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Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: thedoc on 09/01/2014 12:24:13

Title: Why do we have two high tides a day?
Post by: thedoc on 09/01/2014 12:24:13
If the tides are caused by the gravity of the moon, why is there a high tide on the side of the Earth furthest from the moon as well as on the closest side?

Peter Conway

Asked by Peter Conway


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Title: Re: Why are there two high tides a day?
Post by: Pmb on 07/12/2013 00:57:38
The tidal force exerted on the earth causes a stretching effect on the earth. This causes the water to be pulled towards the moon on one side of the earth and pushed away from it on the other. This results in two bulges. As the earth turns each location on earth passes through the two bulges causing two tides.
Title: Re: Why are there two high tides a day?
Post by: David Cooper on 07/12/2013 18:28:09
Land barriers get in the way of that though and make it much more complicated, so it's really a case of oceans and seas sloshing around in resonance with the ideal pattern (for a landless globe that's all sea). The result is that in most places you get two high and low tides a day, though with times that vary from place to place far out of sync with the ideal case (when the two bumps would pass over), and in a few places there are four high and low tides a day instead of two, while in others there is only one high and low tide a day.
Title: Re: Why are there two high tides a day?
Post by: Bored chemist on 07/12/2013 19:08:04
pmb,
How does gravity push water away from the moon?
Title: Re: Why are there two high tides a day?
Post by: A Davis on 08/12/2013 00:09:17
I've never understood how there can be a high tide on each side of the earth, but I do know that the Sun also creates a tide some days they act in opposition and there's no tide at all. Never heard a satisfactory explanation.
Title: Re: Why are there two high tides a day?
Post by: evan_au on 08/12/2013 11:07:57
If you take a single large mass (say the Moon, which has a stronger tidal pull than the Sun due to its closer distance), the Moon's gravity effectively tugs on the center of Earth's mass, ie the center of the Earth follows an ideal elliptical orbit, and this point feels no tide.

Now, if you add a second mass (the Sun), which does not always line up with the Moon:
Title: Re: Why are there two high tides a day?
Post by: David Cooper on 08/12/2013 19:00:43
That's close to it. If you think about every particle of the Earth trying to follow its own orbit round the sun, only a tiny proportion of them are going to be on the line of the orbit that the centre of the Earth follows. The rest want to do other things. The particles on the Earth nearest the sun are orbiting too slowly and want to behave as if they are at aphelion, so their natural course to follow would take them closer in towards the sun and away from the Earth (if the Earth's gravity wasn't holding them back). The particles on the opposite side of the Earth want to behave as if they are at perihelion, so their natural course to follow would be to take them further away from the sun and thereby away from the Earth too.

This same effect can rip comets apart, leaving a line of chunks like the Shoemaker-Levy remnants that slammed into Jupiter one after another.
Title: Re: Why are there two high tides a day?
Post by: Pmb on 08/12/2013 19:30:47
pmb,
How does gravity push water away from the moon?
That’s a great question. I may have interpreted this incorrectly. For the correct description see Newtonian Mechanics by A.P. French (1975) page 533. If you don’t have a copy of this textbook then you can download it from http://bookos-z1.org/book/2033048/a8c794

What I had in mind is different and it’s possible that I confused the two.

Consider a planet or radius R located at the origin of the (x, y, z) coordinate system.

At t = 0 let one object be released from rest at a location along the z-axis at a distance r1 > R. Let a object be released at the same time at a distance r2 > r1. Since r1 is closer to the center of the planet than r2 it will accelerate towards the center faster than r2. Therefore object 1 will accelerate away from object 2. An observer in freefall half way between them will see the objects accelerate away from him. This corresponds two the two particles being forced apart by inertial forces.

Calculation shows that the forces follow a 1/r^3 law. This is similar to how things are working on the earth in the gravitational field of the moon. The actual description is different and is described in French’s text. A good read but hard to understand in my opinion.
Title: Re: Why are there two high tides a day?
Post by: bizerl on 09/12/2013 00:38:36
I've always understood it as this:

On the near side of the Earth, the Moon's gravity is pulling the water away from the Earth so the bulge on the near side is the water that is being pulled towards the Moon more than the rest of the Earth.

On the far side, the Moon's gravity is pulling the Earth away from the water, so the bulge on the far side of the Earth is the water that isn't attracted to the moon as much as the rest of the Earth.

Then there's the sun...

And probably quantum somewhere...  [;D]
Title: Re: Why are there two high tides a day?
Post by: bizerl on 09/12/2013 00:40:55

  • The point on Earth's surface farthest to the Moon is further away than the center of the Earth, and feels a weaker pull than the center of the Earth, so water piles up higher there (a high tide). Alternatively, it could be viewed as if centrifugal force "throws" the water outwards on the far side of the Earth.

I was lead to believe that "centrifugal" was a dirty word in physics?
Title: Re: Why are there two high tides a day?
Post by: Pmb on 09/12/2013 21:29:24

  • The point on Earth's surface farthest to the Moon is further away than the center of the Earth, and feels a weaker pull than the center of the Earth, so water piles up higher there (a high tide). Alternatively, it could be viewed as if centrifugal force "throws" the water outwards on the far side of the Earth.

I was lead to believe that "centrifugal" was a dirty word in physics?
Not at all. It's very important. More so in general relativity then elsewhere.

The centrifugal force is what's called an inertial force defined as any force which is proportional to the mass of the body.

See http://home.comcast.net/~peter.m.brown/gr/inertial_force.htm
Title: Re: Why are there two high tides a day?
Post by: A Davis on 10/12/2013 01:12:25
I think I understand now, so if I drew a perfect elipse representing the earths orbit around the Sun there would be a monthly ripple on the orbit towards and away from the sun every two weeks, does this effect show up in any other measurement say the earths gravity or astronomical measurement.
Title: Re: Why are there two high tides a day?
Post by: David Cooper on 10/12/2013 23:02:34
There's nothing that happens every two weeks that isn't happening all the time. The part of the Earth furthest from the sun is being forced to orbit faster than the course it's following should allow, but its attached to the rest of the Earth and can't move about much. The water on top of it there can move more easily though, so it lifts off slightly, and that's why you get a bulge in the sea there. On the opposite side the opposite applies: the part of the Earth there is being forced to orbit more slowly than the course it's following should allow, so it wants to fall more towards the sun. Again it can't move much, but the water floating around on top of it can, so you get a bulge there too.
Title: Re: Why are there two high tides a day?
Post by: A Davis on 15/12/2013 20:25:19
It does happen all the time and the calculaton is extremely complicated, I've found ten variables in the calculation already there may be more. I was hoping that someone would answer the questions I posed, but I have answered one of them for myself after calculating the shift in the earths center of mass towards the moon.
Title: Re: Why are there two high tides a day?
Post by: David Cooper on 26/12/2013 23:40:31
It struck me last night that I'd only half switched my brain on when posting in this thread, due to pressure of time - I dash through here once a day to see what's new and am always in a hurry to get offline as quickly as possible. What I said before applies to the smaller component of the tide, that part caused by the sun, but I completely forgot to look at the larger component of the tides involving the moon.

The Earth and moon go round each other to some degree, but the point on the line between them which they actually go round is contained within the Earth because the Earth is so much more massive than the moon - it's an 8000 mile diameter vs. a 2000 mile diameter, so that's a ratio of 4:1, with the ratio for the volumes and the apx. masses being 64:1. That means that at the point of balance between them should be about 1/64 of the way from the centre of the Earth to the centre of the Moon, so that'll be 1/64 of 250,000 miles which works out at 3906 miles - that's about a hundred miles below the Earth's surface (beneath the point nearest to the moon). It's doubtless more complicated than that, so feel free to correct these figures if you have better ones.

Importantly though, the Earth must go round this point, even if it's more of a wobble around it rather than what would normally count as an orbit. Again though, the movement of the Earth in carrying out this wobble is perpendicular to the direction to the moon, so again the water on the side nearest the moon will not be moving fast enough to maintain this orbit and will try to follow a different path from the Earth as a whole, leading it to try to fall away slightly from the Earth, while the water on the far side will be moving too fast to maintain this orbit and will try to lift away outwards as a result.

I'm not convinced this explanation for either component of the tide is really any better than the one that says it's down to gravity pulling the water nearest the sun/moon more strongly towards it and pulling the Earth towards it more strongly than the water on the far side [I'll call this the second explanation] - it seems to me that they're just different ways of describing the same thing, with one of them (the second) being declared to be wrong when it may actually be equally correct. After all, if you started with just the moon and Earth in stationary positions about 250,000 miles apart and with neither of them rotating, the water on the Earth nearest to the moon would accelerate more quickly towards the moon and the water on the far side least quickly, so the same bulges would occur through the mechanism described by the second explanation. The first explanation may appear to apply better if you start them off in orbit around each other such that they stay roughly the same distance apart all the time, but it's really the same mechanism with a sideways movement component added in to confuse the issue, thereby leading to more complex explanations being brought in while the simplest explanation is then regarded as wrong, and yet the simplest explanation (the second one) looks as if it is still correct. So I think I've been misled by experts into rejecting it in the past, and now I'm putting it back at the top of my list for explanations of this phenomenon. It's still not the whole story, of course, because continents block the progress of these bumps of water and lead to the tides being completely out of time with these alignments in many locations as the oscillations work their way around seas and oceans, out by six hours in just as many places as it's on time, but the essential driving force which inputs energy into those oscillations will still be the simple one of the second expanation.
Title: Re: Why are there two high tides a day?
Post by: evan_au on 27/12/2013 00:09:16
While force of gravity follows an inverse square law with distance, the tidal force follows an inverse cube law (http://en.wikipedia.org/wiki/Tidal_force#Mathematical_treatment) with distance (being the difference of two points on an inverse square law curve).

This results in the Moon producing a higher tide than the Sun, due to its closer distance.
This is despite the fact that the Sun exerts a much stronger gravitational force on the Earth than the Moon does.
Title: Re: Why are there two high tides a day?
Post by: David Cooper on 28/12/2013 20:52:31
Thanks - that's a good link. It also agrees that the simplest explanation in this case is indeed the right one.
Title: Hear the answer to this question on our show
Post by: thedoc on 09/01/2014 12:24:13
We discussed this question on our  show



Dominic - Tides certainly are caused by the gravity of the moon which is always pulling the Earth very gently towards the moon. Now, the side of the Earth which is faced directly towards the moon is slightly closer to the moon from the middle of the earth. That means it feels a slightly stronger pull because gravity decreases with distance from the moon. And so, that's being pulled more strongly towards the moon and so, you can understand why you get a high tide there. Water is being pulled there more strongly towards the moon.



Now, on the far side, the pull towards the moon is weaker than anywhere else on the Earth, just because it’s further away from the moon. And that means the moon is pulling down on that water on the far side less strongly than elsewhere. And so, it rises up away from the moon in the opposite direction from the moon to form this second high tide. So, you've got two, one on either opposite side of the Earth.



Chris - As the planet turns, it’s turning through both of those bulges of water, so you get high tide number 1, then it takes 12 hours to get round to the other side which is half a rotation, half a day, and there's the second bulge, second high tide.



Dominic - Exactly, so. Those two bulges stay in the same place in space more or less on the line through the Earth to the moon, and the Earth is rotating once every 24 hours, so we move through one of those two bulges every 12 hours as you say.



Chris - And just very briefly, Dominic, the difference between a spring and a neap tide. How does that happen and why?



Dominic - The sun also produces tides. They're about half the strength of the lunar tides. So, that's another signal on top of the lunar tides and sometimes the sun and moon tides coincide and sometimes they don't. They coincide at full moon and new moon, and then you have tides which can be 30%, 40% higher than at other times of the month when the moon and the sun are ninety degrees apart in the sky when you have what are called neap tides which are much lower.



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Title: Re: Why do we have two high tides a day?
Post by: Colmik on 10/01/2014 14:35:50
Evan-au said "the center of the Earth follows an ideal elliptical orbit", but that's not quite true.  The Earth and moon are tied together into a system by gravity, and it is the centre of mass of that system that goes around the sun in a smooth orbit.  Well, if you want to be sniffy about it, there are lots of other influences on this system, and the highest tides of all occur when the Earth, Sun, and all planets are in a line - but that doesn't happen often.  However, ignoring these lesser influences and concentrating on the Earth-moon system, it is the centre of mass of the system that orbits the sun, not the centre of the Earth.  I believe that the centre of mass of the system is, in fact, inside the Earth, but not at its centre.  If the moon is directly over the UK, then the UK is closer to the centre of mass of the Earth-moon than New Zealand is, and whether "centrifugal" is a dirty word or not, it is helpful to think of water close to New Zealand as being thrown out by centrifugal force.

Think of a Scotsman throwing the hammer in the highland games - he is twirling around with the hammer (an iron ball on a chain) being thrown out in front of him, but he is also leaning backward to counter-balance it - and his kilt is being thrown out behind him.  The hammer is one tide, and his kilt is the other.

Incidentally, as every yachtsman knows, the Spring Tides occur about 3 days after new moon and full moon - not on the day itself.
Title: None
Post by: Bill on 04/05/2015 03:07:08
Chris and Dominic seem to be saying the high tides are close to the moon and opposite the moon.  That way of thinking of it is just wrong.  If it were true, then high tides would occur simultaneously along lines of longitude, but in fact they don't.  Have a look at this: http://www.seafriends.org.nz/oceano/tides.htm
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 29/05/2015 08:07:08
1) “Centrifugal” force isn NOT a forbidden word whatsoever: it is just a poorly understood and poorly explained force. I´m not going to deliver now any further explanation about why I say so, perhaps in another post if found convenient.
But please consider what follows: clear examples of real centrifugal forces appear.
2) Why is Earth from pole to pole smaller than between opposite points of the equator?
Because Earth´s rotation originates centrifugal forces which tend to increase equator´s diameter. Not only at oceans: also solid parts “suffer” that kind of force and get deformed. No doubt about it.
3) What happens if we consider now two celestial objects rotating around a common axis? For same reason, ALL material parts of both objects are subjected to centrifugal forces. The further from common axis, the stronger the force. All particles tend to get further from the axis of rotation.
4) What is causing that rotation? This is clear for everybody: apart from certain suitable initial conditions, gravitatory mutual attraction causes the rotation (centripetal force, “twin” force  -rather mother force- of the centrifugal one …).
Frequently, to simplify, this attraction is considered to work between centers of gravity of each of the two considered objects. But it is not actually so.
Making only “half” of that simplification, we can say that EACH material particle of one of the objects is attracted by the other object, even considering particles of equal masses, differently: the further the particle from the other object, the smaller the attraction …
5) Globally, totals of those centripetal and centrifugal forces are equal. Otherwise the distance between the objects would not keep constant.
But their spatial distributions are opposite from each other. At “inner” parts of both objects gravitatory atraction is bigger than centrifugal one, and at “outer” parts centrifugal forces are stronger than gravitatory ones (relative to the other object).
This fact originates deformation of those objects, even of their solid parts. Needless to say sea surface is much easily deformed than solid parts.
6) The Moon/Earth case is a “tricky” one. Most knowledge “sources”, when dealing with Earth´s movements, mention translation (actually an annual rotation around the sun), the daily rotation … and then they mention the several very slow changes of Earth´s axis direction, and similar very long period changes … But I consider that the third most important movement of Earth is its 28/29 days rotation of both Moon and Earth … Moon´s movement wouldn´t be possible without Earth´s one, and this last one causes the 2nd daily Earth´s tide. Few people mention that movement, I think because the axis of rotation is kind of hidden: it crosses the Earth, but NOT through its center of gravity, but at some point between that center and Earth´s surface. By itself this rotation causes bulges at both parts nearer and further to the Moon, but bigger at further one (it is also further from axis of rotation). But attraction from the Moon is bigger at the side nearer to it.
The result is two similar high tides, one “below” the Moon, and the other on the antipodes. There is actually some delay: due to the quick tangential movement of sea because of Earth daily rotation, high tides try to catch up with their theoretical positions but don´t “succeed” …
This last fact has curiously had very big importance in Earth´s history (perhaps even in the beginning of life), especially at its early stages when Moon was much much closer, Earth´s own rotation was much much quicker, and subsequently tides were much much stronger. But this is not the moment to go any further about that.
Title: Re: Why are there two high tides a day?
Post by: PmbPhy on 29/05/2015 08:37:01
Quote from: Bored chemist
pmb,
How does gravity push water away from the moon?
I learned this years ago from Newtonian Mechanics by A.P. French. I don't recall the details of the derivation though. I only make sure that I follow a derivation at least once and then after that I can have confidence in knowing that it's right. I'll scan the derivation in and post it later on today and reread it myself. Until then Wiki explains it nicely. See:
http://en.wikipedia.org/wiki/Tidal_force#Mathematical_treatment

Especially Fig. 4
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 29/05/2015 13:01:27
In Reply #21 a wikipedia link is given. It surprises me the don´t mention the centrifugal forces originated by Earth´s rotation around the Moon/Earth rotation axis, as I exposed at Reply #20.
They just say (in my words) that gravitational (due to the Moon) acceleration at ocean nearer side is bigger than at intermediate solid parts of the Earth, and last one is bigger than at the further side of the Earth. And those facts produce kind of opposite acceleration there, relative to main Earth´s body, causing the second high tide.
I´ve previously heard that argument, but consider it insufficient.
If those three zones of the Earth experience only those gravitational accelerations from the Moon, why ALL of them don´t cause a movement of the Earth towards the Moon?
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 29/05/2015 16:58:10
I scanned the section of Gravity From the Ground Up by Bernhard Schutz which covers ocean tides. It's on my companies website at http://www.newenglandphysics.org/Science_Literature/Journal_Articles/schutz_tides.pdf
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 29/05/2015 21:17:34
In pdf linked in reply #23, I can see that Schutz also mentions only the different gravitatory pulls from the Moon …
I repeat the question on my post #22:
"If those three zones of the Earth experience only those gravitatory accelerations from the Moon, why ALL of them don´t cause a movement of the Earth towards the Moon? “
Tomorrow I´ll try to explain my “theory” of the centrifugal force used in my first post, which I consider necessary to have a correct explanation of the issue. I´m slow when typing in English, and have no time now. But just an analogy that I consider useful:
Imaging an athlete of hammer trow speciality. He or she can´t keep verticality when throwing the hammer. It is necessary to lean a little backwards. Otherwise the hammer could not be rotated. Both the athlete and the hammer will rotate around an axis situated near the forward part of the athlete´s body.
If the hair of the athlete is long and not  fixed by some device, instead of keeping its normal downward direction due to its weight, it will move back and upwards …
That cannot be due to anything similar to what stated by Schutz relative to tides: CENTRIFUGAL force is the cause.
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 29/05/2015 23:59:08
Quote from: rmolnav
In pdf linked in reply #23, I can see that Schutz also mentions only the different gravitatory pulls from the Moon …
Did you read that PDF file carefully? Did you see an error in it?

Why do you believe that centrifugal force plays such an important role in ocean tides?

Quote from: rmolnav
That cannot be due to anything similar to what stated by Schutz relative to tides: CENTRIFUGAL force is the cause.
Let me get this right. You actually believe that tides are caused entirely by centrifugal forces? If so then that's clearly wrong. It can't be correct because the existence of tides corresponds to the position of the moon and centrifugal forces can't account for that. It also can't account for the periodicity of tides either.
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 30/05/2015 00:33:40
Quote from: rmolnav
That cannot be due to anything similar to what stated by Schutz relative to tides: CENTRIFUGAL force is the cause.
You appear to think that tidal forces are caused by centrifugal forces and that's simply not true. That would mean that there'd be no tides since the centrifugal force on matter at any place on earth doesn't change with the time of day like ocean tides do.
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 30/05/2015 00:35:18
Quote from: rmolnav
That cannot be due to anything similar to what stated by Schutz relative to tides: CENTRIFUGAL force is the cause.
You appear to think that tidal forces are caused by centrifugal forces and that's simply not true. That would mean that there'd be no tides since the centrifugal force on matter at any place on earth doesn't change with the time of day like ocean tides do.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 31/05/2015 11:41:39
I know centrifugal force can be considered as an inertial, actually kind of ficticious force … But not only that way.
Imagine a solid spherical object rotating as artificial satellite around Earth. As all we know, the hole object is globally subjected only to one gravitatory centripetal force, which is producing an acceleration. Being that acceleration perpendicular to the object velocity, it doesn´t change its linear velocity, only its direction, in such a way that the object rotates. To simplify we can consider the orbit is circular.
That is possible only with suitable values  of angular speed and distance: those values have to match. The gravitatory force divided by object´s mass has to be equal to the square of the angular speed multiplied by the radius (distance from Earth).
For a given angular speed, centripetal acceleration of that satellite is proportional to the distance, but gravitatory acceleration is inversely proportional to the square of the distance.
If we made the simplification of considering the hole mass of the satellite concentrated in its center of gravity, for a given angular speed there would be a distance suitable to get the object moving as a satellite.
In that case, no centrifugal force would actually be affecting the object. That force, in the fashion I´m  using this concept, derives from action/reaction principle: if Earth is producing a gravitatory, centripetal force on the object, this is also producing another equal opposite force on the Earth. But the object does´t suffer that reaction force.
But the object has a size. Different parts of it are at different distances from Earth. Subsequently gravitational forces are different. But the hole object is kind of obliged to rotate at the same angular speed … New internal forces appear affecting different parts of the object!
Let us imagine the object divided into many thin cylindrical slices (with radius equal to distance to Earth). All parts of each slice rotate with same radius. If there were no internal forces, only a central slice could continue rotating with the given global angular speed, but the rest could not.
All parts of the object interact with contiguous ones, but we can consider central parts of the object “ run the show”: they are rotating at the “ correct” speed, and forcing the rest to keep their pace. 
Each slice experiences the gravitatory force at that distance, and additional internal forces from contiguous slices. The sum of all those forces has to produce the centripetal acceleration which makes it rotate at the given angular speed.
Going from center slice to further side, slices are being forced by contiguous closer ones to rotate with a centripetal acceleration bigger than what only gravity would produce on each slice. That is acheived by inner direct tensile forces, from closer slices to further contiguous ones. And that fills the centripetal acceleration “gaps”. Due to action/reaction principle, further slices produces an opposite and equal force on closer ones. Those forces  are CENTRIFUGAL, not ficticious but real, and affect all those slices of the object.
Something similar happens going from center slice to closer side. In this case gravitatory forces get bigger and bigger than what required to produce the centripetal force at the given angular speed. Further slices force contiguous closer ones to rotate with a  centripetal acceleration smaller than what produced by gravity there. That is achieved by direct internal tensile forces from further slices on closer ones, which is also a CENTRIFUGAL, real force.
Those “ internal” imbalances are the cause of the tendency of any celestial object rotating around another (actually around a common axis of rotation) to get an ovoid shape, causing many interesting phenomena, some of them mentioned by Schutz previously linked work as a pdf. And, as far as I can see, they are also the actual cause of the two Earth´s high tides.
I have other arguments that further “ prove” my (?) theory, related to tide cycle due to the other “couple” Sun/Earth, but this post is already too long …
By the way, Schutz (or whoever has written “Investigation 5.1: Tides and eclipses“ at mentioned work), make a bizarre calculation when comparing Sun and Moon gravitatory effects, that I also find wrong ...
     
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 31/05/2015 11:48:20
Sorry. An "n" is missing at "But the object does**´t suffer that reaction force". It is clear, isn´t it?
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 01/06/2015 09:02:21
After my last posts I´ve found my point can be put in a simpler way:
Each slice is affected by its own gravity, and internal stresses from both nearer and further contiguous slices. The net result of all forces has to produce the centripetal acceleration to make the slice rotate at the given angular speed.
The closer the slice, the bigger the gravitatory force (it increases inversely proportionally to the square of the distance), but the smaller the centripetal force necessary to produce the given angular speed (proportionally to the radius or distance). The dynamic equilibrium of satellite rotation can only be achieved if the net force produced on each slice by contiguous ones (either direct or reaction forces) is in an outward direction: a CENTRIFUGAL force, which affects all slices bar central one, where net force from contiguous slices is null (gravitatory force equal to what necessary to produce the given angular speed).
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 01/06/2015 15:24:02
I HAD NOT SEEN Reply #       :
"You appear to think that tidal forces are caused by centrifugal forces and that's simply not true. That would mean that there'd be no tides since the centrifugal force on matter at any place on earth doesn't change with the time of day like ocean tides doI".

That is utterly erroneous. In any low latitude area, when Moon is almost just over there, even solid earth get deformed, in this case mainly due gravitatory pull from the Moon. And it happens the same at antipodes, there mainly due to centrifugal force. Logically, the deformation is smaller than at open oceans, where "fluidity" of water makes deformation much easier ... Even we ourselves weigh a little less, similar to what happens at the equator, compared to our weight at poles.
Something similar, but smaller, happens at noon and midnight, in relation to Sun/Earth effects. 
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 02/06/2015 03:15:25
Quote from: rmolnav
That is utterly erroneous.
And you are utterly wrong. You're claiming that tidal forces, i.e. ocean tides, are due solely do to centrifugal forces which means that the moon has nothing to do with it because centrifugal forces are not caused by the moon.

You may not like it and you may not understand it but the descriptions in those PDF files are the correct description for tidal forces. There is no doubt about it whatsoever. If you think that those derivations are wrong then you are seriously mistaken.

 You have the derivations right in front of you and they've been there for days now and you've made no attempt to state what's in those derivations that's wrong. I.e. you haven't proved that the derivations and therefore the conclusions are wrong.
Keep it civil - Mod
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 02/06/2015 07:04:07
Reply #31:
"... You're claiming that tidal forces, i.e. ocean tides, are due solely do to centrifugal forces which means that the moon has nothing to do with it because centrifugal forces are not caused by the moon".
Sorry, but i have NOT claimed what you say. What in several ways I have said is that ocean tides are mainly due to an imbalance between gravitatory pull from the Moon (the closer, the higher the pull), and centrifugal forces due to the fact that it is not 100% right that Moon is rotating around Earth: both are rotating around a common barycentral axis, and the further from that axis, the bigger the centrifugal force.
And I´m not promoting my own idea and "not inviting critical debate about it ...". F.e., twice I have invited anybody to answer the question:
"If those three zones of the Earth experience only those gravitatory accelerations from the Moon, why ALL of them don´t cause a movement of the Earth towards the Moon? “
To be honest, I admit that perhaps I"ve been too assertive. and I beg your pardon. It would have been due to the fact that I feel pretty sure about the issue. Logically I know I could be wrong. If any of you think it is so, please kindly show it to me with facts and arguments, not with sweeping, not accurate generalizations. 

Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 02/06/2015 10:11:44
If anybody think I was utterly wrong at #30, as said at #31, please kindly google "tidal earth crust deformation" ...
There are plenty of studies considering a fact that deformation.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 04/06/2015 07:54:58
No reply in two days ... I do hope somebody could be considering my point, without seeing clearly wether I´m right or wrong.
Please kindly compare #29 with what follows. People with some knowledge on calculation of internal stresses of structure materials know that the way I dealt with the issue there is basic on that field.
Another analogy to make it clearer for people not familiar with the matter:
Imaging just a steel cylindrical bar hanging from one of its ends. If we made a horizontal cut, the lower part would fell down. Why it didn´t fall before? Because of internal tensile stresses: upper side of the section was pulling lower one, exactly with a force equal to the weight of lower part of the bar.
If we had made the cut a little lower, we could say the same. In this case the weight of lower part would be a little less: just the weight of the slice between the two cuts.
As the slice is not experiencing any acceleration, the sum of all forces applied to it is null. The sum of internal stresses it suffers from contiguous material, plus its own weight, has to be null.
If we produced any upward acceleration to the hanging point, internal stresses would increase in such a way that the sum of weight of the slice plus stresses from contiguous material would give a net force that divided by slice mass would be equal to the acceleration.
Very similar is what happens in the case of two objects rotating around a common axis. In this case each slice of one of the objects is subjected to gravitational attraction from the other object, and that force plus net forces from contiguos slices, divided by the slice mass, has to be equal to the square of angular speed multiplied by the radius (or distance), actual centripetal acceleration at rotatory movements.
Please kindly go now back to #29 ... As far as I can see, it should be clear for most people. 
   
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 04/06/2015 16:12:04
Quote from: rmolnav
No reply in two days ...
That's because when we see that someone can't understand an argument and/or correctly back up there's then we usually give up. In this case you still haven't done what's required of you and that's to find an error in the derivations that are in those PDF files that I posted which are from various textbooks.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/06/2015 07:36:16
Sorry, but what said in #35 is NOT true. F. e., THREE times I have previously asked:
"If those three zones of the Earth experience ONLY those gravitatory accelerations from the Moon, why ALL of them don´t cause a movement of the Earth towards the Moon? “
It´s you, and rest of readers, who haven´t reply that question. Do you think that question is absurd and doesn´t deserve to be answered?
Within the theory in mentioned PDF, that question could be answered saying that those forces are producing the centripetal accelerations of the rotating movement. But, as I´ve already shown in different ways, in different earth areas gravitational pull from the moon and rotatory centripetal force are NOT equal to each other ... Forces from internal stresses have to be considered.
ANOTHER error, as far as I can see:
In mentioned PDF, tidal forces are considered to be derived  from just the variation of pull from the moon with the distance. That´s why they conclude that although gravitational force decrease to the square of the distance, tidal force decrease to the third power of the distance.
Beeing honest, I have to say I can´t refute their mathematical explanation ... But I insist: stuff different from just gravitatory pull and distance have to be considered to explain tides. Such us internal stresses as I´ve already said, some of them centrifugal. And also what follows.
Even considering a bellow the moon ocean bulge, that could be cause just by pull from the moon, we could call it high tide only in comparison to what happens 90 degrees away, where low tide. But the reason is not just a difference in what they call tidal forces (smaller pull where further from the moon). Even if the pull were the same at both places, tides would happen.
Why? Because main cause of ocean surface shape is OTHER force which has also to be considered: gravitatory one from earth itself. That alone would produce a spherical shape. Adding moon pull bellow it (now I´m "forgetting" inner stresses), the result is kind of less weight of the water there. BUT 90 DEGREES AWAY from there moon´s pull doesn´t actually affect water weight, and we have low tide.
Tides are kind of a dynamic equilibrium that results from MORE forces than just pull from the moon and its variation with the distance ...
 
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/06/2015 10:21:03
A couple of things more.
Firstly, I beg your pardon for my many English errors. Perhaps they make more tiring to read my posts. It takes me long to write them, and when written I am not sufficiently patient to check them carefully before sending ...
I´m afraid PmbPhy didn´t completely read #34.

But what said there is BASIC Physics. Standard Finite Element Methods to calculate structures are based on what said there: ALL internal and external forces affecting each small finite element of any object have to be considered. If not in balance, deformations and/or movements (with acceleration) happens ...
By the way: in #34, for people not familiar with the subject, I considered the hanging bar initially without any acceleration (still, but it could also be in movement at constant speed). Afterwards with an upward acceleration produced by an artificial force applied to the hanging point: internal stresses would increase.
Needless (?) to say that if we let the bar fall down freely, internal stresses would disappear. The unique force which would be affecting each slice would be its weight, with the result of "g" acceleration there ...
The sum of ALL forces affecting ANY part of the object, and the resulting acceleration, has to match. 

 
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 07/06/2015 12:20:58
Two days more and … I´m going to post a “divertimento” on the issue , hoping it may be interesting to some folks.
Nature is far more complex than what seems at first sight. Ocean surface shape depends on not just a few physical phenomena.
I´ll try and list what, as far as I know, affects it, in a decreasing order of importance. I hope not to forget anything. If any disagreement, please let me know, and we can discuss the issue.
1) Gravity from earth itself. If our planet is considered alone, without any movement, without any other celestial object gravitationally affecting it, etc, sea surface “should" be spherical …
2) I say “should” because that is only in theory: gravity is not uniform over sea surface, because amount and density of inside material varies …
That produces sea local level differences much much bigger than tidal differences.
3) Let us “allow” earth daily rotation … Due to centrifugal forces, equator diameter gets bigger than distance from pole to pole.
4) If we now include moon/earth rotation, then we have main part of tides, which we have been discussing. We shouldn´t forget that local conditions (mainly size and shape of continents) produce changes on “theoretical” tides, because water is not completely free to “obey” the forces we have been discussing.
5) Now earth translation (actually rotation around the sun): it produces tides similar to moon related ones, but smaller. These don´t change with the clock. For similar reasons to lunar tides, a high tide happens “bellow” the sun, at noon, and another at antipodes, at midnight. Also with a certain gap due to the fact that the water bulges cannot catch up with their theoretical position …
Needless to say that when lunar tides syncronize with sun ones, we have the strongest actual tides. This happens when full and new moons.
6) Meteorological conditions also affect: atmospheric pressure, winds, etc.
7) I haven´t mentioned water temperature, as if it did´t change with space and time … That isn´t actually so, and the result is that some additional, but relatively small changes happen.
An additional curiosity in this respect. Local increase in temperatures of ocean upper part (f.e., at El Nińo zone) are beeing detected thanks to “gravitational” Physics … The temperature increase produces there tiny bulges that are detected with sophisticated satellites (f.e., GRACE, a couple of satellites), thanks to very very tiny changes in the distance to each other, derived from gravitational changes.
The software they use must be “unbelievable”: how can they tell apart the many different gravitatory existing factors?
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 08/06/2015 00:45:06
rmolnav - Who are you talking to? I lost interest in this thread myself.

I'm curious. If you're just posting to everyone then why don't you modify one thread and put everything in that thread instead of creating multiple posts?
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 08/06/2015 13:13:50
PmbPhy:
I´m posting ideas which intend to answer the asked question: why are there two tides a day?.  Or at least related to the issue, which could be interesting to somebody.

COLMIK (especially for you, but possibly interesting to some other folks):
In relation to something you said in #18, I´ll put another curious physical detail.
You said that so called “spring tides” (strongest ones) do not happen the day when full or new moon, but actually a few days later.
That´s certainly so, and something I had not previously found any explanation to, consistent with rest of “my” theory …
Last night I started to ruminate it in bed. I found following possible explanation.
In some previous posts I´ve mentioned the fact that moon related high tide doesn´t happen when moon is exactly over our meridian, but with some delay. That´s because the top of the tidal bulge can´t catch up with below moon meridian, due to the high linear velocity of earth rotation (40,000 km/24h at equator).
Although main component of attraction between moon and main part of the tidal bulge is vertical, a relatively tiny tangential component does actually occur. Naturally, in both senses (action and reaction principle)
I´m not going to expose now interesting consequences that fact has been having for the moon, which I even discussed some time ago with Neil F. Comins, after seeing a very interesting tv show where he appeared (relative to moon and earth early stages).
But that tangential component of mentioned attraction tries to decrease the delay. Ocean surface quickly moves eastward due to earth rotation, so tidal bulge is east to moon, but always “moving” westwards trying to reach moon meridian.
Something similar happens when considering only sun related part of tides. At noon the sun is over our meridian, and high sun related tide happens. Actually some time later, for reasons similar to what said for moon tides. But now, being the sun much far away, the tangential component of the sun pull on the main part of the bulge must be almost negligible. Subsequently, the angular gap between bulge and sun must be bigger than moon related one.
A couple of days after new moon, the two bulges are on same meridian. Sun is then west to the moon, but the higher delay of its related bulge puts it  same place than moon related bulge. They fully add up and tide coefficient reaches then a maximum.
Any comments would be wellcomed.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 13/06/2015 10:34:48
Looking for some information relative to local solid earth effects on ocean tides, I´ve seen in wikipedia an article titled "Tide".
There the term "centrifugal force" is also kind of forbidden ... But there is a paragraph which has "open my eyes", in a way consistent with "my" theory.
As far as I can see, denial of the existence of any real centrifugal force is a physical error (as I´ve previously explained), which requires a mathematical "trick" to reach an explanation of tides.
Mentioned paragraph says:
"The tidal force produced by a massive object (Moon, hereafter) on a small particle located on or in an extensive body (Earth, hereafter) is the vector difference between the gravitational force exerted by the Moon on the particle, and the gravitational force that would be exerted on the particle if it were located at the Earth's center of mass".
A DIFFERENCE OF A REAL VECTOR (MOON ATRACTION ON A PARTICLE), AND THAT ATRACTION "IF" THE PARTICLE WERE LOCATED SOMEWHERE ELSE IS A PHYSICAL NONSENSE. Sorry. For me it is just a mathematical trick.
That mention of the "Earth´s center of mass" reminds me what I said that on a slice there located actual Moon attraction and rotational movement centripetal force are in balance. But not in the rest, and internal stresses appear. Each slice is supporting its own weight, Moon attraction on it, and internal forces (stresses multiplied by surfaces) from contiguos slices, many of them CENTRIFUGAL.
A dynamic equilibrium is reached on EACH particle if the result of adding up all those vectors divided by the mass of the particle is the actual rotational centripetal acceleration where considered particle is located. Any imbalance would produce internal stress changes and/or additional movement of the particle. 

Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 14/06/2015 22:08:06
Perhaps yesterday I was kind of too assertive when saying what quoted from wikipedia was a “physical nonsense”. I´ll try to explain why I said so.
I do understand a comparison between real location of an earth particle and “if” it were located at earth center of gravity can be made. And a term as “tidal force” or something similar could be defined in relation to that comparison …
After all, that´s similar to what I said that if all mass of a rotating object were at its center of gravity the gravitatory pull from the other object would be equal to rotation centripetal acceleration multiplied by its mass, the centripetal force. And in that case there wouldn´t be any centrifugal force (neither a tide whatsoever). BUT objects do have a size, and then …
Following the idea of the quote, I would say something such as:
“Tidal effect (on a given particle) is the difference between net forces supported by the particle in two scenarios: the real one, and another fictitious with the particle situated at the center of gravity of the earth”.
Considering just differences between moon pull vectors is NOT sufficient. How could the particle physically “feel” that difference? How could it “know” it is in a location away from center of gravity?
The two scenarios differ more than just that:
- The particle at earth center of gravity has a rotational movement in balance: net internal force affecting the particle would be null. But at its actual location that balance doesn´t exist, and the sum of forces affecting the particle, due to internal stresses (centripetal, centrifugal and others) is not null, in general.
- Particle own weight is real at its actual location, but at earth center of gravity would be null.     
In real scenario, the particle does “feel” moon gravitatory pull, also its own weight, and stresses from contiguous particles, and does “know” how to react: with actual centripetal rotational acceleration, which has to be the same across the earth. And, if not in that dynamic equilibrium, with additional movement and/or deformation (which would affect back internal stresses …).
That´s why tides happen, included the usual two high tides a day.
Another important detail. In some posts i have said something like “radius or distance”, in relation to the rotation of a solid object around earth, as artificial satellite. That could also be correct if considering earth rotation around the sun. But in moon related tides, as both moon and earth rotate around an axis distant from earth center less than earth radius, the radius of that earth rotation is much much smaller than the distance to the other object of the couple, the moon. Pull from the moon is inversely proportional to the square of that distance (for different parts of the earth), but centripetal force required for the rotation, for actual angular speed (360ş/plus 27 days), is proportional to the rotation radius (for each earth particle). Differences between different parts of the earth are relatively much much higher in centripetal accelerations (causing earth rotation around barycentral axis), than in moon pulls.
That produces higher imbalances, and higher internal stresses, included centrifugal ones, than if we were considering just distance to the moon differences.
Curious enough, earth part closest to moon is at same side of rotation axis than moon: moon pull (causing that earth rotation) is there in the sense opposite to centripetal acceleration …
That produces centrifugal forces similar to what happens with earth daily normal rotation. But in an asymmetrical way, and much much slower.
That´s why for me the second moon related high tide, at opposite side to moon, is mainly due to centrifugal force. 
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 17/06/2015 10:10:57
Just a short post to put another argument/evidence supporting "my" theory ... as far as I can see.
It is well known that Io, one of the satellites of Jupiter, is internally really hot, even with strong vulcanism that most recent, sophisticated studies say it is due to internal tide friction.
Could all that energy come just from differences across Io in the gravitational pull from Jupiter, being Io´s diameter only 3,643.2 km? For me, NO.
It is true that Jupiter´s pull is really big, but the radius of Io´s rotation around Jupiter is similar to our Moon´s. Relative gravitational differences cannot be big enough to produce such friction.
It is close to Jupiter, and that´s why it needs to rotate along its orbit really quickly, in less than a couple of our days.
Such a high angular speed produces really high imbalances between centripetal acceleration and gravitational pull at different locations. That produces really high internal stresses, and deformation and movements happen quickly, with the result of really high friction.
In the universe there have been satellites which even broke/exploded due to really high internal, tidal stresses. It would be unthinkable that were due just to differences in gravitational pull, without the existence of the imbalances and the internal forces (centrifugal ones included) I´ve been referring to. 
Title: Re: Why do we have two high tides a day?
Post by: evan_au on 18/06/2015 11:22:23
Quote from: rmolnav
In the universe there have been satellites which even broke/exploded due to really high internal, tidal stresses.
The Roche limit (https://en.wikipedia.org/wiki/Roche_limit) is where the gravitational tug of a massive body will disrupt a smaller body that is held together by gravitational forces. Effectively, the tidal forces tear the smaller body apart.

This applies to comets which are "rubble piles" held together only by gravity. However, bodies which are made of solid rock or solid iron will survive these tidal forces at a closer radius than the Roche limit.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 18/06/2015 12:05:33
Thank you, evan_au. I was just drafting something more, but relative to what I said about Io ...
Being honest, I have to say something DOESN´T MATCH.
In relation to what I said about Io, after sending my last post I remembered several times I had previously thought:
"How odd that huge energy dissipation from Io! If tidal friction is so high, why Io has´t got synchronized with Jupiter yet, as our Moon is with Earth? Io must be very young …"
Because tidal friction happens within Earth thanks to its daily rotation. The two normal bulges are continuosly changing location, internal stresses produce deformation and additional movements, and friction occurs. But there is no tidal friction in the Moon ... Its shape has a couple of relatively small bulges, but at fixed locations. It was precisely tidal friction, happening there long long ago, which slowly synchronized it with Earth (in the sense that it is still rotating around its axis, but exactly 360ş in same period of its rotation around Earth).
I said to myself: let us google Io ... And in wikipedia I found that Io is already synchronized with Jupiter !!!
Then, as far as I can see, Io huge friction cannot come from tides, not in “my” model (including internal centrifugal forces), let alone if we consider just local differences in gravitational pull ...
An explanation is given in the article, rather odd for me.
They say it is due to "Laplace resonance". The periods of rotation of Io (the closest) and the two other Jupiter´s satellites (Europa and Ganymede) happen to be proportional to 2,3 and 4 … Periodically they get in line with Jupiter … Well, anyone interested can have a look in the web.
But small celestial objects such as Europa and Ganymede, especially compared to Jupiter, “shoudn,t” affect Io that much. When any of them in line with Io and Jupiter, outer Io bulges must get a little bigger, some deformation happens, and some heat is released … But, can that be sufficient to feed Io´s huge volcanic activity?
Really strange for me!
Any comments would be appreciated.   

 
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 18/06/2015 12:17:14
Sorry. I said "outer Io bulges must get a little bigger ...". I mean both bulges: if an (or two) outer satellite pull outward, Jupiter has to compensate increasing its inward pull ... Otherwise Io would get out of orbit ... Well, that is kind of simplification. Those phenomena must be really complex.
Title: Re: Why do we have two high tides a day?
Post by: alancalverd on 18/06/2015 16:13:01
Who "we"? Southampton has four high tides per day, some Mediterranean ports have almost no tidal range at all.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 20/06/2015 12:12:23
alancalverd, #46
For me, your second question is easier to answer than the first. Firstly I´ll put my answer to the second, and make just a small comment/guess about the first afterwards.
For the sake of simplification, let us consider only the pull from the Moon, and suppose that Moon´s orbit and our equator´s plane were the same. And in open sea, not to have local influences (and not far from the equator)
If we have the Moon over us, we have high tide. Moon´s pull is in opposite direction to water weight, actual (or apparent) density of water decreases (though very, very slightly), and sea surface get deformed due to that. Low tides are 90ş away from high tide location. There Moon pull vector and water weight are perpendicular to each other, and Moon´s pull don´t affect water density at all. In intermediate locations, something also intermediate happens.
From west to east ends of Mediterranean sea there are less than 4,000 km. When Moon is over one end, the above mentioned effect there and at the other end are not too different, and sea surface deformation is small.
Regarding second question, my guess is related to the fact that, you know, we all actually have FOUR high tides: two related to the Moon (below it and at antipodes), and two related to the Sun (at noon and midnight, with some delay I commented about in previous posts). But we only see the result of adding up Moon and Sun effects …
Perhaps in Southhampton, due to its latitude, some local conditions, and the fact that apparent orbits of Moon and Sun in our skies are different, at least in some periods of the year Sun related high tides are not “hidden" by Moon related tides …
That could match with four not uniformly distributed over the day high tides: two at fixed time (1 or 2 hours after noon and midnight), related to the Sun, and the two related to the Moon changing with its location in local skies … And not equally distinguishable over the year, due to changes in apparent orbits of Moon and Sun …
Has Southhampton annual cycle of tides such kind of characteristics?
If not, I can´t guess any other possible explanation ...
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 01/10/2015 10:45:39
Recently I´ve seen a Wikipedia article where arguments similar to mines on #26 and 28 are given, when explaining differences between centripetal and centrifugal forces at different points of a rotating solid. In this case it is kind of a cable/elevator, with a "counterweight" rotating around our planet over the equator ...
I must say I find the "invention" just SCIENCE FICTION, not to be discussed here (they use centrifugal force as if it were a primary force, ready to be used, and that is utterly erroneous). I would say it is quite opposite to considering centrifugal forces not real at all, but BOTH errors.
But the centrifugal forces they are considering are REAL, similar to internal forces I explained are transmitted between different slices of any two objects rotating around its barycenter axis (f.e. Moon/Earth): 
"A space elevator cable rotates along with the rotation of the Earth. Therefore, objects attached to the cable will experience upward centrifugal force in the direction opposing the downward gravitational force. The higher up the cable the object is located, the less the gravitational pull of the Earth, and the stronger the upward centrifugal force due to the rotation, so that more centrifugal force opposes less gravity. The centrifugal force and the gravity are balanced at GEO. Above GEO, the centrifugal force is stronger than gravity, causing objects attached to the cable there to pull upward on it".
https://en.wikipedia.org/wiki/Space_elevator
 
Title: Re: Why do we have two high tides a day?
Post by: ProjectSailor on 01/10/2015 11:34:13
yes the space elevator is science fiction.. however i dont see what it has to do with tides.. which are caused by gravitational attraction from the sun and moon... for something that has been known for centuries, if not millenia, I have a book that accurately tells me in advance the heights of all the tides in a year and the times.. and are accurate within reason (weather effects have a great deal to play in some areas)

If you think we don't understand them at all.. how is this possible? we can even predict tides in areas where the land plays a massive part (southampton is a great example)

you are right in one thought though, centrifugal forces will play a part but since this is constant it will have little effect on TIDES but mostly on CURRENTS but are not the only effect on currents!
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 03/10/2015 12:31:27
ProjectSailor, #49
Thank you. Sorry if I didn´t put it clearly enough.
Regarding your last point, of course currents are due not only to tides, but also to local geographical conditions.
And I brought up the space elevator article just to show the use of centrifugal force concept as a real thing, by scientists, and in a similar way to mine on my posts #20 and following ones, when trying to explain why there are two high tides per day.
You know, there are people who strongly refuse even the use of the term "centrifugal force". They consider it ONLY as an inertial, kind of ficticious force. For them ONLY centripetal force actually exists, producing the centripetal acceleration that changes the direction of the considered object, making it rotate …
Please have a look at first thread posts. F. e., at #9 where it is said:
"I was lead to believe that "centrifugal" was a dirty word in physics?”
For me without the presence of centrifugal forces it is not actually possible to understand the existence of not only a high tide “following” Moon´s location (as you say "caused by gravitational attraction from …), but also another at antipodes (not considering Sun´s effect for the sake of simplicity).
In several posts I tried to explain my ideas, as you can see if some available time (posts 20, 26, 28 and 31 are suggested). There was a rather nasty discussion with “PmbPhy”, some comment/question (#43 and 46), and then long silence (until my #48).
Had you any doubt and/or opposite idea, please kindly post it. A rational discussion is always enriching for both sides.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 13/10/2015 12:26:56
(Sorry I wrote #49 instead of #51 in last post first line)
Listening to something about detection of exoplanets, I´ve met something that should be useful to those who don´t accept “my” theory of Earth´s wobbling due to its rotation, together with our Moon, around their barycentral axis, being that what causes the high tide at the Earth´s side opposite to the Moon (with its subsequent centrifugal forces)
Even very, very distant stars, if a planet is turning around them, experience that wobbling movement. And that is being used trying to detect exoplanets:
 
"...the fact that a star does not remain completely stationary when it is orbited by a planet. It moves, ever so slightly, in a small circle or ellipse, responding to the gravitational tug of its smaller companion".

http://www.planetary.org/explore/space-topics/exoplanets/radial-velocity.html
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 01/11/2015 12:22:34
Apart from the issue of the wobbling I referred to in my last post, I previously exposed the other "leg" of my argument: the reality of centrifugal force.
Several people argued against it. A link to a scientific paper showed that the author even had made a "bizarre" (at least to me) physical and mathematical explanation just because he had apparently forbidden himself the use of the discussed term (centrifugal f.).
I also said (#42) that in a wikipedia page something similar happens ...
Am I wrong, alone against everybody?
No. Recently, when discussing other related item (tidal locking), I found in wikipedia an explanation similarly erroneous to me, which I consider could also be properly exposed considering centrifugal forces ... I decided to check thoroughly more internet pages.
RESULT: even in the wikipedia page titled "centrifugal force", where initially the inertial, kind of ficticious character of the concept is exposed, going down one can find:     
"In another instance the term refers to the reaction force to a centripetal force. A body undergoing curved motion, such as circular motion, is accelerating toward a center at any particular point in time. This centripetal acceleration is provided by a centripetal force, which is exerted on the body in curved motion by some other body. In accordance with Newton's third law of motion, the body in curved motion exerts an equal and opposite force on the other body. This reactive force is exerted by the body in curved motion on the other body that provides the centripetal force and its direction is from that other body toward the body in curved motion"
AS I ALSO SAID, if the body has a relatively big size, and centripetal acceleration is exerted by gravity from other celestial object, different parts of it experience different gravitational forces, not matching with the required centripetal acceleration. The closer the part the bigger the gravitational attraction (to the square of the distance), BUT the smaller the required centripetal acceleration (all parts are obliged to rotate at same angular speed).
That HAS TO BE compensated by internal stresses/forces, all in "couples" action/reaction (following Newton´s third law), being half of them centrifugal, and not ficticious but REAL. And the other half centripetal (in addition to gravitational ones).
The distribution of all those internal forces produces the deformation of the body, being one of the results the two opposite sea surface bulges (high tides), that move around the planet due to its rotation, one always trying to catch up with Moon´s relative position and the other with the point furthest to the Moon (not considering Sun´s effect, for the sake of exposition clearness)
 
Title: None
Post by: Wiz on 28/07/2016 11:39:30
This answer is wrong. The bulge on the opposite side of the earth is caused by a centripetal force created by the earth's rotation, which is in effect trying to throw the oceans off into space.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 03/10/2016 10:56:33
I´m afraid you haven´t carefully read my last post, let alone many others of last year, when I gave many examples and explanations.
Don´t try to read them now. It would be too time consuming.
But please kindly read what I happened to post just yesterday, because last month, once again, another discussion relative to centrifugal force started. And without agreeing in the basics of that concept, it would be useless to discuss the four tides a day subject:
"Imagine you are rotating a weight, with the help of your hand (and wrist) and a string. Somebody has already put this case.
Let us put a dynamometer between weight and string. It will show the centripetal f. that is producing the rotational movement (the dynamometer pulling the weight).
But the ACTION AND REACTION principle says that if that mentioned force exist,  the weight is also pulling the dynamometer with another opposite and equal force. That is a REAL force, and CENTRIFUGAL.
By the way, the same could be said in relation with the knot between the string and the dynamometer ... And this instrument functions with two opposite forces applied at its extremes. At the inner one it would be the centripetal force (the string pulls inwards the dynamometer ), and at the outer one the centripetal force (the weight pulls outwards the dynamometer)"
Should you accept this for me cristal clear question, please let me know. Then I would suggest which of my last year post could more easily make you understand my vision of sea tides, because when the pull is a "tele-pull" (gravity), instead of direct through a string, the application of the action and reaction principle is rather trickier.
Title: Re: Why do we have two high tides a day?
Post by: evan_au on 03/10/2016 23:04:30
Quote from: ProjectSailor
I have a book that accurately tells me in advance the heights of all the tides in a year and the times.. and are accurate within reason (weather effects have a great deal to play in some areas). If you think we don't understand them at all.. how is this possible?
They don't generate those books by calculating the tides from first principles.

Observers use Fourier Analysis, and by observation over a long period they work out the relative frequency, amplitude and phase of each contributor to the tide at a particular port; some observers have identified over 300 contributors (of which the position of Sun and Moon are major contributors), although fewer are relevant at any individual port. By using these historical Fourier coefficients, they can predict future tides.

See: https://en.wikipedia.org/wiki/Theory_of_tides#Harmonic_analysis

Quote from: Wiz
The bulge on the opposite side of the earth
The bulge of ocean water does not stay on the opposite side of the Earth from Moon or the Sun, for two reasons:
1) The North-South continents are in the way, and
2) Even if the Earth had no continents, and was an ocean of uniform depth, calculations suggest it would take 30 hours for this bulge to propagate around the Earth, not the slightly over 12 or 24 hours as observed.

In fact, the tidal bulges (there are several) remain in their own ocean basin, and rotate (roughly) once per day or twice per day, depending on the local geography.

Attraction of Sun & Moon drive these tidal bulges; whether it is primarily driven by the diurnal (roughly 24 hours) or semi-diurnal period (roughly 12 hours) depends on the resonant frequency of the individual ocean basin.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 04/10/2016 11:43:54
#56
I consider you are right that tidal bulges depend on many more facts than Moon and Sun interactions with Earth, and that N-S large continents are very important global factors.
But within E-W large water basins the (I would say) nş1 bulge, due to lunar attraction, logically happens bellow the Moon. But the very high tangential speed due to Earth´s rotation (40,000 km/24h) makes impossible that position, and it puts the bulge some distance from Moon´s vertical, towards East.
And Moons attraction makes that bulge to be always changing its longitudinal position. We could say the bulge is continuosly trying to get a bellow Moon position (not considering coastal effects) 
That´s why, f.e., with new Moon (Sun´s effect adds to Moon´s) there is always a high tide some time after 12:00 solar time.
So I consider that:
"2) Even if the Earth had no continents, and was an ocean of uniform depth, calculations suggest it would take 30 hours for this bulge to propagate around the Earth, not the slightly over 12 or 24 hours as observed"
could not be correct. For mentioned "nş1" bulge it would take 24 h + app. 50 min., this delay due to Moon´s eastward movement in 24 h. (by the way, the same dayly delay of tides in not rare locations).
Title: Re: Why do we have two high tides a day?
Post by: evan_au on 04/10/2016 21:52:51
Quote from: rmolnav
nş1 bulge, due to lunar attraction, logically happens below the Moon
This implies that the nş1 tidal bulge should start from the Eastern end of an ocean basin and move to the Western end, following the Moon.
- But what happens when this tidal bulge do when it reaches the western end? Where does all the water go?

They really didn't know the details, until the arrival of global ocean radar mapping from satellites*.

What actually happens is that if you look on a world map, the tidal bulge circulation tends to go anticlockwise in the Northern hemisphere, and clockwise in the Southern hemisphere. (This might suggest some influence from the Coriolis effect?) The bulge returns to the place where it started.

These tidal bulges circulate around a region in the open ocean that has a very low tidal range, or around an island (there is a particularly nice circulation around New Zealand).

The tidal variation in the Indian ocean actually splits in two, with half going east-to-west, and half going west-to-east. Nothing quite so simple as following the Moon...


If that doesn't work, try www.youtube.com/watch?v=5zi7N06JXD4
*Ocean-watching radar may have been promoted by the desire of the military to detect submarines via gravity waves (gravity waves are much easier to generate and detect than gravitational waves). But the rest of us benefit from having better visibility of weather at sea, and better weather prediction on land.
Title: Why are there two high tides a day?
Post by: chris on 04/10/2016 23:25:50
This answer is wrong. The bulge on the opposite side of the earth is caused by a centripetal force created by the earth's rotation, which is in effect trying to throw the oceans off into space.

What?

A centripetal force acts towards the centre of rotation...
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/10/2016 18:39:56
#59
You are quite right. Centripetal force obliges considered objects to follow a circular path, instead of a straight line.
But those objects (o parts of matter, such as water particles) somehow have to exert an opposite and equal force on the prime mover (the one that causes the centripetal force), following action and reaction principle.
I say "somehow" because there are many ways, depending on each particular case. In the case of the 2nd high tide (opposite to the Moon) is rather tricky.
I would suggest you to read my reply #55, where a very simple case is exposed.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/10/2016 19:17:21
#58
"- But what happens when this tidal bulge do when it reaches the western end? Where does all the water go?"
Tides are kind of very soft waves. High tide "excess" of water is not moving (relative to earth) horizontally, water level basically just oscillate vertically. Forgetting local effects, wether included, that is the result of all acting forces, mainly weight, lunar and solar attraction, and centrifugal forces.
When western basin ends, that physical limitation changes the scenario. Tidal effects continue on solid earth, but deformations are logically much, much smaller. And water tends to go back to a spherical shape.
Then, depending of each case, quite interesting phenomena (as what mentioned by you) can occur. But next day, when when Moon reaches the eastern edge of the basin, high tide and the bulge "chase" of the Moon occur again.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 06/10/2016 11:23:42
#58 (continuation)
We should also keep in mind that we say "bulge", but this term is just a massively exaggerated analogy ... Sea surface deformation is relatively very, very tiny. Even strongest tides are just in the order of app. a couple of tens of meters. Locally it could be consider something big, but that hight decreases very, very slowly from high tide place to low tide places, app. 90 degrees eastward and westward the Moon, along 10,000 km ... (at the equator) (!!!)
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 01/12/2016 12:08:44
Yesterday I sent some comments to a Scientist I am discussing with, relative to cause of sea tides, and centrifugal forces.
As it could interest other people, I am going to put them here.
Previously I have to remember something additional, necessary to understand my arguments.
Many people don´t realize that "the Moon rotates around the Earth" is not 100% exact … Cosmic rotating objects actually rotate in pairs, around their barycenter, kind of center of gravity of whole masses of the pair.
If one of the objects is much more massive, then they are still orbiting about their barycenter, but the “center of mass” is located inside the larger object, at some distance from its C.G. That is the case of Moon "around" the Earth:
"It is quite clear, for me too, that the addition of all forces of attraction from the Moon, what produces is the required centripetal acceleration for the whole of the Earth.
But, as far as I can understand, that is not the whole picture.
I see you yourself have been very close to the "edges" of my stand when saying:
"If you're on the side of the Earth facing away from the Moon, you are being pulled slightly less than the Earth as a whole, so your centripetal acceleration due to the Moon is a bit LOWER".
Please kindly note those places further from the Moon are also further from the axis of rotation. And required centripetal acceleration, for same angular speed as the whole Earth, must be actually HIGHER, proportionally to the radius.
The bulk of our planet kind of obliges both closer and further parts to rotate at same angular speed.
Any portion of Earth, either liquid or solid, if only with the rotation movement, has to satisfy the law that the addition of all force vectors acting ON it, divided by its mass, has to be the required centripetal acceleration.
That only can be reached through additional forces, resulting from internal stresses.
Let us imagine a narrow cylinder connecting closest to and furthest from the Moon Earth surface points (I suppose isotropic material and perfect spherical Earth, for the sake of simplicity), and let us divided it into e.g. 1 m pieces.
Forces acting laterally on those pieces would be equilibrated.
And longitudinally acting forces on each of them would be own weight (attraction from the whole Earth), attraction from the Moon, and tensile or compressive stresses/forces exerted by contiguous pieces. Shear stresses would also be present, but for simplification I won´t take them into consideration.
Let us consider four contiguous pieces A, B, C and D at furthest half of the Earth, A the closest to Moon and axis of rotation.
Between contiguous pieces stresses/forces are opposite but equal (3rd Newton´s Principle).
Required centripetal acceleration of B is slighty bigger than A´s, and slightly smaller than C´s. Considering only changes in comparison with when no rotation, one can deduce that the addition of forces on B (exerted by A and C) has to be centripetal, but smaller than what exerted on C by B and D, because the further from rotation axis, the bigger Moon´s attraction "deficit".
That means that every pair of contiguos pieces are pulling each other, half of those forces inwards (centripetal), and the other half outwards, for me CENTRIFUGAL forces (because they are in the sense of fleeing from a center).
I know that word is considered "politically incorrect" among many Physicists … But, whatever we call them, they are REAL forces, which apparently have been obliterated by many.
And I can´t really understand why! Those forces are present across the universe: planet/satellites, star/planets, twin stars … For me, Inner bulges are due to the higher attraction from the other massive object, but at outer bulges those centrifugal forces are paramount ...
Title: Re: Why do we have two high tides a day?
Post by: Janus on 02/12/2016 18:36:14
In pdf linked in reply #23, I can see that Schutz also mentions only the different gravitatory pulls from the Moon …
I repeat the question on my post #22:
"If those three zones of the Earth experience only those gravitatory accelerations from the Moon, why ALL of them don´t cause a movement of the Earth towards the Moon? “
Tomorrow I´ll try to explain my “theory” of the centrifugal force used in my first post, which I consider necessary to have a correct explanation of the issue. I´m slow when typing in English, and have no time now. But just an analogy that I consider useful:
Imaging an athlete of hammer trow speciality. He or she can´t keep verticality when throwing the hammer. It is necessary to lean a little backwards. Otherwise the hammer could not be rotated. Both the athlete and the hammer will rotate around an axis situated near the forward part of the athlete´s body.
If the hair of the athlete is long and not  fixed by some device, instead of keeping its normal downward direction due to its weight, it will move back and upwards …
That cannot be due to anything similar to what stated by Schutz relative to tides: CENTRIFUGAL force is the cause.

Both the Earth and Moon are in free fall.  So would be an Earth-Moon pair falling directly towards each other.  IOW, if you were to stop both the Earth and Moon in their orbital motion, so that they suddenly started to fall directly towards each, the tidal bulges would not change in shape at that moment.  If the Moon had been falling in towards the Earth from some far distance, at the moment it reached the present Earth-Moon distance, the tides would be the same as they would be for the Moon orbiting at that distance. The relative motion of the Earth and Moon with respect to each other doesn't matter.
Title: Re: Why do we have two high tides a day?
Post by: evan_au on 03/12/2016 06:33:44
Quote from: Janus
If the Moon had been falling in towards the Earth from some far distance, at the moment it reached the present Earth-Moon distance, the tides would be the same as they would be for the Moon orbiting at that distance.
Resonance is one factor in the height of tides - the most extreme tides in the world (eg near Anchorage in Alaska) occur because the resonant frequency of the ocean basin is close to the driving frequency of the Earth's rotation (slightly over 12 hours between high tides).

If the Earth & Moon were falling directly towards each other (without the Earth rotating), these resonances would not occur, and these instances of extreme tide heights would not occur.

Similarly, there are regions in the middle of Earth's oceans which have almost zero tidal range (amphidromic points (https://en.wikipedia.org/wiki/Amphidromic_point)), because the oscillations cancel out at that point.  If the Earth & Moon were falling directly towards each other, these oscillations would not occur, and I expect that the distribution of amphidromic points would be quite different.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 06/12/2016 11:36:51
#65 evan_au
On previous posts I´d already said that local circumstances, resonance mentioned by you included, does affect the hight of tides.
But that doesn´t mean what I said on #63 to be main causes of tides is erroneous ...
I saw both #64 and 65 just some minutes ago. Regarding what said by Janus on #64, I´ll reply at another moment. I need more time, firstly to "ruminate" about it, and after that to write the reply.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 06/12/2016 11:57:45
#66 (Continuation)
Just after posting #66 I´ve realized what you said about resonances has a basic error. Janus, I presume, was "stoping" the rotation of the pair Earth/Moon. But that rotation (around their barycenter, 28/29 day cycle) is not which allows the resonances mentioned by you. It is the DAYLY Earth´s rotation about its own axis !!
Title: Re: Why do we have two high tides a day?
Post by: evan_au on 06/12/2016 19:46:05
Quote from: Janus
The relative motion of the Earth and Moon with respect to each other doesn't matter.
Quote from: rmolnav
the resonances mentioned by you. It is the DAYLY Earth´s rotation about its own axis !!
I agree with you, rmolnav. The Earth's rotation relative to the Moon is the major driver for tides (followed by the Earth's rotation relative to the Sun).
- If the Moon were "far away", it's tidal pull would be small in amplitude, but would occur every 12 hours, as the Earth rotates.
- At its present distance (300,000 km), the Moon's tidal pull is larger than that from the Sun, but it occurs about every 12 hours and 20 minutes, as the Moon has rotated in its orbit in that 12 hours, and it takes a bit longer than 12 hours for the Earth to rotate to it's original position relative to the Moon. This changes the resonances a bit.
- If we imagine the Moon only 100,000 km away, the tidal range would be huge (27x higher = inverse cube law), but the high tides will be a lot longer than 12 hours apart, and will have very different resonances.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 07/12/2016 09:01:57
#64 Janus and #68 evan_au
After "ruminating" Janus´s post last night, a few minutes ago I saw evan_au´s ...
On last one I´ve already seen some details I can´t agree with, but some more time is necessary to prepare and post a well thought reply.
Janus: For a couple of year I´ve been posting tens and tens of replies and/or comments, trying to convey my stand step by step, giving well recognized physical principles I support my statements on, pinning down others´concrete statements I can´t agree with, drawing parallels with several practical cases ... Both here and on  "http://www.thenakedscientists.com/forum/index.php?topic=68025.50", "What is centrifugal force?".
Just as an example, as an introduction to a physical analysis of pendulum´s movement, I posted on last one (#60):
"It seems simple, but there are small but important details that, if not being careful, one can misinterpret facts, or at least confuse others …
That´s why I am now going to consider only the scenario statically (and step by step), without any movement: fixed hanging point, string and weight (I´ll call it W), all in a straight vertical line.
- a) Primary acting force: Earth pulls downwards W.
- b) W does not move. According to 2nd Newton´s principle, the sum of all forces acting on W has to be null.
- c) The unique object that can exert another force on W is the string: it must somehow pull W, with equal but opposite force (watch out: those two opposite forces are not action/reaction forces (3rd principle); they are acting on a unique object, and 3rd principle is about two objects exerting a force on each other).
- d) If the string pulls upwards W, applying now 3rd principle to that pair of objects, we can deduce that W must be pulling downwards the string lower extreme.
- e) That force seems to be a centrifugal one (and centripetal the one mentioned on - c)), but we should keep in mind that if there is no rotatory movement at all, a proper “center” does not actually exist.
- f) And when with movement, we have also to be careful with the term “centripetal force”, because it usually refers to the the radial component of adding up all forces acting on W,  which divided by the mass would give us the “centripetal acceleration” that makes W not to follow a rectilinear trajectory. And in many cases some of those added forces may be in the sense of the “center”, but compensated by others and not producing any acceleration by themselves".
AND YOU, not quoting anything from my last post here but from another one and a half year old, without specifying any concrete steps of my arguments you don´t agree with, and without giving any physical principle to support your stand, JUST SAY:
"Both the Earth and Moon are in free fall.  So would be an Earth-Moon pair falling directly towards each other.  IOW, if you were to stop both the Earth and Moon in their orbital motion, so that they suddenly started to fall directly towards each, the tidal bulges would not change in shape at that moment.  If the Moon had been falling in towards the Earth from some far distance, at the moment it reached the present Earth-Moon distance, the tides would be the same as they would be for the Moon orbiting at that distance. The relative motion of the Earth and Moon with respect to each other doesn't matter"
PLEASE KINDLY note that if I´d just said something is white, you could say I´m wrong, that it is black ...
But It would be OF NO USE, to you, me and any other reader !!   
   
Title: Re: Why do we have two high tides a day?
Post by: Bill S on 07/12/2016 15:47:55
Fascinating link,  Bill. Having lived by the sea - Cornwall,  then East Anglia (UK) - all my life,  I've long been aware that tides were not as simple as the two bulge model suggests, but I've never tried answering my own questions.  There's some food for thought in your link.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 07/12/2016 22:19:46
#68 evan–au
You say:
" If the Moon were "far away", it's tidal pull would be small in amplitude, but would occur every 12 hours, as the Earth rotates".
Sorry but I´m afraid you are mixing things ...
Firstly, there are neither two nor one Moon tidal pull ... Its pull is continuous, no matter how far away the Moon were.
Two so called bulges always happen. And they continuously change their longitudinal position with Earth 24 h. own rotation.
It is clear that the one app. below the Moon is due to higher attraction there, due to less distance to the Moon.
Many people consider the opposite bulge to be also due to similar reason: that part is further from the Moon, pull there is smaller than mean pull to the whole Earth, and water there kind of falls behind ...
I feel pretty sure that those differences in gravitational pulls are not sufficient to explain further high tides. Here and on "http://www.thenakedscientists.com/forum/index.php?topic=68025.50" I have explained many, many times my reasons.
I neither can repeat now all that staff, nor ask you to read it.
Nevertheless, I´ll put just a few condensed lines.
The whole Earth has to rotate around Moon/Earth barycenter at 2PI/app.28 d. angular speed. Moon´s pull on the whole Earth produces the centripetal force for that rotation. That means centripetal force matches distance from C.G of Moon and Earth.
But further Earth parts, on the one hand they are being pulled less by the Moon. And on the other, being required centripetal acceleration proportional to distance to the barycenter, to rotate at same angular speed centripetal force should be higher ...
That "deficit" of centripetal force can only be compensated by internal stresses/forces. The result is that "rows" of material between bulges are being stretched by internal tensile stresses, two opposite between each pair of contiguous particles (according to 3rd Newton´s Principle).
Half of them are inward and compensate centripetal force deficit, and the other half are outward (I consider them to be centrifugal, in the sense of fleeing from a center), and produce both earth deformation and further high sea tides ...
Rest of #68, as far as I can understand, has other errors, but I´ll refer to them tomorrow on another post.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 08/12/2016 12:30:49
#72 (Continuation)
evan_au said on #68:
"At its present distance (300,000 km), the Moon's tidal pull is larger than that from the Sun"
If with "Moon´s tidal pull" you mean its gravitational attraction, that is not right. Even being the Moon much closer, the Sun is so much massive that its pull is higher.
What is actually larger is the centripetal force "deficit" (see #71) at Earth´s parts further from the barycenter, that happens to lie within Earth, app. 2/3 of its radius away from its C.G. That is due to the fact that in Earth´s rotation around the Sun, required centripetal acceleration at furthest part (midnight) is relatively much more similar to centripetal acceleration of the whole Earth. And gravitational pulls are more similar too.
We could say that four "bulges" actually happen: two visible related to the Moon, and two smaller related to the Sun, but not visible because we see sea surface deformation due to both effects added up.
App. twice a month, when the three celestial objects form a line (full or new Moon), "non visible" Sun related bulges are at same longitudinal position as Moon related ones, and we have so called spring tides (in Spanish, literately translated, "live tides"), with tide´s range at its maximum (visible and invisible bulges add up). When instead of in line they are 90ş apart (Moon at one of its quarters), Sun related "invisible" bulges are situated where Moon related low tides, and we have "neap tides" ("dead tides in Spain), with tide´s range at its minimum.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 10/12/2016 11:59:04
Last couple of months I´ve been discussing our "issue", and also the sheer concept of centrifugal force, on "live science". On some of their last web pages they had shown the idea I´m refuting that centrifugal f. is a kind of taboo, something not real …
Some minutes ago I sent them:
It´s funny. Just a few minutes ago I saw that Lauren Cox, more than six years ago, on a page ALSO FROM LIVESCIENCE,
http://www.livescience.com/29621-what-causes-the-tides.html
showed a stand very, very similar to mine (linking cause of tides to REAL centrifugal forces):
"The moon's gravitational pull on the earth is strong enough to tug the oceans into bulge. If no other forces were at play, shores would experience one high tide a day as the earth rotated on its axis and coasts ran into the oceans' bulge facing the moon.
… As the moon circles the earth, the earth moves in a very slight circle too, and this movement is enough to cause a centrifugal force on the oceans.
… This inertia, or centrifugal force, causes the oceans to bulge on the opposite side facing the moon ..."
But I wouldn´t say "This inertia, or centrifugal force ...". It is a real, "inertial" force, in the sense that it is due to 3rd Newton´s Principle, based (as the other two) on inertia physical REALITY."
http://www.livescience.com/52488-centrifugal-centripetal-forces.html
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 13/01/2017 12:28:36
#70 Bill S
Perhaps I should have commented your post before ...
In my mother tongue (Spanish), "fascinante" sounds perhaps too much ...
Anyhow, thank you for what could affect my own interventions. If any doubt or disagreement, please send them. I´ll do my best to try and clear our ideas.
Recently I´ve sent a couple of comments to another related topic. I think it could be of your interest:

"Will we eventually lose the moon?"
https://www.thenakedscientists.com/forum/index.php?topic=36739.msg461127#msg461127 (https://www.thenakedscientists.com/forum/index.php?topic=36739.msg461127#msg461127)
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 15/11/2017 12:22:07
Looking for something else in Youtube, a couple of days ago I found a VERY interesting video, which thoroughly analyzes both tides basic causes (globally considered: Moon, Sun ... the two "standard" bulges), and local tides. Don´t miss it!
Said that, its authors are among the certainly big group of scientists that don´t want even mention centrifugal forces ... Short time ago I sent a comment explaining why I consider they are wrong, as far as that concrete issue is concerned:
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 17/11/2017 12:00:30
Said that, its authors are among the certainly big group of scientists that don´t want even mention centrifugal forces ... Short time ago I sent a comment explaining why I consider they are wrong, as far as that concrete issue is concerned
In my last post I intended to include just the link to the youtube page, but the video "play" engine (?) turns up …
But what don´t appear are sent comments …
What follows is what I sent, not literally (I had a lapsus and had to correct some details)
Both Moon and Sun, causes of tides, are similarly considered in the video, without even mentioning centrifugal forces.
As we know, Moon is main cause (particularly in timing), and Sun tides affect total tides only in their cycling intensity.
But Moon/Earth dynamics is rather trickier than Sun/Earth´s. as far as distribution of forces is concerned. That is why I´ll refer to last one before.
Diagram timed in the video 2:23 is about Moon related bulges. Let us imagine a similar diagram, but with the Sun instead of the Moon. Shown "medium"  gravitational pull (average as if acting on Earth´s C.G.) is rested both at nearer and farther sides, trying to explain the bulges ...
The authors don´t consider those vectors any centrifugal force, they´re just comparing to find kind of "relative bulges" caused by "relative forces". In other words, the usual concept of inertial frame of reference, which I consider applies to distances, velocities and accelerations, but that applying it to forces is an erroneous "trick". But I will not discuss that now any further. 
What´s causing "medium" black vector at C.G.? : the centripetal acceleration required for the rotation. But that REQUIRED (to orbit at each distance) acceleration is NOT uniform across the Earth, horizontally in our diagram. Farther parts of the Earth, to rotate at same angular speed (an average of 2π radians a year), NEED higher centripetal acceleration than closer ones: just the OPPOSITE to what happens with gravitational forces.
ONLY internal stresses can compensate that imbalance ... Thanks to them, the excess of pull on closer parts is transmitted, from right to left in the diagram, to farther parts ...
But according to Newton´s 3rd Motion Law, if at any considered Earth´s section, right part is pulling (towards the Sun) contiguous left part, this one is also pulling the former towards the left, same force but opposite direction: OUTWARDS.
All those internal forces are centrifugal REAL forces, which, together with their "mirror" centripetal ones (the fraction not necessary for the rotation), stretch the Earth, solid parts included.
Logically, that deformation is much bigger where liquid parts, our oceans. And the TWO Sun related bulges ARE DUE TO THAT.
In another post, better tomorrow, I´ll put what Moon related …This way could be better for some of possibly interested folks: t could facilitate a posterior understanding of the trickier case of Moon related bulges.

Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 18/11/2017 12:25:23
ONLY internal stresses can compensate that imbalance ... Thanks to them, the excess of pull on closer parts is transmitted, from right to left in the diagram, to farther parts ...
AS A CONTINUATION of my post of yesterday, yet another analogy, to help understanding.
Yesterday I happened to have to buy a new tyre for my byke. When I had it hanging from my right hand, I felt how heavy it was … and saw some deformation: its circular shape got slightly oval.
What physically was happening is clear: gravity was acting across the tyre, but my still hand was preventing it from getting a g downwards acceleration. My hand was pulling upwards, and total force acting on the tyre was null (forgetting consequences of Earth´s movements …)
But my hand upward force was neither acting on the tyre´s C.G., nor uniformly distributed across the tyre (as gravity was). Similarly to what I said yesterday, through internal stresses my hand´s pull was being transmitted and distributed along the tyre, acting both upwards and downwards (Newton´s 3rd Law), each part of the tyre was still, but the tyre was being stretched vertically.
Then I tried and spinned myself with my arm almost horizontal, the tyre still gripped by my hand, as if I were going to throw it like a disc by an athlete.
I was feeling I had to pull the tyre inwards, the higher the angular speed, the higher the necessary pull. My hand was making rotate actually only what it was touching, but required force was much bigger than if my hand were gripping only some ten cm piece of tyre ...  Rest of the tyre rotated because, almost identical reason as what quoted, "the excess of pull on closer parts was being transmitted, outwards in this case, to farther parts …"
In Earth rotation around the Sun I was talking about yesterday, centripetal force is not initially exerted through physical contact at a spot as in the tyre case…
Gravitational pull is distributed over Earth sphere. Not having a pull "concentration" at any spot, "excesses of pull" are much smaller, but, as explained yesterday, they also happen. And also centrifugal REAL forces appear, due to mentioned Newton´s 3rd Law.
Because of that, deformations due to all those forces (particularly tidal bulges) have to be relatively much smaller than in the tyre case. And that is what really happens. Sun related bulges (by the way, very much exaggerated term …) are only a few meters from high and low tides … 10,000 km apart !! (at the equator).
 
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 19/11/2017 11:45:41
In another post, better tomorrow, I´ll put what Moon related …This way could be better for some of possibly interested folks: t could facilitate a posterior understanding of the trickier case of Moon related bulges.
Here it is ... though a day later due to additional analogy of yesterday (hope useful).
Mentioned video diagram timed 2:23 is misleading: it seems like a static situation … Authors not only ignore any possible centrifugal force: they don´t even mention any centripetal force. If only what shown were the physical reality there, why Earth doesn´t moves toward the Moon?. Because of same reason as Moon doesn´t move towards the Earth: both are actually rotating around their barycenter, and gravitational pulls are causing their centripetal accelerations. 
As that barycenter is situated app. 2/3 Earth´s radius from its center (logically, at Moon side), Earth´s rotation in this "couple dance" is a kind of wobbling, similar to when a child plays with a hula-hoop.
Though Earth daily spinning generates much, much stronger internal stresses and centrifugal forces (some 28/29 times higher angular speed …), it acts always same way: trying and increasing Earth´s diameter at the equator and lower latitude parallels.
That Earth´s wobbling generates internal stresses and deformations which we can perceive added to those mentioned stronger, permanent ones.
Let us imagine video diagram timed 2:23 with a vertical line some 2/3 Earth´s radius at right side of its center: the axis of this rotation.
Earth parts at the left of that axis require a total force per unit of mass proportional to its distance to that axis, in order to get required centripetal acceleration at ach considered spot. But gravitational pull from Moon is the farther the smaller (to the square of the distance to the Moon). That imbalance can be compensated only by internal stresses: parts closer to the axis have to add an inward pull on contiguous farther parts … Then we have Newton´s 3rd Law, and each part pulls inner contiguous one with same but opposite (and REAL) force, that is, outwards, "fllying from a center" (axis of rotation), CENTRIFUGAL …
At Earth parts at right side of that axis, required centripetal forces are smaller, also the farther from the axis, the bigger. BUT now gravitational pulls from the Moon are opposite to required centripetal forces. Only own gravitational Earth´s pull can supply required centripetal forces …That´s similar to what happens due to daily Earth´s rotation: it tends to increase sea level at that spherical segment. And that effect adds to what derived from the obvious fact that there Moon´s gravitational pull is bigger, and a bulge also occurs at that side, due to both reasons ...
And due to Earth´s daily spinnig, those two opposite bulges seem to rotate around Earth, trying to keep in line with Moon. Result: the main part (Sun caused tides have to be added) of our "popular" (but not so well known) tides …

 
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 12/12/2017 11:27:33
Let us imagine video diagram timed 2:23 with a vertical line some 2/3 Earth´s radius at right side of its center: the axis of this rotation.
Earth parts at the left of that axis require a total force per unit of mass proportional to its distance to that axis, in order to get required centripetal acceleration at ach considered spot. But gravitational pull from Moon is the farther the smaller (to the square of the distance to the Moon). That imbalance can be compensated only by internal stresses: parts closer to the axis have to add an inward pull on contiguous farther parts … Then we have Newton´s 3rd Law, and each part pulls inner contiguous one with same but opposite (and REAL) force, that is, outwards, "fllying from a center" (axis of rotation), CENTRIFUGAL …
I wish we ALL had a better knowledge of nature, or at least more imagination ...
I´ve long been arguing in line with what above, and neither myself, nor anybody else so far (unless they kept it "in secret") have imagined what below, and realized that what quoted is WRONG !!
Some people actually said it was wrong, but not for the correct reason. Most of them just considered that centrifugal forces are never real, but only apparent of ficticious ... And I gave many examples of REAL centrifugal forces ...
But recently I learnt that the 28/29 days circular movement of the couple Earth/Moon around their barycenter, though for the Moon is a rotation, for the Earth is NOT: Earth only REVOLVES around mentioned barycenter ...
That implies that all Earth points, center of mass included, follow equal circular paths.
Subsequently, my explanation, that would be valid for the Moon (it rotates at 2π radians/some 28 days angular speed), is NOT for the Earth !!
And possible imbalances between Moon´s pull on each point (either where solid or liquid parts), that could cause centrifugal forces, are not so straightforward to tell (?) ...
I´m trying to sort that out, but not easy ... For all my discussions on the subject, I´ve read many, many articles. But I NEVER saw any reference to what above !!
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 15/12/2017 10:50:49
But recently I learnt that the 28/29 days circular movement of the couple Earth/Moon around their barycenter, though for the Moon is a rotation, for the Earth is NOT: Earth only REVOLVES around mentioned barycenter ...
That implies that all Earth points, center of mass included, follow equal circular paths.
Subsequently, my explanation, that would be valid for the Moon (it rotates at 2π radians/some 28 days angular speed), is NOT for the Earth !!
And possible imbalances between Moon´s pull on each point (either where solid or liquid parts), that could cause centrifugal forces, are not so straightforward to tell (?) ...
I´m trying to sort that out, but not easy ...
As a continuation of #80, I can say that, as far as I can understand, I´ve already sorted it out ...
Another day I´ll further elaborate on the issue, but now I´m going to put here main points.
Curiously, as Earth only revolves around the barycenter, all particles of the Earth follow equal circular paths, its center of mass included, logically at same angular speed (2π radians every some 28 days). So, centripetal accelerations required for the circular movement of each particle are the same in value, but the direction of each one is towards the center of its circular path.
As those accelerations can´t match with Moon´s pull on each particle, there is an imbalance, similar (but more complex) to what in #79. That imbalance only can be countered by internal stresses. If we analyze them along line Moon´s C.G./barycenter/Earth´s C.G., and similarly orientated lines, we find (as in #79) that particles closer to the Moon have to add a pull on contiguous farther parts … And due to Newton´s 3rd Law, each part pulls inner contiguous one with same but opposite (and REAL) force, that is, outwards from Moon.
Whether those last forces should be called "centrifugal" is arguable ... But they, in any case, together with all Moon´s pulls on each particle,  stretch considered line of particles, both where solid Earth, and where water.
Other similarly orientated lines of material suffer similar stretches, but smaller ...
Deformations were solid Earth depend on elasticity. But water logically is more free to move, and that´s the cause of high tide so called "bulges".
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 20/12/2017 07:58:22
Curiously, as Earth only revolves around the barycenter, all particles of the Earth follow equal circular paths, its center of mass included, logically at same angular speed (2π radians every some 28 days). So, centripetal accelerations required for the circular movement of each particle are the same in value, but the direction of each one is towards the center of its circular path.
As those accelerations can´t match with Moon´s pull on each particle, there is an imbalance, similar (but more complex) to what in #79. That imbalance only can be countered by internal stresses. If we analyze them along line Moon´s C.G./barycenter/Earth´s C.G., and similarly orientated lines, we find (as in #79) that particles closer to the Moon have to add a pull on contiguous farther parts … And due to Newton´s 3rd Law, each part pulls inner contiguous one with same but opposite (and REAL) force, that is, outwards from Moon.
I think I¨ve already managed to grasp the actual mechanism of internal stresses in cases, like Earth attracted by Moon, when the object revolves, but doesn´t rotate.
ALL MATERIAL POINTS not only follow equal circular paths, logically at same speed, but even they ALWAYS are at farthest (from Moon) point of their own path …
Then, ALL centripetal forces required for the circular movement of each particle are equal both in value and in direction: ALWAYS parallel to line Earth´s C.G./barycenter/Moon´s C.G., and towards the Moon.
If we imagine the revolving object cut into slices (all its material equally far from Moon), and analyze in detail all forces (gravitational, interactions, and the ones required for their circular movements, ALL PER UNIT OF MASS), we find that mentioned "differential gravitational forces" are what remain kind of "free" (not necessary for the circular movements), and either stretch solid Earth, or move water (and change pressure) towards the Moon (where excessive Moon´s pull), or in the opposite direction (where excessive centrifugal force).
Being all centripetal forces equal, "excessive" pull at closer slices is transmitted outwards (at same  rate as Moon´s pull decreases) ... Between each pair of contiguous slices opposite pulls occur (Newton´s 3rd Law), that could be arguably called centripetal and centrifugal ...
So, "differential gravitational forces" tool (usually called "tidal forces") kind of hides a "chain of transmission" of forces between contiguous particles, centrifugal forces included, what affect the whole revolving object.
I used to think "how on Earth so many people, even scientists, say those differential gravitational forces are the unique deep cause of tides, without even mentioning any circular movement, let alone centrifugal forces ?? ... Centrifugal forces HAVE TO be taken into consideration !!!"
And THEY WERE, but without most people (including me) noticing ... !!!
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 20/12/2017 20:29:49
As a continuation of #82, just a detail to avoid anybody could get mislead.
When talking about position of material points relative to the Moon, we have to keep in mind that what is being analyzed is exclusively the  Moon /Earth "dance" around  their barycenter... The daily spinning of Earth has to be disregarded.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 24/12/2017 18:37:13
If we imagine the revolving object cut into slices (all its material equally far from Moon), and analyze in detail all forces (gravitational, interactions, and the ones required for their circular movements, ALL PER UNIT OF MASS), we find that mentioned "differential gravitational forces" are what remain kind of "free" (not necessary for the circular movements), and either stretch solid Earth, or move water (and change pressure) towards the Moon (where excessive Moon´s pull), or in the opposite direction (where excessive centrifugal force).
Being all centripetal forces equal, "excessive" pull at closer slices is transmitted outwards (at same  rate as Moon´s pull decreases) ... Between each pair of contiguous slices opposite pulls occur (Newton´s 3rd Law), that could be arguably called centripetal and centrifugal ...
So, "differential gravitational forces" tool (usually called "tidal forces") kind of hides a "chain of transmission" of forces between contiguous particles, centrifugal forces included, what affect the whole revolving object.
As most likely not everybody buys what quoted, I´m going to put what follows, that I sent last year to a scientist by email, when discussing on the issue. I´m sure this analogy can help understanding:
"Surely you also played what in Spain we call "el látigo", literally in English the whip: a row of several boys, each left hand grasping right hand of contiguous inner boy … "Inner" because they run around one of them at a center, trying to keep the row in a radial line at a certain equal angular speed.
Let us consider five boys, A, B, C, D, and E, A being on a fixed point.
Let us suppose each boy hand-to-hand distance is 1 m (logically, the same between their C.G.), they weight of each 40 kg, and they rotate at 1 radian/sec.
Centripetal acceleration of E has to be 1x1x4 = 4 m/sec2. Centripetal force, 4x40 =160 newtons. It only can by exerted by right hand of D pulling inwards left hand of E.
Centripetal acceleration of D has to be 1x1x3 = 3 m/sec2. Required centripetal force: 3x40 =120 newtons. But he is "suffering" two opposite pulls: one outwards (centrifugal) exerted by E, of 160 newtons (3rd Newtons´Principle), and another inwards, exerted by C. If net force on D has to be 120 newtons inwards, right hand of C has to pull left hand of D with an inward force of 120+160=280 newtons.
Also due to 3rd Newton´s Principle, that means right hand of C is pulled outwards (centrifugal) by left hand of D with a force of 280 newtons ...
… C : 1x1x2 = 2 m/sec2 … 2x40 = 80 newtons … 280+80= 360 newtons.
… B : 1x1x1 = 1 m/sec2 … 1x40 = 40 newtons … 360+40= 400 newtons.
The result is boys, especially their arms, get stretched.
All those hand-to-hand transmitted forces are similar to what happens between contiguous pieces of material of above mentioned "columns" perpendicular to axis of rotation. Though in last case there is another important "source" of centripetal forces, Moon´s attraction, it can´t exert required forces along those columns, because Moon´decreases with distance, and required centripetal force increases with distance from axis of rotation.
There is also the difference that own Earth´s gravity effect, important for tides, does´t  significantly affect rotation of boys (energy wasted in friction could be reduced to a minimum with boys skating on ice). There are also other centrifugal forces due to daily Earth´s own rotation, which produces a permanent bulge around the equator.  But what exposed above is the real scenario in comparison with if Earth were just not rotating around Moon/Earth.barycenter, with the purpose of analyzing forces caused by that rotation alone.
The result of all those forcers is both a stretch of solid Earth´s parts, and the pseudo-equilibrium of oceans with its "theoretical" spherical shape very slightly changed, with two bulges".
By that time I had not yet learnt that Earth doesn´t actually rotate, only revolves. Then, the comments to the analogy should be changed.
On the one hand, in the game required centripetal forces increase proportional to the radius, but all particles of Earth follow identical circles, what requires same centripetal accelerations ...
On the other hand, in the game only "internal" stresses supply force for centripetal accelerations, but in our real case Moon gravitational pulls (on each Earth´s particle) are main cause of centripetal accelerations. But not uniformily ...
BUT, similarly to the game, between contiguous slices of any imaginary "column" of material parallel to line of centers of G. Moon/Earth, opposite pulls have to turn up to compensate the imbalance, in such a way that the net force acting on each slice (three vectors to add: Moon´s gravitational pull, and two opposite pulls from contiguous particles), divided by its mass, gives the required centripetal acceleration, the same across the Earth as previously said.
Then, centrifugal forces resulting distribution is kind of a "mirror" image of gravitational pull distribution. At Earth C.G. there are not "spare" forces (all Moon´s pull is spent in centripetal acceleration), but at closer and farther points that balance does´t occur: at closer points there is an "excess" of gravitational pull, and in farther points a deficit, what results in "spare" outwards forces ("arguably" centrifugal ... perhaps another day I´ll come back to that).
BOTTOM LINE: the well known diagram with average pull at C.G., bigger vector pull at closer side, and smaller one at further side, which after deducing average bigger vector changes direction to side opposite to the Moon ...
       
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 14/02/2018 08:51:36
Not having seen any further comment to my last posts, I imagine some of you agree with them, but surely many don´t buy my explanations really convinced.
As I continued discussing the topic in other sites, especially youtube pages, I´ve found some nuances to add to my explanations, to try and convey "my" (?) ideas ...
I do agree differences in gravitational forces (so called “differential g.f.”) are necessary condition to have real "tidal forces”, but not sufficient, let alone both concepts should be considered synonymous.
Let us imagine our planet were cut through into two equal parts, by a plane perpendicular to moon-earth line.
The center of mass of closer part would be obviously closer to moon than the farther one, and gravitational force acting on it would be higher.
IF NO OTHER REAL FORCE acted on them (apart from own gravity and internal stresses within each half), they would move towards the moon with different accelerations, following 2nd Newton´s Law.
But, on the one hand, their solid parts are kind of stuck, and they exert internal stresses on each other. And both solid and liquid parts of the two halves are attracted towards earth C.G.
On the other hand, both follow equal circular movements, as earth doesn´t rotate around moon-earth barycenter: it revolves.
Those two identical circular movements require equal centripetal forces, expected to be provided by moon´s pull.
But moon´s pull on closer half is bigger than necessary, and farther one is smaller.
The bulk of moon´s pull on closer half produces its centripetal acceleration. No centrifugal force associated to that fraction: 3rd Newton´s Law tell us closer half exerts an equal pull ON THE MOON. But the excess of moon´s pull on that half is transmitted to the farther half, same way we saw it happens between children at their "whip" game (see last post).
That transmitted pull is precisely what the farther half needs to get same centripetal acceleration, due to the fact that moon´s pull on that part is insufficient.
And  3rd Newton´s Law implies that farther half, apart from its own pull ON THE MOON (which couldn´t be called centrifugal force), also exerts an outward pull ON CLOSER HALF, as part of the reaction to its total centripetal force. That outward force is a real CENTRIFUGAL force: if it were not real, that closer half wouldn´t move the way it does: it would move towards the moon, getting separated from the other half !!!
Whatever the position of the imaginary dividing plane, similar things could always be said.
THOSE internal stresses are the real “tidal forces”: they stretch our planet. And water, being more “free” than solid parts, moves to form the two opposite bulges, as an “infinitesimal” deformation (a few meters of sea level difference some 10,000 km apart) of the spheric shape it would have with only own earth gravity ...
We have to keep in mind that moon´s effect, rather than a dynamic phenomenon, is a succession of static situations. If earth didn´t have its daily rotation, the bulges would form very, very slowly, because moon-earth “dancing” lasts 28/29 days per cycle … And that is what counts, if we analyze only that “dance”. Adding afterwards earth daily rotation, we have the tidal bulges trying to keep in line with moon´s position ...


     
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 19/02/2018 12:39:21
Seen from outside, as my stand refutes  what so many people (scientists included) say, most probably "should be" the wrong one ... But there are several details that could explain those so frequent, total or partially  errors:
1) Many folks don´t keep in mind Newton´s Motion Laws apply also to any part of objects, and internal stresses appear, as explained in my last posts (and long ago, e.g. #28)
2) Many ignore or forget that earth´s C.G. is actually following a continuously "curled" path around the sun: earth-moon barycenter is what follows an elliptical orbit.
And that implies additional curvature of earth parts (apart from what due to daily rotation), what required additional centripetal forces ...
3) And 3rd Newton´s Law reactions to those forces "may" be real centrifugal forces, as explained in my last posts.
But many consider those only as "apparent" and/or fictitious forces, that "could not" cause real effects such as tides. I´ve been refuting that since my first post here, #20 in May 2015 !!
By the way, a few months ago a scientist from NOAA, when discussing about a rather old publication of them about tides, told me:
"The publication you are referring to ... It used terminology of science and forces which were common in the 1950s.  Such as centrifugal force.  Centrifugal force was always an "imaginary force" (not a real / measurable force).  But that type of description made the concepts easier to understand and explain.  That  description and use of centrifugal force continued to be common practice until the 1970-80's.  At that point, the terminology shifted and the textbooks used in grade schools were changed to use a more modern terminology and description of this "effect" being a result of inertia rather than an "imaginary force"."
4) And, curiously, the mathematical operation of subtracting gravitational force at earth C.G. from local g.f. at each place gives correct results ... But NOT because that "artificial" operation is what actually happens ...
As explained in my last post, subtracting at each part the required centripetal force there from gravitational force, we get a kind of spare force, that is transferred from the considered material part to the rest of the earth, as internal stresses.
THOSE forces are what actually cause the tides.
BUT, as earth doesn´t rotate around the barycenter, only revolves, centripetal forces are the same across the whole earth. And as at earth C.G. gravitational force is equal to centripetal force (neither surplus nor deficit there), subtracting g.f. there gives same result as subtracting required centripetal force at each place.
Curiously, as moon is tidal locked to earth, it doesn´t revolve but rotates. There differences between gravitational forces and required centripetal forces are relatively higher, because last ones also vary with distance, in this case proportionally to the distance (in sense opposite to g.f.).
And differential gravitational forces mathematical "trick" would give erroneous values when analyzing tidal forces in the moon.       
Title: Re: Why do we have two high tides a day?
Post by: smoker42 on 22/02/2018 19:44:23
Because of the Moon
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 26/02/2018 07:38:51
That outward force is a real CENTRIFUGAL force: if it were not real, that closer half wouldn´t move the way it does: it would move towards the moon, getting separated from the other half !!!
Sorry. I meant “it would intend to move towards the moon ...” That centrifugal force would be much, much smaller than gravitational attraction between the two halves.
I was more precise before, when saying:
“IF NO OTHER REAL FORCE acted on them (apart from own gravity and internal stresses within each half), they would move towards the moon with different accelerations, following 2nd Newton´s Law”
By the way, even with mutual attraction, those accelerations would also be centripetal, changing the revolving fashion of the two halves. That attraction would initially keep them “together”, but they would slide on each other ...

Title: Re: Why do we have two high tides a day?
Post by: opportunity on 26/02/2018 08:04:43
If the tides are caused by the gravity of the moon, why is there a high tide on the side of the Earth furthest from the moon as well as on the closest side?

Peter Conway

Asked by Peter Conway


                                        Visit the webpage for the podcast in which this question is answered. (http://www.thenakedscientists.com/HTML/podcasts/naked-scientists/show/20140107/)

[chapter podcast=1000585 track=14.01.07/Naked_Scientists_Show_14.01.07_1001836.mp3](https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.thenakedscientists.com%2FHTML%2Ftypo3conf%2Fext%2Fnaksci_podcast%2Fgnome-settings-sound.gif&hash=f2b0d108dc173aeaa367f8db2e2171bd)  ...or Listen to the Answer[/chapter] or [download as MP3] (http://nakeddiscovery.com/downloads/split_individual/14.01.07/Naked_Scientists_Show_14.01.07_1001836.mp3)


As much as I want to read all the posts, the answer is simple.....the earth spins.....so when the pull is strongest with the moon, obviously there's a drag on the Earth, yet to keep the earth on a stable spin there has to be an equal counter-weight on the opposite side.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 26/02/2018 11:20:03
the answer is simple.....the earth spins.....so when the pull is strongest with the moon, obviously there's a drag on the Earth, yet to keep the earth on a stable spin there has to be an equal counter-weight on the opposite side.
The idea of the necessity of a counter-weight to keep stability is correct ... But:
1) The question is actually HOW that necessity is physically fulfilled.
2) If with "spin" you refer to the daily Earth rotation, that movement should be dismissed when analyzing root causes of tides. Moon-Earth couple movement is actually a cycle of 28/29 days. If Earth didn´t have its daily spin, the two tidal bulges would also occur. And the reason of that is what we are discussing.
Logically, as Earth does spin daily, mentioned bulges are continuously trying to keep in line with Moon, and that causes tides, but as a kind of secondary reason.
By the way, Earth daily spin causes the so called equatorial bulge, much, much bigger than the two tidal bulges we are discussing about: equator diameter is more than 40 km bigger than the distance between N and S poles !! Centrifugal forces are involved here too.
And Moon and Sun related tides occur on top of that deformed sphere.
Title: Re: Why do we have two high tides a day?
Post by: opportunity on 26/02/2018 11:32:53
the answer is simple.....the earth spins.....so when the pull is strongest with the moon, obviously there's a drag on the Earth, yet to keep the earth on a stable spin there has to be an equal counter-weight on the opposite side.
The idea of the necessity of a counter-weight to keep stability is correct ... But:
1) The question is actually HOW that necessity is physically fulfilled.
2) If with "spin" you refer to the daily Earth rotation, that movement should be dismissed when analyzing root causes of tides. Moon-Earth couple movement is actually a cycle of 28/29 days. If Earth didn´t have its daily spin, the two tidal bulges would also occur. And the reason of that is what we are discussing.
Logically, as Earth does spin daily, mentioned bulges are continuously trying to keep in line with Moon, and that causes tides, but as a kind of secondary reason.
By the way, Earth daily spin causes the so called equatorial bulge, much, much bigger than the two tidal bulges we are discussing about: equator diameter is more than 40 km bigger than the distance between N and S poles !! Centrifugal forces are involved here too.
And Moon and Sun related tides occur on top of that deformed sphere.

It's worked itself out, right?
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 14/04/2018 11:53:40
It's worked itself out, right?
When I first saw what quoted, I didn´t send any further comment because, rather than a Physics question, it´s kind of a Philosophy one ...
Somehow you are right, because nature works by itself, regardless of our actions and theories.
But, watch out, we are analyzing facts a posteriori. But if gravitational and Newton Motion Laws were different (or the initial conditions), things would have happened differently, and existing  equilibrium could have not been reached: the Moon, if existing as a single object at all, could have either gone away to the space, or fallen back onto the Earth ... Or tides could be quite different, or not be happening at all ...
And it´s quite reasonable to analyze how and why things are actually happening as they are. If everybody have had the attitude your comment implies, sciences would not have gone where they now are, and we could still be in the stone age !! 
Title: Re: Why do we have two high tides a day?
Post by: opportunity on 14/04/2018 13:05:10
I'll take that as a compliment  ;)

I don't have any tickets on myself. Coudos to your "attitude detector".

Apologies though for not explaining the physics of stable spinning objects. I wrongly thought it was already understood.

Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 26/04/2018 18:40:25
I wish we ALL had a better knowledge of nature, or at least more imagination ...
I´ve long been arguing in line with what above, and neither myself, nor anybody else so far (unless they kept it "in secret") have imagined what below, and realized that what quoted is WRONG !!
Some people actually said it was wrong, but not for the correct reason. Most of them just considered that centrifugal forces are never real, but only apparent or ficticious ... And I gave many examples of REAL centrifugal forces ...
But recently I learnt that the 28/29 days circular movement of the couple Earth/Moon around their barycenter, though for the Moon is a rotation, for the Earth is NOT: Earth only REVOLVES around mentioned barycenter ...
That implies that all Earth points, center of mass included, follow equal circular paths.
Subsequently, my explanation, that would be valid for the Moon (it rotates at 2π radians/some 28 days angular speed), is NOT for the Earth !!
And possible imbalances between Moon´s pull on each point (either where solid or liquid parts), that could cause centrifugal forces, are not so straightforward to tell (?) ...
I´m trying to sort that out, but not easy ... For all my discussions on the subject, I´ve read many, many articles. But I NEVER saw any reference to what above !!
I´m quoting that because a couple of weeks ago I saw some surprising prove that even eminent physicists can be (or could have been) as wrong as I used to be, in relation with that question of Earth´s revolving instead of rotating, in its "dance" with the Moon.
If somebody has any doubt about that, please consider what follows.
It´s a little tricky, because of the daily Earth´s spin, which not having anything to do with Moon/Earth dynamics, has to be dismissed.
Imagine our meridian is Greenwich, it is 6:00 p.m., and the Moon is at its zenith (subsequently in first quarter).
Next day at same time the Moon would be some 360ş/24 eastwards. That means that if Earth were not spinning daily, our meridian would be keeping its position relative to far stars, after that time moving in opposition to the Moon.
The same would happen in subsequent days.
So, Earth does revolve around Moon/Earth barycenter, it does NOT rotate ...
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 26/04/2018 20:07:15
Next day at same time the Moon would be some 360ş/24 eastwards.
Sorry for the lapsus. I should have said "... some 360ş/28 ..."
Title: Re: Why do we have two high tides a day?
Post by: Yusup Hizirov on 26/04/2018 20:24:29
If, the tidal hump on Earth was created by the Moon - that would not be an ellipse but a "drop". (The force of gravity is added, not compensated).
What prevents a drop of water hanging on an apple, create two tidal hump?
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 26/04/2018 23:22:57
What prevents a drop of water hanging on an apple, create two tidal hump?
The apple is not rotating around a barycentre with another mass
Title: Re: Why do we have two high tides a day?
Post by: Yusup Hizirov on 27/04/2018 01:46:34
In some places there is only one tide per day, where does the barycenter go?
Can one sea move on a barycenter, and another sea does not?
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 27/04/2018 09:43:49
In some places there is only one tide per day, where does the barycenter go?
Can one sea move on a barycenter, and another sea does not?
That’s like asking if a wheel axle goes!
Clearly from the earth/moon rotational frame, they both rotate around a common barycentre, but from an earthcentic frame the barycentre rotates around the earth.
For the rest I’ll answer in the other thread where you are misunderstanding the papers you quote.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 27/04/2018 11:02:17
In some places there is only one tide per day, where does the barycenter go?
Can one sea move on a barycenter, and another sea does not?
Regarding first question, apart from what said by Colin2B, I´ll only mention that the barycenter is some 2/3 of Earth´s radius off from its center, so "inside" our planet. But it is not something physical, just abstract stuff: the center of mass of total mass of Earth and Moon.
And it is the one which actually follows the elliptical orbit around the Sun, not the Earth´s C.G.
Regarding the second one, you have to keep in mind that we are discussing the root causes of tides, which are astronomical if you like. But due to the different continents and other varying local circumstances, ocean water is not free to move "obeying" only those root causes. In some places resonance and other phenomena may occur, and that can hinder or enhance tides in different fashions.   
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 27/04/2018 11:50:51
What prevents a drop of water hanging on an apple, create two tidal hump?
Let us analyze that carefully ...
A "free" drop of water is roughly spherical. If hanging on (or "from"?) an apple, it actually has TWO kind of bulges.
The upper one, due to surface tension, would be equivalent to sublunar tidal bulge, in this case due to an "excess" of gravity attraction from the Moon, compared to centripetal force required for the circular movement of all water "drops".
And the lower one, due to the weight of all water of the drop, would be equivalent to opposite tidal bulge, in this case due to a deficit of gravity attraction from the Moon, also in comparison to required centripetal force. That defficit has to be compensated by the bulk of Earth´s mass, exerting an additional pull on water there, which, if not sufficient centripetal force, would inertially move outwards ... The equilibrium is reached thanks to 3rd Newton´s Motion Law: centrifugal forces turn up, as reaction to that additional pull from the bulk of the Earth.
That happens also where solid earth (and even in the atmosphere), and all across of the half of the Earth farthest from the Moon ...
Solid parts are stretched, and water, being much more free to move, deformes ocean surface shape until  own water weight "says": that´s enough ...
By the way, similarly to what happens at lower part of the drop hanging from the apple surface. In this case inward pull is water surface tension, and outward pull is water weight ...       
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 27/04/2018 12:08:13
Clearly from the earth/moon rotational frame, they both rotate around a common barycentre,...
Just an important nuance.
The main purpose of my post of yesterday was to insist again that Earth doesn´t rotate around the barycenter: it actually revolves. Paths of points farther from the barycenter are not bigger than those of closer points. The circular path followed by every material Earth point has a radius equal to the distance from Earth´s own C.G. to the barycenter.

Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 27/04/2018 12:26:16
The main purpose of my post of yesterday was to insist again that Earth doesn´t rotate around the barycenter: it actually revolves.
Sorry, I was trying to reply to this post at the same time https://www.thenakedscientists.com/forum/index.php?topic=38127.msg540291#msg540291 and had a slip of the keyboard on terminology.
I agree with you
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 27/04/2018 20:02:26
pmb,
How does gravity push water away from the moon?

Due to the fact that he closer you are the stronger the pull of a gravitating body and the further you are away the less strong. This has the net effect of stretching whatever is in  the field.

And yes, for the most part there are only two tides per day, David is speaking about anomalies due to geometrical formations and that's not part of the two tides per day in oceans and seas. He never seems to mention those parts.

This link is from an authoritative source: https://oceanservice.noaa.gov/education/kits/tides/tides05_lunarday.html
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 27/04/2018 23:21:01
This link is from an authoritative source: https://oceanservice.noaa.gov/education/kits/tides/tides05_lunarday.html
This one has more detail https://tidesandcurrents.noaa.gov/restles3.html
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 28/04/2018 11:43:50
Sorry, I was trying to reply to this post at the same time https://www.thenakedscientists.com/forum/index.php?topic=38127.msg540291#msg540291 and had a slip of the keyboard on terminology.
I agree with you
Don´t worry: even without any keyboard slip, it´s easy to use the verb "to rotate" instead of "to revolve". But I wanted to insist on that detail, because the problem comes when somebody makes some dynamics analysis of it without keeping that in mind: the two cases are quite different.
Curiously, even the NOAA site linked by PmbPhy is not quite correct in that respect. They say: 
"The lunar day is 50 minutes longer than a solar day because the moon revolves around the Earth in the same direction that the Earth rotates around its axis".
They use both terms:
1) " ... the moon revolves around the Earth ..." But the Moon actually ROTATES, more exactly around the Moon/Earth barycenter, because it is tidal locked to our planet.
2) "... the Earth rotates around its axis". That´s correct, because the "revolving" of the Earth I previously referred to is  the some 28 day one around the barycenter. And they refer to its daily rotation. I prefer to say "spins" instead of "rotates" though ...
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 28/04/2018 12:26:43
Don´t worry: even without any keyboard slip, it´s easy to use the verb "to rotate" instead of "to revolve". But I wanted to insist on that detail, because the problem comes when somebody makes some dynamics analysis of it without keeping that in mind: the two cases are quite different..
Yes, I’m aware of the technical difference, but I think the 2 get confused in common usage, spin would be clearer.
I generally prefer the barycentre explanation in the link I gave because it’s very clear where the forces and relative motions are occuring

I have removed the alternative theory posts to new theories as I think it is unfair go clutter up your thread.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 28/04/2018 12:28:54
Due to the fact that he closer you are the stronger the pull of a gravitating body and the further you are away the less strong. This has the net effect of stretching whatever is in  the field.
You and me have discussed this topic many times, and you always insist on the idea of differential gravitational forces, as if they were the only cause of the stretching, and eventually the tides.
But that "net effect", the result of a mathematical operation our minds can make, is insufficient: farther matter cannot "know" how big moon gravitational pull on closer matter is (in order to react to that "net" gravity), unless you consider internal stresses caused by the field of centripetal and centrifugal forces, originated by Earth´s revolving around the barycenter.
Colin2B has linked another work from NOAA. I learnt about it some months ago, but preferred not to bring it here because they go even beyond my stand, in relation to the use of the term
 "centrifugal force". I even contacted NOAA by mail, and one of them sent me a few mails.
I´ll further comment on that in another post replying to Colin2B post. But not today, because I´m afraid that, though they have gone in my line (considering centrifugal forces), they´ve gone "too far" ... And it will take long time to write something  sufficiently clear.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 28/04/2018 12:35:04
This one has more detail https://tidesandcurrents.noaa.gov/restles3.html

Colin2B has linked another work from NOAA. I learnt about it some months ago, but preferred not to bring it here because they go even beyond my stand, in relation to the use of the term
 "centrifugal force". I even contacted NOAA by mail, and one of them sent me a few mails.
I´ll further comment on that in another post replying to Colin2B post. But not today, because I´m afraid that, though they have gone in my line (considering centrifugal forces), they´ve gone "too far" ... And it will take long time to write something  sufficiently clear.
I´ll try to have it ready for tomorrow ...
Apart from that, I don´t quite understand what you mean when saying:
I have removed the alternative theory posts to new theories as I think it is unfair go clutter up your thread.
Perhaps I didn´t see some posts ...
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 28/04/2018 16:51:09
Don’t worry about it, I like the method because it is clear on where the forces and relative positions of earth moon are. I can live with their terminology, I think it was written before the modern trend of explaining via inertia and frame dependant forces.

The other thread/posts I was refering to is https://www.thenakedscientists.com/forum/index.php?topic=73127.0
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 30/04/2018 12:42:55
This one has more detail https://tidesandcurrents.noaa.gov/restles3.html
As I said two days ago, I saw that NOAA work last december. Previously I had seen another, also from NOAA, not completely in agreement with my stand. Then I contacted them, and one of its scientists and myself kept emailing each other for a couple of weeks. He was the one who sent me that link, and also another much more thorough:
https://tidesandcurrents.noaa.gov/publications/Tidal_Analysis_and_Predictions.pdf
though the analysis of root causes of tides are similarly long. The bulk of the much longer one is in relation with geographical local factors all across the world.
Then was when I posted #80 and 81. For me it was an "eureka" moment, especially for the "revolving" feature of Earth movement, in its necessary "dance" with the Moon to keep it in its orbit.
I saw that fact made compatible, as far as results are concerned, my "centrifugally influenced" stand, the one of so many who talk about only differential gravitational forces, and also the stand of those NOAA experts, which use the concept of centrifugal force beyond Newton´s Motion Laws.
I´m glad you "can live with their terminology", but others such as PmbPhy could say that to say total centrifugal force at Earth C.G. is equal to total Moon´s gravitational pull is wrong, because Earth is experiencing a centripetal acceleration ... They´d say that centrifugal force there wouldn´t be real ...and I agree with them. Perhaps they would bring up the issue of non-inertial systems of reference, what NOAA people don´t mention ...
And NOAA people also use "centrif. f." across the whole Earth, in a similar fashion, as one of the causes of tides ... But "fictitious" forces can´t produce real effects !!
I´m afraid all that is rather confusing. And that´s why I discussed with them. But no clear conclusion !!
I´m considering the possibility of copying here all those arguments of both parts, for anybody who could be interested.


Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 01/05/2018 11:51:04
I can live with their terminology, I think it was written before the modern trend of explaining via inertia and frame dependant forces.
I haven´t found the date of the NOAA "short" work, but the long one is actually dated 2007 ...
They mention centrifugal force as something kind of inherent to circular motion, without neither specifying which agent produces it, nor even mentioning terms such as centripetal acceleration and force ...
I´m afraid that is going to keep the big confusion existing out there, even among many physicists.
That problem is the one I first detected, and fight against, when I met this thread ... years ago. But to no avail so far !!
By the way, one of the first things I was told by the mentioned NOAA scientist was:
"The publication you are referring to is "Our Restless Tides", a 10-page pamphlet developed in the 1950's to provide a basic description of the forces which create the tides.  It's intended audience were the grade school children and adults of that time.  It used terminology of science and forces which were common in the 1950s.  Such as centrifugal force.  Centrifugal force was always an "imaginary force" (not a real / measurable force).  But that type of description made the concepts easier to understand and explain.  That  description and use of centrifugal force continued to be common practice until the 1970-80's.  At that point, the terminology shifted and the textbooks used in grade schools were changed to use a more modern terminology and description of this "effect" being a result of inertia rather than an "imaginary force".
RESULT OF INERTIA ... I do agree: Newton´s Motion Laws come from inertia !!
"THIS EFFECT" ... Centrifugal forces, in the cases when they are REAL, are not just an "effect" or tendency due to inertia ... Affected objects "tend" to keep their velocity vector due to their inertia, centrifugal FORCE is derived from inertia, but it is something more than that "effect" ...
   
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 03/05/2018 22:23:46
I can live with their terminology
I´m quoting that again just to add that, as the core of our current discussion is the interpretation of the term "centrifugal force", and existing confusion, I´ve passed my previous post (#112) to the specific thread on the matter:
https://www.thenakedscientists.com/forum/index.php?topic=68025.new#new,
post #167 there.
It has induced several replies, some surprising because they show also a confusion among some of our "colleagues", the thread OP included ...
That made me send copies of the texts of a couple of emails sent by me to institutions with a very clear confusion on the issue. One of them the very Britannica dictionary.
I suggest anybody interested to have a look there.
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 23/05/2018 17:53:14
If the tides are caused by the gravity of the moon, why is there a high tide on the side of the Earth furthest from the moon as well as on the closest side?

Peter Conway

Asked by Peter Conway


                                        Visit the webpage for the podcast in which this question is answered. (http://www.thenakedscientists.com/HTML/podcasts/naked-scientists/show/20140107/)

[chapter podcast=1000585 track=14.01.07/Naked_Scientists_Show_14.01.07_1001836.mp3](https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.thenakedscientists.com%2FHTML%2Ftypo3conf%2Fext%2Fnaksci_podcast%2Fgnome-settings-sound.gif&hash=f2b0d108dc173aeaa367f8db2e2171bd)  ...or Listen to the Answer[/chapter] or [download as MP3] (http://nakeddiscovery.com/downloads/split_individual/14.01.07/Naked_Scientists_Show_14.01.07_1001836.mp3)



See: http://www.newenglandphysics.org/other/Taylor_tides.pdf
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 24/05/2018 11:57:20
See: http://www.newenglandphysics.org/other/Taylor_tides.pdf
We have long discussed this issue ...
What said in linked site could not actually considered erroneous, but it´s insufficient. No clue is given to following question:
How "on Earth" could a portion of water far from Earth C.G. "know" the value of how much it would be pulled by the Moon if, instead of its actual position, it were situated where Earth´s C. G., and be able of deducting that vector from Moon´s pull in its actual position?
Because, as far as I can understand, material reacts only to the forces really exerted on it, either by CONTIGUOUS material (push, pull, hydraulic pressure ...) or "tele-forces" (such as gravity)
As early as #37 I refuted what you have said on #36 (rather than on #35, as erroneously said there).
No reply from you then ...
I´ve sent lots of posts, and though not many replies, I know there are people who agree with me (e.g. Colin2B).
Logically I can´t ask you to read now so many posts, but let me suggest you to have a look at #86 ...
Any further comment would be appreciated.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 26/05/2018 10:48:31
Don’t worry about it, I like the method because it is clear on where the forces and relative positions of earth moon are. I can live with their terminology ...
Hallow!
Just in case you haven´t recently read:
https://www.thenakedscientists.com/forum/index.php?topic=68025.new#new
I suggest you to have a look at, especially from #181 on.
I honestly think I´ve already grasp the ideas and method of my NOAA "friends", but I need some time to elaborate and send a clearly explaining post ...
PmbPhy intervened, and in my reply there is a kind of clue to what came to my mind in the eureka moment I mention:
"In particular, the centrifugal force is an inertial force which is observed in rotating (non-inertial) frames (a quote from PmbPhy post)
I´m afraid that is rather confusing ... What do you mean with "... which is observed in ..."?
Does it mean it is a real force "observed" by us? Or, in that "particular" case, is it rather a kind of tricky tool WE apply to bring a "fictitious" situation (an accelerating frame of reference) back to reality (with all REAL effects which have disappeared in that frame of reference)?

 
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 27/05/2018 03:04:20
Quote from: rmolnav
We have long discussed this issue ...
And I explained it when I first mentioned it or was asked. The earth or moon doesn't "know" anything since they don't think. :)

Surrounding all planets there is a gravitational field whose strength decreases as 1/r[sup2[/sup]. IT's always there just like blades of grass are always on a gold course. A person or moon who which is in free fall in the earth's gravitational field will experience different forces on their body. If falling feel first then the feet will have a greater force on them than the force on his head. If will feel to the man falling that his body is being pulled apart. Same thing with the oceans on earth due to the moon's gravitational field.

If you're wondering why I didn't reply to something you posted its because I hold that you ignored what I posted. In particular the derivations for tidal forces and ocean tides and you refusing to prove that they are wrong, Homey don't play that.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 27/05/2018 07:41:08
A person or moon who which is in free fall in the earth's gravitational field will experience different forces on their body. If falling feel first then the feet will have a greater force on them than the force on his head. If will feel to the man falling that his body is being pulled apart
I thought it was clear that when I mentioned "think" it was just a rhetoric question ...
But a really "free fall" is only possible for material points ...
If a man is falling feet first, as you say his feet will have a greater force on them than the force on his head ... But both ends are not really in free fall, because they are kind of tight together through the body. Both have to get equal acceleration, and that implies that the body is somehow transmitting information between its parts ... The bulk of the body, accelerating according to gravitational field at body´s c.g., "forces" (pulls) upper parts not to let them fall behind, and pulls back lower parts, "forcing" them not to accelerate more than the c.g. ...
Those forces (certainly originated by the not uniform gravitational field), and their reactions (3rd Newton´s Law), are the direct cause of the "pulling apart" mentioned by you. It is not just a "feeling" due to the gravitational field gradient !!
That rather obvious idea has been present in all my posts, but, honestly, I´m afraid you feel too sure your rather "excluding" theory is the right one, and you´ve never taken "mine" worthy to consider. 
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 27/05/2018 16:15:45
Quote from: rmolnav
But a really "free fall" is only possible for material points ...
That is incorrect. The term "free fall" is defined to mean "subject only to the force of gravity."

I have to stop here since I/m extremely busy with something else and you keep making false accusations about me, i.e. ad hominems. I don't tolerate that so I won't.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 27/05/2018 18:17:51
On #119 it is said:
"The term "free fall" is defined to mean "subject only to the force of gravity."
I know ... But if we are being that precise (considering even the quite negligible difference between gravitational field in the vertically falling man´s feet and head), we should say the man as a whole is in free fall ... but strictly speaking only the horizontal section through his c.g. would actually be in free fall. The rest of his body would be subject not "only to the force of gravity", BUT ALSO to the internal stresses this morning I referred to, and have continuously been referring to since my very first post here, #20 ... three years ago !!
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 28/05/2018 07:56:34
...we should say the man as a whole is in free fall ... but strictly speaking only the horizontal section through his c.g. would actually be in free fall
I´m quoting that from my last post, because, even Britannica.com people (who, as most people do, ONLY consider the varying gravitational field as cause of tides, something I consider ERRONEOUS, or at least insufficient), in this "free fall" subject, surprise, surprise, do agree with me:
"Gravitational forces are never uniform, and therefore only the centre of mass is in free-fall. All other points of a body are subject to tidal forces because they move in a slightly different gravitational field. The Earth is in free-fall, but the pull of the Moon is not the same at the Earth’s surface as at its centre; the rise and fall of ocean tides occur because the oceans are not in perfect free-fall"
https://www.britannica.com/science/free-fall-physics 
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 28/05/2018 09:11:07
...Britannica.com people (who, as most people do, ONLY consider the varying gravitational field as cause of tides, something I consider ERRONEOUS, or at least insufficient) ...
Many may think "What a stubborn man, he continues to go against most people ... after three years !!).
I already said, on #86:
"Seen from outside, as my stand refutes what so many people (scientists included) say, most probably I "must be" the wrong one ... But there are several details that could explain those so frequent, total or partially  errors: ..."
and then I listed several reasons.
Now I´ll bring up a real case which justifies my stand. We have to take things (at least as far as tides are concerned) with a pinch of salt, even if they come from folks MSc in Physics ...
I´ve seen really surprising errors. E.g.: please kindly have a look at:
http://ipho.org/problems-and-solutions/1996/IPhO_1996_Theory.pdf
3rd problem is about tides … Mathematically seems to be correct … But, leaving aside the centrifugal force issue, its solution has a basic Physics error …
They analyze Earth movement around Earth-Moon barycenter as if it ROTATED (farther points following a bigger circular path), but EARTH ACTUALLY REVOLVES around the barycenter, and ALL its material points follow circular paths with identical radius !!
The dynamics of those movements are utterly different: different distribution of required centripetal forces, different mismatch between them and Moon´s gravitational field, and therefore different distribution of internal stresses that compensate imbalances (real centrifugal forces included) ...
So, any solution to the that way given problem would have nothing to do with reality ... 
And, apparently, neither one of the organizer, nor one among the participants, detected it !!
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 08/06/2018 08:16:09
... even Britannica.com people (who, as most people do, ONLY consider the varying gravitational field as cause of tides, something I consider ERRONEOUS, or at least insufficient) ...
I´m quoting that again in order to add something more related to that.
I´ve been discussing that  with more people ... One recently sent a link with an article from some science institution, where according to him (I couldn´t read it because the link didn´t work) only gravitational differences were mentioned (neither centrifugal forces, nor something else related to circular movement). So, he said, "my" theory was wrong ...
What follows was my last reply:
"Another article, now form NASA, where internal stresses related to satellite-planet  dynamics (with circular movements) are mentioned, not just "tidal forces" only due to gravitational differences:
" … Hurford and his team believe the discrepancy can be resolved if Enceladus’
rotation rate is not uniform – if it wobbles slightly as it rotates. Enceladus' wobble,
technically called "libration," is barely noticeable. "Cassini observations have ruled
out a wobble greater than about 2 degrees with respect to Enceladus' uniform
rotation rate," says Hurford.
… Depending on whether the wobble moves with or against the movement of
Saturn in Enceladus' sky, a wobble ranging from 2 degrees down to 0.75 degrees
produces the best fit to the observed warmest zones," said Hurford.
The wobble also helps with the heating conundrum by generating about five times
more heat in Enceladus’ interior than tidal stress alone, and the extra heat makes
it likely that Enceladus' ocean could be long-lived, according to Hurford ...
The wobble is probably caused by Enceladus' uneven shape. "Enceladus is not
completely spherical, so as it moves in its orbit, the pull of Saturn's gravity
generates a net torque that forces the moon to wobble," said Hurford."

https://www.nasa.gov/mission_pages/cassini/whycassini/cassini20100708-b.html
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 09/06/2018 12:04:00
I´ve been discussing that  with more people ...
Another reply I sent refuting that idea that ONLY differential gravity counts:
"You say:
1) "The antipodean bulge, even though it seems to move away from the surface of the earth, is actually accelerating towards the moon in free-fall".
Wrong! That bulge is subject NOT ONLY to Moon´s gravity (free-fall requirement), but also to rest of Earth gravity, its own weight …
3) "Another problem with the centrifugal explanation of tides is that they OPPOSE stretching rather than CAUSING it.  A rigid solid is an example of the strongest internal centripetal-centrifugal force pairs, and yet it exhibits no tidal response at all, assuming perfect rigidity".
Wrong too! At least in REAL cases, because what do you mean with "perfect rigidity" ?? And with "no tidal response at all" ??
However small the elasticity of a real object might be, if it has a circular movement centripetal forces will be required to cause it … And if the source of those forces can´t supply EXACTLY the required centripetal force at each location (either in a case such as the "whip" children game, or one such as Earth revolving, or Moon rotating), contiguous material elements (or other gravity source) will have to give or take the difference, because 2nd Newton´s Law applies.
That originates internal stresses which, please DON´T FORGET, now due to 3rd Newton´s Law, they are exerted on each other of considered contiguos elements, with OPPOSITE senses … And THAT is precisely the root cause of the stretching !!
Sorry if what follows might be considered kind of "too" honest, but to say "Another problem with the centrifugal explanation of tides OPPOSE stretching rather than CAUSING it" shows not only a Physics misunderstanding, but also a rather serious Logics one !!
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 15/06/2018 11:48:50
I think it may be interesting to some of you to have a look at what this very morning I sent to same "youtuber" mentioned on my previous post:
"As I already told you, I contacted the author of the interesting article linked by you. He replied, initially with a rather radical attitude, and he continued to say ONLY gravitational gradient matters, and that forces related to Earth revolving don´t interviene in tide generation ... After a few pairs of mutual replies, yesterday I sent him:
 "... In any case, the point I´m bringing up doesn´t require sophisticated calculus, as it is in the realm of the "interface" between inanimate nature (physical objects and their visible movement) and our intelligent mind. But I wouldn´t call that to get phylosofical : before any calculation we have to interpret correctly what we see or experience.
You say: "… ask yourself "Are forces "real"? No, they are mathematical tools for describing how material bodies interact with each other and how they move". I fully agree with you. But they can be directly exerted by an object A on another (or a part of another) B either by contact (internal stresses if parts of a body), or as "tele-forces" such as gravity … BUT for that last type of forces only their masses and distance between each other count, NOT stuff such as "the force that A would exert on B if B were situated where G" (whatever G could be). That is something we can handle, but nature can´t react to, if not present any other kind of hidden "information mechanism".
That is the reason why I asked: "How "on earth" a certain material part of our planet, which can "feel" Moon's gravity at its location, as well as own Earth's gravity there (its weight), but "knows" neither how far from Earth's C.G. it is, nor Moon's gravity there, could react to so called "tidal forces" (differences between gravity vectors at those two positions) ?…"
After writing that I realized you also say in one of mentioned articles:
"The only real forces that act on the body of the earth are:
- The gravitational forces between each part of the earth and every other part, and the gravitational forces on parts of the earth due to the moon, sun, and the nearly negligible forces due to more distant bodies in the solar system.
- Internal tensile forces within the materials of the earth".
Those phrases are very close to my stand:
Internal forces (stresses) - either tensile, compressive, or shear stresses - are exerted between contiguous parts, and a part of them works as follows.
Each part (B) of the Earth is "forced" by the rest (all other parts) to move together, in our case of revolving with no rotation, with identical centripetal force. Where Moon´s (A) pull doesn´t match with that required centripetal acceleration, rest of the Earth gives or takes the difference, through contiguous part direct contact. That difference can be "moon-wards" or outwards.
A correct application of 3rd Newton´s Law would oblige us to separate those two sources of centripetal force on B, because a fraction is exerted by the Moon (A), and another by the part of the Earth which is not B … B would be exerting back a pull on the Moon, an another on the rest of the Earth, through contiguous parts (internal stresses). Those pair of pulls on each other (B, and the rest of the Earth) are, one of the pair "moon-wards" (part of total centripetal force), and the other outwards, "centrifugal" in its more general sense, and "real" in the sense that is according to Newton´s 3rd Law, not just a mathematical tool linked to the chosen reference system …
THOSE internal forces (difference between gravitational force and required centripetal force), as far as I can understand, are the one which directly generate all tidal effects.
Another thing is that, as in the Earth case centripetal force is constant, and at Earth C.G. it is exactly in balance with Moon´s pull there, what the so called standard model does (to subtract Moon´s pull on B if it were at Earth C.G., from Moon´s pull at B actual position), gives equivalent results.
 But that doesn´t mean that model is THE 100% correct one (and no other including those forces linked to the revolving movement). As I already said, it wouldn´t work in an analysis of Earth related Moon tides, or in the analysis of tidal effects in any of the billions of celestial objects that are tidal locked to another in their rotation around their barycenter" TO THAT he replied with a short email piece, saying that he couldn´t write a longer reply due to a family problem, BUT he said my take  IS INTERESTING ... though he added:
"I don't think it is helpful to think of tidal bulges as being caused by tensile forces in earth or water. Rather, dNewtonian dynamics forces earth materials to positions that bring those internal tensile forces to equilibrium. The gravitational field gradient provides the differential force that reshapes earth and water. The tensile forces limit that process”. Much, much “flexible” attitude !
I´ll let him “rest” for a few days, and then I´ll insist with some analogy with some examples he himself explains on his articles ...
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 16/06/2018 10:57:11
... he added:
"I don't think it is helpful to think of tidal bulges as being caused by tensile forces in earth or water. Rather, dNewtonian dynamics forces earth materials to positions that bring those internal tensile forces to equilibrium. The gravitational field gradient provides the differential force that reshapes earth and water. The tensile forces limit that process”. Much, much MORE “flexible” attitude !
(edited).
Surely he didn´t read my post carefully, because he says it is interesting, but he insists in what I refuted: "differential force" is an idea of our minds, NOT something inanimate objects can directly react to !!
Internal stresses change due to the mismatch between required centripetal force (constant in Earth case) and Moon´s pull across our planet (with a "gradient"), as I previously explained. Where solid Earth a stretch occurs, and where water pressure distribution changes, and that IS what actually "reshapes water" ...

Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 17/06/2018 11:53:15
As a continuation of my two previous posts ...
I´m going to reply what I referred to as follows, but not yet because "our man´s" wife was going to have a knee surgery, and he preferred to leave our discussion "stand by" ...
"On the one hand, I´m glad you find "my take" interesting. But on the other I´m kind of sad, because you found necessary to send me Newton´s 3rd Law "properly stated" ... Do you think I had used it wrongly ?.
In "my case" I was very careful with its application ... But it cannot be applied exactly as originally stated, because three "bodies" interviene, rather than just two: Moon (A), a certain part of Earth (considered portion B), and the rest of our planet not included in B.
When a cart is pulled by two horses, if accelerating and supposing no friction losses, 2nd and 3rd Newton´s Laws apply:
- Added pull of the horses on the cart is equal to the mass of the cart multiplied by the acceleration it gets.
- The cart is pulling back EACH horse with same size pull each horse is exerting on the cart.
By the way, I´ve seen you yourself explain a really similar case using EXACTLY same arguments of my take (your article "Tidal misconceptions")
Under the subtitle "Weight" you say:
"Weight may be defined as the force required to keep an object at rest relative to its surroundings. This definition is consistent with most colloqual interpretations of the word (surprise!).
The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v˛/R (the centripetal force). Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation".
TAHT CASE is quite similar to "mine", when Earth is revolving around the barycenter. A man on Earth surface where antipodal bulge builds would also experience ALL those changes, due to a lightening equally caused by its circular movement (though much, much smaller).
Unique difference: now we have two "horses" pulling instead of only one ...
IT´S ALSO CLEAR that one of the "horses" (the Moon) pulls Earth massive parts there (B) less than it does on closer parts, but what that actually causes is that the other "horse" (massive parts of Earth not being B) HAS TO pull more, kind of "supply" the difference. And corresponding reaction according to Newton´s 3rd Law, equal but opposite (therefore outward - "centrifugal" in its general sense) and exerted by B ON REST OF EARTH, is also bigger than at closer parts.
And, as you say for equatorial bulge, I can say:
 "This is the reason for antipodal bulge of the Earth due to its revolving around Moon-Earth center of mass".
You also say:
"I don't think it is helpful to think of tidal bulges as being caused by tensile forces in earth or water" ...
It´s not a question about being "helpful" or not ... As far as I can understand, and in line with what you yourself say in the equatorial bulge case, root causes of tidal bulges are several: Varying Moon´s pull, Earth own pull on each of its parts, initial conditions that gave tangential speed to Earth ... and nature motion laws (basically due to the so called "inertia"), as Newton stated.
CHANGES in internal forces where solid Earth (and corresponding stretching), and in pressure distribution where water (and its corresponding pilling up at opposite areas) are nature´s answer to above mentioned causes. They are actually the tides themselves, rather than their cause.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 19/06/2018 11:34:34
As a continuation of my two previous posts ...
I´m going to reply what I referred to as follows, but not yet because ...
As a further continuation to my last posts, just to say I´ve already sent mentioned message, adding what follows:
"By the way, apart from nature itself (and its own motion laws), mentioned
tangential speeds could be considered the actual cause of tides, whatever the energy source that made those speeds possible.
Without those initial speeds, and Moon-Earth "dance" originated by them, tides wouldn´t have been happening for billions of years as they have !"
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 24/06/2018 11:07:18
As a further continuation to my last posts ...
I got a rather sad reply from our man, the author of the very interesting works I referred previously to:
https://www.lockhaven.edu/~dsimanek/scenario/tides.htm#centrip
https://www.lockhaven.edu/~dsimanek/scenario/centrip.htm
He is 81, his wife 80, the surgery went well, but they have no other family help, and among other things he told me:
"So I'll be quite busy helpling her for a few weeks, so don't expect a reply for a while"
After that I asked him if he has some kind of team right hand colleague which could continue our discussion. I feel pretty sure I could get them at least accept some of the errors in the above linked works ... But no answer so far.
Apart from what I said from nine to five days ago, he said some absurd things, I think mainly due to the fact that he frequently makes titanic efforts to avoid the use of anything related to centrifugal (or even simply "outward") forces, and/or Earth revolving ... Kind of viscerally biassed reasoning ! And he critics many other works biassed ...
E.g. ("exam" ):
"Question 10: The picture (not copied here) and text below are from the NOAA-NOS website. Your tax dollars at work to propagate misconceptions (!!)
Gravity and inertia are opposing forces acting on the earth's oceans, creating tidal bulges on opposite sides of the planet. On the "near" side of the earth (the side facing the moon), the gravitational force of the moon pulls the ocean's waters toward it, creating one bulge.
On the far side of the earth, inertial forces dominate, creating a second bulge.
Identify the specific misconceptions in the picture and the text. 
Answer 10. The picture suggests that the near bulge is only (or "mainly", isn´t it) due to gravitation, the other one only due to "inertial forces". The text speaks of "inertial forces", without saying that such a term has no meaning (??) except in a non-inertial coordinate system. The phrase "pulls the ocean waters toward it" implies "motion toward it" (or "deformation", does´t it?). The moon exerts gravitational forces on the far side bulge not much smaller than on the near side, and if these forces are "pulling" toward the moon on the near side, they are also pulling toward the moon on the far side. No mention is made of that (but they say "inertial forces dominate" !!.)
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 27/06/2018 12:10:20
I´ll continue to refute what dealt with on my last posts.
One of first paragraphs of “Misconceptions about tides” (“lockhaven.edu-dsimanek” previously linked sites), says:
"One of the few books that clearly defines "tide" at the outset is The Planetary System by Morrison and Owen [1966]: "A tide is a distortion in the shape of one body induced by the gravitational pull of another nearby object." This is definition (2) above. It clearly says that tides are the result of gravitation, without any mention of rotation of the earth”.
That last comment seems to show the author is biassed: he starts considering rotation (“revolving”, rather) has nothing to do with tides …
With that fixed idea, his reasoning may be kind of distorted, reaching erroneous conclusions !!
Further comments we can do about it:
1) If only "the gravitational pull of another nearby object” (the Moon in our case) counted, Moon and Earth would have fallen onto each other long, long ago …
2) I know standard Physics says: but they both rotate/revolve around the barycenter, and that prevent them to fall onto each other !!
3) Right. Therefore, without that rotation/revolving we wouldn´t have had the tides we´ve had for billions of years !!
4) And as Moon´s gravitational pulls on farthest hemisphere ocean water “units" are insufficient to cause required centripetal acceleration at their place, those water “units" tend to follow in the direction of their tangential speed (logically, due to the “inertia” phenomenon). Part of own Earth pull on that water has to be “used” to produce that centripetal acceleration deficit.
5) The author himself says on https://www.lockhaven.edu/~dsimanek/scenario/centrip.htm
that (explaining why on the equator we weigh less than at higher latitudes, being our weigh maximum at both poles): 
"The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v2/R (the centripetal force). Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation".
EXACTLY THE SAME happens (why it wouldn´t? …) to water further from the Moon due to Earth revolving around Moon-Earth barycenter, because its weight there has same direction and sense as required centripetal force.
And that water lightening there, together with the displacement of water from rest of further hemisphere (caused similarly, but with different angles between weight and required centripetal forces) is actually the reason of the antipodal bulge formation !!
The varying Moon´s pull also intervienes, but only causing previously mentioned centripetal force “deficit” !!
Another thing is that, as Earth revolves without rotation (in its “dance” with the Moon), required centripetal force is constant across it. And as at Earth c.g. it is in balance with Moon´s pull there, the “fields” of gravitational differences, and the one of differences at each place between required centripetal force and gravity, are equivalent !!
That fact keeps kind of hidden to many the error of thinking Earth revolving has nothing to do with tides formation … The so called “standard model”, considering only gravitational gradient, would give erroneous results when analyzing e.g. Earth tidal effects on the Moon, or on any of the many billions of celestial objects tidal locked to another !!   
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/07/2018 11:23:55
I´ll continue to refute what dealt with on my last posts.
They author of the articles I´m referring to also says:
“ … Once they launch into the rotating coordinate mode and start talking about centrifugal forces … They also forget that the non-uniformity of moon's gravitational field over the volume of the earth is alone sufficient to account for both tidal bulges, bulges that would be essentially the same if ... the earth and moon were not moving relative to each other”.
HE is the one who apparently forgets that hypothetical case is NOT possible … unless some other external force were countering the gravitational attraction between both celestial bodies !!
Let us “play” with our imagination. Some “possible” (?) cases could be:
1) A kind of “cosmic" superman were able to keep them separated their current distance, pushing them outwards on their inner sides. Earth would not get tidal bulges but quite the opposite: it would become slightly flatter (internal compression stresses increase)
Similar situation, but pulling outwards from their outer sides (where strong solid parts in the Earth side ...). Kind of bulges would build, but outer one would be quite different, depending on the way our superman could  "grasp" the Earth … (internal tensile stresses increase)
Some “ miraculous” outward forces were applied directly on Moon and Earth respective centers of mass, countering mutual gravity attraction … Earth inner hemisphere would get an ovoid shape, kind of bulge (internal tensile stresses increase), but outer hemisphere would experience the opposite type of deformation: it would get slightly flater (internal compression stresses increase).     
THEREFORE to get our real bulges it is paramount the way Moon-Earth mutual attraction is countered to keep actual distance … And the real way is their rotation/revolving around their barycenter !! THAT EARTH REVOLVING implies inertia related outward forces, quite an intervening fact in bulges formation, as explained on my last post.
So, to say “… bulges that would be essentially the same if ... the earth and moon were not moving relative to each other” is erroneous, or at least pointless, because it compares something real with something impossible ...
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/07/2018 11:28:55
Sorry there should be:
"2)" before "Similar situation, but pulling outwards ..."
"3)" before "Some “ miraculous” outward forces were applied ..."
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 07/07/2018 15:21:19
When discussing with somebody else things said on the very interesting articles I´m lately referring to:
https://www.lockhaven.edu/~dsimanek/scenario/tides.htm#centrip
https://www.lockhaven.edu/~dsimanek/scenario/centrip.htm
he asks me what below.
I´ve seen many people somehow have rather confused ideas about daily manifested tides (does Earth daily spinning cause tides ??), when Moon-Earth "dance" movements have almost a month period ...   
That´s why I´ve decided to send this post.
I answered what follows.
"Your question:
" ...(I had previously said)  DAILY Earth spinning causes the "permanent" equatorial bulge, but it has NOTHING to do with Moon-Earth dynamics ... Does this mean that you also agree that the earth's spinning has nothing to do with tides?"
My answer:
I chose carefully my words. There is a trick is in the way you put the question ... The earth's spinning has nothing to do WITH THE ROOT CAUSE OF TIDES, but it does with the way they manifest ...
When we say "tides", in its global sense (local changes of sea level can have other causes), we refer to the cyclical change of sea level related mainly to Moon position in its monthly rotation. To analyze their actual causes, we have to disregard the daily spinning of our planet (imaging we check the position of the bulges everyday at same time). That way we would have the bulges, always in line with the Moon, "moving" around our planet in an app. 28 days "two-parts" cycle ...
THAT IS the basic phenomenon of tides.
BUT, as Earth does spin daily, its meridian in line with the Moon changes continuously, and though bulges maintain its position relative to the Moon, they continuously change meridian too. And Moon related tides SEEM to have an app. 12+12 h. cycle ...
BUT THAT SPINNING doesn´t cause the tides, though it causes the so called equatorial bulge, easily explainable as done on  “lockhaven.edu-dsimanek” previously linked sites, when analyzing how we weigh less on lower latitudes, especially on the equator: 
"... Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation".
ANOTHER THING is that the monthly revolving of the Earth in its "dance" with the Moon, also causes similar inertia related effects (not to use tricky concept "centrifugal forces"), which actually are one of the causes of tides, especially on the hemisphere further from the Moon, where antipodal bulge appears ...
On closer hemisphere, though those inertial effects have also an outward direction, as Moon´s pull there is higher, the later prevails and the better understood bulge builds towards the Moon !!
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 15/07/2018 12:15:22
For those not interested on a thorough explanation of the two daily hight tides, A REALLY SIMPLE one is what said by NOAA, "question" nş 10 as on:
https://www.lockhaven.edu/~dsimanek/scenario/tides.htm#centrip 
"Gravity and inertia are opposing forces acting on the earth's oceans, creating tidal bulges on opposite sides of the planet. On the near side of the earth (the side facing the moon), the gravitational force of the moon pulls the ocean's waters toward it, creating one bulge.
On the far side of the earth, inertial forces dominate, creating a second bulge
" (see figure on linked site).
All that is RIGHT, though not complete. Across the oceans both moon´s gravitational forces and inertial forces act ... BUT we have to understand the term "inertial forces" as "inertia related forces", NOT in the restricted sense "dsimanek" article author says:
" The text speaks of "inertial forces", without saying that such a term has no meaning except in a non-inertial coordinate system" ...
WHATEVER the reference system, the revolving of earth around the barycenter makes ALL its parts follow identical circular paths. That implies a "tendency" (due to INERTIA) of water to follow the tangent ... On the far side that kind of lightens water, as here several times commented in comparison to the the very author´s explanation of equatorial bulge.
Why "on earth" we, as NOAA people does, could not call those water weight lightenings "inertia related forces", or even "inertial forces" ??
NOAA people also say "On the far side of the earth, inertial forces dominate”…  QUITE RIGHT too,  because there moon´s pull is smaller.
They could also have said "On the near side of the earth, gravitational forces dominate" (there moon´s pull is bigger than "inertial forces"), though they only say "On the near side of the earth (the side facing the moon), the gravitational force of the moon pulls the ocean's waters toward it, creating one bulge".
That is why I consider all that is RIGHT, though not complete ... "Dsimanek" author critic could only be explained as a biassed attitude due to his visceral opposition to anything different to his idea that ONLY gravitational gradient intervienes on bulge formation … IT DOES, BUT IT IS ONLY ONE OF THE SIDES OF THE COIN !!
It could not be otherwise. The absence of those "inertial" forces (not to use the even more controversial term "centrifugal"), would imply no circular movement of the Earth around the barycenter, and Earth and Moon could not have kept their actual "dance" for billions of years !!

 
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 16/07/2018 21:22:45
You really need to condense all that down into a book (and then get someone to translate it into readable form).
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 22/07/2018 11:56:08
You really need to condense all that down into a book (and then get someone to translate it into readable form).
I´m glad you say so, hopping it implies you´ve found things I´ve said are worth a book (though you exaggerate ...) !
But, you know, I type with only two fingers, and not being English my mother tong it would take me “eons” …
Though since my first post here (#20), more than three years ago, I´ve sent a huge number of posts. That moment I could not even imagine I would have to write so much. And I´ve also sent lots of posts to other “The naked scientists” threads, especially "What is centrifugal force?"
I think a big problem has been the fact that many people feel very sure about the correctness of things said by many physicists, for instance what I´ve been refuting lately:
https://www.lockhaven.edu/~dsimanek/scenario/tides.htm#centrip
To go “against the grain” is harder than I thought!
And, once and again, I´ve had to send my reasons, putting  them in different ways, looking for success.
The main issue has been the concept of centrifugal force, since my very first post. Many physicists restrict its use to when in an “artificial” scenario, “existing” only inside our minds: the so called “non inertial system of reference” …
But, however we could agree to call them, “outward” forces do exist whenever a massive object follows a curved path, which produce REAL physical effects. They are not just a tendency to follow the tangent, e.g. as we can “feel” when in a car ...
Unless people understand and agree with that, the explanation of the existence of the high tide further from the Moon will be erroneous.
By the way, lately I´ve been ruminating on that concept, "not inertial system of reference”, and how it is most frequently handled … I´ve found what I consider “logical” flaws, but need some more time to prepare a post to properly try and convey that idea. It could be a paramount reason of the enormous existing confusion on both centrifugal force and antipodal bulge issue ...
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 23/07/2018 00:08:11
The reasons I suggested that you should condense it all into a book are as follows:-

(1) You've practically written a short book here on the subject already in terms of volume.

(2) If it was turned into a proper book, it would have to be organised into chapters delivering everything in the right order and with no repetition of any of the content and with anything extraneous removed.

(3) If it was a book, it would have a contents list and index making it easy to find the parts of most interest to any specific reader.

(4) If it was a book, it would have an introduction which would set the scene for everything else that follows, and that would make it easier for people to work out why you think centrifugal force has anything to do with tides. I still haven't seen the connection.

If you took away the sun, moon and other planets, the tides would completely disappear even though the Earth's rotation and the centrifugal/centripetal force would continue to apply in full.

If you only had the Earth and the moon and the Earth didn't rotate, you would have no centrifugal force, but you would still have tides: two of them per month.

Due to the lack of an index and proper structure though, I can't find the right place to start reading to get that vital explanation of the part of your argument that I've somehow managed to miss. How does centrifugal/centripetal force do anything other than create a static bulge (all round the equator)? And if you have identified a real effect rather than an imaginary one, how big a percentage would you put on it to represent its role in causation of tides?

Edit:-

I think I've found what I was looking for. In the case where the moon rotates round an Earth that isn't spinning, the moon would move the Earth about in a circle over the course of a month, and that would, I think you're suggesting, create centrifugal force. But does it count as centrifugal force if the Earth isn't rotating? Perhaps it doesn't, but I don't want to rule out an effect on the basis of it being misnamed. The Earth is being caused to move - to accelerate, and to change direction as it does this, which might lead someone to think that the water will be flung out away from the Earth when the Earth changes direction. But the acceleration force is gravity, and that gravity is acting on the water as well as the Earth, so it's all being accelerated in the same direction. The only thing that varies is the strength of that acceleration, and it's stronger on the closer material and weaker on the more distant material. That is not an addition to the normal cause of tides - it is precisely the normal cause of tides and it is not centrifugal/centripetal force.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 23/07/2018 11:36:40
In the case where the moon rotates round an Earth that isn't spinning, the moon would move the Earth about in a circle over the course of a month, and that would, I think you're suggesting, create centrifugal force. But does it count as centrifugal force if the Earth isn't rotating?
Thank you for your ideas about a possible book, but, frankly, I´m afraid it would be too hard a task for me.
What quoted is very important. I´ll answer your question with an analogy.
Imagine we have a bowl with water on the flat surface of a table. Let us make it slowly follow a circular path, trying to maintain the angular speed, and avoid any resonance situation.
The water level will rise at outer side of the bowl, against the water weight ... Here the bowl inner wall "forces" outer water layer to follow a circular trajectory (centripetal force), and as per 3rd Newton´s Motion Law water will "push" outward on bowl inner circular surface ... The same happens between contiguous cylindrical layers of water ... Those outward forces, whether we call them centrifugal forces or not, are quite real, and the result is that water pressure increases outwards, and that makes the water rise there (kind of high tide).
In the case of moon-earth monthly dance, certainly due to mutual gravitational forces (but also to suitable initial speed vectors of both celestial objects), disregarding daily earth spinning, earth is in a similar case to the water bowl analogy.
Now there are no walls to force inwards the water, but own water weight (much, much greater) makes the job keeping farther water "connected" to the rest of the oceans.
But that "tendency" of that water to follow the tangent, kind of lightens its weight, as very well explained on previously linked site:
https://www.lockhaven.edu/~dsimanek/scenario/centrip.htm
though here is for the daily earth spinning case, what forms the so called equatorial bulge.
And, whatever distribution of moon´s pull across our planet, that water weight (pull from rest of the planet) lightening does intervene on tides, particularly on farther hemisphere bulge formation


Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 23/07/2018 20:42:10
Imagine we have a bowl with water on the flat surface of a table. Let us make it slowly follow a circular path, trying to maintain the angular speed, and avoid any resonance situation.
The water level will rise at outer side of the bowl, against the water weight ... Here the bowl inner wall "forces" outer water layer to follow a circular trajectory (centripetal force), and as per 3rd Newton´s Motion Law water will "push" outward on bowl inner circular surface ... The same happens between contiguous cylindrical layers of water ... Those outward forces, whether we call them centrifugal forces or not, are quite real, and the result is that water pressure increases outwards, and that makes the water rise there (kind of high tide).

The water bulges up at one side of the bowl as you move it in a circle, but it does the exact opposite at the other side of the bowl - not a bulge, but a dip. That doesn't match up with the behaviour of tides.

Imagine that we only have the Earth and the moon, and that the Earth rotates at a speed that keeps the same side facing the moon at all times. What's would be the result? No tides. What you would see is the water bulging up towards and away from the moon, and that bulging is not driven by centrifugal force in any way, but solely by greater gravitational attraction to the moon on one side and less gravitational attraction on the other.

Once you have that picture in your head, you can then create tides by rotating the Earth at any different speed (including zero rotation). Introduce the sun and you have a second pair of bulges which can add to or subtract from the bigger bulges caused by the moon.

If there was a centrifugal effect from the Earth being sent in a circle by the moon, what would happen? There would be an amplification of the height of the tide on one side of the Earth and a reduction of it on the opposite side, but no such effect is measured, and no such effect should apply either. Let me explain why no such effect should apply.

With the moon moving the Earth in a small circle, the moon is directly applying its gravity both to the Earth and to the water - there is no further transfer of that force needed from Earth to water as it is already being applied directly to both. The case of a bowl of water being moved around by you is radically different - you move the bowl and the bowl transfers force to the water. If you were able to apply force to every atom of the water and bowl directly, the water would not slosh around or bulge at all. You could, if you had access to such a force, accelerate a bowl of water sideways in an instant to 100m/s and decelerate it back to 0m/s a moment later, and not a drop would spill - all the water would stay where it is in relation to the container throughout this manoeuvre. Gravity works like that - a falling bowl of water doesn't leave the water behind, but the water falls with it. This is very different from the equivalent in zero G (in a space station) where you could accelerate the bowl "downwards" away from its content by pulling it, leaving the water behind. The moon pulls both the Earth and water at the same time, pulling on every part of it with almost the same force, and the only reason you get two bulges from this is that the force declines over distance as it spreads out further away from the moon.
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 24/07/2018 11:53:21
The water bulges up at one side of the bowl as you move it in a circle, but it does the exact opposite at the other side of the bowl - not a bulge, but a dip. That doesn't match up with the behaviour of tides.
You’re correct if thinking of earth without land masses. What is being described here is the basin effect eg North Atlantic which results in a rotating tidal wave around a node.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 24/07/2018 12:20:15
Imagine that we only have the Earth and the moon, and that the Earth rotates at a speed that keeps the same side facing the moon at all times. What's would be the result? No tides.
I need more time to properly refute several things on your post ... For now, just something about what quoted.
That imaginary case is the real case of the moon, because it is tidal locked to our planet.
And there tides, in the exact way we have them here, certainly doesn´t happen, but tidal effects do, including a stretching, kind of bulges.
Moon is rotating around the barycenter, and centripetal (mainly earth pull) and centrifugal forces do happen at different parts of it. And also own moon gravitational forces.
At closer moon half, earth attraction prevails. At further moon half, earth pull is insufficient to supply the centripetal force necessary for the rotation ... Own moon gravity has to supply the deficit, and all those further parts kind of lighten, due to same reasons explained on:
https://www.lockhaven.edu/~dsimanek/scenario/centrip.htm
"The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v2/R (the centripetal force). Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation"
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 24/07/2018 13:49:40
The water bulges up at one side of the bowl as you move it in a circle, but it does the exact opposite at the other side of the bowl - not a bulge, but a dip. That doesn't match up with the behaviour of tides.
I started with your 2nd paragraph, but that does´t mean I agree with 1st one !
In the first place, an analogy is just that, an analogy. Scenarios can´t be identical.
In my analogy, we were supposed to move the bowl with our hands ... But if ("forgetting" the mass of the bowl), we imagined the bowl water molecules pulled horizontally by some "magic" magnet (attracting water), following a circular path similarly to moon-earth case, and forces were properly balanced, we would also see water rising at closer bowl side (magnet pull would prevail at that side), as well as the one at further side. Now it would be smaller than when we move the bowl with our hands (for identical radius and angular speed), because water pressure there would rise due to only the excess of centrifugal force, in comparison to the smaller magnet pull there ...
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 24/07/2018 19:35:03
The water bulges up at one side of the bowl as you move it in a circle, but it does the exact opposite at the other side of the bowl - not a bulge, but a dip. That doesn't match up with the behaviour of tides.
You’re correct if thinking of earth without land masses. What is being described here is the basin effect eg North Atlantic which results in a rotating tidal wave around a node.

The issue under discussion in this particular case is the imagined centrifugal effect of the Earth moving in a circle once a month as the moon goes around it, which, if there was such an effect acting, would lead to one high tide per day on a planet with no continents and even sea depth all the way round.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 24/07/2018 21:39:40
Moon is rotating around the barycenter, and centripetal (mainly earth pull) and centrifugal forces do happen at different parts of it. And also own moon gravitational forces.

The whole moon is being pulled by the Earth's gravity, and that removes the opportunity for centripetal force to apply. For centrifugal/centripetal force to apply, you'd need to have something like a string connecting the Earth and moon to apply the force instead of gravity, and that string would only apply the force at one point with all the rest of the material feeling as if it's being pulled outwards relative to it. That would produce one bulge if it attached to the centre, but it would produce two bulges if it was attached to the nearest point, and given that the gravity is highest at the closest point, something not unlike that is actually happening, so I might just be about to change my mind on this whole issue...

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At closer moon half, earth attraction prevails. At further moon half, earth pull is insufficient to supply the centripetal force necessary for the rotation ... Own moon gravity has to supply the deficit, and all those further parts kind of lighten, due to same reasons explained on:

I can see now that you could regard this as centrifugal/centripetal force on the basis that the change in gravitational strength does produce a situation similar to the moon and Earth being connected by a string, except that there's an elastic string connecting to every atom which applies less force to the atoms that are furthest away and more force to the ones that are nearer. You then have pull forces acting within the moon and planet to hold them together, these being provided by locally generated gravity, so nothing feels stressed other than by the changing deformation as they rotate. So, on that basis, tides arguably do have something to do with centripetal force and can be said to be entirely caused by it. It all comes down to whether we decide to class this as centripetal or not. Up until now, I would certainly not have done, but now I'm not so sure. It will come down to definition, and the distinctions may be arbitrary.

Imagine a case where there's a planet and a "moon", but there is no orbiting - the planet is being pulled along on a string in a straight line by a powerful spaceship and is accelerating, while the moon is being pulled along behind it by gravity such that it keeps up and maintains separation. The moon's gravity is causing two bulges in the planet's ocean, but there is no rotation involved in the system at all, so are the bulges driven by centripetal force in a case such as this one? No. When you get "flung back" by acceleration in a car, that is also like centripetal force, but without any rotation. Ultimately, all these cases are the same phenomenon, and the primary cause is the greater phenomenon, not dependent on rotation. It is caused by different amounts of force being applied to different material in different places.

If we accelerate an object in a straight line by pulling it on a string, we create an equivalent of swinging it round and round on the end of a string. In the latter case, it's centripetal force. In the former, it isn't, and yet it's really the same thing. Centripetal/centrifugal force is a subset of the greater phenomenon.

With a moon going round a planet, or both of them going round the same point, we have the same relationship to the case where the planet is being pulled by a spaceship while the moon is pulled along behind it by gravity. The greater phenomenon is the cause of both cases, but in the former case it could be attributed to the subset of that greater phenomenon called centripetal. However, there may be a problem with the definition of centripetal if it excludes "gravitational string", and I've no idea whether using forces as "string" has ever been used in the definition of centripetal to settle the argument, either by including or excluding its role. Everything thus rests on a technical definition and it's a matter of luck as to who is officially right.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 25/07/2018 11:12:44
Imagine a case where there's a planet and a "moon", but there is no orbiting - the planet is being pulled along on a string in a straight line by a powerful spaceship and is accelerating, while the moon is being pulled along behind it by gravity such that it keeps up and maintains separation. The moon's gravity is causing two bulges in the planet's ocean, but there is no rotation involved in the system at all, so are the bulges driven by centripetal force in a case such as this one? No
You yourself answered your question: No.
But, I´d rather say "no" to the very question: if the powerful spaceship would be making the earth accelerate, and with the moon back at current distance, earth deformation would depend on the way the spaceship force were transmitted to earth:
1) If somehow it were transmitted pulling on earth c.g., we wouln´t have two bulges: forward hemisphere would get flatter.
2) If the spaceship pushed on earth back side, the whole earth would get flatter (if pushing through a sufficiently big flat surface): no bulges at all.
3) Only if pulling from forward earth parts, we would get two bulges. But forward bulge would not be due only to less gravitational pull from more distant moon, but also to the very spaceship direct pull ...
We could even draw a parallel with our real case. In your imaginary case, the agent which causes moon-earth distance doesn´t diminish, is the mega-spaceship ... In the real case, similarly the agent is a kind of "hidden" force ("centrifugal" f., or at least "outward" force), due to inertia and the fact that all earth parts try to follow the tangent, and moon-earth dance, through invisible "gravity strings", is continuously "bending" the trajectories of all those earth parts ...
Without mentioned dynamic effects of that earth-moon dance, we could not even be here discussing the issue, because they were paramount in the very history of our planet evolution and the beginning of life !
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 25/07/2018 20:26:08
But, I´d rather say "no" to the very question: if the powerful spaceship would be making the earth accelerate, and with the moon back at current distance, earth deformation would depend on the way the spaceship force were transmitted to earth: ...

I didn't design that bit of the system well enough, and you've taken it in the wrong direction, missing the point as a result.

Let's have the spaceship apply an even force to the whole of the Earth and all the water on it with a tractor beam so that the Earth simply accelerates and the moon accelerates after it. In this case, there are two bulges in the water with one ahead and the other behind. The bulges in this case are not caused by any kind of rotation, but the mechanism for forming them is the same as it is in the real case of the Earth with the moon orbiting it - it's the difference in the moon's gravitational attraction acting on the Earth as it reduces with distance. The important point is that the cause occurs in a straight line case with no rotation, so it is not caused by centripetal force. Take away the spaceship and the moon will fall into the Earth, and as it does so there will be an amplification of the two bulges, again not caused by centripetal force. This clearly shows the real mechanism behind the tides.

In the case of the Earth with the moon orbiting it, rotation is present and is part of the mechanism for maintaining the bulges indefinitely (with the orbit making sure the Earth and moon don't collide), but it is not that rotation that causes the bulges - they are already there from the mere proximity of Earth and moon and the reduction of gravitational force over distance. I now think I was wrong about it being a matter of technical definition as to who is right - centrifugal/centripetal force is not part of the cause of tides, but merely plays a peripheral role.

I said before that we could see gravity as being like strings linking every atom of the Earth and sea to the moon, but with them being elastic so that some can pull with more force than others. The difference in strengths of pull in different places leads to the deformation, and that makes it similar to a case where a string is attached to the nearest point and not to any other, leading to it seeming possible that it could be equivalent to the case of a ball being swung round in circles on the end of a string. But if you stop the ball going round on the string (do this in deep space where there's no gravity worth speaking of), it just sits there at the end of the string with no deformation. If you do the same thing with the elastic gravity "strings" linking to every atom of Earth and sea from the moon, they continue to pull and the bulges are maintained when the rotation is removed. The rotation therefore did not cause the bulges, so their cause cannot be attributed to centripetal/centrifugal force.

Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 27/07/2018 11:17:04
Let's have the spaceship apply an even force to the whole of the Earth and all the water on it with a tractor beam so that the Earth simply accelerates and the moon accelerates after it. In this case, there are two bulges in the water with one ahead and the other behind. The bulges in this case are not caused by any kind of rotation, but the mechanism for forming them is the same as it is in the real case of the Earth with the moon orbiting it - it's the difference in the moon's gravitational attraction acting on the Earth as it reduces with distance
Your model has really improved … Now both bulges would be identical to real, natural ones (disregarding effects due to sun and other celestial objects) …
But in that scenario bulges wouldn´t be due to "the difference in the moon's gravitational attraction acting on the Earth as it reduces with distance” as you say, because massive stuff can react only to forces exerted on it, either gravity from more or less distant matter, or direct pulls or pushes (even share stresses) exerted by contiguous material. It cannot directly react to gravitational attraction differences between positions thousands of km apart !! How “on earth” could it ??
The behind bulge would actually be due to the fact that there backward pull from moon, at any given position, is bigger than forward pull from the spacecraft. And ahead bulge would be due to the fact that there moon backward pull is smaller than forward pull from the spaceship !! 
Your new model makes possible to also improve the parallel I drew: now we would be exerting on our planet an artificial, uniform and outward (relative to moon-earth couple) force with the spaceship, which is actually replacing the also uniform and outward force that, in the real case with earth revolving around the barycenter, is exerted by nature. How? … As I previously said, moon-earth dance is continuously bending the trajectory of each earth molecule (forcing it to be circular). ALL earth material points are continuously at farthest distance from moon (within its respective circular trajectory), what implies centripetal and centrifugal forces inherent to that revolving are all parallel to the straight line between earth and moon centers of mass.
Take away the spaceship and the moon will fall into the Earth, and as it does so there will be an amplification of the two bulges, again not caused by centripetal force.
That is another prove of the impossibility of having the tides we have had for billions of years without mentioned inertial effects due to the revolving of earth and the rotation of moon, both logically around their common center of mass.
What quoted would not happen if, exactly at the moment of disconnection of the spaceship, somehow we could give earth and moon the “tangential” real velocity vectors they got in nature. The barycenter would follow its previous straight trajectory, at the speed acquired until that moment, and earth-moon couple would dance again, the earth with the two bulges that in your model were artificially originated ... 
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 28/07/2018 01:00:20
Your model has really improved … Now both bulges would be identical to real, natural ones (disregarding effects due to sun and other celestial objects) …
But in that scenario bulges wouldn´t be due to "the difference in the moon's gravitational attraction acting on the Earth as it reduces with distance” as you say, because massive stuff can react only to forces exerted on it, either gravity from more or less distant matter, or direct pulls or pushes (even share stresses) exerted by contiguous material. It cannot directly react to gravitational attraction differences between positions thousands of km apart !! How “on earth” could it ??

What's the problem with this being the mechanism? Gravity's strong enough to make the moon orbit the Earth, and the moon's gravity is strong enough to make the Earth wander round in a circle. Why shouldn't the difference in the strength of the moon's gravity as it falls off over distance also be enough to raise the sea a little on the nearest and furthest sides? If there was no difference in the gravitational strength over distance, each body would act on the other just like the spaceship's tractor beam with no bulges being generated in the ocean at all.

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The behind bulge would actually be due to the fact that there backward pull from moon, at any given position, is bigger than forward pull from the spacecraft.

If you don't think the fall off in gravitational strength over distance is enough to account for the tides, the moon would apply it's backward pull to the whole Earth and its ocean and pull them all backward equally rather than doing so more strongly on the nearest part (creating a bulge there) and less strongly on the furthest part (creating a bulge there too).

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And ahead bulge would be due to the fact that there moon backward pull is smaller than forward pull from the spaceship !!

You're describing the same mechanism as mine while claiming mine is wrong and yours is right - it is the fall off in gravitational strength that is doing the job of creating the bulges.

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Your new model makes possible to also improve the parallel I drew: now we would be exerting on our planet an artificial, uniform and outward (relative to moon-earth couple) force with the spaceship, which is actually replacing the also uniform and outward force that, in the real case with earth revolving around the barycenter, is exerted by nature. How? …

The spaceship is only needed in the linear case to stop the moon catching the Earth and colliding with it. If we have the moon orbiting the Earth instead, they won't collide, so we can do away with it. We don't need it though anyway to see the mechanism for tides because it's all there on show until they collide, and the height of the bulges grows as they near each other.

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As I previously said, moon-earth dance is continuously bending the trajectory of each earth molecule (forcing it to be circular). ALL earth material points are continuously at farthest distance from moon (within its respective circular trajectory), what implies centripetal and centrifugal forces inherent to that revolving are all parallel to the straight line between earth and moon centers of mass.

You haven't taken in what I said at the end of my previous post. If you stop the Earth and moon (and all the water of the ocean) in an instant by applying uniform forces to each body, what happens to the bulges? They don't disappear, but are maintained (and will grow as the two bodies accelerate towards each other). If you are whirling a ball round your head on a string and then apply uniform forces to eliminate the movement in the same way, what happens to the bulges? They disappear as the ball springs back into a sphere. In the case of the ball on the string, the centripetal force has been removed, so the bulges disappear. In the case of the Earth and moon, there is no centripetal force to remove, so the bulges remain - they are caused by something else.

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Take away the spaceship and the moon will fall into the Earth, and as it does so there will be an amplification of the two bulges, again not caused by centripetal force.
That is another prove of the impossibility of having the tides we have had for billions of years without mentioned inertial effects due to the revolving of earth and the rotation of moon, both logically around their common center of mass.

The only difference is that an orbit prevents a collision, so the system continues to function for much longer.

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What quoted would not happen if, exactly at the moment of disconnection of the spaceship, somehow we could give earth and moon the “tangential” real velocity vectors they got in nature. The barycenter would follow its previous straight trajectory, at the speed acquired until that moment, and earth-moon couple would dance again, the earth with the two bulges that in your model were artificially originated ...

If the spaceship can tow the Earth in the manner I described, there is no reason why the same advanced technology couldn't apply forces to Earth (including ocean) and moon to send them in the right directions to establish an orbit, but the bulges are caused by the same mechanism before and after these changes - it is not the orbit that produces them.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 30/07/2018 10:34:20
What's the problem with this being the mechanism? Gravity's strong enough to make the moon orbit the Earth, and the moon's gravity is strong enough to make the Earth wander round in a circle. Why shouldn't the difference in the strength of the moon's gravity as it falls off over distance also be enough to raise the sea a little on the nearest and furthest sides?
Imaging three tennis ball size pieces of earth stuff, one were earth c.g. (E),  another just under the moon (M), and the other at the antipodes (A).
If we had them without the rest of our planet, they would be free to react to forces such as moon gravity … M would accelerate the most, and A the least. Distances between M and E, and between E and A, would increase, though M and A don´t even “know” moon pull on E. And certainly without needing any rotation or revolving.
But in the real case, M and A neither are free whatsoever, nor “know” how much E is pulled by the moon, let alone that material stuff “knows” how to subtract pull vectors, in order to directly react to gravitational differences !!
Our planet own gravity, many, many million times grater than moon gravity differences, together with rigidity where solid parts, “forces” them to move basically together … That causes internal stresses, what also depends on the way the planet globally reacts to the pull of the moon.
As you say, "the moon's gravity is strong enough to make the Earth wander round in a circle”, but that movement also requires suitable initial “tangential” speeds of both earth and moon. And those speeds, and their continuous change of direction due to the rotation/revolving, also affect internal stresses, what inevitably intervenes on tides.
You cannot say tides are due only to moon pull differences. The ways M and A react to the pull of the moon at their different distances depend on the let us say  degree of freedom the rest of the planet allows them …
If, e.g., we analyzed tidal effects at the moon, we would find that the answer of the whole moon to earth¨s gravity is different than the opposite we are dealing with. Moon is tidal locked to earth, and further parts of it require bigger centripetal acceleration than closer ones, for its actual rotation. But earth only revolves around the barycenter: all its parts require identical centripetal force.
Therefore, distribution of internal stresses due to those not equal “ global” movements is utterly different, and that also affects tides. 
A single cause such as gravitational differences can´t be the unique reason of tides, because  in mentioned two cases the field of total force acting on each unit of mass of the earth are the result of differently caused individual forces ...
If you don't think the fall off in gravitational strength over distance is enough to account for the tides, the moon would apply it's backward pull to the whole Earth and its ocean and pull them all backward equally rather than doing so more strongly on the nearest part (creating a bulge there) and less strongly on the furthest part (creating a bulge there too).
That´s logically flawed. What I say is that gravitational differences can´t be the UNIQUE cause of tides, that they intervene on the formation of tides, but only as one of the sides of a more complex “coin”.
And it is absurd to deduce from that, if it were right, "the moon would apply it's backward pull to the whole Earth and its ocean and pull them all backward equally …”
You haven't taken in what I said at the end of my previous post. If you stop the Earth and moon (and all the water of the ocean) in an instant by applying uniform forces to each body, what happens to the bulges? They don't disappear, but are maintained (and will grow as the two bodies accelerate towards each other).
We could imagine bizarre situations applying artificial forces to change the scenario, and get something similar to actual tides for a short period of time … But that wouldn´t mean our billions of year old natural tides are due only to causes identical to the ones at those situations ... 
If you are whirling a ball round your head on a string and then apply uniform forces to eliminate the movement in the same way, what happens to the bulges? They disappear as the ball springs back into a sphere.
That is erroneous, because outer parts of the ball require bigger centripetal force than inner ones (the ball rotates, as moon´s case …).
But, curiously, inner bulge would be bigger than outer one, because internal tensile stresses (interactions between contiguous material), caused also by centripetal and centrifugal forces, kind of accumulate inwards ...
To make the bulges disappear you would need to apply opposite forces varying proportionally to the radius (required centripetal force: ω˛r)
And if we made the ball rotate with a sling (pushing inwards on its outer side), instead of bulges the ball would get slightly flatter. As before, required centripetal force would be bigger at outer side, but the now compression internal stresses would be also bigger there, because they accumulate outwards …
All that shows how much the distribution and type of forces causing the circular movement affect internal stresses, and logically deformations and tidal effects.
In the case of the Earth and moon, there is no centripetal force to remove, so the bulges remain - they are caused by something else.
That is also erroneous! The whole earth, in order to be able of making the moon rotate, is revolving around the barycenter, similarly to the waist movement of a child playing “hulla-hoop” …
And centripetal forces (and centrifugal ones) necessarily have to be present, they affect internal stresses, and due to that also affects the dynamics of tides.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 30/07/2018 10:41:36
(required centripetal force: ω˛r)
Sorry ... That´s actually the required centripetal acceleration. It could also be considered as centripetal force ... but per unit of rotating mass !!
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 30/07/2018 17:55:05
But in the real case, M and A neither are free whatsoever, nor “know” how much E is pulled by the moon, let alone that material stuff “knows” how to subtract pull vectors, in order to directly react to gravitational differences !!

The Earth's material applies a gravitational pull upon itself, pulling it into a ball. Its rotation causes an equatorial bulge and polar flattening. The moon's gravitational pull then adds to this and causes two bulges on opposite sides. They all add together and none of the material knows or cares where the gravity comes from that is acting on it - it simply moves to where the total force takes it.

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As you say, "the moon's gravity is strong enough to make the Earth wander round in a circle”, but that movement also requires suitable initial “tangential” speeds of both earth and moon. And those speeds, and their continuous change of direction due to the rotation/revolving, also affect internal stresses, what inevitably intervenes on tides.

If you apply an even force to something, you don't get any such effect. you only get the tidal bulges because it is not an even force, but one that falls off over distance. The bulk of the force on each point is the same and generates no stresses - it's only the excess at the near side and deficit at the far side that causes them.

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You cannot say tides are due only to moon pull differences. The ways M and A react to the pull of the moon at their different distances depend on the let us say  degree of freedom the rest of the planet allows them …

The Earth's tidal bulges are precisely caused by the differences caused by the moon and sun's gravity. All the other distortions which these add on top of are not included as tidal effects. As the Earth rotates, the tidal bulges move, but the other distortions remain (ignoring the stresses caused by the mass of water changing as the bulges move - where I live, the land I'm standing on bends downwards measurably at high tide under the weight of the mass of water).

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If, e.g., we analyzed tidal effects at the moon, we would find that the answer of the whole moon to earth¨s gravity is different than the opposite we are dealing with. Moon is tidal locked to earth, and further parts of it require bigger centripetal acceleration than closer ones, for its actual rotation. But earth only revolves around the barycenter: all its parts require identical centripetal force.

It is not centripetal force - the force remains in place when rotation is removed. The moon is tidally locked to the Earth by the same mechanism of the Earth's gravity falling away over distance. The moon has no ocean to produce bulges, but it has the equivalent set in stone in its mass distribution, thereby allowing its rate of spin to be locked at one rotation per orbit. Because its orbit isn't circular, it doesn't rotate at a constant rate, but wobbles, and the corrections are made by the gravitational tidal forces. Because the Earth's gravity is much stronger than the moon, the difference in force across the moon is bigger than the difference in force across the Earth, so if the moon had an ocean it would have much bigger tide bulges. It doesn't though, so all we have is a wobbling rock which is continually moving away from its average alignment and being corrected back towards it, with stresses generated as a consequence.

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A single cause such as gravitational differences can´t be the unique reason of tides, because  in mentioned two cases the field of total force acting on each unit of mass of the earth are the result of differently caused individual forces ...

The tidal part is entirely driven by the gravitational force from the other body decreasing over distance. Any other forces acting are not contributing to tides - the tides are acting on the top of those.

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If you don't think the fall off in gravitational strength over distance is enough to account for the tides, the moon would apply it's backward pull to the whole Earth and its ocean and pull them all backward equally rather than doing so more strongly on the nearest part (creating a bulge there) and less strongly on the furthest part (creating a bulge there too).
That´s logically flawed.

No - your understanding of it may be logically flawed, but what I said is correct. If the moon's gravitational force acts so evenly across the whole Earth that it isn't responsible for the tides, there will be no bulges produced by placing the two bodies close to each other (while they fall together), and, incidentally, you also lose spaghettification of objects falling into black holes. If we put the two in orbit around each other instead, again there will be no tidal bulges because the moon's gravitational force will be too even across the whole Earth and its ocean to produce them. There is no centripetal force to make up the difference because acceleration by an even force field does not produce centripetal force. Remember the glass of water on a desk being accelerated sideways by such a field from zero to 100m/s in an instant and back to zero again the next - the water would not slosh about at all, but would sit with a flat surface at the top throughout. The same applies if you move this glass of water round in a circle by applying an even force field to it, and you should realise that when you repeat the straight line version of this experiment to move the glass north, then east, then west, then south in quick succession, each time doing so by applying an even force field to it. The glass moves round the corners of this square without the water moving relative to it at all. No centripetal forces are acting in a case like this. To send the glass round in a circle at a constant speed of 100m/s, all we need to is apply two force fields to it at 90 degrees to each other, each one accelerating it up to 100m/s on one axis, then decelerating it to zero on that same axis, then to -100m/s on that same axis, then back to zero again. No sloshing of the water - the surface remains flat at all times. That is what happens with the moon and Earth's gravity when they act on each other, except for the crucial difference that the strength of that gravity diminishes with distance, and that's what generates the bulges in the ocean.

(Incidentally, the bulges on an Earth with no land and equally deep water everywhere would be tiny - we only have big bulges where they run up against coasts. I've heard that at Fiji there is no noticeable tide at all.)

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What I say is that gravitational differences can´t be the UNIQUE cause of tides, that they intervene on the formation of tides, but only as one of the sides of a more complex “coin”.

If you take away the gravitational difference caused by it diminishing over distance, you will have no tides at all. The complexities that you are imagining do not contribute to the tides, even though centripetal bulge does add to to the degree to which the sea bulges away from a spherical shape - the tides are just the distortions added to that already-distorted surface and are not the total distortion away from a sphere.

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And it is absurd to deduce from that, if it were right, "the moon would apply it's backward pull to the whole Earth and its ocean and pull them all backward equally …”

You may find it absurd, but it is correct regardless. I refer you back to the glass of water experiment with the even force field being applied - no bulges and no sloshing. You simply aren't modelling this correctly in your head, and that's why you're imagining absurdities where there are none.

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We could imagine bizarre situations applying artificial forces to change the scenario, and get something similar to actual tides for a short period of time … But that wouldn´t mean our billions of year old natural tides are due only to causes identical to the ones at those situations ... 

These bizarre situations are designed to pin down precisely which parts of a system are responsible for which results. The bulges are caused by precisely the same thing, and that is the difference in gravitational pull over distance in accordance with the inverse square law. The only reason tides continue in a relatively constant way (in terms of bulge size) for millions of years is that the moon's orbit prevents it being ended by a collision. If we were to put the moon into an orbit where it comes much more directly towards the Earth, then swings round it and heads back out to the far point again, we would see the bulges in our ocean grow in size as the moon sweeps in towards us, then they'd diminish as the moon leaves, and if you then change the orbit again to make it a straight line leading to a collision, the bulges would grow in a near-identical way as the moon approaches even though all the rotation has been removed from the system. And we know this, because if the force was evenly applied instead, there would be no bulges generated in any of these cases, just like the glass of water being moved around by even force fields.

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If you are whirling a ball round your head on a string and then apply uniform forces to eliminate the movement in the same way, what happens to the bulges? They disappear as the ball springs back into a sphere.
That is erroneous, because outer parts of the ball require bigger centripetal force than inner ones (the ball rotates, as moon´s case …).

Then stop the movement with a force that acts more strongly on the furthest out part of the ball - the aim is simply to stop the orbit to reveal the truth about what was causing the distortion. When you stop the orbit of the ball on the string, it pings back to a spherical shape. When you do the equivalent with the Earth and moon, no such return to spherical shape would occur because the cause of the distortion is still applying in full.

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To make the bulges disappear you would need to apply opposite forces varying proportionally to the radius (required centripetal force: ω˛r)

To make the bulges disappear from the ball, it is sufficient to stop it moving and have the string go slack.

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All that shows how much the distribution and type of forces causing the circular movement affect internal stresses, and logically deformations and tidal effects.

What it shows is how centripetal force acts on things, and you're the one proposing that it causes tides. It doesn't have any role in causing tides though, so none of these complications apply.

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In the case of the Earth and moon, there is no centripetal force to remove, so the bulges remain - they are caused by something else.
That is also erroneous!

The error is not in what I said, but in your understanding.

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The whole earth, in order to be able of making the moon rotate, is revolving around the barycenter, similarly to the waist movement of a child playing “hulla-hoop” …
And centripetal forces (and centrifugal ones) necessarily have to be present, they affect internal stresses, and due to that also affects the dynamics of tides.

That's where your entire misunderstanding comes from. You have recognised a similarity between two systems and made the mistake of imagining that they must work the same way, but they don't. An even force field being applied to something does not generate any stresses in it and does not distort it. This makes it possible to move something in a circle without anything being flung outwards or pulled in by centrifugal/centripetal force.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 31/07/2018 19:25:22
I consider you talk a load of rubbish ! ...
I´m not going to refute all that. It´d be useless. I thought we could "solve" our Physics differences, but I see we have also a quite different mind set, and I can´t agree with your Logics. That is much more difficult to solve ...
Just one question. You say:
"If you take away the gravitational difference caused by it diminishing over distance, you will have no tides at all. The complexities that you are imagining do not contribute to the tides, even though centripetal bulge does add to to the degree to which the sea bulges away from a spherical shape - the tides are just the distortions added to that already-distorted surface and are not the total distortion away from a sphere".
You seem to refer ("already-distorted surface") to the equatorial bulge caused by daily earth spinning.
That is caused by inertial effects. Water, with a linear speed tangent to equator (if exactly there), tends to follow the tangent. But own earth gravity obliges the water to follow a circular path.
In this case, without any significant gravity differences, we have a relatively huge permanent bulge, because centripetal/centrifugal forces are proportional to the square of angular speed, some 29 times bigger than in the moon-earth "dance"
Somewhere you say tides occur on top of that permanent equatorial bulge. Quite right.
But, why earth daily spinning can cause that permanent equatorial bulge, and (according to you) the some 29 day revolving of the earth around the barycenter (please don´t mix with daily earth rotation !!), besides actual gravity forces on each location, can´t intervene in tides formation ??
In this last earth movement, water (and solid parts too) are also somehow obliged to follow circular paths, instead their inertial tendency to follow the tangents ... Therefore, centripetal and centrifugal forces are also present, and they do intervene on the 29 day cycle of moon-éarth dynamics, and on lunar related tides !!
There is an image:
main-qimg-851d9284749b378191c8ae87e4e2e4c2
that can help guys understand better moon-earth dance, especially earth revolving around the barycenter.
By the way: the image has a detail a little bit erroneous ... Any guess?

Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 31/07/2018 19:29:24
There is an image:
main-qimg-851d9284749b378191c8ae87e4e2e4c2
that can help guys understand better moon-earth dance, especially earth revolving around the barycenter.
By the way: the image has a detail a little bit erroneous ... Any guess?
A Google page appears. Please click on images ...
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 31/07/2018 21:51:58
I consider you talk a load of rubbish ! ...

That's why the world's in such a mess - the people who talk the most sense aren't recognised by the masses who can't work out what makes sense and what doesn't, so all the decisions are made by majorities of people with faulty understandings.

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I´m not going to refute all that. It´d be useless.

To try to correct something correct is indeed a fruitless endeavour (unless you realise that it is correct).

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I thought we could "solve" our Physics differences, but I see we have also a quite different mind set, and I can´t agree with your Logics. That is much more difficult to solve ...

If you aren't prepared to break down the system into component mechanisms properly to find out precisely what causes what, how are you ever going to know if you're right or wrong? I'm just trying to help you do that. The key thing for you to understand is that when an even force field is applied to an object that's free to move, every single particle in it will be pulled with equal force, so this adds no stress within that object at all. You could use such an even force field to make the object travel in a circle, and all the particles would respond exactly the same amount and move in exactly the same direction without generating any stress at all. That's what an orbit essentially does using gravity (except for the fall off in strength of gravity over distance, without which there would be no tides). That's all you need to understand to see why centripetal force is not part of the mechanism.

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But, why earth daily spinning can cause that permanent equatorial bulge, and (according to you) the some 29 day revolving of the earth around the barycenter (please don´t mix with daily earth rotation !!), besides actual gravity forces on each location, can´t intervene in tides formation ??

The daily spinning has already had its full effect - the Earth's rotation causes a distortion and that distortion is fairly constant, only diminishing very gradually over time as the rotation slows. The revolving around the barycentre is irrelevant for the reason set out in my previous paragraph - an even force being applied gives no scope for stresses to be generated in the material to which it is applied if all that material is free to move. The Earth and moon are both free to move, so you cannot put any stress into them by applying an even force to either of them, and that means you can't distort them that way - no bulges.

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In this last earth movement, water (and solid parts too) are also somehow obliged to follow circular paths, instead their inertial tendency to follow the tangents ... Therefore, centripetal and centrifugal forces are also present, and they do intervene on the 29 day cycle of moon-éarth dynamics, and on lunar related tides !!

The reason you think I'm talking nonsense is that you have put a faulty idea into your model of reality which is causing you to make mistakes whenever you try to build on top of it. None of the material of the Earth of moon feels as if it's travelling round anything in a circle - it all feels as if it's going in a straight line (ignoring spin). Imagine a case where the spin is removed. All the material of the moon is following a circular path which is exactly the same shape and size for ever single particle of the moon, and all the material of the Earth is following a much smaller circular path which is exactly the same shape and size for every single particle of the Earth. If you imagine this orbiting system is maintained by an even gravitational force being applied to each body by the other, you have no means whatsoever of generating tidal bulges. You incorrectly suppose that the mere fact of these bodies moving along circular paths will try to fling material outwards in the same way as happens with a ball on a string being whirled round your head, but that is based on a fundamental misunderstanding where you equate two different mechanisms and assume that the stresses generated by one must also be generated by the other, but they aren't. There is no location on a non-spinning planet or moon where the orbital path of that body would be detectable by any apparatus designed to measure stress if the gravitational force was applied evenly.

I very much doubt you'll believe that even now, so I'll give you even more help. Imagine four rockets in a square formation near a little asteroid with insufficient gravitational strength for them to orbit it in any useful amount of time. If we want the four rockets to travel round it in a circle while maintaining their square formation and without rotating that formation, what do we do? We make them all put out the same amount of thrust in the same direction at all times, so they maintain the same distances to each other and the square maintains its alignment with the stars while it travels round and round the asteroid. The square does not bulge. Now compare that with a non-spinning moon and non-spinning planet with an even gravity field being applied by each to the other, and you have exactly the same thing - no bulges appear because every particle has the same force applied to it from the other body (and applied in the same direction). We have the two bodies moving on circular paths without centrifugal/centripetal force having any ability to act.

Now replace the even gravity field with one that falls away in strength over distance, and what happens? Tidal bulges suddenly appear, and we can see very clearly the mechanism that drives them. Take away the orbits and the bulges are maintained by the mere proximity of the two bodies alone, revealing that the bulges are not caused by the circular paths. If you were to take the moon further away from the Earth and then release it to let it fall onto the Earth, the bulges in the ocean would grow as the moon falls towards the Earth, and at the point where the moon and Earth are 250,000 miles apart, these bulges would be the same size as they are when the moon is orbiting the Earth at that distance.

You can lead a horse to water, but you can't make it drink. I have taken you to the water and there's nothing more that I can do for you if you aren't able to drink.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 01/08/2018 13:58:30
As I did yesterday, I´ll quote only one paragraph from all that rubbish, today the first one.
That's why the world's in such a mess - the people who talk the most sense aren't recognised by the masses who can't work out what makes sense and what doesn't, so all the decisions are made by majorities of people with faulty understandings.
You are wrong even on the “social” history of the issue.
What actually happens is the other way around. Your model is even called by many the standard model, because many people like you, physicists included, keep in their minds erroneous ideas already corrected by the few … My stand here, since my very first post #20 on 29/05/2015 (!!), has most frequently gone against the grain, always defending the importance of centrifugal force on the existence of tides, especially on the hemisphere furthest from the moon.
But I´m not alone, what in its case would clearly mean I was the wrong. In several moments I even contacted eminent scientists, directly by email, to discuss some details or ask some questions. And I discovered interesting details of the huge confusion that exists, and has existed, out there ...
Just as an example, the first reply from a NOAA scientist was: 
"Based on the topics and points of discussion, you are obviously someone with a high level of science, physics, and mathematical education, training and understanding.
Unfortunately, this level of knowledge and understanding can sometimes work against you - as you look for a "deeper" level of explanation, or do not understand why some concepts that you understand are not included in the descriptions.
The publication you are referring to is "Our Restless Tides", a 10-page pamphlet developed in the 1950's to provide a basic description of the forces which create the tides.  It's intended audience were the grade school children and adults of that time.  It used terminology of science and forces which were common in the 1950s.  Such as centrifugal force.  Centrifugal force was always an "imaginary force" (not a real / measurable force).  But that type of description made the concepts easier to understand and explain.  That  description and use of centrifugal force continued to be common practice until the 1970-80's.  At that point, the terminology shifted and the textbooks used in grade schools were changed to use a more modern terminology and description of this "effect" being a result of inertia rather than an "imaginary force”.
[/b]
I´m afraid your model is somehow “contaminated” by ideas which are several decades out of date, what added to your to me "absurd" logic make you say all that rubbish !!
By the way, without entering the amazing ideas of Einstein on deep nature of gravity, we should not forget he started from his idea that inertia and gravity are the same (I´d personally rather say “almost” the same …), derived from their almost identical effects …
Once again, how gravity forces could intervene on the formation of tides, but inertial ones (such as centrifugal forces inherent in circular movements) could not ??
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 01/08/2018 19:11:17
As I did yesterday, I´ll quote only one paragraph from all that rubbish, today the first one.

What rubbish? I've analysed the system for you and I've done so correctly, illustrating each point carefully. Just labelling it "rubbish" isn't adequate. You need to pick it apart and show where it's broken, but you'll find that you can't because it isn't broken. When I tell you that an even gravitational field applied to an object acts equally upon every single particle of that object, I mean exactly that, and I'm also right. Such a field does not generate stresses in that object if the object can move freely - the object simply accelerates in the direction the force makes it move in and it does so without taking on any distortion. That makes it possible for the object to move in a circular path without any tidal bulges because there is no centripetal/centrifugal force involved at all. The way to produce the bulges is to change it from an even force to one that varies over distance (as it spreads out away from the source), and that is how tides are generated. It's that simple, but you are incapable of understanding that because you are determined that your explanation must be right and that any rival explanation therefore must be wrong because it opposes yours. You're shackled in your thinking by your inability to reject an incorrect idea that you are so emotionally tied to that you cannot let it go, no matter how clearly it is shown to you that you're wrong.

Now, you could make that same accusation to me in an attempt to defend your position, but I'm not emotionally tied to this - I don't care what wins out as the cause of tides other than wanting that winner to be the correct answer, and I approach everything else in science and every other field on the exact same basis - you should never develop an emotional bond to a theory, but be ready to ditch it in an instant as soon as a fatal flaw appears with it. That means that I test the proposed mechanisms to see if they really do what's claimed of them. If you look back a few posts (to reply #144), you'll see that this is absolutely true - I shifted position radically when I realised why you thought centrifutal/centripetal force had a role in tides because I suddenly saw such a strong parallel between the two systems (ball swinging round in circles at end of a string, and planet/moon held in orbit by gravity acting like invisible strings) that I thought the difference in the proposed mechanism might be an artificial one caused by an inadequate technical definition of centripetal/centrifugal, and I was happy to think that you might therefore be right. However, when I wrote my next post (reply #146), I ruled that out because I realised that there was a key difference between the two cases in that the bulges remain in the gravity case when the rotation is removed, thereby proving that it cannot be driven by centripetal/centrifugal force. That was the point where the case was proved, and everything since then is about you failing to get that point while you remain superglued to a disproved position.

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You are wrong even on the “social” history of the issue.

No I'm not - I see monkeys making bad decisions practically everywhere. There's a reason why China is heading for world domination, because they have people with a significantly higher level of expertise making decisions there without having to pander to people of lesser minds. They still make woeful decisions most of the time, of course, because even those people with higher expertise are poor thinkers, but they are better than average, and that adds up to a big economic advantage over time.

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What actually happens is the other way around. Your model is even called by many the standard model, because many people like you, physicists included, keep in their minds erroneous ideas already corrected by the few …

You don't have to look far around this forum to find that I do not follow the herd. When it comes to gravity, I am now leaning most strongly towards the idea of it not being a force - it is just a slowing of the speed of light in the proximity of mass, as that's all it takes to change the course of objects (and of light), and to change their speed of travel (so there is no need for it to curve space in the way proposed by GR). When I refer to gravity as being a force in this thread, it doesn't mean that I believe it is a force, but the way a force acts would create the same results as the more complicated explanation without getting bogged down in unnecessary technical details as to the actual mechanism by which gravity produces orbits. (In the same way, a believer in GR might also use the idea of gravity being a force in this discussion rather than going into unnecessary complications about curved space and the orbiting body moving in a straight path at all times through non-Euclidean Spacetime.)

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My stand here, since my very first post #20 on 29/05/2015 (!!), has most frequently gone against the grain, always defending the importance of centrifugal force on the existence of tides, especially on the hemisphere furthest from the moon.

Lovely, but who cares about the grain? I don't care how much support (or how little) an idea gets from troops of monkeys. The only thing that matters is whether it's right or not, and to find that out it's necessary to test the idea properly to see where it breaks (or doesn't). Your idea breaks at the point where the tidal bulges remain after the orbital rotation is removed from the system, thereby revealing that the rotation from the orbit had absolutely no role in producing the bulges, with the result that the cause cannot be attributed to centripetal/centrifugal force.

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But I´m not alone, what in its case would clearly mean I was the wrong. In several moments I even contacted eminent scientists, directly by email, to discuss some details or ask some questions. And I discovered interesting details of the huge confusion that exists, and has existed, out there ...

There are many eminent scientists who are wrong on many things. Nothing can be taken on trust on the basis of authority - everything has to be tested for what it is itself without biasing the results by factoring in dogma and other baggage.

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Just as an example, the first reply from a NOAA scientist was: 
"Based on the topics and points of discussion, you are obviously someone with a high level of science, physics, and mathematical education, training and understanding.

That is a standard, polite wording. The thought going through the mind of the author of such phrases runs along the lines of, "How can I get rid of this tiresome pest while pretending to respect him?"

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The publication you are referring to is "Our Restless Tides", a 10-page pamphlet developed in the 1950's to provide a basic description of the forces which create the tides.  It's intended audience were the grade school children and adults of that time.  It used terminology of science and forces which were common in the 1950s.  Such as centrifugal force.

It is common in education to provide false explanations of things in an attempt to simplify them, so the higher you go with your studies, the more times you will have such explanations thrown out for you to be replaced with better false explanations, and if you get far enough you might eventually be lucky enough to be see some true explanations.

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Centrifugal force was always an "imaginary force" (not a real / measurable force).  But that type of description made the concepts easier to understand and explain.  That  description and use of centrifugal force continued to be common practice until the 1970-80's.  At that point, the terminology shifted and the textbooks used in grade schools were changed to use a more modern terminology and description of this "effect" being a result of inertia rather than an "imaginary force”.[/i]

Here, we see that the scientist may not have a firm understanding of the tides himself though - all he's doing here is trying to get rid of the idea of centrifugal force by referring to the real mechanism that is behind centrifugal force, but he hasn't realised that this isn't the mechanism behind the tides.

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I´m afraid your model is somehow “contaminated” by ideas which are several decades out of date, what added to your to me "absurd" logic make you say all that rubbish !!

I can't work out if that's still part of what the scientist said to you, or if you're saying it to me. What I'm giving you is direct analysis where I look at the role of rotation (the orbit) and see what happens when that orbit is removed. The tidal bulges remain, so the rotation cannot have a role in producing them, and that means centipedal/centrifungal force has no role in causing tides.

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By the way, without entering the amazing ideas of Einstein on deep nature of gravity, we should not forget he started from his idea that inertia and gravity are the same (I´d personally rather say “almost” the same …), derived from their almost identical effects …

Einstein saw a parallel and read too much into it, just as you have done. SR and GR both fail to function correctly in accordance with their specifications - all simulations of them have to cheat as a result in order to maintain the pretence that the model works. But we don't need to go into that side issue - it is sufficient to prove the case that the tidal bulges remain when there is no rotation at all.

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Once again, how gravity forces could intervene on the formation of tides, but inertial ones (such as centrifugal forces inherent in circular movements) could not ??

Centrifugal forces are not inherent in all circular movements - that's where you're continuing to fail to understand this. You are trying to have them act in a case where they do not act. A ball whirling round your head on the end of a string feels a force - the whole of it feels the force. A moon going round a planet would not feel any force acting on it at all if the force between them was even, acting with the same strength on each particle (and applying in the same direction to each). It's only when you make that force uneven that you get distortion, and that distortion is not dependent on orbiting - the exact same distortion applies in a straight line system.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 02/08/2018 14:50:02
There are many eminent scientists who are wrong on many things. Nothing can be taken on trust on the basis of authority - everything has to be tested for what it is itself without biasing the results by factoring in dogma and other baggage
When I said I directly contacted several scientists to discuss details of the issue, I meant I did tell them the things they had said I didn´t agree with, or at least I didn´t understand ... And what I´ve been telling you is the result of both my initial ideas, and all those discussions, some of them really fruitful ... It´s not just "trust on the basis of authority" !! I do know "There are many eminent scientists who are wrong on many things" …
Though I wonder if you really think a scientist who ...
" ... spent most of his career in NOAA and much of his time working on tide related
problems as a specialty even while tackling jobs with a much broader scope ...
"
and held positions at NOAA such as:
"Chief Scientist of the National Ocean Service; Director of the Coast Survey Development Laboratory; Director of the World Data Center for Oceanography; Principal Investigator for the NOAA Global Sea Level Program; and head of the U.S. national tides and currents program (in a earlier organizational form before it became CO-OPS)",
if being clearly on my side (about the discussed issue), instead of on yours, still he could be wrong on the very core of the issue ...
And what you also said:
"Just labelling it "rubbish" isn't adequate. You need to pick it apart and show where it's broken, but you'll find that you can't because it isn't broken”
should not be applied to my case. I´ve been refuting many single things said by you (#136,138,141,142,145,147,149,152… - only then i started mentioning the word “rubbish”! ), but you keep on repeating same things, without any change connected to what I´d previously said … It seems you have not read all my posts !!
By the way, you said too:
"The key thing for you to understand is that when an even force field is applied to an object that's free to move, every single particle in it will be pulled with equal force, so this adds no stress within that object at all" ...
I´d rather say, as I´ve said many times, but with other words:
"The key things for you to understand are that” …, even if moon´s pull were uniform across our planet,
1) earth, as it´s not “free to move” (moon and earth c.g.s have to keep their distance, and the barycenter has to follow its orbit around the sun), it would experience additional internal stresses ...
2) if due to initial conditions earth parts were somehow following not rectilinear paths, centripetal and centrifugal forces would be occurring ...
3) if those paths were as they actually are (circumferences with ALL earth parts continuosly at maximum distance from the moon), inertia would tend to make water follow the tangent, earth own gravity would avoid it … and we would get the further hemisphere bulge ... as I said even if moon pull there were the “uniform” average !!   
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 02/08/2018 17:42:01
Hi David,

I agree with you about the bulges, but it is not clear how a small vertical motion in the middle of the oceans can transform into huge ones on the coasts. That's what tsunamis do, but they are considered as a fast wave that breaks on shallow waters, so they can break in any direction from the impact point, while the two opposing tidal bulges are only going west and could only break on eastern coasts if they were a wave, which is not what is observed. I didn't read all the thread, so maybe you already talked about that, but if not, can you give me your explanation please?

Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 02/08/2018 20:04:52
Though I wonder if you really think a scientist who ... if being clearly on my side (about the discussed issue), instead of on yours, still he could be wrong on the very core of the issue ...

It would be fully possible for a scientist who's an expert on tides to have a gap in his knowledge about the fundamental cause of them (or for his knowledge to be built upon an error which makes no importance to the science that he's doing). It isn't clear from your quote though whether he has such a gap or fault, because what you quoted was simply him objecting to centrifugal force being regarded as a real force, so he wasn't actually addressing the issue of the cause of tides.

If you want to find out who's right about the fundamental cause of tides, you need to produce a model that replicates tides when you run a simulation of them, and that model has to be compatible with reality rather than contradicting it. If you apply an even force to an object in a simulation and the object is free to move, it won't distort - you can send it round a circular path with such a force being applied from the centre and there will be no bulges produced. You can also simulate the way the bulges grow as the moon moves towards the Earth on a straight path and show that no rotation is necessary to generate them. I can do this in my head without difficulty, but I know that this is a skill which is not shared by everyone, and indeed that only a tiny minority can do it well.

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And what you also said:
"Just labelling it "rubbish" isn't adequate. You need to pick it apart and show where it's broken, but you'll find that you can't because it isn't broken”
should not be applied to my case. I´ve been refuting many single things said by you (#136,138,141,142,145,147,149,152… - only then i started mentioning the word “rubbish”! ), but you keep on repeating same things, without any change connected to what I´d previously said … It seems you have not read all my posts !!

The reason I keep repeating things is that you keep repeating the same incorrect things instead of fixing your mistakes, so I keep trying to help you recognise where you're wrong by spelling things out more and more clearly for you. I can't force you to recognise an error if you're incapable of recognising it though, so there is no guarantee that you will ever understand this stuff. It appears that running simulations with precision in your head is not something you can handle, so perhaps we need to move to using computer programs to run the models openly instead so that it's harder for you to go on denying reality.

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...even if moon´s pull were uniform across our planet,
1) earth, as it´s not “free to move” (moon and earth c.g.s have to keep their distance, and the barycenter has to follow its orbit around the sun), it would experience additional internal stresses ...

Of course it's free to move! The Earth's sitting in space without being moored to anything - it will simply accelerate towards anything that applies a gravitational force to it. The Earth, moon, sun, other planets, etc. are all falling through space without any strings tied to them - they are free to go wherever they're flung. They don't have to keep their distance from each other, but simply fall uncontrolled, but fortunately, if they miss each other at closest approach, they naturally produce orbits which keep repeating the same pattern.

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2) if due to initial conditions earth parts were somehow following not rectilinear paths, centripetal and centrifugal forces would be occurring ...

Why? With an even force applying, there is no centrifugal force involved because every particle is accelerated the same amount in the same direction, thereby ensuring that no stress is generated.

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3) if those paths were as they actually are (circumferences with ALL earth parts continuosly at maximum distance from the moon), inertia would tend to make water follow the tangent, earth own gravity would avoid it … and we would get the further hemisphere bulge ... as I said even if moon pull there were the “uniform” average !![/b]

Why should water follow the tangent when the Earth doesn't follow the tangent? Why are you applying one rule to the Earth and a different rule to the water sitting on it?
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 02/08/2018 20:18:48
but it is not clear how a small vertical motion in the middle of the oceans can transform into huge ones on the coasts.

The Earth's rotation moves the bulges towards the coasts which block their way, so they bunch up there. It's not nearly that simple though, as I'll explain further down.

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That's what tsunamis do, but they are considered as a fast wave that breaks on shallow waters, so they can break in any direction from the impact point, while the two opposing tidal bulges are only going west and could only break on eastern coasts if they were a wave, which is not what is observed.

These bulges are tiny and don't add up to the size of tsunamis. What you actually get is resonance in ocean basins and seas, and in different parts of the world you can get no tides at all, one tide per day, two tides per day, four tides per day, ... 128 tides per day. Think about a young child on a swing - you can push them each time they come back to you, or you can push them every second time, but they swing at the same rate regardless - the basin determines how many high tides a day it can handle, and the rotation of the earth puts in regular nudges to maintain the oscillation.

Now let's think about the size of the bulges. To make a big bulge, you have to move a lot of water, but we simply don't see any significant amount of water rushing around as a bulge crosses an ocean. The bulge hardly shows up at all. What must actually happen is that you get a reduction of pressure as the sea tries to lift, but water is so incompressible that it can change in pressure a lot without any measurable change in volume. These "bulges" thus manifest themselves as pressure changes rather than actual bumps. These pressure changes then serve to drive currents involved in the tides that we see at coasts where the water does move about and change in height a lot, but this water is only moving at a few miles per hour and not at the speed of the "bulges" which travel at the speed of an aeroplane.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 02/08/2018 20:50:55
The Earth's rotation moves the bulges towards the coasts which block their way, so they bunch up there. It's not nearly that simple though, as I'll explain further down.
Wouldn't they only bunch on the east coasts of the emerged lands if it was the case?

What you actually get is resonance in ocean basins and seas
If I understand well, the tidal wave would get amplified in a basin the same way a sound wave is amplified in a trumpet. If the trumpet or the basin is longer, the frequency is lower, and vice versa. It makes sense, but it's hard to imagine. It means that if we were to begin such a cycle, the first one would give a low tide, and the following ones would give higher and higher tides until friction losses equalize the gravity nudges. I think it also answers my first question.
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 03/08/2018 08:30:24
Wouldn't they only bunch on the east coasts of the emerged lands if it was the case?
No, various reasons. Firstly the pull of the moon/sun system passes over (or the earth rotates from under, depending on how you view it) and releases the bulge so it can fall back. Secondly, the sloshing backwards and forwards can due to the distance result in a wave superposition that results in the waves performing a seesaw movement, a form of resonance.  Also the movement of the bulge isn’t straight across a basin west-east but Coriolis effect turns the movement to the right (northern hemisphere). For example in the English Channel when the bulge moves west to east you get slightly higher tides on the south (french) coast and then on the north (english) when it goes back east to west. Motion in a basin is more difficult to explain without pictures, but it results in a bulge that rotates counter clockwise around a null point - look up amphidromic circulation and dynamic theory of tides for details.   

If I understand well, the tidal wave would get amplified in a basin the same way a sound wave is amplified in a trumpet. If the trumpet or the basin is longer, the frequency is lower, and vice versa. It makes sense, but it's hard to imagine.
This only happens in some very specific places where the geology works to produce a resonance - look up Bay of Fundy where length of bay causes a resonance with 12 hr tides hitting the entrance.
In many other places eg Atlantic the bulge is amplified by the wave hitting shallow water of continental shelf just like waves on a beach increasing in ht as the water shallows. It is also amplified due to amphidromic effects as the further it is from the null point, the higher the bulge will be.
As David says, the geography is responsible for the specific local tides and these will deviate from the equilibrium 2 bulges/day eg Solent has a double peak high tide due to water entering first the west then the east entrance.

However, as David pointed out in an earlier post, I think this is moving away from the objectives of the OP which is to discuss the specifics of the moon/earth effects of the equilibrium theory and the terminology used.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 03/08/2018 17:37:02
More thoughts:-

Going back to the mention of tsunamis, they are started at a point location and then spread from there carrying all the energy in a compact wave, but tidal "bulges" are generated across thousands of miles rather than at a single point, so you don't have the same intensity of impact at the coast - it's a much gentler business even if it involves a lot more total energy.

On the (undiscussed) issue of how much of the energy is newly added by the moon and sun's pull and how much might be a remnant of resonance from previous cycles, the variation between spring and neap tides shows that a lot of energy must be lost from the system on each cycle because there is a big variation depending on how the bulges generated by the moon and sun combine.

Note to rmolnav:-

If your mechanism was correct about the centripetal/centrifugal effect flinging the water out in two bulges at opposite sides as the Earth rotates around the barycentre, wouldn't that generate a single bulge on the side furthest out from the barycentre with the sea lowest on the side nearest to it? That would lead to most places having one tide per day instead of two. (You may have discussed this earlier, but due to the quantity of your writing here, it's hard to find out in a reasonable length of time.)
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 03/08/2018 21:47:58
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...even if moon´s pull were uniform across our planet,
1) earth, as it´s not “free to move” (moon and earth c.g.s have to keep their distance, and the barycenter has to follow its orbit around the sun), it would experience additional internal stresses ...
Of course it's free to move! The Earth's sitting in space without being moored to anything - it will simply accelerate towards anything that applies a gravitational force to it. The Earth, moon, sun, other planets, etc. are all falling through space without any strings tied to them - they are free to go wherever they're flung. They don't have to keep their distance from each other, but simply fall uncontrolled, but fortunately, if they miss each other at closest approach, they naturally produce orbits which keep repeating the same pattern.
You keep repeating erroneous things, I´m afraid that due to basic misconceptions …
If an artificial satellite is orbiting around the earth, it is due to suitable speed vector perpendicular to earth pull, what makes it follow an elliptical path. The whole earth pull is the centripetal force, and the satellite (considered as a whole) neither experiences nor exerts any centrifugal force. It´s reaction to earth pull (3rd Newton Law) is just to pull back on earth … what practically doesn´t affect our planet …
Earth-moon dynamics is something quite different, especially as far as our planet is concerned. Moon not only pulls earth, what makes its linear speed change direction (proportionally to the pull)… The moon also keeps moving in the opposite sense, what makes earth mentioned “answer” (the continuous  "bending” of its intended rectilinear path, due to inertia) “insufficient” … Earth path gets more curved than if it were going to orbit around a static moon: it is a much smaller circumference, and around the barycenter instead of around the moon ...
That´s why I consider that, rather than talking about “free falls” within a given gravitational field, it is more realistic to consider earth-moon interactions as if they were part of a single extended object, with no massive parts apart from where both celestial objects actually are, and rotating/revolving about the common center of mass, maintaining the distance between them due to their dynamic equilibrium.
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2) if due to initial conditions earth parts were somehow following not rectilinear paths, centripetal and centrifugal forces would be occurring ...
Why? With an even force applying, there is no centrifugal force involved because every particle is accelerated the same amount in the same direction, thereby ensuring that no stress is generated.
Repeating erroneous things does´t make them right !!
I don´t know how old are you (I´m 74), but you seem to have forgotten basic Physics boys usually learn when teenagers ...
Without a centripetal force an object can´t get a curved trajectory whatsoever !!
If you “ … use such an even force field to make the object travel in a circle”, that would be insufficient: you also need a certain initial speed of the object, and the field force has to have a component perpendicular to initial speed… As soon as that field started acting on the object, that very field force (or part of it) would become the centripetal force, whatever caused that field !! 
And if, for any reason, at some parts of the object the required centripetal force doesn´t match with the field force there, interactions with the rest of the object appear, that is, internal stresses !!
The bottom line would be that those internal stresses, exerted on considered part by contiguous material, and/or by part of gravitational pull exerted by own earth, would have to compensate the mismatch between moon pull there and required centripetal force (in one or in the opposite sense).
Those internal stresses would include "moonward" forces (centripetal), and outward forces (centrifugal, in a non restricted sense of the term).
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3) if those paths were as they actually are (circumferences with ALL earth parts continuosly at maximum distance from the moon), inertia would tend to make water follow the tangent, earth own gravity would avoid it … and we would get the further hemisphere bulge ... as I said even if moon pull there were the “uniform” average !![/b]
Why should water follow the tangent when the Earth doesn't follow the tangent? Why are you applying one rule to the Earth and a different rule to the water sitting on it?
Please, don´t twist my words. I didn´t say “water (on earth) follow the tangent”, but "inertia would tend to make water follow the tangent”, and exactly the same happens to earth as a whole … But neither of them is fully free, and their dynamics features (movement, deformation, inertia related reaction forces …) depend on each “scenario” ...
And, as water is relatively more "free" to move than earth solid parts, as I´ve said many times, bulges happen, as a result of both actual moon pull at each location, and also previously mentioned inertial effects !!
As I´ve also said many times, the "information" of differential moon pulls, between not contiguous parts of the earth, can exist in our intelligent minds, but cannot reach and directly be exerted on those parts !!
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 03/08/2018 22:52:31
Going back to the mention of tsunamis, they are started at a point location and then spread from there carrying all the energy in a compact wave, but tidal "bulges" are generated across thousands of miles rather than at a single point, so you don't have the same intensity of impact at the coast - it's a much gentler business even if it involves a lot more total energy.
Yes, the wavelengths are quite different. The way a tsunami is generated produces a wave pulse of a crest preceded by a trough which is why the sea recedes a considerable distance from the beach just before the crest hits. Again the shallowing causes the crest to increase in size.

the variation between spring and neap tides shows that a lot of energy must be lost from the system on each cycle because there is a big variation depending on how the bulges generated by the moon and sun combine.
Water creates very high damping in this system.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 04/08/2018 00:25:45
You keep repeating erroneous things, I´m afraid that due to basic misconceptions …
If an artificial satellite is orbiting around the earth, it is due to suitable speed vector perpendicular to earth pull, what makes it follow an elliptical path. The whole earth pull is the centripetal force, and the satellite (considered as a whole) neither experiences nor exerts any centrifugal force. It´s reaction to earth pull (3rd Newton Law) is just to pull back on earth … what practically doesn´t affect our planet …

You're failing to recognise the key difference. When you stop a ball on the end of a string from going round and round just above the ground and just let it sit on the ground at the end of the string, the deformation disappears. If you do the same with an orbiting object, the deformations remain because they are not driven by the orbit. That is why centrifugal/centripetal force cannot validly be regarded as the cause of tides - they remain in full without centrifugal/centripetal force acting at all.

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Earth path gets more curved than if it were going to orbit around a static moon: it is a much smaller circumference, and around the barycenter instead of around the moon ...

It hardly moves at all, but it doesn't matter how much it's moved by gravity, that doesn't make any force that can generate bulges. The bulges only occur because of the change in gravitational strength over distance.

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That´s why I consider that, rather than talking about “free falls” within a given gravitational field, it is more realistic to consider earth-moon interactions as if they were part of a single extended object, with no massive parts apart from where both celestial objects actually are, and rotating/revolving about the common center of mass, maintaining the distance between them due to their dynamic equilibrium.

If you're going to ignore existence of bulges during free falls, you're making it harder for yourself to see the key point that the are not caused by the rotation. All the orbit does is enable the Earth and moon to continue to operate for a long length of time without colliding. The bulges are not caused by the path that either body is moving in. If you want to create a bulge, you could actually do it by grabbing the Earth and forcing it to move in a straight line while the water (free to move) tries to continue to follow the orbit.

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2) if due to initial conditions earth parts were somehow following not rectilinear paths, centripetal and centrifugal forces would be occurring ...
Why? With an even force applying, there is no centrifugal force involved because every particle is accelerated the same amount in the same direction, thereby ensuring that no stress is generated.
Repeating erroneous things does´t make them right !!

Then you need to learn to stop doing it.

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I don´t know how old are you (I´m 74), but you seem to have forgotten basic Physics boys usually learn when teenagers ...

It should not be classed as centrifugal/centripetal force in an orbiting system because the force continues to act in full if you stop the rotation and is therefore entirely independent of that rotation, rendering the rotation incidental. (Physics got rid of the term centrifugal force because it was an incorrect way of looking at things, and if it is still common to use centripetal force in the description of orbits, that is another error which should be corrected - it is only acceptable as an analogy.) Most importantly though, the rotation has no role whatsoever in forming a bulge.

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Without a centripetal force an object can´t get a curved trajectory whatsoever !!

I don't accept that it's centripetal force other than as an analogy. That's entirely a side issue though, because the only point that matters here is that whatever we call it, it doesn't generate bulges It is not the mechanism for tides.

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If you “ … use such an even force field to make the object travel in a circle”, that would be insufficient: you also need a certain initial speed of the object, and the field force has to have a component perpendicular to initial speed… As soon as that field started acting on the object, that very field force (or part of it) would become the centripetal force, whatever caused that field !!

Same irrelevant disagreement. The important point is that it doesn't generate bulges.

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And if, for any reason, at some parts of the object the required centripetal force doesn´t match with the field force there, interactions with the rest of the object appear, that is, internal stresses !!

If you apply an even force, it will match everywhere and there will be no stresses. The mechanism for tides is that the force is not even, as is demonstrated by the straight-line case with no rotation.

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The bottom line would be that those internal stresses, exerted on considered part by contiguous material, and/or by part of gravitational pull exerted by own earth, would have to compensate the mismatch between moon pull there and required centripetal force (in one or in the opposite sense).
Those internal stresses would include "moonward" forces (centripetal), and outward forces (centrifugal, in a non restricted sense of the term).

The stresses are caused by the force not being even, and as I keep pointing out, they continue to apply if the rotation is eliminated, thereby revealing that they are not centrifugal/centripetal by any viable definition at all. If you have a planet with an ocean travelling in a straight line through space with no bulges in the ocean as there is no stress applied to it, if a spaceship then uses a tractor beam to cause the path of the planet to curve by applying an even force to it, no bulges appear despite the change in direction. If no change occurs between the beam not being applied and being applied, no part of the direction change can have anything to do with generating bulges.

Yes, this is repetition, but it's necessary because it's correct and you continually refuse to accept reality.

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3) if those paths were as they actually are (circumferences with ALL earth parts continuously at maximum distance from the moon), inertia would tend to make water follow the tangent, earth own gravity would avoid it … and we would get the further hemisphere bulge ... as I said even if moon pull there were the “uniform” average !![/b]
Why should water follow the tangent when the Earth doesn't follow the tangent? Why are you applying one rule to the Earth and a different rule to the water sitting on it?
Please, don´t twist my words. I didn´t say “water (on earth) follow the tangent”, but "inertia would tend to make water follow the tangent”, and exactly the same happens to earth as a whole … But neither of them is fully free, and their dynamics features (movement, deformation, inertia related reaction forces …) depend on each “scenario” ...

I'm not twisting anything. If you imagine that you have centrifugal force involved in making the Earth and moon orbit each other (in their dance around the barycentre) and you also have it involved in making the water of the ocean go along with the Earth, you have no mechanism for inertia to move the water any differently than the Earth. Your idea that the water should follow the tangent should naturally lead to the Earth needing to follow the tangent too, and because it isn't tied to the moon, it would be free to do so. You have failed to provide a mechanism for the water to behave differently from the Earth other than by applying different rules to each. The Earth is governed by gravity such that it follows a curved path, while the water is magically not governed by the same gravity such that it tries to follow the tangent. NO - both will follow the same path (unless the gravitational force being applied is not an even one, which is of course the case, and it's that unevenness that causes the tides).

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And, as water is relatively more "free" to move than earth solid parts, as I´ve said many times, bulges happen, as a result of both actual moon pull at each location, and also previously mentioned inertial effects !!

The water and Earth are exactly as free as each other to follow the same path as each other. You only get bulges from an uneven force, and then what you get is a change in pressure in the water (which we're calling a bulge) with stresses also generated right through the Earth.

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As I´ve also said many times, the "information" of differential moon pulls, between not contiguous parts of the earth, can exist in our intelligent minds, but cannot reach and directly be exerted on those parts !!

There is a gravitational force being applied to each particle of Earth and sea from the moon, and it tries to move each of those particles. Because the force isn't evenly applied, it pulls at some with greater strength than others, and that's what generates the bulges. Simple as that - not driven by any rotation.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 04/08/2018 13:52:19
@ Molnav,

In case it hasn't been presented, here is my two cents picture of the phenomenon.

Its uneven acceleration at different distances from the pulling body that creates the bulges. Put three balls side by side, accelerate them at three different rates in the same direction, and the distance between the balls is going to increase with time. That's what creates the bulges in the case of the earth, and what prevents them to increase forever is the earth's own gravity that also accelerates the bulges towards one another.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 04/08/2018 14:15:49
Put three balls side by side, accelerate them at three different rates in the same direction, and the distance between the balls is going to increase with time.
What quoted is right ... but not what you say afterwards.
If you have a look at my #149, you´ll see that, using same analogy, I said:
"Imaging three tennis ball size pieces of earth stuff, one were earth c.g. (E),  another just under the moon (M), and the other at the antipodes (A).
If we had them without the rest of our planet, they would be free to react to forces such as moon gravity … M would accelerate the most, and A the least. Distances between M and E, and between E and A, would increase, though M and A don´t even “know” moon pull on E. And certainly without needing any rotation or revolving.
But in the real case, M and A neither are free whatsoever, nor “know” how much E is pulled by the moon, let alone that material stuff “knows” how to subtract pull vectors, in order to directly react to gravitational differences !!
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 04/08/2018 14:52:44
I regret to have to say: more, and more, and more rubbish ...
One of the many differences between our stands is that you say ONLY differential pull causes the bulges, but I don´t say ONLY inertial effects related to earth revolving can cause the bulges ...
And with your absurd logic, if you find another scenario where deformations similar to the bulges occur without any curved movement, you wrongly deduce your stand is the right one!!
E.g.:
The bulges are not caused by the path that either body is moving in. If you want to create a bulge, you could actually do it by grabbing the Earth and forcing it to move in a straight line while the water (free to move) tries to continue to follow the orbit.
Whatever the imagined way you did that, it would be a different case, where same reason (inertia) would have similar effects because you change the speed vector of a massive body ...
But that doesn´t mean those inertial effects wouldn´t happen changing only speed vector direction, what is continuously happening to the revolving earth !!
As I thought yesterday, perhaps it would be good for you (and for many others, I´m afraid ...) to go back to basics.
If e.g you google "phisics.ohio-state.edu" and go to "Dynamics of Uniform Circular Motion" (Chapter 5, 5.3 Centripetal Force), you can read (apart from many other things and formulas):
"...Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration.
The centripetal force is the name given to the net force required to keep an object moving on a circular path
".

Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 04/08/2018 15:30:00
What quoted is right ... but not what you say afterwards.
Your example is more complicated than mine even if it is similar, so I will keep mine and add gravitation to it instead. Let my three balls be accelerated by the earth's gravitation while falling directly to it, and add mass to them so that they are also accelerated towards one another. We can give them a large mass so that they stay together while falling, then reduce it so that they get away from one another just a bit, or reduce it so much that they get away from one another a lot. The just a bit case represents the tides' case, with no input from forces due to rotation.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 04/08/2018 16:13:03
"...Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration.
The centripetal force is the name given to the net force required to keep an object moving on a circular path".
There is certainly a motion produced by the earth in the direction of the moon and vice versa, but this motion is not driven by a rigid link like a rope for instance, so contrary to the tangential one, it is not purely inertial. I have a theory that says particles must execute small steps to justify their inertial motion in space, but those steps cannot justify their gravitational motion, so some other steps have to be executed by the gravitating particles towards one another, and those steps have to be driven by the information exchanged between the two gravitating bodies, contrary to my inertial steps that are driven by the information exchanged between bonded particles. The information exchanged between gravitating particles nevertheless transforms immediately into steps similar to my inertial motion's steps, so that there is no more force involved than for a body on inertial motion. Force develops only when the steps between bonded particles get out of sync, which is far from being the case for orbiting particles since they have all the time they need to adjust to their orbital changes without getting out of sync.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 04/08/2018 18:40:02
But in the real case, M and A neither are free whatsoever, nor “know” how much E is pulled by the moon, let alone that material stuff “knows” how to subtract pull vectors, in order to directly react to gravitational differences !!

Picture three massive balls, M, E and A, which attract each other using gravity. They are aligned in a straight line with where we're going to place the moon. M and A are both sitting on springs, so they don't contact E directly, but they are indirectly sitting on it, and each is sitting at height h over the surface of E. Now we introduce the moon into the system. The three balls begin to accelerate towards the moon, and each tries to go at its own speed, but the local pull on M and A from E feels stronger than the pull from the moon, so they will remain close to E. What happens though is that M settles to a different altitude slightly higher than h above the surface of E because part of the Earth's pull on it is cancelled out by the extra pull on M that E isn't getting (that extra being the result of E not being pulled with quite as much force as M). A also settles to an altitude slightly higher than h above the surface of E because E is being pulled towards the moon more strongly than A is. Those are the tidal bulges (and in this case they are actual bulges rather than just pressure reductions).

As the three balls fall towards the moon, the difference in force between the moon and each successive ball grows stronger, so the altitudes of M and A above the surface of E will increase. None of the balls need to know anything - they simply react directly to the forces being applied to them. This is the mechanism behind the tides and it does not depend on rotation from orbits.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 04/08/2018 19:30:38
I regret to have to say: more, and more, and more rubbish ...

Then you need to learn to stop posting it.

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One of the many differences between our stands is that you say ONLY differential pull causes the bulges, but I don´t say ONLY inertial effects related to earth revolving can cause the bulges ...

Practically all the bulge involved in tides is produced by the mechanism I've described. Any potential additions to it will be trivial, but I haven't seen you propose anything that could serve as an addition. You magically have water wanting to follow the tangent while the Earth follows the curve in a case with an even force being applied, and that's plain bonkers (loco). In reality, for water to want to follow the tangent, it would have to have zero force applied to it by gravity from the other body, and that certainly isn't going to happen. What we actually have is a lesser force being applied by the distant body's gravity at the far side and a greater one at the near side, and those directly pull up the bulge on one side and pull the Earth away from what becomes the bulge at the other side. The orbital path is irrelevant and has no role in the causation, and you can tell that this is the case because the bulges are the same size at that separation distance whether there is an orbit or not. It is a massive error to attribute the bulges to a rotation when they are identical to the ones that are there without the rotation - the rotation clearly cannot have a role if it makes no difference to them. The biggest question that remains here for this thread though is how many more pages of posts will it take before you recognise that simple fact.

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And with your absurd logic, if you find another scenario where deformations similar to the bulges occur without any curved movement, you wrongly deduce your stand is the right one!!

It is an extension of the same scenario. The moon's orbit is growing in size, and some day it will become more linear, eventually reaching a point where it collides with the Earth (if the sun hasn't swallowed both of them first). At no point during this transition from near-circular orbit to more linear orbit and ultimately to a nearly straight line path to impact does the mechanism for the bulges change - it remains the exact same mechanism throughout.

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E.g.:

The bulges are not caused by the path that either body is moving in. If you want to create a bulge, you could actually do it by grabbing the Earth and forcing it to move in a straight line while the water (free to move) tries to continue to follow the orbit.

Whatever the imagined way you did that, it would be a different case, where same reason (inertia) would have similar effects because you change the speed vector of a massive body ...
But that doesn´t mean those inertial effects wouldn´t happen changing only speed vector direction, what is continuously happening to the revolving earth !!

Like I said, it's the same case and the mechanism doesn't change. The orbit aspect of it is irrelevant other than as a means to delay the collision. Going back to the three balls with springs holding M and A off the surface of E, the height of the balls above the surface of E is dependent on one single factor, and that is the distance to the moon (because it's that distance that determines how much extra force is applied to M over that applied to E, and how much more is applied to E over that applied to A.

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As I thought yesterday, perhaps it would be good for you (and for many others, I´m afraid ...) to go back to basics.

It would be a good idea if you took your own advice. Picture a centrifuge used to separate out a mixture of heavy and light material. It goes round and round to throw the material outwards, and the heaviest material ends up collecting together at the outside. Now picture a scientist with a poor grasp of physics who, in the interests of reducing maintenance and energy costs, tries to make a centrifuge using a black hole to make the container move round in the same size of circle at the same speed as in the earlier centrifuge. The container whirls round and round this circuit just like the other one, but the material inside it is weightless, so it simply doesn't separate out - it does not serve as a centrifuge, and the reason for that is that there is no force on the content generated by the rotation: no centrifugal/centripetal force is acting on it. It's the same with the water sitting on the Earth - there is no centrifugal/centripetal force acting on it either.

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If e.g you google "phisics.ohio-state.edu" and go to "Dynamics of Uniform Circular Motion" (Chapter 5, 5.3 Centripetal Force), you can read (apart from many other things and formulas):
"...Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration.
The centripetal force is the name given to the net force required to keep an object moving on a circular path
".

It doesn't matter how much justification you can find for calling it centripetal force. This is a grey area in which some usages of the word aren't fully rational. What is absolutely certain though is that centrifugal/centripetal force is not the cause of the bulges because the bulges remain when the rotation is removed. It is plain wrong to attribute them to a rotation which is not causing them.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 04/08/2018 22:29:20
I have a question about the tides that has not been answered yet, so I might as well post it here.

While the earth and the moon execute their common orbital trajectory with the sun, their common barycenter goes at the right speed all the time, but not their respective gravity centers: those are too fast when they transit on the far side, and too slow when they transit on the near side. They should thus behave as if they were at an aphelia on the far side, and at a perihelia on the near one, slowing down on their orbital trajectory with the sun and getting away from it progressively during the first half of their far side transit (and then doing the inverse during the second half), and accelerating and getting closer to the sun during the first half of their close side transit (and doing the inverse also during the second half), what should progressively increase and decrease the distance between them during their transit. I never heard of such an observation, but I can't find a flaw in the logic. Would it be too small to be observable?

Now if we apply this mechanism to the tides, we find that it would coincide with the height they get in the middle of the oceans, and that it would also coincide with the horizontal speed they get near the shores. This time, it may be observable, but how to distinguish it from the effect of the other mechanism if it only adds to it?
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 04/08/2018 23:45:14
While the earth and the moon execute their common orbital trajectory with the sun, their common barycenter goes at the right speed all the time, but not their respective gravity centers: those are too fast when they transit on the far side, and too slow when they transit on the near side.

They would be too fast or slow if they weren't also pulling against each other. If the moon is furthest away from the earth at the same time as it's directly between the sun and Earth, it will be moving relatively slowly such that if the Earth suddenly disappeared, the moon would then follow a new orbit that would take it closer to the sun. The Earth can't just disappear though, so the moon is held up by it and accelerated back up to speed as it continues to orbit around it. While the moon is between the Earth and sun, the Earth is further out and moving faster than the barycentre, so if the moon suddenly disappeared, the Earth would follow a new orbit that would take it further away from the sun. Again though, the presence of the moon slows it back down and all the speeding up and slowing down is balanced out.

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They should thus behave as if they were at an aphelia on the far side, and at a perihelia on the near one,

A body moves fastest at perihelion and slowest at aphelion, but here we have them slower when closer to the sun and faster when further from it.

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... slowing down on their orbital trajectory with the sun and getting away from it progressively during the first half of their far side transit (and then doing the inverse during the second half), and accelerating and getting closer to the sun during the first half of their close side transit (and doing the inverse also during the second half), what should progressively increase and decrease the distance between them during their transit.

If you imagine a case where the moon's orbit round the Earth is circular, it won't accelerate relative to the Earth at all, but it will be accelerating relative to the sun throughout the outer part of its orbit (when the Earth is nearer to the sun than the moon is), and decelerating relative to the sun throughout the inner part (when it is nearer to the sun than the Earth is). If you're imagining that the moon's greater movement towards the sun during part of its orbit round the sun will lead to the sun accelerating it, that won't happen because the sun's pull on the moon has already been applied to set a course for the Earth and moon collectively to travel around it, so it can't then add some more to accelerate the moon even more towards it - the movement of the moon inwards is driven by the Earth, and the Earth's gravity serves to decelerate that inward movement as soon as the moon is nearer the sun than the Earth is.

The place where a complication comes in is that the sun's pull on the moon is stronger when the moon is between it and the Earth, and weaker when the moon's furthest out, so that must have some impact on the shape of the orbit, but I certainly can't model that in my head - it needs to be simulated in a computer to see what the effect of that is, although exaggerated proximity to the sun in such a simulation would likely be necessary to make that effect visible to the eye.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/08/2018 10:59:08
Let my three balls be accelerated by the earth's gravitation while falling directly to it, and add mass to them so that they are also accelerated towards one another. We can give them a large mass so that they stay together while falling, then reduce it so that they get away from one another just a bit, or reduce it so much that they get away from one another a lot. The just a bit case represents the tides' case, with no input from forces due to rotation.
Main lack of freedom in the tides´real case is due to the fact that M and A are pulled by own earth gravity with forces some billion times bigger than the differential moon´s pull, NOT to rigidity when they are part of a solid, compact object ...
Your "redesigned" case leaves them free, though initially with a pseudo-equilibrium between pulls between each other, and earth´s gravity. Quite a different scenario !!
 
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/08/2018 13:57:42
I need more time to analyze all said here in last 24 hours!
But going again to basics, just some short comments ...
When I started here more than three years ago, I said "Centrifugal force is not a forbidden word whatsoever" ... I couldn´t imagine that, if not forbidden, it is so a controversial term.
Mainly because in some moment physicists decided to define it as the "fictitious" force applied to the object following a curved path if we analyze the movement using a so called "non inertial frame of reference" ... Any other "outward" force, in the direction of "flying" (sense of "fugal") from a "center", shouldn´t be called centrifugal force, according to them ...
O.K. It´s just a question of terminology. If the force does exist, and produces their effects according to laws of Physics, it does´t matter much the term we use.
But in my wildest dreams I could have thought I would have to discuss about the term "centripetal force".
Yesterday I posted what said on a "ohio.state.edu" page:
"The centripetal force is the name given to the net force REQUIRED to keep an object moving on a circular path".
Similar things are said on a lot of other sites ...
But one can read now:
 
It doesn't matter how much justification you can find for calling it centripetal force. This is a grey area in which some usages of the word aren't fully rational
Could you please send some link, from a University or similarly reliable source, where one could see that is not another absurd idea of yours ??
There could be some not fully rational usages of the term ... Absurd things are said by many people. But do you really think that term is a "grey area"??
I sincerely imagine you would find it crystal clear area, if it reinforced your fixed idea that ONLY differential pull from moon (apart from sun´s similar effects) counts for tides !!
By the way, the so called "non inertial frame of reference" is something I have in my back log (?), I mean some day I have to try and convey some own ideas about it, surely controversial ...
But I´m sure it won´t be easy at all ... seeing that we can´t agree on even really basic terms !! 
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 05/08/2018 14:43:21
By definition, a force is felt when we try to accelerate a massive object, and it is so when we throw a ball, swing it in circles, or hold it against gravity. If there is no force felt, then there is no force by definition. In free fall, no force is felt, and whether we are rotating or not is unimportant since we can't feel any anyway. That's why I prefer using the idea that, when no force is felt and we still have a motion, then that motion must depend on an internal mechanism that is not yet elucidated. The curved space/time mechanism is not physical so it can't be studied, but the idea that this kind of motion may be driven by an exchange of information can: we just have to study the way light could be exchanged between two bonded particles during their acceleration, and move them so that they stay on sync all the time.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/08/2018 17:25:22
Picture three massive balls, M, E and A, which attract each other using gravity. They are aligned in a straight line with where we're going to place the moon. M and A are both sitting on springs, so they don't contact E directly, but they are indirectly sitting on it, and each is sitting at height h over the surface of E. Now we introduce the moon into the system. The three balls begin to accelerate towards the moon, and each tries to go at its own speed, but the local pull on M and A from E feels stronger than the pull from the moon, so they will remain close to E.
That´s right …
But those springs would act as a kind of “information” transmission chain, what in real case is done as internal stresses, transmitting action/reaction forces between contiguous material, a real “mechanism” you´ve rejected several times … You said many times just differential gravity is the mechanism !
But if we count with internal stresses, we have to include not only those, but the ones originated by revolving related inertia ...
The springs would also slightly change their lengths due to the revolving of the earth, even if moon attraction were the same on M and on A …
M and A, at their maximum distances from the moon (within their own revolving trajectories), with tangential speeds perpendicular to line between moon´s and earth´s centers, would require a centripetal force in both cases towards the moon. Otherwise they could not follow their circular paths.
Same happens on E, where the centripetal acceleration is provided by moon´s pull there: they exactly match with each other.
Where M moon´s pull is stronger than required centripetal force: part of it does accelerate M towards the moon (the required centripetal acceleration for revolving of M), and the rest is what actually causes the bulge there.
But where A moon´s pull is smaller, insufficient for the centripetal force necessary to make it follow its circular path. Own earth´s pull has to "supply" the difference. Water kind of become slightly lighter, and a bulge builds also there, same way as previously explained for the formation of the equatorial bulge:
"The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v2/R (the centripetal force). Now compare these two cases. On the non rotating earth, man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation”.
(https://www.lockhaven.edu/~dsimanek/scenario/centrip.htm)
 
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 05/08/2018 20:30:09
Here is an exert from the page on "Tidal misconceptions (https://www.lockhaven.edu/~dsimanek/scenario/tides.htm#centrip)" that I found on your last link Rmolnav:

Quote from: Lockhaven.edu
We shall argue that the "tidal bulges", which are the focus of attention in many textbooks, are in fact not due to rotation, but are simply due to the combined gravitational fields of the earth and moon, and the fact that the gravitational field due to the moon has varying direction and strength over the volume of the earth.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/08/2018 21:02:26
In reality, for water to want to follow the tangent, it would have to have zero force applied to it by gravity from the other body, and that certainly isn't going to happen
It is more politically correct not to use terms such as rubbish ... O.K.
But how could I call that nonsense? Back to basic again:
1) If water is revolving as a part of the earth, it has a velocity vector.
2) Due to 1st Newton´s Motion Law, its inertial "tendency" is not to change that vector (size or direction), unless a force were applied.
3) If the gravity from the other body were zero, the water would actually "follow" the tangent, rather than just "want to" !!
4) It is precisely the not null force due to gravity, basically from the other body, what is required to overcome mentioned inertia: that vector divided by considered mass will give the acceleration vector necessary to change velocity vector ...
5) Its component perpendicular to initial speed, or part of it if in "excess", will be the centripetal acceleration, absolutely necessary for the water to follow a curved path ...
6) For further details, please have a look at my recent post of a few hours ago ...
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/08/2018 21:33:56
If there is no force felt, then there is no force by definition. In free fall, no force is felt, and whether we are rotating or not is unimportant since we can't feel any anyway
But our case is not actually a perfect free fall ...
I´m rather tired. Let me just copy what I sent to somebody else a couple of months ago:
"Encyclopaedia Britannica:
"Free-fall, in mechanics, state of a body that moves freely in any manner in the presence of gravity. The planets, for example, are in free-fall in the gravitational field of the Sun ... Gravitational forces are never uniform, and therefore ONLY the centre of mass is in free-fall. ALL OTHER POINTS of a body ARE SUBJECT to tidal FORCES because they move in a slightly different gravitational field. The Earth is in free-fall, but the pull of the Moon is not the same at the Earth’s surface as at its centre; the rise and fall of ocean tides occur because the OCEANS ARE NOT IN PERFECT FREE-FALL"
You may say: OK, but those "tidal forces" are kind of imaginary, just the differences in gravitational field ...
But they are quite REAL, internal stresses within Earth´s material. Otherwise they COULD NOT produce real effects such as tides !!
Gravitational field is something abstract, which exist only in our minds, let alone gravitational field "differences" ... Those concepts are useful for OUR calculations (if well done ...), but material stuff ONLY can react to real forces, either gravitational (always exerted directly on considered stuff), or forces exerted through direct contact by contiguous material elements (pulls, pushes, hydraulic pressure ...)"
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/08/2018 21:46:21
Here is an exert from the page on "Tidal misconceptions" that I found on your last link Rmolnav:
Quote from: Lockhaven.edu
We shall argue that the "tidal bulges", which are the focus of attention in many textbooks, are in fact not due to rotation, but are simply due to the combined gravitational fields of the earth and moon, and the fact that the gravitational field due to the moon has varying direction and strength over the volume of the earth.
I know that, and I discussed the issue directly with the author, via email.
Initially he kept his stand, but later he said he found my take "interesting" ... But he had a familiar problem (he is 81, and his wife 80), and told me he had at least to take a break in our discussion.
Were you interested, I did put here most of the texts of interchanged pieces of email (#125 to #134).
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 05/08/2018 22:15:00
O.K. It´s just a question of terminology. If the force does exist, and produces their effects according to laws of Physics, it does´t matter much the term we use.

Quite right - there's no difficulty in understanding what centrifugal means and in converting to a more correct alternative. Both descriptions map to the same events easily enough. What matters though is that you try to make sure no one develops an incorrect causal understanding of what's going on by thinking there's an outward force being generated by rotation rather than an inward one - whichever wording they use, they must still visualise the force in the right direction. In the same way, when you use the word centripetal in a place where I consider it invalid, you should not turn your sloppy use of the word centripetal into a belief that rotation has a role in generating tidal bulges.

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But in my wildest dreams I could have thought I would have to discuss about the term "centripetal force".

Then you need to broaden your mind. Science is a process that leads (generally) towards improved understanding and to repeated corrections in the way things are described and labelled. There are two very different things going on with rotation and force, and a clear distinction should be made between them to prevent people like you from tripping up over them by mistaking one for the other.

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Could you please send some link, from a University or similarly reliable source, where one could see that is not another absurd idea of yours ??

I am always more interested in actual science than error-ridden authorities. I haven't seen anyone in science support my position (primarily because I haven't looked for that) - what I'm saying is based 100% on what I see when I look directly at the physics involved in this. If a force is generated by rotation, that is clearly centripetal force - a force that comes into play because of the rotation. If a force persists when the rotation is removed, it cannot have been generated by the rotation and has therefore should not, in my opinion, be labelled as centripetal force. It already has a better name as what it actually is: in this particular case gravitational force. If everyone wants to call it centripetal force regardless, that's up to them, but they should not let their sloppy labelling contaminate their understanding of the physics.

The point under discussion in this thread is the cause of the bulges, and the bulges persist when the rotation is removed. You attempt to attribute the bulges to centripetal force, but you're making two errors in the process. (1) The first is that you're playing a game of words - you want centripetal force to be involved because the Earth and moon are following curved paths, and you see cases where authorities have used labels in a particular way which backs your position (where centripetal has either been applied to orbits or appears to include them without specifically stating that it does so), but it is always better to look directly at the physics and use that as a guide to how labels ought to be defined. A force attributed to rotation which (the force) persists when the rotation is removed should not be given a name that implies that it is caused by rotation. (2) Your second error, having brought "centripetal force" into a situation where it doesn't belong causally, is to assert that the rotation drives the bulges on the basis that the rotation necessarily involves centripetal force and centripetal force will necessarily make the water on the outside try to follow the tangent while the Earth tries to follow the curve. In reality, both things will try to follow the same curve if the gravitational force is applied evenly because they are in free fall. What then locks you into your incorrect understanding though is that the gravity isn't evenly applied, so it does allow the water to try to follow a different curved path from the curved path the Earth follows, and a "bulge" appears which fits neatly with that explanation. However, the bulge is exactly the same as the bulge that would be there without the rotation, which shows that it is not caused by the attempt to move along a different path.

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There could be some not fully rational usages of the term ... Absurd things are said by many people. But do you really think that term is a "grey area"??

To me it isn't particularly grey, but other people make it so. If they want to attribute it to the rotation purely on the basis that a curved path is involved, even though it is not caused by the rotation, I see that as a clear error.

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I sincerely imagine you would find it crystal clear area, if it reinforced your fixed idea that ONLY differential pull from moon (apart from sun´s similar effects) counts for tides !!

Given that the differential pull does account for it in full (and persists with the rotation not present), what is the point of proposing some other cause that cannot add anything to the actual cause?

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By the way, the so called "non inertial frame of reference" is something I have in my back log (?), I mean some day I have to try and convey some own ideas about it, surely controversial ...
But I´m sure it won´t be easy at all ... seeing that we can´t agree on even really basic terms !!

We don't need to agree on the use of the term centripetal. The only important task is for us to find agreement on what causes the bulges. If you are determined to call it centripetal force but are prepared to accept that the rotation has no role in generating the bulges, then that will be good enough, but you need to realise that you will be extending the meaning of the word centripetal into an even more controversial area by applying it where there is no rotation at all.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 05/08/2018 22:39:42
But those springs would act as a kind of “information” transmission chain, what in real case is done as internal stresses, transmitting action/reaction forces between contiguous material, a real “mechanism” you´ve rejected several times … You said many times just differential gravity is the mechanism !

Those springs are equivalent to the pressure changes where the "bulges" are. The greater pull on M makes it rise higher above the surface of E, and the mechanism for that is differential gravity.

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But if we count with internal stresses, we have to include not only those, but the ones originated by revolving related inertia ...

The rotation doesn't add any such extra stresses because E is in free fall.

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The springs would also slightly change their lengths due to the revolving of the earth, even if moon attraction were the same on M and on A …

If you're talking now about the Earth spinning and creating a bulge towards the equator and flattening towards the poles, those are not relevant to tides. If we have a series of balls sitting on springs sitting round the equator of E, all of them will find a height caused by E's spin, then they will gain extra height as they pass through positions M and A and lose all that extra height when half way in between those locations. The extra height above the surface of E that these balls have due to E's spin compared with the height they would be at if E didn't spin is caused by centrifugal force (inertia), but this is not a component of the tidal variation.

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M and A, at their maximum distances from the moon (within their own revolving trajectories), with tangential speeds perpendicular to line between moon´s and earth´s centers, would require a centripetal force in both cases towards the moon. Otherwise they could not follow their circular paths.

A is held by E's gravitational force and only rises a little because the moon is pulling on A less strongly than E. This gravitational force is applied independent of the rotation and should not be attributed to the rotation.

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Same happens on E, where the centripetal acceleration is provided by moon´s pull there: they exactly match with each other.

Again, this force is independent of the rotation.

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Where M moon´s pull is stronger than required centripetal force: part of it does accelerate M towards the moon (the required centripetal acceleration for revolving of M), and the rest is what actually causes the bulge there.

Again you're attributing a force to rotation which is not caused by the rotation. The bulge is caused directly by the greater strength of gravitational pull applied to M.

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But where A moon´s pull is smaller, insufficient for the centripetal force necessary to make it follow its circular path. Own earth´s pull has to "supply" the difference. Water kind of become slightly lighter, and a bulge builds also there, same way as previously explained for the formation of the equatorial bulge:

No - it's a different cause. The equatorial bulge is caused by rotation. The tidal bulges are not.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 05/08/2018 23:05:02
In reality, for water to want to follow the tangent, it would have to have zero force applied to it by gravity from the other body, and that certainly isn't going to happen
It is more politically correct not to use terms such as rubbish ... O.K.
But how could I call that nonsense? Back to basic again:

It's your nonsense - you claimed it should try to follow the tangent while the Earth should follow its orbit.

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3) If the gravity from the other body were zero, the water would actually "follow" the tangent, rather than just "want to" !!

Indeed. The argument was about an even force being applied to cause a non-rotating Earth to follow a curved path, and with such an even force being applied, the water would follow the exact same curved path and not produce a bulge. This reveals that the bulge cannot be driven by the involvement of a curved path, but by the difference in gravitational strength that actually occurs over distance - the real cause is that it not an evenly applied force.

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5) Its component perpendicular to initial speed, or part of it if in "excess", will be the centripetal acceleration, absolutely necessary for the water to follow a curved path ...

You are still trying to attribute the cause to a rotation that is not the cause - the bulge would be there without the rotation's involvement and it is a mistake to try to pin a different cause to it than the actual cause of the bulges. In the first page of this thread, I incorrectly pinned the cause on the same thing as you, but I switched away from that explanation when I realised that the bulges would remain if the rotation was removed - it is not possible for there to be two different causes for the exact same thing. One of the proposed causes is the actual cause, while the other is merely an apparent cause which is easy to mistake for the real one.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 05/08/2018 23:22:58
Let me just copy what I sent to somebody else a couple of months ago:
"Encyclopaedia Britannica:
"...ALL OTHER POINTS of a body ARE SUBJECT to tidal FORCES because they move in a slightly different gravitational field. ...

It doesn't make the mistake there of attributing it to rotation, so what it says is correct.

I've thought of a new static system to use as an example for you which ought to settle the matter. Imagine a stationary Earth with two moons held away from it on opposite sides, each supported by a long, strong rod. These moons would generate tidal bulges. If we rotate the system, an equatorial bulge will also develop in addition to the existing bulges. If we rotate this system at a speed where the Earth's gravity will keep the moons in orbit around it, we can remove the rods and the bulges remain. If we stop the Earth spinning, the equatorial bulge disappears. The tidal bulges remain throughout though, even when we get to this beautiful point where the Earth is both stationary and not rotating. The tidal bulges in this system cannot be caused in any way by the Earth's movement because it has none. Your explanation for the causes of tides is thus shown more clearly than ever to be horribly wrong.

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Gravitational field is something abstract, which exist only in our minds, let alone gravitational field "differences" ... Those concepts are useful for OUR calculations (if well done ...), but material stuff ONLY can react to real forces, either gravitational (always exerted directly on considered stuff), or forces exerted through direct contact by contiguous material elements (pulls, pushes, hydraulic pressure ...)"

Nice - so how do you account for the bulges in the system I've just set out for you? Good luck with that one!
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 06/08/2018 13:50:20
Imagine a stationary Earth with two moons held away from it on opposite sides, each supported by a long, strong rod.
To illustrate the difference between the two kinds of effects, we can also imagine accelerating the rotation of the earth until some of the surface starts orbiting, and then completely stop the rotation while part of the surface is still orbiting. What is orbiting got at a certain height and speed because of rotation, but it goes on staying at the same height and speed without it.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 06/08/2018 14:02:22
I know that, and I discussed the issue directly with the author, via email.
Initially he kept his stand, but later he said he found my take "interesting"
Tell him to join our discussion so that we can defend him from you. You should be ashamed not to respect old people like him! (just kidding :0)
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 06/08/2018 19:04:42
To illustrate the difference between the two kinds of effects, we can also imagine accelerating the rotation of the earth until some of the surface starts orbiting, and then completely stop the rotation while part of the surface is still orbiting. What is orbiting got at a certain height and speed because of rotation, but it goes on staying at the same height and speed without it.
I have to prepare with more time my reply to D. C., not only in connection to that ...
Just a couple of things now:
1) The most important: when I´ve mentioned that effect, saying even "permanent" equatorial bulge, and usually adding something like "due to earth daily spinning", I was just saying that due to similar inertial reasons, earth app. monthly revolving around moon-earth barycenter should deform oceans surface, on its own right ... And I even said, at least once, something like "please don´t mix the two movements", because it´s not the first time D. C. seems to think I connect the moon related tides to earth daily spinning !!
Therefore, thank you for your words, but the difference is so obvious, that no additional cases need to be added "to illustrate" it ... The problem is that D.C. and me seem to speak in different languages !!
2) If you ruminated your scenario, let alone if you made some maths, you would find that, before centrifugal forces made some of the earth surface stuff "levitate", earth would kind of explode ...
Logically, equatorial bulge would previously get higher and higher, and also wider. Even with smaller radius at higher latitudes, those forces (proportional to ω˛r), would increase dramatically.
And the required angular velocity to orbit at earth surface, I don´t remember any figure, but is really huge !! 
 
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 07/08/2018 11:44:10
Instead of refuting the lot of new absurdities I see in your last posts, i´m going to refer only to one, that could be the root of all your "problems" ...
I posted here the basics of "centripetal force", linking the "physics.ohio-state.edu" site I took them from. To no avail as far as your ideas are concerned.
I also explained, step by step and very carefully, the application of that basic concept to our case (see #181, particularly 4 and 5). Also to no avail: your reply shows you have not caught back even a pinch of that stuff ... which perhaps you have forgotten ...
And the day before yesterday, you broke your own "mark":
A force attributed to rotation which (the force) persists when the rotation is removed should not be given a name that implies that it is caused by rotation.
In other paragraphs you also refer to "forces caused (or not) by rotation", or to "forces caused (or not) by certain movement", as if we could say some forces are caused by some movements, and others are not ...
But things are the other way around !! Movements cannot cause any force ... Forces do cause movements (or change them) !!
Your ideas on Physics are flawed even at the very "basics of the basics" level ...
No wonder you say things such as "It doesn't matter how much justification you can find for calling it centripetal force. This is a grey area in which some usages of the word aren't fully rational" ... and so on.
By the way, you could not send us any link where we could see where you took that from, because it´s just what YOU have deduced when reading many things ...
I wonder, couldn´t it rather be a problem of your sight, or of your mind-set ??
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 07/08/2018 19:59:16
Instead of refuting the lot of new absurdities I see in your last posts, i´m going to refer only to one, that could be the root of all your "problems" ...
I posted here the basics of "centripetal force", linking the "physics.ohio-state.edu" site I took them from. To no avail as far as your ideas are concerned.

I can't find your link. However, all the definitions I've found of centripetal force make it very clear that rotation is involved, as in movement following a curve. Most definitions also refer specifically to it being a circular path. None of them suggest that centripetal force can apply to a straight line course. When it comes to the tides, they are caused by forces acting in a straight line and the rotation is incidental. That's all there is to it. You are the one posting absurdities by calling things absurdities that are fully valid and can be demonstrated in full through simulations.

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I also explained, step by step and very carefully, the application of that basic concept to our case (see #181, particularly 4 and 5). Also to no avail: your reply shows you have not caught back even a pinch of that stuff ... which perhaps you have forgotten ...

"4) It is precisely the not null force due to gravity, basically from the other body, what is required to overcome mentioned inertia: that vector divided by considered mass will give the acceleration vector necessary to change velocity vector ..."

If you wrote that in Spanish it might be easier to understand.

"5) Its component perpendicular to initial speed, or part of it if in "excess", will be the centripetal acceleration, absolutely necessary for the water to follow a curved path ..."

But in both cases you're just building upon the same old error - the bulges are fully accounted for by differential gravity without any rotation being required.

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And the day before yesterday, you broke your own "mark":
A force attributed to rotation which (the force) persists when the rotation is removed should not be given a name that implies that it is caused by rotation.
In other paragraphs you also refer to "forces caused (or not) by rotation", or to "forces caused (or not) by certain movement", as if we could say some forces are caused by some movements, and others are not ...
But things are the other way around !! Movements cannot cause any force ... Forces do cause movements (or change them) !!

If a ball is sitting on the end of a rope and is not moving relative to the other end of the rope, there is no force acting through the rope. If you send the ball round in circles though, that movement generates force in the rope. The movement is the cause. The movement of the ball generates the force in the rope - the force from the rope does not set the ball moving. If you're looking for a primary cause for the centripetal force, it's not just the movement of the ball though as the ball could already be moving in a straight line before hooking onto the end of the rope. Once it's hooked onto the rope, the ball's movement then generates a force in the rope, and then the force in the rope changes to course the ball is travelling in. The cause of the force being generated in the rope is thus a combination of the movement of the ball AND the ball being locked to the end of the rope. The force being generated in the rope follows that. Causation has a strict order and you are violating it.

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Your ideas on Physics are flawed even at the very "basics of the basics" level ...

Not at all - you are the one who's mauling physics here because you've learned a stack of incorrect rules which you are applying as if they come from a deity. The physics itself should be your guide and not the mistakes of your teachers.

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No wonder you say things such as "It doesn't matter how much justification you can find for calling it centripetal force. This is a grey area in which some usages of the word aren't fully rational" ... and so on.

I say it because when something is already explained in a straight line case, that explanation holds in a rotating case and the rotation is irrelevant. The ball on the end of the string only generates a force in the rope when it's going round on the end of the rope. Take the orbit out of it and the force in the string disappears. In the gravity case though, the force continues to act in full when the orbit is removed, and that tells you that it is a radically different case.

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By the way, you could not send us any link where we could see where you took that from, because it´s just what YOU have deduced when reading many things ...
I wonder, couldn´t it rather be a problem of your sight, or of your mind-set ??

Everything you need is here - there is no need to follow links to fake authorities. I draw directly upon physics and reason, and I only state what I know is backed up by physics (and when I also know it could be simulated on a computer to show the mechanisms in action and demonstrate that they do exactly what I claim).
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 07/08/2018 21:45:04
If a ball is sitting on the end of a rope and is not moving relative to the other end of the rope, there is no force acting through the rope. If you send the ball round in circles though, that movement generates force in the rope. The movement is the cause. The movement of the ball generates the force in the rope - the force from the rope does not set the ball moving. If you're looking for a primary cause for the centripetal force, it's not just the movement of the ball though as the ball could already be moving in a straight line before hooking onto the end of the rope. Once it's hooked onto the rope, the ball's movement then generates a force in the rope, and then the force in the rope changes to course the ball is travelling in. The cause of the force being generated in the rope is thus a combination of the movement of the ball AND the ball being locked to the end of the rope. The force being generated in the rope follows that. Causation has a strict order and you are violating it.
Again and again, utterly twisting actual, basic Physics. I´ll leave it there without any further comment !!
Please, could somebody else post what he or she thinks of that ??
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 07/08/2018 22:21:19
Again and again, utterly twisting actual, basic Physics.

What I said conforms 100% to physics, whether basic or otherwise. You are the one twisting physics if you imagine that the force in the rope is the primary cause. The primary cause is the ball's movement in a direction which will then lead to the force being generated in the rope due to the rope being tied to the ball. That comes first. The force generated in the rope then applies to the ball and changes its course.

And don't lose track of the real point under discussion here either. A ball on the end of a string where nothing is moving does not produce a force in the string. A ball that's swinging round in a circle on the end of a string does produce a force in the string. When you stop the ball, the force in the string is lost.

Now do the same thing with an object orbiting a planet with the planet's gravity holding it in that orbit. The movement of the object does not generate the force that holds it in orbit. If you stop the object, the force does not disappear, but continues to act in full and will drag the object down onto the planet.

I don't know which part of that you're incapable of understanding, or how you can have got your thinking into such a tangled state that you can't acknowledge that this is correct and instead repeatedly refer to it (like every other rational, correct statement I supply you with) as rubbish, nonsense, an absurdity, etc. Please switch your brain on properly and stop wasting my time.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 08/08/2018 11:57:01
What follows is mainly intended not for D. C., but for others. Because he doesn´t accept the authority of eminent physicts (if against what he supposes to "see" just observing physical facts), and most likely he won´t accept even definitions from "eminent" dictionaries, such as Merriam Webster:
1) movement: the act or process of moving;  especially  : change of place or position or posture.
2) to move: to change the place or position of / to cause to go or pass from one place to another with a continuous motion.
3) force : an agency or influence that if applied to a free body results chiefly in an acceleration of the body and sometimes in elastic deformation and other effects.
From which part of those definitions could we deduce that "movements can cause forces" ??
As I previously said, couldn´t rather be that either D. C. sight, when observing physical facts, or the way his mind processes information from his eyes, have some type of problem ??
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 08/08/2018 19:25:59
From which part of those definitions could we deduce that "movements can cause forces" ??

Think about the ball on the end of a string going round and round and the force generated in that string. What happens when you stop the ball? The force in the string disappears - a force stops the ball, and the change of movement of the ball then changes the force in the string (eliminating it in this case). What happens if you start the ball moving again? The force in the string reappears as a result of the ball moving. You clearly can't start the ball moving creating the force in the string - the movement of the ball comes first. Clearly another force will have to start the ball moving, but that's a different issue - we're only concerned with the cause of the force in the string, and the force that starts the ball moving can also be detached from the system by having the ball move in a straight line for a long time with no force being applied to it after the initial shove. The ball then hooks onto the string and bingo: the force in the string is now generated. The cause of that force is the movement of the ball and it being locked to the end of the string. What's so hard about understanding that?

Another example of movement causing a force is where something collides with something else - the force does not cause that movement, but the movement and collision with something else then causes the force to be generated (before the force modifies the movement).

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As I previously said, couldn´t rather be that either D. C. sight, when observing physical facts, or the way his mind processes information from his eyes, have some type of problem ??

No - all that's happened here is that you've revealed a fundamental error in your understanding of physics (and causation). If you were to program simulations and tried to reverse the causality of things, your programs would fail to function correctly. If you simulate a force in the string to try to make the ball move at 90 degrees to the application of that force, it will fail to do what you want - the ball will be pulled directly towards the centre instead. You have to start by applying a force to move the ball in the right direction, and then you have the ball's movement generate the force in the string as a result.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 10/08/2018 13:40:36
That stubbornness of yours doesn´t deserve further explanations, but I´ll continue for the sake of the interest of others ...
Your investigations, if any, are quite the opposite to Sherlock Holmes´s … He would never say the first found suspect was the murderer, without a thorough analysis of all available details !
Despising loftily Newton´s Laws, and what said today by eminent physicists, and definitions given on recognized dictionaries, you just observe what happens (or just seems to happen), and conclude:
If you send the ball round in circles though, that movement generates force in the rope. The movement is the cause. The movement of the ball generates the force in the rope - the force from the rope does not set the ball moving
Think about the ball on the end of a string going round and round and the force generated in that string. What happens when you stop the ball? The force in the string disappears - a force stops the ball, and the change of movement of the ball then changes the force in the string (eliminating it in this case).
That is utterly erroneous both as:
1) physical reality: the centripetal force doesn´t actually disappear ,
2) as logic (?) conclusion: one thing is “correlation” and quite another “causation” (also “ basics of basics” in its realm) 

Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 10/08/2018 14:29:29
1) physical reality: the centripetal force doesn´t actually disappear
Are you talking of the force you need to develop to hold to your idea while it is being swung around? (just kidding :0)
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 10/08/2018 19:04:55
Quote from: rmolnav on Today at 13:40:36
1) physical reality: the centripetal force doesn´t actually disappear
Are you talking of the force you need to develop to hold to your idea while it is being swung around? (just kidding :0)
I´m not sure of your intentions with that, but if the hammer an athlete is going to throw were suddenly stopped by a physical obstacle, since that very moment centripetal force (previously changing the direction of the hammer movement), would change to accelerate linearly the weight and the wire "inwards", and even the athlete would move back and fall down ... Clear prove that the centripetal force was being exerted by the athlete on inner extreme of the wire, transmitted section by section through the wire, and finally to the weight through outer extreme of the wire !! 
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 10/08/2018 23:39:36
That stubbornness of yours doesn´t deserve further explanations, but I´ll continue for the sake of the interest of others ...

There are two stubborn people here: one is stubbornly right and the other is stubbornly wrong.

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Your investigations, if any, are quite the opposite to Sherlock Holmes´s … He would never say the first found suspect was the murderer, without a thorough analysis of all available details !

If Holmes was analysing this, he would tackle the task in a rational way, as I do.

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1) physical reality: the centripetal force doesn´t actually disappear ,

The force is completely gone - the rope is slack. What sense does it make to claim that a force is still there when it's no longer present?

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2) as logic (?) conclusion: one thing is “correlation” and quite another “causation” (also “ basics of basics” in its realm)

That is precisely your problem - you are being misled by correlation and cannot determine the actual causation. In the case of a ball on the end of the string, the movement of the ball causes the force to be generated in the string. In the case of an object orbiting a planet, the force is not caused by the movement of the object, and this shows up when we stop the object moving and then watch it fall down onto the planet as the force continues to act.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 11/08/2018 00:02:09
...if the hammer an athlete is going to throw were suddenly stopped by a physical obstacle, since that very moment centripetal force (previously changing the direction of the hammer movement), would change to accelerate linearly the weight and the wire "inwards", and even the athlete would move back and fall down ... Clear prove that the centripetal force was being exerted by the athlete on inner extreme of the wire, transmitted section by section through the wire, and finally to the weight through outer extreme of the wire !! 

In the case of a ball going round on a string attached to a pole which is turn anchored to the Earth, if the ball is stopped (perhaps by hitting the same bat that set it moving in the first place), 99.99% of the force in the string may disappear, the only remnant being caused by a little flexing of the pole (which will have been pulled towards the ball throughout rather than away, which is quite different from the case of the hammer-thrower). A new force will then be generated in the string as the ball falls, again generated by the movement of the ball which tries to take it where the string can't let it go.

Your hammer-throwing case is more like two balls going round and round on opposite ends of a string. To eliminate the force in an instant, you'd have to stop both of them moving simultaneously because the movement of both is generating the force, although just stopping one of them would substantially reduce the force in the string, and if you were to record the transmission of the change in force in the string by putting a series of detectors along it, you would find that the change starts at the end where the stopped ball is and it races from there to the other end, probably at the speed of sound in the string. That movement of change reveals the route of causation, and it's the very opposite of the one you've asserted.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 13/08/2018 18:42:06
As I´ve already said several times, I´m afraid It´s useless to carry on discussing with you …
In the case of a ball going round on a string attached to a pole which is turn anchored to the Earth, if the ball is stopped (perhaps by hitting the same bat that set it moving in the first place)
It would be quite a spectacle to see you hitting a ball hanging from a string and pole tens of times (or hundreds) trying to make it get and follow a horizontal circular path, without any resting interval (otherwise the ball would descend).
In any case, if somehow we could made the ball properly rotate that way, the movement of the ball is logically necessary for its rotation: no centripetal force could make a null speed vector change direction !! But that doesn´t mean the movement “causes” the centripetal force !!
You ignore (or forget, or don´t understand and handle wrongly…) other details of the scenario such as the concept of inertia, the action and reaction principle, that to have a string tight two opposite forces are required, etc, etc.
A prove of that is:
... a little flexing of the pole (which will have been pulled towards the ball throughout rather than away,
Don´t you realize that, if the pole has "been pulled towards the ball”(as you correctly say), that pull cannot be a “centripetal” force caused by the ball movement, because it has an outward direction ??
I would call it centrifugal force, in a singular flavor of the concept, because in that case we could say it is an “active” force (action), and the centripetal one is a “reactive” force (reaction), when usually the opposite happens: centrifugal force is an inertial reaction to an “active” centripetal force …
I´m afraid that, as you consider centripetal force a “grey” area, adding those details could be a “black” area for you …
In any case, I´ll carry on for the sake of the interest of others:
Hitting a ball produces a transference of momentum, always through forces. At the very initial instant some deformations (of the ball and the bat) occur, what produces opposite pushes on each other, which then change both speed vectors … The ball gets a speed, and its inertia tries to make it go straight … If a string attached to a pole prevent that to happen, what the ball´s inertia does (its “movement”, in your words) is to tighten the string: it pulls outward the string outer end, and the string inner end will pull on the pole … The pole will react exerting an equal but opposite force on the string (3rd Newton´s Motion Law), a centripetal force … THAT centripetal force makes the rectilinear movement of the ball change into circular movement, that is, it causes the rotational movement … Quite the opposite of what you say !!

Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 13/08/2018 21:27:51
As I´ve already said several times, I´m afraid It´s useless to carry on discussing with you …

That's because I'm right and you're wrong, but you're determined that wrong is right.

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In the case of a ball going round on a string attached to a pole which is turn anchored to the Earth, if the ball is stopped (perhaps by hitting the same bat that set it moving in the first place)
It would be quite a spectacle to see you hitting a ball hanging from a string and pole tens of times (or hundreds) trying to make it get and follow a horizontal circular path, without any resting interval (otherwise the ball would descend).

You specialise in making errors and irrelevant objections. I don't know if you've ever heard of Swingball which has a ball on a string that you hit with a bat, the other end being attached to the top of a pole, but the high speed of the ball takes it round and round in circles very nicely, and if you let the ball hit the bat while the bat's moving away from it at the right speed you can take all the speed out of the ball (at which point it will accelerate (from stationary) downwards and swing in towards the pole, though this part of the action is a side issue  we're only really interested in what happens when the ball stops going round horizontally). The small amount of descent on each circuit due to gravity is also something you should be able to ignore when focusing on the physics in question, but you seem incapable of tuning out all the noise, and that's likely why you struggle so much to get to a pure understanding of any individual part of any mechanism.

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In any case, if somehow we could made the ball properly rotate that way, the movement of the ball is logically necessary for its rotation: no centripetal force could make a null speed vector change direction !! But that doesn´t mean the movement “causes” the centripetal force !!

Of course the movement causes the force. Stop the movement and the force disappears. Start the movement and the force returns. The force doesn't cause the movement, but is caused by the movement. Each change from movement to non-movement or the opposite leads to a change in force in the string starting at the ball end and transmitting to the pole end at the speed of sound in the string (and you could record that by putting sensors in the string to monitor its local tension - though note that this is just the small initial increase in tension and not the one that ramps it up by many magnitudes when the centripetal force kicks in). That is the course the causality takes.

You have the idea that forces cause movement and that movement can't cause force, but both are possible and both are common. A force can start something moving, and the movement can then lead to a different force being generated. In the case of a ball on a string going round in circles, the first force sets the ball moving, that movement then generates the centripetal force in the string, the centripetal force then modifies the movement of the ball, changing the direction of its travel.

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You ignore (or forget, or don´t understand and handle wrongly…) other details of the scenario such as the concept of inertia, the action and reaction principle, that to have a string tight two opposite forces are required, etc, etc.

There are plenty of ways to apply transformations to my description of events to turn it into other correct wordings, but the existence of other correct wordings does not make my wording wrong. Most importantly though, my wording has causation working in the right direction - opposite to the incorrect direction that you assert for it. You want the centripetal force to cause the movement of the ball (even though all it does is modify the movement of the ball).

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A prove of that is:
... a little flexing of the pole (which will have been pulled towards the ball throughout rather than away,
Don´t you realize that, if the pole has "been pulled towards the ball”(as you correctly say), that pull cannot be a “centripetal” force caused by the ball movement, because it has an outward direction ??

Again you blind yourself with side issues. None of that negates the fact that centripetal force is acting in the string and pulling on the ball, and when the ball is stopped, that centripetal force disappears 100% as a result of the movement being removed. There is some force in the string left over, but I never called that remnant force centripetal force, so why are you making out that I did? It is a force that comes from the flex in the pole.

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I would call it centrifugal force, in a singular flavor of the concept, because in that case we could say it is an “active” force (action), and the centripetal one is a “reactive” force (reaction), when usually the opposite happens: centrifugal force is an inertial reaction to an “active” centripetal force …

That's fine, but when the centripetal force has gone entirely (when the ball's been stopped), that centrifugal force has also gone. All we have left is the force from the flex in the pole, and that is neither centrifugal nor centripetal.

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I´m afraid that, as you consider centripetal force a “grey” area, adding those details could be a “black” area for you …

The questionable part of the application of the word is where you're labelling gravitational pull as centripetal force. I don't consider that valid because the gravitational force continues to act in full when the orbiting object is stopped.

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The ball gets a speed, and its inertia tries to make it go straight … If a string attached to a pole prevent that to happen, what the ball´s inertia does (its “movement”, in your words) is to tighten the string: it pulls outward the string outer end, and the string inner end will pull on the pole … The pole will react exerting an equal but opposite force on the string (3rd Newton´s Motion Law), a centripetal force … THAT centripetal force makes the rectilinear movement of the ball change into circular movement, that is, it causes the rotational movement … Quite the opposite of what you say !!

Now I understand why you're getting this wrong. It's because you've got a lot of that right. There is one force applied by the ball to the string which spreads from ball to pole at the speed of sound (revealing the causation, which you ignore), and then another force is applied in the opposite direction to equalise it, starting at the pole and then ending at the ball, again spreading at the speed of sound, and it's this latter force that's called centripetal. The centripetal force then modifies the movement of the ball, but crucially it does not cause the movement of the ball - it merely modifies that movement.

The chain of causation is this: the ball moves --> it applies a force to the string --> centripetal force is generated in the string (the opposite way). With any chain of causation, if you to remove an intermediate item in the chain, what's left behind remains true, so we get: the ball moves --> centripetal force is generated in the string. We never get: centripetal force is generated in the string --> the ball moves. The ball was already moving. What we do have though is: centripetal force is generated in the string --> the movement of the ball is modified. You're conflating the movement of the ball and the modification of the movement of the ball, and that's where you're tripping up. Like I keep telling you, when you stop the ball, the centripetal force disappears as a result, and when you start the ball moving again, the centripetal force reappears as a result. The direction of causality is ball to pole. The direction of response is pole to ball, and this response then causally drives the modification of the movement of the ball.

And why does this matter? Again, let me focus your attention on the actual argument in question. With a ball on a string, centripetal force is generated by the movement of the ball. With an orbiting object, the force keeping it in orbit continues to apply in full when the ball has no movement - the movement of the orbiting object did not generate the gravitational force. No amount of diversion tactics will overturn these key facts.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 14/08/2018 15:29:23
My two cents comment again! :0)

If we tie the ball to its far end through a hole in it, it will flatten instead of stretching, and if we tie it at many places with many strings, it will neither stretch nor flatten. For the ball to get deformed by the centripetal force the same way gravitation deforms the seas, the force on its close end has to be stronger than the force on its far end, but if we tie two balls to the same string and rotate them at the same time, on the contrary, the force on the closer one will be weaker than the force on the farther one. The equation is F=w˛r, and w˛ is the angular velocity, so if r increases, the force automatically increases since w˛ is the same for both balls, which means that if gravitation would work this way, the seas would flatten instead of stretching.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 14/08/2018 18:57:09
Not necessary to refute that lot of errors. Anybody with a minimum of Physics education can see them.
But just a couple of things.
The questionable part of the application of the word is where you're labelling gravitational pull as centripetal force. I don't consider that valid because the gravitational force continues to act in full when the orbiting object is stopped.
What absurd a phrase last one! ... When did I label "gravitational pull as centripetal force". A gravitational pull (caused by a celestial object) can become a centripetal force if another object has a suitable speed vector to get an elliptical path, thanks precisely to that pull. E.g.: the pull exerted by earth on the moon ...
And, if for any reason the inertial tendency of the moon disappeared (e.g.: due to a sudden change of its velocity), the path would change, and the term "centripetal" could have no sense any more (if no proper rotation, no proper "center" where the pull were pointing at ...).
But as long as the rotation exists, the gravitational pull is a centripetal force !! 
The centripetal force then modifies the movement of the ball, but crucially it does not cause the movement of the ball - it merely modifies that movement.
Don´t twist my words. I´ve always said that the centripetal force causes the rotational movement, not just the "movement". And also that an initial speed vector, whatever the way it was originated, is logically necessary too !!
The chain of causation is this: the ball moves --> it applies a force to the string --> centripetal force is generated in the string (the opposite way).
Please kindly put it more clear:
1) "... centripetal force is generated ..." What do you mean with "generated" ??
2) "(the opposite way)" Opposite to what, to the by you previously mentioned force, or to the direction I say is the correct one ?

Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 14/08/2018 22:23:24
Not necessary to refute that lot of errors. Anybody with a minimum of Physics education can see them.

It doesn't do you any good repeatedly referring to correct things as errors - you just make yourself look more and more incompetent.

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But just a couple of things.
The questionable part of the application of the word is where you're labelling gravitational pull as centripetal force. I don't consider that valid because the gravitational force continues to act in full when the orbiting object is stopped.
What absurd a phrase last one! ... When did I label "gravitational pull as centripetal force".

Are you reallyy incapable of working out what I'm referring to from the context of this conversation? You're calling it centripetal force in the case of a moon orbiting a planet.

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A gravitational pull (caused by a celestial object) can become a centripetal force if another object has a suitable speed vector to get an elliptical path, thanks precisely to that pull. E.g.: the pull exerted by earth on the moon ...

And the reason I class that as a different category is that the force is not generated by the movement, which is why I question whether it should be called centripetal force at all. I've made that clear a good half dozen times now, but you don't take anything in. And this is just a side issue which you keep trying to divert things onto to hide the reality that the fundamental cause of the tides is gravitational force and not centripetal force. That is the big question under discussion here, and I have shown you the answer. The rest is just you engaging in diversion tactics because you don't like the answer.

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And, if for any reason the inertial tendency of the moon disappeared (e.g.: due to a sudden change of its velocity), the path would change, and the term "centripetal" could have no sense any more (if no proper rotation, no proper "center" where the pull were pointing at ...).
But as long as the rotation exists, the gravitational pull is a centripetal force !! 

We're looking for the fundamental answer of the cause of tides, and that's gravitational pull and the difference in its strength over distance - not centripetal force. Your explanation is highly misleading.

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The centripetal force then modifies the movement of the ball, but crucially it does not cause the movement of the ball - it merely modifies that movement.
Don´t twist my words. I´ve always said that the centripetal force causes the rotational movement, not just the "movement". And also that an initial speed vector, whatever the way it was originated, is logically necessary too !!

Why have you spent so much time then attacking what I've said about the movement of the ball causing the force to be generated if you've understood all along that it does exactly what I said it does? What's your game now? You realise you've been wrong, so you try to rewrite history as to what's happened in this conversation? Wll, the actual history of events is all still there for people to check. Tal vez, la problema es que no entiende Ingles bastante bien.

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Please kindly put it more clear:
1) "... centripetal force is generated ..." What do you mean with "generated" ??

Y aqui habemos un ejemplo de eso - no entiende la palabra "generate". I mean exactly what the word says: if something is generated by something, it means it comes into existence because of that something. If a force is applied to something and causes it to move, the force generates the movement. If a movement causes a force to appear, it generates that force.

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2) "(the opposite way)" Opposite to what, to the by you previously mentioned force, or to the direction I say is the correct one ?

The opposite way to the force being applied by the ball - the ball pulls one way on the string and the pole pulls the other. As the pole's pull on the string provides resistance, the ball pulls its force on the string higher and higher and the pole reacts by matching it. The force from the ball end (the cause) comes first, and the centripetal force from the pole end (the result) is in every part a response driven by the cause.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 15/08/2018 10:06:56
Oxford Dictionary
"Centripetal force":
Noun.
A force that acts on a body moving in a circular path and is directed towards the centre around which the body is moving.
Example sentences
‘Newton analysed the motion of bodies in resisting and non-resisting media under the action of centripetal forces.’
‘Now, if we set a cylindrical space station in rotational motion at a certain angular velocity, anything that moves with it will need a centripetal force to keep it rotating with the station.’
‘Armed with his conception of material space, which was the location of gravity, combined with the constant action of the centripetal force of universal gravitation, Toland believed he had accounted for all phenomena in the universe.’
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 15/08/2018 19:37:08
Oxford Dictionary
"Centripetal force":
Noun.
A force that acts on a body moving in a circular path and is directed towards the centre around which the body is moving.
Example sentences
‘Newton analysed the motion of bodies in resisting and non-resisting media under the action of centripetal forces.’

Lovely - you have a justification for using the word centripetal with gravity. That doesn't negate the fact that there are two different cases which need to be differentiated between: (1) a case where centripetal force is generated by the movement of the thing going round [and where it disappears if the object is stopped], and (2) a case where centripetal force is not generated by that movement but is 100% independent of it [and where it continues to act if the object is stopped]. This is why I don't like the use of the word centripetal in the latter case, but if other people are happy to use it with such semantic ambiguity, that's fine, just so long as they understand the two different cases and don't allow it to contaminate their understanding of things like tides.

In the case of tides, the movement is irrelevant because the gravitational force and the bulges remain if the movement is removed, and this reveals that the cause of the tides is not centripetal force. By claiming it is, you mislead people into thinking that the movement is fundamentally a part of the cause, but it is incidental to it.

A case with a ball on a string being distorted out of shape as it swings round in circles is very different - the distortions there are actually caused by centripetal force. Stop the ball moving, and the force disappears to prove it.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 16/08/2018 19:22:27
... a case where centripetal force is generated by the movement of the thing going round [and where it disappears if the object is stopped]
Again:
1) In cases where a string, rope, wire or chain transfer the centripetal force to the rotating object, ONLY if there is an active supply of energy at the inner end of the linking device, could the object keep rotating for some time. "Your" cases are only transient movements.
2) In any case, what you "see" in those cases has a certain reality, but you interpret it wrongly.
I wish not to have to mention "centrifugal force": you handle the term "centripetal" so erroneously, that it will be much tougher to discuss with you "centrifugal" one ...
What you see, rather than "the movement causes the centripetal force", could be expressed "the movement originates a CENTRIFUGAL force", because both the inertia of the object (tending to follow the tangent), and the centripetal force, are required for the rotation of the object.
As I said some days ago:
Hitting a ball produces a transference of momentum, always through forces. At the very initial instant some deformations (of the ball and the bat) occur, what produces opposite pushes on each other, which then change both speed vectors … The ball gets a speed, and its inertia tries to make it go straight … If a string attached to a pole prevent that to happen, what the ball´s inertia does (its “movement”, in your words) is to tighten the string: it pulls outward the string outer end, and the string inner end will pull on the pole … The pole will react exerting an equal but opposite force on the string (3rd Newton´s Motion Law), a centripetal force … THAT centripetal force makes the rectilinear movement of the ball change into circular movement, that is, it causes the rotational movement … Quite the opposite to what you say !!
This is why I don't like the use of the word centripetal in the latter case, but if other people are happy to use it with such semantic ambiguity
Come on! It is not just "other people" ... I could post here a huge amount of dictionaries and Physics texts using the term same way. The problem is yours, not:
...that's fine, just so long as they understand the two different cases and don't allow it to contaminate their understanding of things like tides.
The very common confusion on tides, rather than from the term "centripetal force" (clear for almost everybody) comes from the concept of "centrifugal force".
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 16/08/2018 21:54:22
1) In cases where a string, rope, wire or chain transfer the centripetal force to the rotating object, ONLY if there is an active supply of energy at the inner end of the linking device, could the object keep rotating for some time. "Your" cases are only transient movements.

If a ball is attached by a string to a massive lump of lead in deep space and is set moving round it in circles (with the massive object rotating to avoid tangling), this would keep going round and round forever. The generation of centripetal force would be continuous, and it's generated by the movement. Remove all the rotation and the centripetal force disappears. There is no active supply of energy at the inner end.

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2) In any case, what you "see" in those cases has a certain reality, but you interpret it wrongly.
I wish not to have to mention "centrifugal force": you handle the term "centripetal" so erroneously, that it will be much tougher to discuss with you "centrifugal" one ...

The misinterpretations are all yours, but you seem incapable of recognising that. Time and time again you produce errors which you never acknowledge when your errors are made clear to you - you just go on claiming to be right and keep accusing me of making errors (which I haven't made). It's a pantomime.

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What you see, rather than "the movement causes the centripetal force", could be expressed "the movement originates a CENTRIFUGAL force", because both the inertia of the object (tending to follow the tangent), and the centripetal force, are required for the rotation of the object.

There are many valid wordings and descriptions, but the movement causes the centripetal force to be generated regardless of how you word it. Removing the movement removes the centripetal force, and reintroducing it restores it. Straightforward cause and effect. Removing the movement in an orbiting system though and the gravitational force does not vanish.

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As I said some days ago:
Hitting a ball produces a transference of momentum, always through forces. At the very initial instant some deformations (of the ball and the bat) occur, what produces opposite pushes on each other, which then change both speed vectors … The ball gets a speed, and its inertia tries to make it go straight … If a string attached to a pole prevent that to happen, what the ball´s inertia does (its “movement”, in your words) is to tighten the string: it pulls outward the string outer end, and the string inner end will pull on the pole … The pole will react exerting an equal but opposite force on the string (3rd Newton´s Motion Law), a centripetal force … THAT centripetal force makes the rectilinear movement of the ball change into circular movement, that is, it causes the rotational movement … Quite the opposite to what you say !!

You won't gain anything by repeating that. The direction of causation is exactly as I stated regardless of how much irrelevant detail you want to go into, and all my descriptions are correct. If you're saying the opposite to me, then you must be wrong. We both appear to agree that the centripetal force causes the rotational movement (the change in movement of the ball), but the movement of the ball itself is what causes the the centripetal force to appear and the centripetal force merely modifies it by changing the direction of travel of the ball (while the ball moves at a constant speed).

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This is why I don't like the use of the word centripetal in the latter case, but if other people are happy to use it with such semantic ambiguity
Come on! It is not just "other people" ... I could post here a huge amount of dictionaries and Physics texts using the term same way. The problem is yours, not:

This is a side issue about the use of the word centripetal in relation to gravity. I simply made the point that I consider it to be an unwise usage of the word because the force exists regardless of the rotation. That's a perfectly reasonable point for me to make, and all the more so when the result of this usage leads to you making such errors in your understanding of tides. It is an area of labelling which needs to be cleaned up in order to remove ambiguity. I can't help it though if other people keep tying themselves to a mess.

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...that's fine, just so long as they understand the two different cases and don't allow it to contaminate their understanding of things like tides.
The very common confusion on tides, rather than from the term "centripetal force" (clear for almost everybody) comes from the concept of "centrifugal force".

The confusion here has nothing to do with misunderstandings related to the use of the term centrifugal force. We don't need the word or idea in the discussion at all. The confusion here is with centripetal force where you are attributing tides to a force related to rotation even though the tides are actually caused directly by differential gravity and do not depend on orbiting. You are trying to drag the rotation in as a fundamental part of the cause, and that is plain wrong. The rotation and centripetal force have no place in the fundamental explanation.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 18/08/2018 18:39:57
#210 David Cooper
I could easily refute many of those paragraphs, erroneous either to Physics or to Logics (or to both) … But I know it´d be useless, and I won´t do it.
Instead I´m going to refute what seems to be the very “foundation” of your stand. And using same arguments that constitute your kind of pseudo-bible, that keeps you despising many scholars´s statements, and twisting basic Physics laws.
You´ve said a lot of times something like: if we “remove" the revolving movement of earth, the bulges remain … So, that movement cannot be causing the tides.
I already did refute a similar idea of yours (#145), in relation with your idea of a mega-spaceship causing bulges. But you didn´t get it, and said I had taken your case in the wrong way !!
Let us forget the spaceship, and have just moon and earth.
If, somehow, we "remove" the revolving, both earth and moon will accelerate towards each other … Though only a transient movement, bulges would remain (and increase) until the collision. OK. Therefore, revolving was not causing them (YOUR deduction).
IN A QUITE SIMILAR WAY, we could reason as follows:
If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)
Therefore, if with the differential moon´s gravity maintained, outer positive bulge doesn´t exist, the moon differential gravity cannot be the cause of outer bulge !!
Same way you´ve “reasoned” lot of times … Either both are right, or both are wrong …
Any guess ??
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 19/08/2018 01:17:15
I could easily refute many of those paragraphs, erroneous either to Physics or to Logics (or to both) … But I know it´d be useless, and I won´t do it.

You could try to, but you'd fail.

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You´ve said a lot of times something like: if we “remove" the revolving movement of earth, the bulges remain … So, that movement cannot be causing the tides.
I already did refute a similar idea of yours (#145), in relation with your idea of a mega-spaceship causing bulges. But you didn´t get it, and said I had taken your case in the wrong way !!

You didn't refute it, and you clearly didn't get it either - you just did your usual thing of confusing yourself with extraneous factors which I didn't go into (you should have been able to rule them all out by imagining the Earth being pulled along by an evenly-applied tractor beam from the spaceship). Maybe you'd see it more clearly if you switch the moon and Earth over so that the moon's being pulled by the spaceship instead and the Earth is accelerating after it, but never able to catch up. Now the only factor is the moon's pull on the Earth and the differential gravity causing the two tidal bulges which persist for the longterm with no rotation being involved.

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Let us forget the spaceship, and have just moon and earth.
If, somehow, we "remove" the revolving, both earth and moon will accelerate towards each other … Though only a transient movement, bulges would remain (and increase) until the collision. OK. Therefore, revolving was not causing them (YOUR deduction).

Indeed - they were not caused by rotation or centripetal force.

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IN A QUITE SIMILAR WAY, we could reason as follows:
If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)
Therefore, if with the differential moon´s gravity maintained, outer positive bulge doesn´t exist, the moon differential gravity cannot be the cause of outer bulge !!
Same way you´ve “reasoned” lot of times … Either both are right, or both are wrong …
Any guess ??

The removal of rotation does not remove the tidal bulges - they remain, but clearly the Earth has to be free to accelerate towards the moon for this to be the case. If you hold the Earth in position, then you have a different result where a single bulge will appear on the side nearest the moon. With the ball on the string, when you stop the rotation, the ball simply sits there with all the distortion removed - there is no continuation of a force trying to pull it towards the centre of the now-lost rotation because that centripetal force has gone, unlike in the gravity case where it continues to apply in full. If you want to apply a force though to maintain the distortion on the ball you can do that, but it wouldn't be centripetal force - this new force would produce the same distortion if the ball's still free to move as it had on it when it was going round in circles, and if you held the ball so that it couldn't move, it would produce a different distortion more like the one in the case where you hold the Earth and don't let it accelerate towards the moon. But, start the ball moving round in circles again and what happens? The centripetal force is restored and the extra force you added to the system will now shorten the string and pull the ball in, because adding that force was cheating. The key difference remains: the gravitational pull is not lost when the rotation stops in one system, but the centripetal force in the string disappears when the rotation stops in the other system. One of these forces is generated by the rotation, but the other isn't, and the one that isn't (i.e. gravity) continues to produce the tidal bulges.

I have no expectation that all of that won't go straight over your head like everything else and you'll fool yourself into imagining that you can refute it, but perhaps it will register with other people who read this.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 20/08/2018 19:23:32
The removal of rotation does not remove the tidal bulges - they remain, but clearly the Earth has to be free to accelerate towards the moon for this to be the case. If you hold the Earth in position, then you have a different result where a single bulge will appear on the side nearest the moon
There we are! (surely it´s a lapse of yours, and you meant "the removal of the rectilinear movement" ...)
It´s unbelievable that you consider absolutely necessary to let the earth free to get the rectilinear accelerated movement towards the moon (to have the second bulge), but time and again you say that (as it is in the real case) the curved accelerated movement of the earth (its revolving around the moon-earth barycenter), has nothing to do with the formation of the outer bulge, and that it is not necessary at all !!
They are two different ways inertia manifests itself, but due to same basic Physics laws ...
In the first case, outer hemisphere massive stuff kind of "falls behind", because it is not sufficiently pulled by the moon to get the rectilinear acceleration of the whole earth. The whole earth stretches, and behind bulge also appears.
In the real case, outer bulge equally requires the revolving movement, because outer hemisphere massive stuff is not sufficiently pulled by the moon to get the required centripetal acceleration for the revolving movement. Internal stresses (in both senses, inwards and outwards) appear, earth stretches too, and outer bulge builds too ...
But with your basic Physics misconceptions, I´m afraid you´ll carry on not accepting that !!
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 20/08/2018 21:41:41
The removal of rotation does not remove the tidal bulges - they remain, but clearly the Earth has to be free to accelerate towards the moon for this to be the case. If you hold the Earth in position, then you have a different result where a single bulge will appear on the side nearest the moon
There we are! (surely it´s a lapse of yours, and you meant "the removal of the rectilinear movement" ...)

I meant exactly what I said. Rewording it makes no difference.

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It´s unbelievable that you consider absolutely necessary to let the earth free to get the rectilinear accelerated movement towards the moon (to have the second bulge), but time and again you say that (as it is in the real case) the curved accelerated movement of the earth (its revolving around the moon-earth barycenter), has nothing to do with the formation of the outer bulge, and that it is not necessary at all !!

There's nothing unbelievable about it - if the Earth and moon fall together along a straight line, the tidal bulges are both there, and will grow in size for several days until the collision. The other movement that we have with the Earth and moon at 90 degrees to the line between them is irrelevant to the cause of those bulges.

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They are two different ways inertia manifests itself, but due to same basic Physics laws ...
In the first case, outer hemisphere massive stuff kind of "falls behind", because it is not sufficiently pulled by the moon to get the rectilinear acceleration of the whole earth. The whole earth stretches, and behind bulge also appears.

That indeed is the cause of the tidal bulges - differential gravity with the Earth and moon being accelerated towards each other and without any need of the orbit.

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In the real case, outer bulge equally requires the revolving movement, because outer hemisphere massive stuff is not sufficiently pulled by the moon to get the required centripetal acceleration for the revolving movement. Internal stresses (in both senses, inwards and outwards) appear, earth stretches too, and outer bulge builds too ...
But with your basic Physics misconceptions, I´m afraid you´ll carry on not accepting that !!

The revolving aspect can be ignored. The pull of the Earth and moon upon each other in a straight line is all you need for the tidal bulges. The sideways movement is simply an addition to it (which conveniently prevents a collision), and having that  incidental sideways component of movement involved is not any justification for claiming that the tidal bulges are generated by centripetal force. At any point in time you have two vectors for the movement of the Earth or moon (with one running along the imaginary line between the two bodies and the other at 90 degrees to it), and two vectors for the acceleration being applied by each body on the other, again with one acting along the invisible line between the two bodies and the other perpendicular to it. The first of these acceleration vectors is never zero but the latter is always zero. It doesn't matter a damn what what the perpendicular action is - the bodies accelerate towards each other (without necessarily getting any closer together) and the material on each body that's closest to the other body is pulled more strongly while the material furthest away is pulled less strongly, and that is all that's involved in the production of the bulges. The perpendicular movement adds nothing to them. All you're doing is claiming that centripetal force causes the tidal bulges on the basis that there is normally orbital rotation involved, but that's s nonsense as the same bulges exist when you remove that orbital movement and reduce things to a straight line case. That's why physicists don't agree with you.

The only case you might have would be if you were to claim that the straight-line case is a special case of rotation and that the word "centripetal" should include it too as part of its remit, in which case when you slam the brakes on in a car while it's moving in a straight line, you are thrown forward and centripetal force acts on you via the seatbelt. If you want to make that argument, I'll be happy to back you up and join the call to include that in the definition of centripetal so that it is no longer associated solely with cases involving curved paths. However, even then, if you insist on classing gravity as centripetal force (and also logically apply the new definition of centripetal to straight-line action), you then have people jumping up in the air and being pulled straight back down by centripetal force. Do all the physics books have to be rewritten to say that centripetal force causes things to fall rather than gravity? No - the fundamental explanation of why things fall is gravity, just as the fundamental cause of the tides is differential gravity. And you also still have the key difference between two cases with things moving in circular paths where in one case the rotational movement generates the centripetal force in the string, and in the other case the rotational movement doesn't (and where the force in the invisible string, gravity, continues to act in full regardless of whether the object is orbiting or not). Even if you class both of these as centripetal and don't consider one case to be more of an analogy, they are radically different beasts in nature which should not be given the same label.

Your attempt to attribute tidal bulges to centripetal force remains at odds with physics though with the way the word centripetal is currently defined (with curved paths being a requirement), and that should be sufficient cause in itself for you to desist. The straight-line case produces the same bulges and demonstrates that you are plain wrong on the whole issue.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 24/08/2018 18:54:01
#214 David Couper
What a lot of absurdities … and false statements !!
In the first place, I neither ever:
… have claimed "that centripetal force causes the tidal bulges” (or other equivalent expressions of you)
… let alone have "classified gravity as centripetal force” (!!)
Again, not even in my wildest dreams could I have thought I was going to have to explain something such as what follows … But I´ll do it, rather than because of you (you seem to need a change of some of your mind “chips”), not to let unanswered all that rubbish, and for the sake of the interest of others.
In relation to a noun such as “force”, one thing is its nature, or its essence, and quite another its let us say function, or its not essential characteristics, what usually is expressed through adjectives.
A force can be big, small, permanent, variable, pulling or pushing contact f., “at a distance f." (not sure the correct English name) … as well as centripetal or centrifugal, a possible additional quality in specific cases, NOTHING TO DO with “classifying gravity as centrifugal force” !!
Different types of forces (or components of forces) can act as centripetal forces, as long as their directions are perpendicular to the affected object trajectory, and in the sense of a “center” (of the circle if circular movement, or the center of curvature if other curved paths).
Gravitational force can act as centripetal force, as long as what above occurs.
E.g.: the parabolic trajectory of a cannon ball is due to the gravitational force action (and the initial speed vector): its tangential component changes the size of the speed, and its component perpendicular to speed vector, acting as a centripetal force, curves the trajectory, not letting the ball to follow its inertial tendency to continue straight …
I previously posted:
"If e.g you google "phisics.ohio-state.edu" and go to "Dynamics of Uniform Circular Motion" (Chapter 5, 5.3 Centripetal Force), you can read (apart from many other things and formulas):
"...Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration.
The centripetal force is the name given to the net force required to keep an object moving on a circular path".
But you just answered:
"It doesn't matter how much justification you can find for calling it centripetal force. This is a grey area in which some usages of the word aren't fully rational (!!)"
 And when I posted what Oxford Dictionary says:
"A force that acts on a body moving in a circular path and is directed towards the centre around which the body is moving.
Example sentences
‘Newton analysed the motion of bodies in resisting and non-resisting media under the action of centripetal forces.’
‘Now, if we set a cylindrical space station in rotational motion at a certain angular velocity, anything that moves with it will need a centripetal force to keep it rotating with the station.’
‘Armed with his conception of material space, which was the location of gravity, combined with the constant action of the centripetal force of universal gravitation, Toland believed he had accounted for all phenomena in the universe’,
you just commented
 “... a case where centripetal force is not generated by that movement but is 100% independent of it [and where it continues to act if the object is stopped]. This is why I don't like the use of the word centripetal in the latter case, but if other people are happy to use it with such semantic ambiguity, that's fine, just so long as they understand the two different cases and don't allow it to contaminate their understanding of things like tides".
No wonder you said centripetal force is a grey area.
I repeat: you draw wrong conclusions (100% right for you !!) from a rather singular and arguable case (the movement of a ball-string-fixed pole around the later, which according to you causes a "centripetal force”, whatever you mean with that term), and are unable to see in nature things happen differently.
By the way, I was going to leave what related to my supposed claim "that centripetal force causes the tidal bulges” for another post, because unless we previously clarify what you actually mean with the term, any discussion would be an additional waste of time …
But just one thing now: what I have said several times is that CENTRIFUGAL forces are ONE of the causes of the bulge more distant from the moon, rather the opposite to what you seem to think I say !!.
As early as in one of my very first posts here (#24), more than three years ago, I already said:
"Imaging an athlete of hammer trow speciality. He or she can´t keep verticality when throwing the hammer. It is necessary to lean a little backwards. Otherwise the hammer could not be rotated. Both the athlete and the hammer will rotate around an axis situated near the forward part of the athlete´s body.
If the hair of the athlete is long and not  fixed by some device, instead of keeping its normal downward direction due to its weight, it will move back and upwards …
That cannot be due to anything similar to what stated by Schutz relative to tides: CENTRIFUGAL force is the cause"
But other times I tried and avoided the use of such a controversial term, and said, as on post #213:
“… They are two different ways inertia manifests itself, but due to same basic Physics laws ...
In the first case (linear accelerated movement), outer hemisphere massive stuff kind of "falls behind", because it is not sufficiently pulled by the moon to get the rectilinear acceleration of the whole earth. The whole earth stretches, and behind bulge also appears.
In the real case, outer bulge equally requires the revolving movement, because outer hemisphere massive stuff is not sufficiently pulled by the moon to get the required centripetal acceleration for the revolving movement. Internal stresses (in both senses, inwards and outwards) appear, earth also stretches, and outer bulge builds too …”
But you have in your head many absurdities about the pretty simple concept of “centripetal force”, and have proved unable to grasp it … I can´t imagine what there can be in your mind in relation to the rather tricky concept of “centrifugal force” ...

 
 

 
 
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 24/08/2018 22:14:13
#214 David Couper
What a lot of absurdities … and false statements !!
In the first place, I neither ever:
… have claimed "that centripetal force causes the tidal bulges” (or other equivalent expressions of you)
… let alone have "classified gravity as centripetal force” (!!)

What you have consistently done is deny the correct explanation whenever it's been put to you, and you've harped on and on about centripetal force instead, so forgive me for failing to understand that you actually agree with the correct explanation and for not recognising that it's a mere illusion that you've spent weeks arguing against it.

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"If e.g you google "phisics.ohio-state.edu" and go to "Dynamics of Uniform Circular Motion" (Chapter 5, 5.3 Centripetal Force), you can read (apart from many other things and formulas):
"...Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration.
The centripetal force is the name given to the net force required to keep an object moving on a circular path".
But you just answered:
"It doesn't matter how much justification you can find for calling it centripetal force. This is a grey area in which some usages of the word aren't fully rational (!!)"

That's all well and lovely, but what you shouldn't be doing is claiming that the force acting on the object is centripetal rather than gravity when it's gravity that's doing the job and any justification for labelling it as "centripetal" is an irrelevance. What you're also repeatedly doing is ignoring the issue in question and obsessing over other arguments which you're then misrepresenting by taking quotes out of context. I've made it very clear multiple times that physics should be making a clear distinction between centripetal force that's generated by a movement and "centripetal" force that's there regardless of the movement because it is not generated by it, and that this distinction is crucial to the central point of our argument because when the movement stops, the force continues to act in full, demonstrating that the cause of tidal "bulges" should not be attributed to centripetal force. That is all there is to it. You lost that argument, but you're incapable of admitting it, and all you're doing now is playing games to hide the fact that you lost it.

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And when I posted what Oxford Dictionary says:

I told you long ago that if you want to call it "centripetal" in cases where the force is not generated by the movement, that's fine - there are plenty of definitions that agree with you. All I said was that I think it's a mislabelling issue in physics which should be tidied up in that it ignores an ambiguity, and that this ambiguity is deeply damaging in that it's causing people like you to make massive mistakes in your understanding. But this is all a separate issue from the one in question in this thread. I don't like sloppy labelling systems, but no amount of attacking me on the basis of established usage over my disagreement with the way labels are used will alter the fact that the cause of tidal "bulges" is straight-line differential gravity. You still refuse to acknowledge the point that when the movement stops, the force continues to act and the cause of tides is revealed to have no dependency on centripetal force.

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I repeat: you draw wrong conclusions (100% right for you !!) from a rather singular and arguable case (the movement of a ball-string-fixed pole around the later, which according to you causes a "centripetal force”, whatever you mean with that term), and are unable to see in nature things happen differently.

So you're still unable to understand that bit. It's all perfectly simple: the movement of the ball generates the centripetal force in the string. Stop the ball and the centripetal force disappears. Start it moving again and the centripetal force returns. That is how it happens in nature. Do the same thing with orbiting objects and you get a very different result. I don't know what it is that stops you from understanding this simple thing, but that inability is your key problem.

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But just one thing now: what I have said several times is that CENTRIFUGAL forces are ONE of the causes of the bulge more distant from the moon, rather the opposite to what you seem to think I say !!.

I may well have got the wrong impression about how much you attribute the "bulges" to the actual cause and how much you attribute it to centripetal/centrifugal force, but I did ask you early on what proportion of it you imagined to be caused by the latter and you simply ignored the question. None of that really matters though, because centripetal/centrifugal force isn't even one of the causes. The cause is straight-line differential gravity alone.

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As early as in one of my very first posts here (#24), more than three years ago, I already said:
"Imaging an athlete of hammer trow speciality. He or she can´t keep verticality when throwing the hammer. It is necessary to lean a little backwards. Otherwise the hammer could not be rotated. Both the athlete and the hammer will rotate around an axis situated near the forward part of the athlete´s body.
If the hair of the athlete is long and not  fixed by some device, instead of keeping its normal downward direction due to its weight, it will move back and upwards …
That cannot be due to anything similar to what stated by Schutz relative to tides: CENTRIFUGAL force is the cause"

It that example, centripetal/centrifugal force can validly be be described as the cause of the equivalent of tidal forces because it stops acting if you remove the movement that generates it. With gravity and orbits though, that isn't the case - stop the movement and the "bulges" remain.

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But other times I tried and avoided the use of such a controversial term, and said, as on post #213:
“… They are two different ways inertia manifests itself, but due to same basic Physics laws ...
In the first case (linear accelerated movement), outer hemisphere massive stuff kind of "falls behind", because it is not sufficiently pulled by the moon to get the rectilinear acceleration of the whole earth. The whole earth stretches, and behind bulge also appears.

Fine - that's straight-line differential gravity in action on an object that's free to move, so why all the disagreement if you actually do understand the correct mechanism?

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In the real case, outer bulge equally requires the revolving movement, because outer hemisphere massive stuff is not sufficiently pulled by the moon to get the required centripetal acceleration for the revolving movement. Internal stresses (in both senses, inwards and outwards) appear, earth also stretches, and outer bulge builds too …”

In the real case, straight-line differential gravity is all it takes - there is no need to try to manufacture extra factors for it. Take two cases for example where there is no perpendicular movement to the gravitational force: place the planet and moon the same distance apart in each case, but in one case have them moving towards each other and in the other case moving apart (along a straight line). Do we get different bulges at that moment where they are that certain distance apart? No. The same acceleration force is being applied, and it's the same amount of difference in acceleration force being applied across the moon/planet from one side to the other, leading to the same size of "bulges" (pressure differences). Then consider a third case where there is perpendicular movement and at that moment no movement away from or towards the other body. Is there any difference to the "bulges"? No. The sideways movement is completely irrelevant as it has no role in forming the bulges. Consider all other angles at which the planet and moon might be moving relative to each other, and ask the same question: are the "bulges" different at a given separation in any of these cases? No. They are not.

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But you have in your head many absurdities about the pretty simple concept of “centripetal force”, and have proved unable to grasp it … I can´t imagine what there can be in your mind in relation to the rather tricky concept of “centrifugal force” ...

The absurdities are all on your side. I clearly understand the physics better than you do - you are using labels in ways that confuse you by making you think two radically different systems are the same, and the result is that you drag something into an explanation of tides that doesn't belong there. All I've done is try to help you stop confusing yourself with labels so that you'll look at the actual causation instead, but you just can't do it. The only thing that matters here is the cause of the bulges, and centripetal/centrifugal force is an irrelevance to them because they are determined by straight-line differential gravity. The size of the "bulges" is determined by the distance between the moon and Earth regardless of how they're moving relative to each other at the time. There is no magical addition to their size by the relative movement of the two bodies - it's all determined by the straight-line acceleration.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 25/08/2018 14:19:53
If I may,

Molnav, the force applied to a large body made out of particles cannot be treated the same as the force applied to each of its particles taken separately, and you seem to think so. In the case of a body, we need an anchor to hold it, but we can't hold a particle with an anchor. The moon and the earth are not anchored to one another, instead, it is each particle from each of the bodies that is linked to all the particles of the other body by means of the gravitational force, and that force depends on distance, which is not the case with a string. If we would anchor all the particles of the moon to all the particles of the earth by means of strings, there would be no bulge, but since the gravitational force weakens with distance, the closer particles are pulled more strongly than the farther ones and it creates bulges.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 29/08/2018 19:05:32
#216 and #217
Most people will be able to see which side the absurdities are on. So, to refute once more many of them is unnecesary … But:
So you're still unable to understand that bit. It's all perfectly simple: the movement of the ball generates the centripetal force in the string. Stop the ball and the centripetal force disappears. Start it moving again and the centripetal force returns. That is how it happens in nature (??). Do the same thing with orbiting objects and you get a very different result ...
Another day with more time I´ll come back to what quoted, which seems to be one of the roots of that unsolvable puzzle of wrong ideas deep in your mind. In line with what I said about how things happen when we hit a ball attached through a string to a pole (#202)
"Hitting a ball produces a transference of momentum, always through forces. At the very initial instant some deformations (of the ball and the bat) occur, what produces opposite pushes on each other, which then change both speed vectors … The ball gets a speed, and its inertia tries to make it go straight … If a string attached to a pole prevent that to happen, what the ball´s inertia does (its “movement”, in your words) is to tighten the string: it pulls outward the string outer end, and the string inner end will pull on the pole … The pole will react exerting an equal but opposite force on the string (3rd Newton´s Motion Law), a centripetal force … THAT centripetal force makes the rectilinear movement of the ball change into circular movement, that is, it causes the rotational movement … Quite the opposite of what you say !!"
By the way, as usual you replied with more absurdities:
"Now I understand why you're getting this wrong. It's because you've got a lot of that right. There is one force applied by the ball to the string which spreads from ball to pole at the speed of sound (revealing the causation, which you ignore), and then another force is applied in the opposite direction to equalise it, starting at the pole and then ending at the ball, again spreading at the speed of sound, and it's this latter force that's called centripetal. The centripetal force then modifies the movement of the ball, but crucially it does not cause the movement of the ball - it merely modifies that movement".
I have another case in mind, perhaps easier to understand, but I have to draft it very carefully, to diminish the possibilities of you to misunderstand and/or twist my words !!
But now just another thing.
... the force applied to a large body made out of particles cannot be treated the same as the force applied to each of its particles taken separately, and you seem to think so. In the case of a body, we need an anchor to hold it, but we can't hold a particle with an anchor. The moon and the earth are not anchored to one another, instead, it is each particle from each of the bodies that is linked to all the particles of the other body by means of the gravitational force, and that force depends on distance, which is not the case with a string. If we would anchor all the particles of the moon to all the particles of the earth by means of strings, there would be no bulge, but since the gravitational force weakens with distance, the closer particles are pulled more strongly than the farther ones ...
OK. But it is equivalent to consider each particle of the earth linked by means of a somehow elastic string to moon´s c.g., certainly with tensions (gravity pull) the shorter the string, the stronger.
But you seem to forget that each earth particle is also linked much more strongly to the rest of the earth, to its c.g. if you wish: its own weight.
Being that hundreds of thousand times bigger than moon´s pull (at earth´s surface), we could say all earth particles are kind of anchored to earth´s c.g., rather than simply “linked”. Though also through not completely rigid strings (especially when water particles).
Due to that, all those particles are not actually free to accelerate as answer to moon´s pull whatsoever … They have a certain freedom though. The way the whole earth moves is paramount to the type and degree of freedom our planet allows each of individual particles …
The whole planet “forces” each of its particles to move “together”, and the distribution of inherent internal stresses depends on the movement of the whole, not only on the differential gravity from moon. And those internal stresses (their change due to moon-earth rotation/revolving) are the ultimate cause of tides.
If we were analyzing tidal effects of earth pull on the moon, internal stresses there would be relatively higher, because not only gravity decreases to the square of the distance: required centripetal force (mω˛r) increases proportionally to the distance. Therefore the imbalance between local pull from earth (which ought to cause each particle rotation), and the required centripetal force, increases more with the distance …Or, if you wish, the further the particle, the higher the centrifugal force.
That´s why to say earth revolving (and forces inherent in that type of movement) doesn´t intervene on tides is utterly erroneous ...   

 
 
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 29/08/2018 22:12:31
#216 and #217
Most people will be able to see which side the absurdities are on. So, to refute once more many of them is unnecesary …

Indeed they will - it's a straight-line force being applied by gravity where the perpendicular movement is irrelevant.

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"Hitting a ball produces a transference of momentum, always through forces. At the very initial instant some deformations (of the ball and the bat) occur, what produces opposite pushes on each other, which then change both speed vectors … The ball gets a speed, and its inertia tries to make it go straight … If a string attached to a pole prevent that to happen, what the ball´s inertia does (its “movement”, in your words) is to tighten the string: it pulls outward the string outer end, and the string inner end will pull on the pole … The pole will react exerting an equal but opposite force on the string (3rd Newton´s Motion Law), a centripetal force … THAT centripetal force makes the rectilinear movement of the ball change into circular movement, that is, it causes the rotational movement … Quite the opposite of what you say !!"

Your interpretation skills are woeful - nothing in that is in conflict with what I said, until you say "Quite the opposite to what you say". You're just playing silly games to divert attention away from the fact that you've lost the argument, in this case by refusing to understand what I mean by "the movement of the ball". The ball is only moving because it's been hit by a bat - that input of force is what sets it moving, and that movement causes the centripetal force to be generated, and that force in turn modifies the movement of the ball. You are then labelling that modified movement as "the movement of the ball" and are attacking me for claiming that that generates the centripetal force, but that is clearly not what I claimed at any time. When I refer to the movement of the ball, I'm referring to the fact that the ball is moving and not to how it's moving. The ball's movement generates the centripetal force and the centripetal force then modifies the way the ball moves. That is what I have said throughout, and it remains fully correct. Stop the ball and the centripetal force disappears.

To help you understand this mind-numbingly simple point, think about it in terms of kinetic energy. The bat hitting the ball gives the ball kinetic energy. That kinetic energy stays with the ball as it goes round and round the pole, and it leads to forces being generated in the string, one of which is the centripetal force that changes the direction the ball is moving in, but it doesn't add or subtract kinetic energy from the ball - it merely changes the direction the ball's moving in. So, when I (rightly) say the movement of the ball generates the centripetal force, I'm referring to the movement of the ball that results from this energy being added to it, and that movement then generates centripetal force which doesn't change the energy that the ball is carrying other than the direction that energy's moving in. When I stop the ball, I remove that energy from it and the centripetal force disappears. At no time does the centripetal force change the amount of movement energy that the ball is carrying, and that's what I'm referring to when I say that it does not cause the movement - all it does is modify the movement by changing the direction the ball is moving in.

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By the way, as usual you replied with more absurdities:

No; I replied with more correct statements which you consider to be absurdities because you are highly irrational and totally deluded about your thinking capabilities.

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The whole planet “forces” each of its particles to move “together”, and the distribution of inherent internal stresses depends on the movement of the whole, not only on the differential gravity from moon. And those internal stresses (their change due to moon-earth rotation/revolving) are the ultimate cause of tides.

The rotation is a separate issue which merely makes the part near the equator stick out more than it does towards the poles - that rotation is practically constant and the bulge it produces has no role in the tides. If the moon didn't exist, the Earth would still be that shape because of its own rotation, and that is not part of tides so you should not be treating it as a component of tides. Your second sentence is plain ridiculous - the "moon-earth rotation/revolving" part of it is the objectionable bit as it does not cause the tides on the Earth - the ultimate cause of tides is the differential gravity with its straight-line acceleration.

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If we were analyzing tidal effects of earth pull on the moon, internal stresses there would be relatively higher, because not only gravity decreases to the square of the distance: required centripetal force (mω˛r) increases proportionally to the distance. Therefore the imbalance between local pull from earth (which ought to cause each particle rotation), and the required centripetal force, increases more with the distance …Or, if you wish, the further the particle, the higher the centrifugal force.

The straight-line acceleration already provides all the force difference involved without any need to bring centripetal force into it.

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That´s why to say earth revolving (and forces inherent in that type of movement) doesn´t intervene on tides is utterly erroneous ...

The earth's revolving certainly does impact on the way tides present themselves. If the Earth was rotating once a month, permanent bulges would form on two opposite sides and they would then grow and shrink a little throughout each month as the moon gets closer and further away. The moon's orbit not being circular would also make the bulges move east and west a bit because the Earth's rotation wouldn't track the moon's position in the sky perfectly. The Earth's actual rotation prevents such bulges from forming (because the water can't move fast enough to keep up), so we just have tiny pressure changes acting over a huge area and adding enough energy to generate oscillations that provide the tides that we see. The cause of it all though is the straight-line differential gravity and not centripetal force. The rotation of the Earth means that the direction straight-line differential gravity is applied in changes all the time, but that isn't a centripetal force issue. And as I keep pointing out, you can remove all rotation from the system and the "bulges" (pressure differences) remain in place, which is what shows that centripetal force does not cause them. Not only does it not cause them, but it doesn't even add to them.

If I was to write a computer program to simulate this and show the forces applying on the surface of the Earth and moon, I would do it as follows. Let's name three points of each body: EC is the centre of the Earth, EN is the point on the surface of the Earth nearest the moon, EF is the point on the surface of the Earth furthest from the moon, while MC, MN and MF are the equivalent points on the moon. To calculate the tidal force on the Earth at point EN, I'd simply find the difference between gravitational force applied to EC and EN, while for  point EF I'd find the difference between the gravitational force applied to EC and EF. Having done that, I could subtract the results from 10 to get the force that would be felt by objects standing on the surface at those points (measured in units called Newts). This is based on 10 being the number of Newts that would be recorded at the equator if the moon wasn't present, but note that the centripetal force aspect is already accounted for in that figure. (At the poles, the value would be higher than 10 because there is no centripetal force acting on them to suppress it.) I would not add any extra centripetal force to those numbers in the way you seek to do because it has already been fully accounted for - we should certainly not deal with this rotation-driven equatorial bulge by factoring it in afterwards and trying to pass it off as part of the tides.

However, you appear to want to make a centripetal addition to the results, so I want to know how you imagine that it should apply so that it can be added to the program. We'll then be able to see whether you're double-counting it (or just counting it late and mistaking it for part of the tides), and I also want to see how big/small an impact this addition would have on the results. I will write the program then, and we'll be able to put your maths into it to replace mine to see how our methods compare. This will help us home in on the truth of the matter, and everyone will be able to see it all in action. Oh, and I've actually cheated by writing half the program in advance, so it may be ready by tomorrow.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 30/08/2018 15:51:48
OK. But it is equivalent to consider each particle of the earth linked by means of a somehow elastic string to moon´s c.g., certainly with tensions (gravity pull) the shorter the string, the stronger.
No need for the strings to be elastic; if they pull more, the particles tied to them will simply accelerate more and the strings will have to shorten to follow the motion of the particles if they are free to move, but if they are not, for instance if the earth follows a circular orbit around the moon and if it is not rotating with regard to the moon, then all the distances that have shortened will stay the same and the strings will not have to shorten anymore.

But you seem to forget that each earth particle is also linked much more strongly to the rest of the earth, to its c.g. if you wish: its own weight.
There is a difference between the action the rotation of a planet and the action the orbital motion of the same planet has on its own particles, and I'm not sure that you figure it out the way I do, so here is a mind experiment about that. If we accelerate the earth's rotation until its surface begins to orbit, the equatorial bulge will get a lot more important than what it is, but once some of the surface will be orbiting, we can slow down the earth and thus reduce the bulge without affecting what is already orbiting. Now, if we then succeed to accelerate at the same time all the orbiting particles again, we won't get a bulge like the equatorial one, but instead, all those particles will follow an elliptic trajectory whose perigee will be located at the place they were accelerated, which means that they will all begin to slow down and move away from the earth for a while until they get at their apogee, and then they will start accelerating again and move closer to the earth until they get at their perigee. Accelerating a circular orbital motion does not produce an equatorial bulge, it produces a pulsation, and it is so because orbiting bodies are free to move from one another radially. It is that liberty that produces the tides in this case, not the rotation.

The water molecules of the seas are free to move radially a bit, and they do so while pulsating a bit as if they were accelerating and decelerating continuously on their orbital trajectory with the moon. David doesn't agree with me on that one but I still think that the tides have something to do with the two faces of the earth not being at the right speed on their orbital trajectory with the moon. If I'm right, it means that there might be no tides if the earth was rotating clockwise instead of counterclockwise, because its two faces could then move at close to their right speed on their orbital trajectory with the moon during their transit, which would incidentally have nothing to do with the centripetal force that produces the equatorial bulges either. That mechanism would not change the way David explains the bulges though, because suddenly stopping the orbital motion of the earth without stopping its clockwise rotation would immediately produce some. That mechanism looks weird because it's new but think of it this way: why would the particles of the earth change their trajectory to produce bulges while they are actually orbiting at the right speed all the time on their orbital trajectory with the moon? After all, it is the difference in the centripetal acceleration at different distances that produces the tides, and it is precisely the same difference that produces the orbital speeds, no?
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 30/08/2018 19:29:02
Le Repteux
Thank you.
I have to go more slowly through your post, as well as through David´s.
But just one thing now.
You, and David too, mention earth rotation and equatorial bulge ...
I´ve said many times that (some replying David´s posts), to analyze properly moon-earth dynamics (and the resulting some 28 days tidal complete cycle), we have to disregard completely the daily earth spinning, which has nothing to do with the moon ...
Earth is revolving around moon-earth barycenter, rather than rotating, let alone rotating around the moon !! The dynamics of that movement differs from the case of a proper "orbiting" of a satellite around a planet, or of a planet around the sun ...
The cycle we see (some 24.8 h. complete cycle) is certainly also due to earth daily spinning (the 0.8 h. due to daily moon change of position), but to mix that with the permanent equatorial bulge caused by that spinning is frequent cause of misunderstandings !!
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 30/08/2018 20:53:06
I pointed out before that there are multiple versions of the straight-line case which produce the exact same "bulges". If a moon and planet are moving apart, or towards each other, or just on the point where they're stopping moving apart and are about to start moving towards each other, the "bulges" are identical if the distance between the moon and planet are the same in each case. Relativity shows us that this is the case, because the acceleration force is the same in all three cases - if you are co-moving with the planet (or moon) each time, you will expect it to behave the same way in all of these situations, and it will do.

The same applies to a case where there is perpendicular movement involved - we can add it to all three of the previous cases, and what happens? We use a frame of reference where we're co-moving with the planet /moon as before, and we get the exact same "bulges" produced as in the cases where there is no perpendicular movement. Relativity tells us that it must always behave the same way - the only thing producing the "bulges" is the straight-line differential gravity.

What we do have though is a necessary coincidence in that when there is a component of perpendicular movement, we see things behaving as if they're trying to follow paths that match up to the ones that would be predicted for them by analysing things on the basis that centripetal force is involved, and that is what some people trip over when they're looking at this - it matches up so well that they are fooled into thinking that centripetal force must be the cause, but it's a mere illusion. The real explanation is much simpler - the straight line acceleration does the whole job, and that's why you get the exact same "bulges" when the perpendicular movement is completely removed. The perpendicular component of movement is demonstrably a complete irrelevance to the cause.

(This is radically different to the case of a ball on a string where the perpendicular movement is essential to the formation of the bulges because that movement generates the centripetal force that generates the bulges - in that case, the cause can be attributed to centripetal force and by extension to the perpendicular movement of the ball. With orbiting objects, stopping the movement doesn't remove the force that's generating the "bulges" and the persistence of the "bulges" with the same force still acting in full tells you that the cause of the bulges is not centripetal force - it is merely being labelled as centripetal in cases where there is perpendicular movement, and some people then make the mistake of using that label as part of their description of the cause. That's the same kind of mistake as claiming that a boxer whose hand has been cut off would have blood pouring out of the end of his arm because there's no fist on the end of it - the use of the word fist is fundamentally wrong there as it should be referred to as a hand.)
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 31/08/2018 00:19:34
I promised a program, and here it is. I might modify it further, but it's got enough done to make it worth sharing now:-

http://www.magicschoolbook.com/science/tides.html (http://www.magicschoolbook.com/science/tides.html)

Note how the size of the "bulges" (or more accurately, the strength of the tidal forces) is not dependent on the direction and speed of travel of moon or planet of given masses and sizes, but is determined by one factor alone - distance between the two.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 31/08/2018 10:34:29
As I´ve already said, to discuss with you is useless, among other things because you have so fixed your (wrong) ideas that you are unable to see and/or understand what said to you.
On my post #152 I told you:
"Just one question. You say:
"If you take away the gravitational difference caused by it diminishing over distance, you will have no tides at all. The complexities that you are imagining do not contribute to the tides, even though centripetal bulge does add to to the degree to which the sea bulges away from a spherical shape - the tides are just the distortions added to that already-distorted surface and are not the total distortion away from a sphere".
You seem to refer ("already-distorted surface") to the equatorial bulge caused by daily earth spinning.
That is caused by inertial effects. Water, with a linear speed tangent to equator (if exactly there), tends to follow the tangent. But own earth gravity obliges the water to follow a circular path.
In this case, without any significant gravity differences, we have a relatively huge permanent bulge, because centripetal/centrifugal forces are proportional to the square of angular speed, some 29 times bigger than in the moon-earth "dance"
Somewhere you say tides occur on top of that permanent equatorial bulge. Quite right.
But, why earth daily spinning can cause that permanent equatorial bulge, and (according to you) the some 29 day revolving of the earth around the barycenter (please don´t mix with daily earth rotation !!), besides actual gravity forces on each location, can´t intervene in tides formation ??
In this last earth movement, water (and solid parts too) are also somehow obliged to follow circular paths, instead their inertial tendency to follow the tangents ... Therefore, centripetal and centrifugal forces are also present, and they do intervene on the 29 day cycle of moon-éarth dynamics, and on lunar related tides !!"
And a couple of days ago you said:
If the Earth was rotating once a month, permanent bulges would form on two opposite sides and they would then grow and shrink a little throughout each month as the moon gets closer and further away. The moon's orbit not being circular would also make the bulges move east and west a bit because the Earth's rotation wouldn't track the moon's position in the sky perfectly. The Earth's actual rotation prevents such bulges from forming (because the water can't move fast enough to keep up), so we just have tiny pressure changes acting over a huge area and adding enough energy to generate oscillations that provide the tides that we see.
THERE WE ARE !! ... What in bold is what is actually happening ... !! (though earth "revolves", rather than "rotates", in its dance with the moon)
After so many posts, haven´t you realized even that basic fact ??
How can you have been so sure of your stand, if you continue to mix two quite different movements, and keep considering that when I say earth-moon dance, in particular earth revolving around the barycenter, I´m not referring to that but to earth daily spinning ??
Because, trying yo avoid confusion, I´ve usually been very careful even reserving the term "spinning" for the a day period movement (though "rotation" could also be used), the term "revolving" for the 28/29 d. period movement of earth, and "rotation" for the equal period movement of the moon ... But to no avail, as far as your "blinded" (at least in this case) mind is concerned !!
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 31/08/2018 19:07:38
I promised a program, and here it is. I might modify it further, but it's got enough done to make it worth sharing now:-

http://www.magicschoolbook.com/science/tides.html
Nice simulation as usual David, and since simulations represent real motions, Rmolnav, you should be able to make one that represents what you think. When we halt the inertial motion of the moon on David's one, it falls directly to the earth, but the bulges do not disappear, whereas they should disappear in yours since you think that they need centrifugal motion to appear. I had another idea that may help to illustrate the difference between gravitation and inertia as far as orbital motion is concerned: if we applied a centrifugal motion vector to the moon as a whole when it orbits, its bulges would follow the same motion its CG would follow, and reversing that vector to bring the moon back on its trajectory wouldn't add to them, which means that such a vectorial motion cannot create them.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 31/08/2018 19:35:47
No need for the strings to be elastic;
You are right, but I meant they should not have a quite fixed length, to allow the particles some "freedom" to move ... Analogies aren´t usually perfect!
Molnav, the force applied to a large body made out of particles cannot be treated the same as the force applied to each of its particles taken separately, and you seem to think so. In the case of a body, we need an anchor to hold it, but we can't hold a particle with an anchor. The moon and the earth are not anchored to one another, instead, it is each particle from each of the bodies that is linked to all the particles of the other body by means of the gravitational force, and that force depends on distance, which is not the case with a string. If we would anchor all the particles of the moon to all the particles of the earth by means of strings, there would be no bulge, but since the gravitational force weakens with distance, the closer particles are pulled more strongly than the farther ones and it creates bulges.
Sorry I answered that thinking it had been said by D.C. ...
If we accelerate the earth's rotation until its surface begins to orbit, the equatorial bulge will get a lot more important than what it is, but once some of the surface will be orbiting, we can slow down the earth and thus reduce the bulge without affecting what is already orbiting
Apart from the fact that you are referring to the "spinning" of the earth (clearly out of my stand on moon related tides), you already posted that idea (#188 ), and I replied:
" ...1) The most important: when I´ve mentioned that effect, saying even "permanent" equatorial bulge, and usually adding something like "due to earth daily spinning", I was just saying that due to similar inertial reasons, earth app. monthly revolving around moon-earth barycenter should deform oceans surface, on its own right ... And I even said, at least once, something like "please don´t mix the two movements", because it´s not the first time D. C. seems to think I connect the moon related tides to earth daily spinning !!
Therefore, thank you for your words, but the difference is so obvious, that no additional cases need to be added "to illustrate" it ... The problem is that D.C. and me seem to speak in different languages !!
2) If you ruminated your scenario, let alone if you made some maths, you would find that, before centrifugal forces made some of the earth surface stuff "levitate", earth would kind of explode ...
Logically, equatorial bulge would previously get higher and higher, and also wider. Even with smaller radius at higher latitudes, those forces (proportional to ω˛r), would increase dramatically.
And the required angular velocity to orbit at earth surface, I don´t remember any figure, but is really huge !!"
I could make other comments to your post of yesterday (not agreeing necessarily), but surely you have already seen that many of the things you and D.C. are saying  are out of my stand and what I am discussing, because earth spinning has nothing to do with moon-earth dynamics ...
It only changes the perception we have of the real monthly tidal cycle, and produces a small misalignment of the bulges, unable to catch up with sublunar (and antipodal) exact point, due to the high spinning speed (some 28 times higher than moon-earth dance, in terms of angular speed, and being earth radius app. 50% bigger than the radius of earth revolving !!).
All that supposing the barycenter were still, disregarding earth-moon pair revolving around the sun, to make simpler the analysis.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 31/08/2018 22:37:16
As I´ve already said, to discuss with you is useless, among other things because you have so fixed your (wrong) ideas that you are unable to see and/or understand what said to you.

As usual, you accuse others of having your faults.

Quote
But, why earth daily spinning can cause that permanent equatorial bulge, and (according to you) the some 29 day revolving of the earth around the barycenter (please don´t mix with daily earth rotation !!), besides actual gravity forces on each location, can´t intervene in tides formation ??

Because if the force being applied was evenly applied rather than diminishing over distance, there would be no tides in a system with two bodies orbiting each other. The whole planet and all the water sitting on it would be pulled equally strongly towards the moon and the whole moon would be pulled towards the planet in the same way with no tidal forces acting at all. With a spinning planet, there is "centrifugal" force "flinging" material out into a bulge round the planet centred on the equator. With two orbiting ones (not spinning) and without differential gravity, there would be no such centrifugal force trying to fling any material in any direction that the planet it's sitting on isn't also moving in (and no equivalent using the preferred equivalent description with the word centripetal). The water would simply sit on the planet and move with it, matching it in acceleration at all times. I told you this long ago, but you never take it in, and that's the root of your entire problem. There is no centripetal/centrifugal factor to add in for tides - it's all done directly by straight-line differential gravity (and you can see this mechanism at work in the computer program).

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In this last earth movement, water (and solid parts too) are also somehow obliged to follow circular paths, instead their inertial tendency to follow the tangents ... Therefore, centripetal and centrifugal forces are also present, and they do intervene on the 29 day cycle of moon-éarth dynamics, and on lunar related tides !!"

If you were to apply a force that doesn't diminish over distance and applied it in the same direction to all parts of the planet/moon, and if the planet/moon isn't spinning, there would be no difference in the paths that any parts of the planet/moon follow - they would follow paths through space of exactly the same shape as each other. I've told you this before, but you obviously didn't get it and I don't imagine you'll get it now either. However, all it would take is a little computer program to test it and find that I'm right. Do I need to write one for that too to get simple realities through to you? With gravity, there is no flinging of stuff caused by the orbit of a planet, which is why you shouldn't be misled by anyone attaching the label "centripetal force" to it.

All you have is differential gravity acting on it and a straight-line acceleration which makes the nearest parts try to accelerate more in that direction and the furthest parts try to accelerate least, all happening in such a way that the component of perpendicular movement is a complete irrelevance. However, the result of this straight-line acceleration does lead to the material behaving in a manner that appears to fit with your explanation. The problem with your explanation though is that the mechanism for this still depends on differential gravity to make the material behave in that way, and so your explanation, if stated fully, has to include the real explanation of differential gravity, and the extra bit you're adding about centripetal force being involved is an unnecessary addition to it which is invalidated by straight-line cases. A massive object like Jupiter can tear comets apart through tidal forces even if they are heading directly towards it. Centripetal force is not part of the cause of tides.

Quote
And a couple of days ago you said:
If the Earth was rotating once a month, permanent bulges would form on two opposite sides and they would then grow and shrink a little throughout each month as the moon gets closer and further away. The moon's orbit not being circular would also make the bulges move east and west a bit because the Earth's rotation wouldn't track the moon's position in the sky perfectly. The Earth's actual rotation prevents such bulges from forming (because the water can't move fast enough to keep up), so we just have tiny pressure changes acting over a huge area and adding enough energy to generate oscillations that provide the tides that we see.
THERE WE ARE !! ... What in bold is what is actually happening ... !! (though earth "revolves", rather than "rotates", in its dance with the moon)

The word "rotates" means "revolves". I can assure you that the bit in bold is not what is actually happening - the Earth rotates nearly thirty times a month (I forget whether it's 28 or 29 times). You appear to have interpreted what I said in a very wayward manner, perhaps also thinking that I'm referring to the size of the tides varying through the month, but I wasn't referring to that at all. I was talking about an Earth rotating (revolving) once a month such that someone standing on the Earth with the moon overhead would be able to watch the moon there the whole time (when the sun doesn't appear too close to it in the sky). This illustrates the point that English isn't your first language and that you may be misunderstanding all manner of other things that I've said too.

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After so many posts, haven´t you realized even that basic fact ??

If you actually understood the text you were reading, you wouldn't be mistaking it for the Earth and moon as they actually are - I was describing a very different scenario.

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How can you have been so sure of your stand, if you continue to mix two quite different movements, and keep considering that when I say earth-moon dance, in particular earth revolving around the barycenter, I´m not referring to that but to earth daily spinning ??

I haven't been mixing them at all. Again, your interpretation skills are the root of your problem. The reason you think I'm talking rubbish is that your translation software is rubbish.

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Because, trying yo avoid confusion, I´ve usually been very careful even reserving the term "spinning" for the a day period movement (though "rotation" could also be used), the term "revolving" for the 28/29 d. period movement of earth, and "rotation" for the equal period movement of the moon ... But to no avail, as far as your "blinded" (at least in this case) mind is concerned !!

I've been very careful in my wording too, but I can't prevent you from mistranslating everything.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 31/08/2018 22:52:53
...and since simulations represent real motions, Rmolnav, you should be able to make one that represents what you think.

That's exactly what I had in mind when I wrote my program. If he wants to provide his maths, I can replace my maths with his in the function called "run()" - the rest of the program should work fine for both versions (so long as the right values are assigned to the right variables for the force felt at the near and far points of both bodies). If he does this, he will soon find that any calculations he makes involving centripetal force are completely superfluous.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 01/09/2018 18:25:25
The word "rotates" means "revolves".
Not for many scientists:
"revolve / rotate
In ordinary speech these two words are often treated as interchangeable, though it’s “revolving credit account” and “rotating crops.” Scientists make a sharp distinction between the two: the earth revolves (orbits) around the sun but rotates (spins) around its axis".
(Prof. Brians, Washington University)
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 01/09/2018 22:01:44
That's interesting - I wasn't aware that any of them restricted the use of words in such an artificial way. RPM, for example, stands for revolutions per minute - not rotations per minute. In this discussion though, you should always be able to work out which meaning is intended by the context - if one interpretation makes sense and the other doesn't, you should go with the one that makes sense instead of choosing the one that doesn't and then launching into an attack based on your incorrect interpretation.

Edit: on looking back, I notice that I did actually use the word "rotate", and you then told me I should have used the word "revolve" and then attacked what I'd said on the basis that you wanted it to mean "revolve" (orbit) rather than "rotate". That's a weird thing to do when it makes full sense the way I said it and is plain bonkers the way you wanted to interpret it.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 02/09/2018 18:39:35
Quote
And a couple of days ago you said:
Quote from: David Cooper on 29/08/2018 22:12:31
If the Earth was rotating once a month, permanent bulges would form on two opposite sides and they would then grow and shrink a little throughout each month as the moon gets closer and further away. The moon's orbit not being circular would also make the bulges move east and west a bit because the Earth's rotation wouldn't track the moon's position in the sky perfectly. The Earth's actual rotation prevents such bulges from forming (because the water can't move fast enough to keep up), so we just have tiny pressure changes acting over a huge area and adding enough energy to generate oscillations that provide the tides that we see.
THERE WE ARE !! ... What in bold is what is actually happening ... !! (though earth "revolves", rather than "rotates", in its dance with the moon)

The word "rotates" means "revolves". I can assure you that the bit in bold is not what is actually happening - the Earth rotates nearly thirty times a month (I forget whether it's 28 or 29 times). You appear to have interpreted what I said in a very wayward manner, perhaps also thinking that I'm referring to the size of the tides varying through the month, but I wasn't referring to that at all. I was talking about an Earth rotating (revolving) once a month such that someone standing on the Earth with the moon overhead would be able to watch the moon there the whole time (when the sun doesn't appear too close to it in the sky). This illustrates the point that English isn't your first language and that you may be misunderstanding all manner of other things that I've said too.
You stubbornly keep ignoring that, IN ALMOST A MONTH, apart from those 28/29 cycles of earth daily spinning, and apart from a little less than a 29/365 of the complete orbit around the sun, our planet does complete a cycle of its revolving around the barycenter ...
Disregarding the daily spinning, which has nothing to do with moon-earth dynamics, we can look at the moon e.g. midnight (when visible), and write down its angular position in its apparent trajectory.
And do the same next day ... We´ll see there has been a change towards the east: THAT is the real movement of the moon. And earth, in its dance with the moon, ALWAYS at the other side of the barycenter, will have followed the same angle in its revolving around the barycenter.
By the way, I posted a link to a google image which shows that "dance" in three different positions, very interesting, but I didn´t see any further comment here ... Did you see it? I said:
(#152 and #153)
"There is an image:
main-qimg-851d9284749b378191c8ae87e4e2e4c2
that can help guys understand better moon-earth dance, especially earth revolving around the barycenter.
By the way: the image has a detail a little bit erroneous ... Any guess?
A Google page appears. Please click on images …"
 
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 02/09/2018 18:55:20
I can assure you that the bit in bold is not what is actually happening - the Earth rotates nearly thirty times a month (I forget whether it's 28 or 29 times). You appear to have interpreted what I said in a very wayward manner, perhaps also thinking that I'm referring to the size of the tides varying through the month, but I wasn't referring to that at all.
I could not have thought that, because it is cristal clear, at least to me, that the size of the tides changes over the month as a consequence of adding sun-earth dynamic effects to moon-earth´s, that continuously vary in its relative position (apart from other minor changes).
And, as I´ve said many times, I´m usually discussing only moon-related tides, for the sake of simplicity.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 02/09/2018 21:12:58
You stubbornly keep ignoring that, IN ALMOST A MONTH, apart from those 28/29 cycles of earth daily spinning, and apart from a little less than a 29/365 of the complete orbit around the sun, our planet does complete a cycle of its revolving around the barycenter ...

Ignoring it? Hardly - I've responded to it repeatedly. Do I really have to do it yet again? The Earth goes round in a circle (= revolves round the barycentre), but as I keep telling you, if you apply an even force you can make something follow a circular path without any material being flung off it because all the material follows the same shape of path - I told you long ago why you cannot make a functional centrifuge using an evenly-applied force to cause something to move in a circle. It's only because there is differential gravity that you get some material "wanting" to follow different paths and that's caused by straight-line acceleration force differences rather than it being flung out by centrifugal force, as demonstrated by the case where all perpendicular movement is removed and the same straight-line acceleration force differences lead to the same tidal effect.

Quote
By the way, I posted a link to a google image which shows that "dance" in three different positions, very interesting, but I didn´t see any further comment here ... Did you see it? I said:
(#152 and #153)
"There is an image:
main-qimg-851d9284749b378191c8ae87e4e2e4c2

It didn't look like a viable link (and still doesn't), but I also didn't need to see an image of this dance, so I didn't follow it up. If you look at my computer program (assuming you have a machine with a screen big enough for it to work), you'll see the dance played out there, and it's greatly exaggerated because I've started with a moon only 1/8 the mass of the planet that it's going round.

Quote
that can help guys understand better moon-earth dance, especially earth revolving around the barycenter.
By the way: the image has a detail a little bit erroneous ... Any guess?
A Google page appears. Please click on images …"

If you can produce an actual link to it, I'll be happy to have a look.

I could not have thought that, because it is cristal clear, at least to me, that the size of the tides changes over the month as a consequence of adding sun-earth dynamic effects to moon-earth´s, that continuously vary in its relative position (apart from other minor changes).

Indeed it does, so you have some excuse for failing to interpret correctly what was said, but I did use the word "rotate" and it meant exactly that - one rotation in a month.

Quote
And, as I´ve said many times, I´m usually discussing only moon-related tides, for the sake of simplicity.

I wasn't aware that I was discussing anything other than tides relating to the moon (either caused by it or caused to it), apart from a couple of references to the sun's involvement, but that was many pages back.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 03/09/2018 11:40:16
174 Le Repteux
Looking for something else, yesterday I saw this post of you, which for some reason I had not made any comment on:
"I have a question about the tides that has not been answered yet, so I might as well post it here.
While the earth and the moon execute their common orbital trajectory with the sun, their common barycenter goes at the right speed all the time, but not their respective gravity centers: those are too fast when they transit on the far side, and too slow when they transit on the near side. They should thus behave as if they were at an aphelia on the far side, and at a perihelia on the near one, slowing down on their orbital trajectory with the sun and getting away from it progressively during the first half of their far side transit (and then doing the inverse during the second half), and accelerating and getting closer to the sun during the first half of their close side transit (and doing the inverse also during the second half), what should progressively increase and decrease the distance between them during their transit. I never heard of such an observation, but I can't find a flaw in the logic. Would it be too small to be observable?"
I´m afraid there is a flaw: what you say is valid for a proper orbiting movement of an object, only "feeling" gravitational pull from the other "central" object, and its inertial "tendency" to maintain its initial speed vector (that manifest itself as the controversial "centrifugal force") ...
But earth c.g. speed vectors caused by its revolving around the barycenter have to be added ...
During its far side transit (new moon +/- a week), earth revolving around the barycenter speed vectors have components same sense as orbiting around sun ... Resulting actual speed increases during the first half, and decreases afterwards.
And during the near side (full moon +/- a week), the opposite happens and real speed decreases initially, and increases afterwards.
The size of tides also "oscillates" due to that, being bigger when new and full moons, and smaller when earth changes from closer side to further side or the opposite.
But it´s tough to try and figure out the real dynamics of the combined movement, because apart from what said and the change in distance from sun (and gravity pull from it), also the curvature of the actual path of earth´s c.g. varies: farther parts are less curved than closer parts ...
That implies bigger radius of curvature of further parts of the path, and smaller of closer parts ...
And don´t forget: centrifugal forces are proportional to v˛/r (being v the tangential speed, and r the radius of actual curvature of the trajectory in each position).
   
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 03/09/2018 18:47:55
The Earth goes round in a circle (= revolves round the barycentre), but as I keep telling you, if you apply an even force you can make something follow a circular path without any material being flung off it because all the material follows the same shape of path
Well, I thought you disregarded earth revolving because you have said (last 29th):
"IF the Earth was rotating once a month ..."
In any case, what quoted is erroneous ... Perhaps material won´t "fly off", but hatever way you could "make something follow a circular path", ALL its massive parts "tend" to follow straight at constant speed ... The basic phenomenon of inertia in Physics !!
And, depending on how the whole body moves, that inertia manifests itself in a way or another, usually as centrifugal force.
But I´m afraid you can´t grasp that, due to the mess you have in your mind even with the much more clear concept of centripetal force ...
Regarding the not working "link":
"main-qimg-851d9284749b378191c8ae87e4e2e4c2"
surely you must know the default solution is to copy and paste it on your search engine, isn´t it?

Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 04/09/2018 11:08:45
David Cooper #163
Looking for something else, yesterday I saw what follows, included by you as "Note for rmolnav" on a reply to other, that I had not seen before, or at least have not made any comment on ... And it is directly connected to what on my recent posts:
"If your mechanism was correct about the centripetal/centrifugal effect flinging the water out in two bulges at opposite sides as the Earth rotates around the barycentre, wouldn't that generate a single bulge on the side furthest out from the barycentre with the sea lowest on the side nearest to it? That would lead to most places having one tide per day instead of two."
Now you are "almost" completely right ...
In fact, IF ONLY that inertial effect acted on our planet, the sublunar area, on the side nearest to the barycenter, would show even a "negative" bulge ... That would be due to the revolving (instead of rotating) of earth: ALL parts follow identical circles, and are ALWAYS at maximum distance from the moon (within each circular path). Therefore, centrifugal forces are ALWAYS in the same sense of kind of "flying" from the moon ...
But I haven´t EVER said that mechanism is the UNIQUE cause of tides ... If we add the (so bravely defended by you) effects of varying moon gravity, we have:
1) Sublunar area. When adding two opposite vectors, the biggest prevails: we have a real, "positive" bulge.
2) Antipodal area. Also two opposite vectors to be added. But now, because moon pull there is smaller, inertial effects (centrifugal forces) prevail: also a "normal" bulge builds !!
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 04/09/2018 18:36:36
 
1) Sublunar area. When adding two opposite vectors, the biggest prevails: we have a real, "positive" bulge.
2) Antipodal area. Also two opposite vectors to be added. But now, because moon pull there is smaller, inertial effects (centrifugal forces) prevail: also a "normal" bulge builds !!
In this case, "positive" means due to gravitation, and "normal" means due to inertia, which means that if the rotation would suddenly stop, the inner bulge would increase and the outward one would decrease. But it is not what the law of gravitation means: it means that both ends of free falling bodies are stretched due to a differential acceleration. Your idea seem to be challenging the law of gravitation. Is it?
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 04/09/2018 18:46:00
The Earth goes round in a circle (= revolves round the barycentre), but as I keep telling you, if you apply an even force you can make something follow a circular path without any material being flung off it because all the material follows the same shape of path
... In any case, what quoted is erroneous ... Perhaps material won´t "fly off", but hatever way you could "make something follow a circular path", ALL its massive parts "tend" to follow straight at constant speed ... The basic phenomenon of inertia in Physics !!

That's where you keep making the basic error. What I said (the thing you quoted) is fully correct. The whole point is that with an evenly-applied force you can make something follow a circular course with every particle following the exact same shape of curve. Picture a group of rockets in deep space moving along a straight line. They all fire their engines at the same moment, pointing them to the side. All of them follow the same shape of path, maintaining the same configuration with the distances between each rocket and the ones around it not changing at all, but they follow a curved path. Evenly-applied gravity would have the same effect as that, accelerating each rocket at the same time the same amount in the same direction, but no one in the rockets would feel any force from this, unlike in the case where they're using rocket engines to make the turn. With gravity, there is no added centrifugal force to consider. What leads to the tidal "bulges" is that the force isn't evenly applied, but lessens over distance, and that makes a difference to the straight-line acceleration force being applied. There is not extra force needing to be applied to that.

Quote
But I´m afraid you can´t grasp that, due to the mess you have in your mind even with the much more clear concept of centripetal force ...

It's you that's repeatedly failing to grasp the point in question. Time and time you reject correct statements that I've made because you don't understand the basic physics.

Quote
Regarding the not working "link":
"main-qimg-851d9284749b378191c8ae87e4e2e4c2"
surely you must know the default solution is to copy and paste it on your search engine, isn´t it?

Search uses a lot of energy and is best avoided when there's nothing to be gained from looking at an image that illustrates something you already understand in full.

Quote
In fact, IF ONLY that inertial effect acted on our planet, the sublunar area, on the side nearest to the barycenter, would show even a "negative" bulge ... That would be due to the revolving (instead of rotating) of earth: ALL parts follow identical circles, and are ALWAYS at maximum distance from the moon (within each circular path). Therefore, centrifugal forces are ALWAYS in the same sense of kind of "flying" from the moon ...

And the part you still don't understand is that there would be no flinging of material with an evenly-applied force of gravity. What makes the bulge is differential gravity, and that covers the whole thing in one go - different strengths of straight-line acceleration towards the moon.

Quote
But I haven´t EVER said that mechanism is the UNIQUE cause of tides ... If we add the (so bravely defended by you) effects of varying moon gravity, we have:
1) Sublunar area. When adding two opposite vectors, the biggest prevails: we have a real, "positive" bulge.
2) Antipodal area. Also two opposite vectors to be added. But now, because moon pull there is smaller, inertial effects (centrifugal forces) prevail: also a "normal" bulge builds !!

The straight-line acceleration under differential gravity does 100% of the job for (1) and (2), so any addition to (2) through your centripetal mechanism will have to be of zero size, rendering the role of that mechanism entirely imaginary.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/09/2018 15:09:16
In this case, "positive" means due to gravitation, and "normal" means due to inertia
Sorry if my words could somehow lead you to what you say ...
As I also said, the two bulges are due to the ADDITION of both causes: moon´s gravitation (always towards the moon) and inertial effects (always in the opposite sense) ...
And both resulting bulges are in same sense as defined on a dictionary (distance to earth c.g. increases) ...
In 1) I said "positive" to underline it is the contrary to "negative" (distance to earth c.g. decreases), as I had previously called it for when not counting moon´s pull (not mentioned by D.C.)
In 2) I said "normal" (I could have said "standard"): just a "bulge", what we usually call a bulge ...
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 05/09/2018 16:31:52
Quote from: rmolnav on Yesterday at 06:08:45

    1) Sublunar area. When adding two opposite vectors, the biggest prevails: we have a real, "positive" bulge.
    2) Antipodal area. Also two opposite vectors to be added. But now, because moon pull there is smaller, inertial effects (centrifugal forces) prevail: also a "normal" bulge builds !!
You seem to have missed my point, so I will present it differently. 1) means that the vector that prevails is due to gravitation, and 2) means that the vector that prevails is due to rotation, so if we stop the rotation, the prevalence of the inertial vector in 2) should disappear, so the bulge should also disappear, and it is not what we can expect if we apply the gravitation formula to a free falling body.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/09/2018 19:02:26
1) means that the vector that prevails is due to gravitation, and 2) means that the vector that prevails is due to rotation, so if we stop the rotation, the prevalence of the inertial vector in 2) should disappear, so the bulge should also disappear, and it is not what we can expect if we apply the gravitation formula to a free falling body.
OK. D.C. and me already discussed that on #211, 212, 213 ...
I had said:
"Let us forget the spaceship, and have just moon and earth.
If, somehow, we "remove" the revolving, both earth and moon will accelerate towards each other … Though only a transient movement, bulges would remain (and increase) until the collision. OK. Therefore, revolving was not causing them (YOUR deduction).
IN A QUITE SIMILAR WAY, we could reason as follows:
If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)
Therefore, if with the differential moon´s gravity maintained, outer positive bulge doesn´t exist, the moon differential gravity cannot be the cause of outer bulge !!
Same way you´ve “reasoned” lot of times … Either both are right, or both are wrong …"
Then he argued that, when he argued we had to let earth free to fall, meaning (I presume) that when saying "remove" the rotation is not to stop earth completely, just its circular movement ...
That I said:
"It´s unbelievable that you consider absolutely necessary to let the earth free to get the rectilinear accelerated movement towards the moon (to have the second bulge), but time and again you say that (as it is in the real case) the curved accelerated movement of the earth (its revolving around the moon-earth barycenter), has nothing to do with the formation of the outer bulge, and that it is not necessary at all !!
They are two different ways inertia manifests itself, but due to same basic Physics laws ...
In the first case, outer hemisphere massive stuff kind of "falls behind", because it is not sufficiently pulled by the moon to get the rectilinear acceleration of the whole earth. The whole earth stretches, and behind bulge also appears.
In the real case, outer bulge equally requires the revolving movement, because outer hemisphere massive stuff is not sufficiently pulled by the moon to get the required centripetal acceleration for the revolving movement. Internal stresses (in both senses, inwards and outwards) appear, earth stretches too, and outer bulge builds too ..."
I did mention neither "centripetal forces", nor "centrifugal forces", to avoid those terms so misunderstood by D.C. ...
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 06/09/2018 11:35:43
With gravity, there is no added centrifugal force to consider.
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But I haven´t EVER said that mechanism is the UNIQUE cause of tides ... If we add the (so bravely defended by you) effects of varying moon gravity, we have:
1) Sublunar area. When adding two opposite vectors, the biggest prevails: we have a real, "positive" bulge.
2) Antipodal area. Also two opposite vectors to be added. But now, because moon pull there is smaller, inertial effects (centrifugal forces) prevail: also a "normal" bulge builds !!

The straight-line acceleration under differential gravity does 100% of the job for (1) and (2), so any addition to (2) through your centripetal mechanism will have to be of zero size, rendering the role of that mechanism entirely imaginary.
WITHOUT KNOWING correctly what "centripetal force" is, it is IMPOSSIBLE for you to grasp the Physics of what we are discussing ...
I already told you:
"In relation to a noun such as “force”, one thing is its nature, or its essence, and quite another its let us say function, or its not essential characteristics, what usually is expressed through adjectives.
A force can be big, small, permanent, variable, pulling or pushing contact f., “at a distance f." (not sure the correct English name) … as well as centripetal or centrifugal, a possible additional quality in specific cases, NOTHING TO DO with “classifying gravity as centrifugal force” !!
Different types of forces (or components of forces) can act as centripetal forces, as long as their directions are perpendicular to the affected object trajectory, and in the sense of a “center” (of the circle if circular movement, or the center of curvature if other curved paths).
Gravitational force can act as centripetal force, as long as what above occurs.
E.g.: the parabolic trajectory of a cannon ball is due to the gravitational force action (and the initial speed vector): its tangential component changes the size of the speed, and its component perpendicular to speed vector, acting as a centripetal force, curves the trajectory, not letting the ball to follow its inertial tendency to continue straight" …
... but to no avail, as far as your erroneous mind set is concerned !!
You have serious problems not only in Physics and Logics (as I told you before), but also even in the very use of your own language ...
Donald Trump "essence" is a "man". As he is also an "american citizen", he could be elected as U.S. president, and during his term in office he is having that "function", that job ...
He is both a "man" (essence), and U.S. president (his main function).
Similarly, the essence of "gravitational force" is, as we all (?) know,  that it exists between any two massive objects (or parts of objects), pulling them towards each other, directly proportional to the their masses, and inversely proportional to the square of separating distance.
"Centripetal force" is not the "essence" of anything, it is a "function" any force (or one of its components), and particularly gravitational force, can have as long as it acts on an object, perpendicularly to its linear speed, and therefore making its trajectory to bend ...
Therefore, there is no need of "any addition to (2) through your centripetal mechanism" ... Same real moon gravity pull at each location has to have the function of centripetal force. And if the required centripetal force (mω˛r) doesn´t match with gravitational pull at a given location, internal stresses and/or the considered massive piece own weight has to supply the difference, inwards or outwards ... (2nd Newton´s Motion Law has to be satisfied)
And "... with gravity, there is no added centrifugal force to consider" ... Sorry, there is rubbish !!
We can´t choose whether to consider to "add" a centrifugal force or not ... Centrifugal forces are  inertial effects which are ALWAYS present, as long as there is a curved trajectory ...
 
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 06/09/2018 18:45:59
The "bulge" nearest the moon is caused by straight-line differential gravity, and so is the "bulge" furthest from the moon. My computer program demonstrates that in action. There is nothing to add to the bulges from the orbital path that isn't already accounted for in full by the straight-line acceleration. That is all there is to it.

What you need to do (if you want to go on disagreeing) is produce your maths to replace mine in the program so that you can show some kind of role for your imagined mechanism that would allow it to add some amount to the far "bulge" that isn't just an addition of zero. Alternatively, you can try to use your mechanism for the far bulge and mine for the near bulge, and then find some magical way to stop my mechanism doubling your far bulge. Good luck with that!
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 07/09/2018 10:24:01
The "bulge" nearest the moon is caused by straight-line differential gravity, and so is the "bulge" furthest from the moon. My computer program demonstrates that in action. There is nothing to add to the bulges from the orbital path that isn't already accounted for in full by the straight-line acceleration. That is all there is to it.
1) "... strait-line differential gravity" ... "already accounted for in full by the straight-line acceleration."
Several times you have also said something about not necessary to add anything relative to a movement perpendicular to above mentioned "straight-line" ...
You seem to forget revolving (and rotation) are kind of two-dimension movements: velocity is certainly perpendicular to that straight-line, but both required (to curve the trajectory) centripetal acceleration, and inherent inertial centrifugal force, are parallel to it ... And that gravity pull at a given location is what massive earth stuff can "feel", not any "differential gravity" ... All or part of that moon gravitational pull HAS TO BE "USED" to make each particle revolve, acting as centripetal force and producing a centripetal acceleration, NOT any "straight-line acceleration" whatsoever ... That only would happen with no revolving at all (zero initial speed component perpendicular to gravitational field direction).
Therefore, to disregard completely those facts is erroneous !!
2) "My computer program demonstrates that in action" ...
Computers can´t "demonstrate" anything whatsoever ... It all depends on the inputs and the program, and on our correct (or erroneous) analysis of the output ... 
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 07/09/2018 15:40:27
If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)
Therefore, if with the differential moon´s gravity maintained, outer positive bulge doesn´t exist, the moon differential gravity cannot be the cause of outer bulge !!
Supporting the orbiting earth to avoid that it falls on the moon if we stop its orbiting speed is similar to what would be happening to the earth surface if we would stop its rotation: the earth surface would simply lose its equatorial bulge, and the orbiting earth would lose its gravitational ones. Now if we would reaccelerate the rotation of the two systems progressively, an equatorial bulge would build up progressively on both systems until they reach their orbiting speed, and stopping the speed at that moment would not erase the bulge immediately, it would erase only when they would stop orbiting because they are getting supported again, so it seems that, after all that talk, you may still be right about the two kinds of bulges being similar. The bulges that belong to free falling systems would then only be a free fall particularity. For instance, supporting the earth with a cup shaped support while we would progressively decelerate its orbital speed would progressively erase the tides, and they would automatically reappear as soon as we would remove the support to let the earth free to fall on the moon again. I'll let that idea age a bit in my mind until David comments it. If it is right though, programming such a support in David's simulation and stopping the speed should simply erase the bulges, and removing the support after a while so that the system free falls again should build up new ones progressively.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 07/09/2018 18:46:47
I'll let that idea age a bit in my mind
Thank you. I usually say I have to further "ruminate" (rumiar in Spanish) ... or "chew it over".
Please kindly be careful when saying things such as:
- "... if we stop its orbiting speed is similar to what would be happening to the earth surface if we would stop its rotation"
-" ... orbiting earth would lose its gravitational ones "
 Not quite clear what you are referring to ... because we have two earth "orbits" (around sun and around moon-earth barycenter), their two respective "revolvings", and its daily "rotation" or "spinning" ...
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 07/09/2018 19:15:27
I usually use the words "rotating" and "orbiting" to distinguish the two kinds of rotation, and I was talking of the earth/moon system, so I thought I didn't have to specify it. More importantly, tell me if what I said looks similar to what you mean.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 07/09/2018 21:08:26
You seem to forget revolving (and rotation) are kind of two-dimension movements: velocity is certainly perpendicular to that straight-line, but both required (to curve the trajectory) centripetal acceleration, and inherent inertial centrifugal force, are parallel to it ...

Choose the right frame of reference and it becomes clear that the perpendicular movement is irrelevant. I explained that to you before, but you made no comment.

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And that gravity pull at a given location is what massive earth stuff can "feel", not any "differential gravity" ...

Differential gravity simply means that further-away parts of it have less gravity acting on them, so they accelerate less in the direction of that straight-line force. That does the whole job.

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All or part of that moon gravitational pull HAS TO BE "USED" to make each particle revolve, acting as centripetal force and producing a centripetal acceleration, NOT any "straight-line acceleration" whatsoever ... That only would happen with no revolving at all (zero initial speed component perpendicular to gravitational field direction).

For every angle you can imagine it moving at, there's a frame of reference that shows that it reduces to straight-line acceleration with differential gravity.

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Therefore, to disregard completely those facts is erroneous !!

The "bulges" are fully accounted for by the straight-line acceleration. The perpendicular component is irrelevant.

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Computers can´t "demonstrate" anything whatsoever ...

Of course they can - they can show you the consequences of the maths that you program into it, and I programmed it to work on straight-line acceleration with differential gravity.

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It all depends on the inputs and the program, and on our correct (or erroneous) analysis of the output ...

You can check the source code and the routine called "run()" where that maths is applied. Your job is to provide your alternative maths for that routine to run on. If you get it right, it should produce the same end result, but with a lot of superfluous number crunching.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 07/09/2018 22:17:58
For instance, supporting the earth with a cup shaped support while we would progressively decelerate its orbital speed would progressively erase the tides, and they would automatically reappear as soon as we would remove the support to let the earth free to fall on the moon again. I'll let that idea age a bit in my mind until David comments it. If it is right though, programming such a support in David's simulation and stopping the speed should simply erase the bulges, and removing the support after a while should build up new ones progressively.

I could add a "hold" button (or modify the "pause" one) to halt the moon and Earth and keep them in place while their gravity continues to act on each other. This would lead to the water on the far side of the Earth being pulled down more strongly than it is at the equator half way between the near and far sides (where the water would be pulled sideways), but it would make no difference to the force on the water on the near side. Hold the Earth like that for a while though and the water will flow to the near side and build up there into an actual bulge. If the planet's spinning though, the water won't have time to build up at the near side, but the application of those forces would doubtless lead to interesting tides: the Earth would fall 38 miles towards the moon in a day if the orbital motion was removed, so the water between the far and near points would perhaps move somewhere close to ten miles eastward in twelve hours and then ten miles west over the next twelve, which adds up to a lot of sloshing about. That would be quite disruptive to the rail network, I suspect.

Edit: that 38 miles a day figure is way out - that's for the moon falling towards the Earth, so the real figure should be closer to half a mile a day.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 07/09/2018 22:26:06
Supporting the orbiting earth to avoid that it falls on the moon if we stop its orbiting speed is similar to what would be happening to the earth surface if we would stop its rotation: the earth surface would simply lose its equatorial bulge, and the orbiting earth would lose its gravitational ones
Last "lost" would not happen if only daily "rotation" stopped: at any location there would be two high tides and two low tides a month ...
But you continue:
Now if we would reaccelerate the rotation of the two systems progressively ...
when before you had said:
" what would be happening to the earth surface if we would stop its rotation" (only one "system" ??).
Sorry, but I find rather difficult to be sure one is "imaging" same case you are actually trying to convey ...
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 07/09/2018 22:42:16
Differential gravity simply means that further-away parts of it have less gravity acting on them, so they accelerate less in the direction of that straight-line force
Again, those parts where less gravity don´t accelerate less in that direction than others ... That would require them to increase their velocity in that direction less than others, what obviously doesn´t happen, because the velocity in that direction of all parts keeps being zero !!
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 08/09/2018 15:17:11
I could add a "hold" button (or modify the "pause" one) to halt the moon and Earth and keep them in place while their gravity continues to act on each other. This would lead to the water on the far side of the Earth being pulled down more strongly than it is at the equator half way between the near and far sides (where the water would be pulled sideways), but it would make no difference to the force on the water on the near side. Hold the Earth like that for a while though and the water will flow to the near side and build up there into an actual bulge.
I was suggesting a cup shaped support to avoid the tides to build up on the near side. I knew that we needed a glass shaped one to avoid any displacement from the far side to the near one, but this way, we can't imagine the far side tides as easily. What I'm trying to show is that if the orbiting earth was supported the way the rotating one is supported, which is everywhere except at its surface, stopping the orbiting speed would erase its orbiting bulges the same way stopping the rotating earth would erase its equatorial bulge, and then letting it fall towards the moon (or accelerating it to its orbital speed again) would build up the free fall bulges again, but we could also only give it enough speed to get an equatorial bulge instead of a free fall one, and then increase that speed until it becomes a free fall one. As I said, from this analysis, it looks as if the free fall and equatorial bulges were only particularities of gravitational bulges.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 08/09/2018 15:42:32
Supporting the orbiting earth to avoid that it falls on the moon if we stop its orbiting speed is similar to what would be happening to the earth surface if we would stop its rotation: the earth surface would simply lose its equatorial bulge, and the orbiting earth would lose its gravitational ones
Last "lost" would not happen if only daily "rotation" stopped: at any location there would be two high tides and two low tides a month ...
But you continue:
Now if we would reaccelerate the rotation of the two systems progressively ...
when before you had said:
" what would be happening to the earth surface if we would stop its rotation" (only one "system" ??).
Sorry, but I find rather difficult to be sure one is "imaging" same case you are actually trying to convey ...
Read my first quote again: I said there that if both orbiting and rotational speeds were stopped, then both the equatorial and gravitational bulges would disappear. I'm comparing two kinds of rotation to see if they are equivalent, so when I say orbital speed, I'm talking of the orbital rotation, and when I say rotational speed, I'm talking of the non orbital one.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 08/09/2018 19:35:09
Read my first quote again: I said there that if both orbiting and rotational speeds were stopped, then both the equatorial and gravitational bulges would disappear. I'm comparing two kinds of rotation to see if they are equivalent, so when I say orbital speed, I'm talking of the orbital rotation, and when I say rotational speed, I'm talking of the non orbital one.
You didn´t actually mention "both", but said one case was similar to other, not both "stops" together ...
In any case, I prefer not to consider earth is orbiting the moon, let alone it is in a "free fall" ... As moon is also moving (and much more than earth), the dynamics of moon-earth dance (always at same distance from each other) should be rather considered as a kind of "spinning" around the barycenter of a unique extended two-parts body (though earth only revolves).
That way the dance effects are really similar to the daily spinning of earth, but with a some 28 days complete cycle ...
According to one of the NASA scientists I referred to some weeks ago:
"As the earth and moon whirl around this common center-of-mass, the centrifugal force produced is always directed away from the center of revolution. All points in or on the surface of the earth acting as a coherent body acquire this component of centrifugal force".
And, as I´ve already said, at closer to moon hemisphere stronger moon´s pull prevails. And at further hemisphere, where moon´s pull is smaller, centrifugal force prevails.
That is why "we have two high moon-related tides a month", and due to the daily earth spinning, we perceive them twice a day (on top of the permanent equatorial bulge, that has nothing to do with moon-earth dynamics).
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 08/09/2018 20:39:29
Differential gravity simply means that further-away parts of it have less gravity acting on them, so they accelerate less in the direction of that straight-line force
Again, those parts where less gravity don´t accelerate less in that direction than others ... That would require them to increase their velocity in that direction less than others, what obviously doesn´t happen, because the velocity in that direction of all parts keeps being zero !!

I expect people to be able to correct the wording for themselves and then to recognise the idea that the text is intended to communicate (and which all intelligent readers will automatically take from it). Let me reword it for nitpickers. Differential gravity simply means that further-away parts of it have less gravity acting on them so they lag behind a little under that acceleration, this resulting in a lower pressure there, while the nearer parts have more gravity acting on them, so they do the opposite, reacting more quickly than the planet as a whole and again generating a lower pressure there. The accelerations are simultaneous and equal in practise, but they wouldn't initially be simultaneous if there was no moon and you suddenly introduced the moon into the system - the lags would occur while the lower pressures are established, and then the acceleration is equalised.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 08/09/2018 20:49:15
I was suggesting a cup shaped support to avoid the tides to build up on the near side. I knew that we needed a glass shaped one to avoid any displacement from the far side to the near one, but this way, we can't imagine the far side tides as easily.

The cup would have to be more of an enclosing sphere with a little hole in the top.

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What I'm trying to show is that if the orbiting earth was supported the way the rotating one is supported, which is everywhere except at its surface, stopping the orbiting speed would erase its orbiting bulges the same way stopping the rotating earth would erase its equatorial bulge,

There are no orbiting bulges to erase, so it's hard to follow what you're trying to do.

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and then letting it fall towards the moon (or accelerating it to its orbital speed again) would build up the free fall bulges again, but we could also only give it enough speed to get an equatorial bulge instead of a free fall one, and then increase that speed until it becomes a free fall one. As I said, from this analysis, it looks as if the free fall and equatorial bulges were only particularities of gravitational bulges.

Well, I can't follow it well enough to work out what that analysis is doing. (The whole issue is already resolved too, so I'd rather work on other things like LaFreniere's work and just leave rmolnav to his delusions.)
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 08/09/2018 20:51:58
And, as I´ve already said, at closer to moon hemisphere stronger moon´s pull prevails. And at further hemisphere, where moon´s pull is smaller, centrifugal force prevails.

On the side nearest the moon, stronger moon's pull prevails, and on the side furthest from the moon, weaker moon's pull prevails. One single mechanism acting on both sides.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 09/09/2018 11:23:23
Differential gravity simply means that further-away parts of it have less gravity acting on them so they lag behind a little under that acceleration, this resulting in a lower pressure there, while the nearer parts have more gravity acting on them, so they do the opposite, reacting more quickly than the planet as a whole and again generating a lower pressure there. The accelerations are simultaneous and equal in practise, but they wouldn't initially be simultaneous if there was no moon and you suddenly introduced the moon into the system - the lags would occur while the lower pressures are established, and then the acceleration is equalised.
What you explain would be a pretty good analysis ... only if there were no earth-moon revolving/rotation , and earth were actually accelerating along a rectilinear path towards the moon ...
But nature keeps moon-earth separation fairly constant, thanks precisely to that "dance" ... which also produces other dynamical effects that you loftly keep despising ...
At earth c.g. there is a balance between moon´s gravitational pull, and centrifugal force inherent in the revolving of earth around the barycenter. But that only happens at a section there (all points at same distance as earth c.g. to moon c.g.).
And, I repeat:
 
... at closer to moon hemisphere stronger moon´s pull prevails. And at further hemisphere, where moon´s pull is smaller, centrifugal force prevails.
... and water from each hemisphere "piles up" (though relatively very, very little) respectively at sublunar and antipodal areas !!
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 09/09/2018 14:38:17
For clarity purposes, I need to specify that I changed my mind. At first, I agreed with David's differential pulling on free falling bodies, and now, I agree with Rmolnav about orbital rotation being involved in the tides. Please, reread that post (https://www.thenakedscientists.com/forum/index.php?topic=49715.msg553256#msg553256) and tell me if you understand it differently.
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 09/09/2018 17:01:17
I agree with Rmolnav about rotation being involved in the tides.
Yes, you might like to look at the NOAA site https://tidesandcurrents.noaa.gov/restles1.html
They take input from survey vessels and buoys all over the world and the barycentre model is the one that most closely matches the observations. It takes account of both differential gravity and rotation.
You have to be careful comparing different frames because although centrifugal force can appear to disappear in some, a rotating earth is not an inertial frame.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 09/09/2018 19:00:33
For clarity purposes, I need to specify that I changed my mind. At first, I agreed with David's differential pulling on free falling bodies, and now, I agree with Rmolnav about orbital rotation being involved in the tides. Please, reread that post (https://www.thenakedscientists.com/forum/index.php?topic=49715.msg553256#msg553256) and tell me if you understand it differently.
Thank you ... I´m going to reread that post carefully, and reply tomorrow.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 09/09/2018 19:09:46
Yes, you might like to look at the NOAA site https://tidesandcurrents.noaa.gov/restles1.html
They take input from survey vessels and buoys all over the world and the barycentre model is the one that most closely matches the observations. It takes account of both differential gravity and rotation.

You've found a site which makes the exact same mistake as rmolnav, and that mistake comes from making an artificial distinction between the two sides. The mechanism is exactly the same for both and it makes no rational sense to say that differential gravity has a greater role on the near side and less of a role on the other - it has exactly the same role on both sides. The centrifugal explanation is a rival one which also would have exactly the same role on both sides if it was the real mechanism, and it's important to note that the numbers that it produces are identical to the ones produced by the straight-line differential gravity explanation. What's happened on that site (and with rmolnav) is that the two explanations are being mixed in a way that they should not be - either you should go for one and stick to it for both sides or you go for the other. It is a monumental error to apply one mechanism to one side and a different one to the other when both sides are governed by the exact same cause.

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You have to be careful comparing different frames because although centrifugal force can appear to disappear in some, a rotating earth is not an inertial frame.

At any given moment, the Earth and moon are effectively under straight-line differential gravity, and that produces all the right numbers. These forces apply in every billionth of a second, and every quintillionth too - the orbit can be completely ignored. Furthermore, the clincher is that the forces are the same at any given moon-Earth distance regardless of whether the directions they are moving in at the time.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 09/09/2018 19:24:36
For clarity purposes, I need to specify that I changed my mind. At first, I agreed with David's differential pulling on free falling bodies, and now, I agree with Rmolnav about orbital rotation being involved in the tides. Please, reread that post (https://www.thenakedscientists.com/forum/index.php?topic=49715.msg553256#msg553256) and tell me if you understand it differently.

I see the point you were trying to make there now. The reason people are tripping up on this though is precisely that the two explanations produce the same numbers, so it's easy to become confused. If you want to use the centrifugal explanation, you should use it for both sides and recognise that it doesn't have a more significant role on one side than the other. When you get to a case though where two bodies are moving directly towards each other, there is no centrifugal force, but the tidal forces continue to act in full. This, as I've pointed out before, could be considered to be a case of centrifugal force in which no revolution is involved, but if this was accepted as centrifugal force, that then means that when a car breaks in a straight line, you would be able to say that you are thrown forwards by centrifugal force. The word is not normally used that way, but logically it should be because the cause is the same, so the word would have to be given an improved definition to include that case. (The same applies to any alternative wordings involving centripetal force.) That would provide a greater excuse for attributing the tides to centrifugal force, but it is still highly misleading because considering the perpendicular component is entirely superfluous.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 09/09/2018 19:31:06
I agree with Rmolnav about rotation being involved in the tides.
Yes, you might like to look at the NOAA site https://tidesandcurrents.noaa.gov/restles1.html
They take input from survey vessels and buoys all over the world and the barycentre model is the one that most closely matches the observations. It takes account of both differential gravity and rotation.
You have to be careful comparing different frames because although centrifugal force can appear to disappear in some, a rotating earth is not an inertial frame.
Thank you.
As you know, some time ago we commented that NOAA work, which had been kindly mailed to me by one of the authors, with whom I had been discussing the issue ... And months ago I (and you too) copied here not only the link, but also some of what we interchanged ...
But recently I have not posted the link again, preferring to discuss myself basic Physics with D. C. ...  In any case in #155 I copied one of the things that NOAA scientist told me:
""The publication you are referring to is "Our Restless Tides", a 10-page pamphlet developed in the 1950's to provide a basic description of the forces which create the tides.  It's intended audience were the grade school children and adults of that time.  It used terminology of science and forces which were common in the 1950s.  Such as centrifugal force.  Centrifugal force was always an "imaginary force" (not a real / measurable force).  But that type of description made the concepts easier to understand and explain.  That  description and use of centrifugal force continued to be common practice until the 1970-80's.  At that point, the terminology shifted and the textbooks used in grade schools were changed to use a more modern terminology and description of this "effect" being a result of inertia rather than an "imaginary force”...
but D.C., loftly deposing mentioned NOAA scientist, replied:
"Here, we see that the scientist may not have a firm understanding of the tides himself though - all he's doing here is trying to get rid of the idea of centrifugal force by referring to the real mechanism that is behind centrifugal force, but he hasn't realized that this isn't the mechanism behind the tides”
I still have to post some ideas regarding the issue of reference systems, which I think is also misunderstood by many ... But they are not easy to convey !!
They are in line with what I already replied to PmbPhy (?), in the specific thread "What is centrifugal force?" ... You can find it easily, because I think it is the last one, though it was months ago ...
 
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 10/09/2018 11:38:55
I'm comparing two kinds of rotation to see if they are equivalent, so when I say orbital speed, I'm talking of the orbital rotation, and when I say rotational speed, I'm talking of the non orbital one.
But you had said:
"... an equatorial bulge would build up progressively on both systems until they reach their orbiting speed, and stopping the speed at that moment would not erase the bulge immediately, it would erase only when they would stop orbiting because they are getting supported again ..."
Sorry but I find it rather confusing:
1) "equatorial bulge ... on both systems" ? ... Moon related bulges happen on low latitude areas, but "equatorial" adj. should be reserved for the "circular" permanent bulge due only to daily spinning ...
2) "... their orbiting speed" ? : you use "orbiting" for the two systems ... ??
3) "...stopping the speed" ... Which speed ??
I suggest you to imagine the moon-earth pair as if it were a single object, but with null mass where there is no real material ...
That "single" object has a some 28 days period "spinning" around its c.g. (the barycenter).
That generates centrifugal forces on all material points, both on moon and on earth.
The ones affecting earth, although the barycenter is some 1/3 earth radius inside it, are all in the sense opposite to moon, because earth only revolves around the barycenter (it is not tidal locked to the moon).
ALL THAT, together with opposite sense gravity field due to the moon, results in what I said last time a couple of days ago:
Quote from: rmolnav on 08/09/2018 19:35:09
... at closer to moon hemisphere stronger moon´s pull prevails. And at further hemisphere, where moon´s pull is smaller, centrifugal force prevails.
... and water from each hemisphere "piles up" (though relatively very, very little) respectively at sublunar and antipodal areas !!
IF WE NOW ADDED earth daily spinnig (within that "imagined" single object) , we would have:
That is why "we have two high moon-related tides a month", and due to the daily earth spinning, we perceive them twice a day (on top of the permanent equatorial bulge, that has nothing to do with moon-earth dynamics).
If somehow what we had were ONLY the daily spinning, there are two possibilities:
1) If somehow earth is keeping at same actual distance from the moon: bulges would depend on the way we kept earth completely still ... In any "imaginable" case, further hemisphere would not deform "outwards" (unless we applied the so called "fictitious" forces on each material point, but It is better not to enter that question now ...)
Equatorial bulge would form independently ...
2) If letting the earth free to accelerate towards the moon:  then for some tens of hours, apart from the permanent equatorial bulge, we would certainly have tides similar to real ones (increasing in size) ...
But, as I already said, that would be due to both differential gravity, and the addition of a manifestation of inertia different to when the real case. Then it is when the explanation of D.C. could be "almost" completely correct:
 
Quote from: David Cooper on 07/09/2018 21:08:26
Differential gravity simply means that further-away parts of it have less gravity acting on them, so they accelerate less in the direction of that straight-line force
Though they actually would "try" to accelerate less, but are "forced" to accelerate the average due to the much bigger own earth gravity and compactness where solid earth ... That would produce internal stresses and changes in water pressure distribution > > deformation and tides ...
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 10/09/2018 14:26:36
These forces apply in every billionth of a second, and every quintillionth too - the orbit can be completely ignored.
If I use my small steps to analyze the orbital motion, then the orbital steps have two components: the inertial one brings the particle away from the orbital path a bit, and the gravitational one puts it back on track at the end. Of course, those two components are executed in the same time, but I see no way to get a curved trajectory on a screen without using the inertial one. I describe my mind experiment differently below in my answer to rmoldav, hoping it will be clearer.

But you had said:
"... an equatorial bulge would build up progressively on both systems until they reach their orbiting speed, and stopping the speed at that moment would not erase the bulge immediately, it would erase only when they would stop orbiting because they are getting supported again ..."
Sorry but I find it rather confusing:
1) "equatorial bulge ... on both systems" ? ... Moon related bulges happen on low latitude areas, but "equatorial" adj. should be reserved for the "circular" permanent bulge due only to daily spinning ...
I agree with your explanation and the NOAA one, so even if you don't seem to understand what I said, I think we simply say the same thing but with different words. I'm trying to properly describe a system where the earth and the moon are united by a long arm, and where their orbital speed is slowed down progressively. Both of them will then progressively be supported by the arm, and the two opposed tidal bulges of each body will progressively become a single inertial (or equatorial) outward one because the two inward ones will be supported by the arm. I think that this mind experiment shows that the two kinds of bulges (inertial and tidal) are equivalent, don't you agree?
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 10/09/2018 18:03:57
The correct explanation is found in texts on classical mechanics. I can probably find them online if anybody wants to read them. They explain it in detail. I.e. why the tidal force on a particle is away from the earth on the moon's side and  when its in the opposite side away from the earth as well, giving rise to the two ocean tides. The addition of the force due to the sun causes spring and neap tides.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 10/09/2018 18:35:45
The correct explanation is found in texts on classical mechanics
Well, surely there must be "correct texts" on classical mechanics (but I´ve also seen some erroneous ...)
And also specific books on tides, as the one I´ve just now referred to on another "thenakedscientist" thread:
Dr. Bruce Parker is the author of the more than 300 pages book linked below, and he spent most of his career in NOAA and much of his time working on tide related problems as a specialty even while tackling jobs with a much broader scope. Positions he held at NOAA included: Chief Scientist of the National Ocean Service; Director of the Coast Survey Development Laboratory; Director of the World Data Center for Oceanography; Principal Investigator for the NOAA Global Sea Level Program; and head of the U.S. national tides and currents program (in a earlier organizational form before it became CO-OPS).
Among his awards are the NOAA Bronze Medal, the Department of Commerce Silver and Gold Medals, and the Commodore Cooper Medal from the International Hydrographic Organization.
Dr. Parker is presently a Visiting Professor at the Center for Maritime Systems at the Stevens Institute of Technology.
 Dr. Parker has written many papers on tidal subjects, some of which are included in the References section of this book, as well as many tidal analysis programs, some still being used in some form in CO-OPS.
He also had the privilege of organizing the program for the International Conference On Tidal Hydrodynamics in 1988 and editing the book that resulted.
Dr. Parker received his Ph.D. in physical oceanography from The Johns Hopkins University, and prior to that an M.S. in physical oceanography from the Massachusetts Institute of Technology, and a B.S./B.A.in biology and physics from Brown University ...
AND HE SAYS:
"At the center of the Earth there is a balance between gravitational attraction (trying to pull the Earth and moon together) and centrifugal force (trying to push the Earth and moon apart as they revolve around that common point).
At a location on the Earth’s surface closest to the moon, the gravitational attraction of the moon is greater than the centrifugal force of the Earth (moving around the center of the revolving Earth-moon system).
On the opposite side of the Earth, facing away from the moon, the centrifugal force is greater than the moon’s gravitational attraction. In a hypothetical ocean covering the whole Earth with no continents (see Figure 2.8) there will be two tidal bulges resulting from these imbalances of gravitational and centrifugal forces, one facing the moon (where the gravitational force is greater than the centrifugal force), and one facing away from the moon (where the centrifugal force is greater than the gravitational force).
Logically, he also analyses thoroughly local effects of tides all over the world ...
https://tidesandcurrents.noaa.gov/publications/Tidal_Analysis_and_Predictions.pdf
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 10/09/2018 21:46:09
"...there will be two tidal bulges resulting from these imbalances of gravitational and centrifugal forces, one facing the moon (where the gravitational force is greater than the centrifugal force), and one facing away from the moon (where the centrifugal force is greater than the gravitational force)."

That's a better wording of it, but the "centrifugal" part is superfluous as it's all down to straight line differential gravity, as shown by a case where a moon heads straight towards a planet.

It's time to start tidying up here, so here are some things to consider to make sure you've understood what's going on:-

(A) Imagine a straight line of snooker balls set up in space. Introduce a planet some distance away in line with them and what happens? The nearest ones accelerate towards it more quickly and the furthest ones accelerate more slowly, so they spread out more. This is straight-line differential gravity in action.

(B) Repeat that, but this time tie each ball to the ones adjacent to them and introduce the planet as before. The nearest ball tries to accelerate most and the furthest ball least, and forces appear in the tethers instead of the balls moving apart.

(C) Keep the two end balls, but replace the rest with one big, lightweight sphere the size of a moon, but with very little mass so that we can ignore its gravitational pull on the two balls. If the line of balls was the same length as the length of this new setup, the forces in the tethers to the end balls are the same as they were in the long-string-of-balls version of the experiment.

(D) Now lets put more balls on tethers all the way round the equator of this sphere. The sphere does not rotate, but it is orbiting a planet. Each of the balls is following a circular path the same shape as every other ball, but the differential gravity leads to times when their tethers have higher tension - most strongly at the nearest side to the Earth with another peak tension at the furthest side, though this peak tension is slightly lower than the nearside one. With evenly applied gravity, there would be no tension in any of the tethers, but we don't have evenly applied gravity.

(E) Now let's think about the way the nearside ball and farside ball are moving. The farside ball is being moved along by the sphere at a speed higher than it should be moving for it to follow the curve that it's being forced to follow by being tethered to the sphere, so it looks as if it will try try to lift out away from the sphere, and this appears to be generating that tension peak. In contrast, the nearside ball is being moved along at a speed lower than it should be for it to follow the curve that it's being forced to follow by being tethered to the sphere, so it too looks as if it will try to lift out away from the sphere, and this appears to be generating that tension peak. This leads to the idea that centrifugal force is important for generating tidal forces (but it's an illusion).

(F) The straight-line differential gravity approach accounts for the tides without relying on rotation in the system. The question though is whether this remains the only mechanism in play when rotation is introduced. Let's consider the nearside ball which isn't moving fast enough to try to follow the path it's being forced to follow by its being tethered to the sphere. Is that an extra force trying to send it away from the sphere? The only force that's acting on this ball is the gravity from the planet, so it can't be an extra force - it has already been accounted for in full by the straight-line differential gravitational force which is generating all the tension in the tether. If there was any extra force flinging it inwards (towards the planet), it would be flung in more strongly than in the non-revolving case, despite having sideways movement that the stationary case lacks, which, if centrifugal force had a role in changing the amount of force applied to the ball, would make it move towards the planet less forcefully rather than more, but the reality is that no additional force is added to the ball in any direction. The same must be the case at the farside ball - there can be no extra force involved, and so the entire action is already accounted for in full by the straight-line differential gravitational force being less strong there. We already know this to be the case though from looking at the business of frames of reference - the sideways component of movement is clearly shown by that to be an irrelevance.

(G) Now let's turn our attention to the role of the barycentre, because we need to see if it does something special to the nearside "bulge" as a result of the barycentre being inside the Earth. The Earth orbits the barycentre in one sense, but in another sense it doesn't - it actually orbits the moon, but the moon keeps moving, so the Earth's orbit round the moon is continually modified as the moon moves to new locations. If you could lock the moon in position, the Earth would not stay close to its current barycentre, but would accelerate towards the moon instead and would go there in a very narrow elliptical orbit, possibly even hitting the moon. (I'll need to check the numbers to find that out, but the Earth is moving at under 40mph [edit: I said 20 originally, but it's still just an very rough approximation] on its path round the barycentre, so it's bound to go very close to the moon.)

At any moment in time, the Earth is following a curved path which for a short time would be shared with the path it would follow if the moon was fixed in position, and that's why I said that the Earth can be said to be orbiting the moon rather than the barycentre. Once you realise that, you can replace the moon with a planet of a thousand times the mass and a thousand times as far away [edit: that should say the square root of a thousand, so make it 33 1/3 times as far away], and this will make the Earth follow the same curve (for a short time). The barycentre therefore has no causal role in the tidal forces generated on the Earth's seas at any moment in time - it is a phantom entity which doesn't apply force to anything. The nearest point on the Earth to the moon has a force applied to it from the moon and not from the barycentre, just as the far point does. Replace the moon with the planet a thousand times more massive and distant and you can see that this is the case - the barycentre cannot be involved as any part of the mechanism for the tidal forces. Proximity of the nearside point of the Earth to the barycentre therefore does not lead to differential-gravity having a greater role for the near "bulge" and further distance out from the barycentre for the farside point does not lead to centrifugal force having a greater role for the far "bulge" - both "bulges" are simply driven directly by straight-line differential gravity.

Let's look at the quote again: "...there will be two tidal bulges resulting from these imbalances of gravitational and centrifugal forces, one facing the moon (where the gravitational force is greater than the centrifugal force), and one facing away from the moon (where the centrifugal force is greater than the gravitational force)."

This explanation contains misleading, superfluous content. The tidal forces result from differential gravity alone as the perpendicular component of travel has no causal role. It's highly misleading because it encourages people to imagine that the perpendicular movement has an important role, but it has none - the perpendicular motion is incidental, and the "evidence" for the proposed mechanism described in (E) is actually a necessary consequence of differential gravity rather than a cause. The only force involved is gravitational - dividing it into "gravitational" and "centrifugal" is an artificial distinction which has no place in the explanation of tidal forces.
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 10/09/2018 22:28:42
The correct explanation is found in texts on classical mechanics. I can probably find them online if anybody wants to read them. They explain it in detail. I.e. why the tidal force on a particle is away from the earth on the moon's side and  when its in the opposite side away from the earth as well, giving rise to the two ocean tides. The addition of the force due to the sun causes spring and neap tides.
I can’t immediately remember which text it’s in, but I assume you mean like this:
http://burro.astr.cwru.edu/Academics/Astr221/Gravity/tides.html
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 11/09/2018 01:17:13
The correct explanation is found in texts on classical mechanics. I can probably find them online if anybody wants to read them. They explain it in detail. I.e. why the tidal force on a particle is away from the earth on the moon's side and  when its in the opposite side away from the earth as well, giving rise to the two ocean tides. The addition of the force due to the sun causes spring and neap tides.
I can’t immediately remember which text it’s in, but I assume you mean like this:
http://burro.astr.cwru.edu/Academics/Astr221/Gravity/tides.html
No. That page is incorrect. I used to make the same mistake myself, even in my first post in this thread. I'll post the ones used at MIT.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 11/09/2018 10:27:21
If you don´t agree with what quoted by me from the "super-expert" Dr. Parker, you had better refuting his paragraphs, rather than proposing new scenarios difficult for you to grasp, due to your difficulties as far as understanding basic Physics laws is concerned ...
E.g.:
 
(C) Keep the two end balls, but replace the rest with one big, lightweight sphere the size of a moon, but with very little mass so that we can ignore its gravitational pull on the two balls. If the line of balls was the same length as the length of this new setup, the forces in the tethers to the end balls are the same as they were in the long-string-of-balls version of the experiment.
Why are you so sure?
Many things change:
1) Ball-big sphere-ball mass is smaller than the full string of balls.
2) The planet pull would diminish proportionally to mentioned mass has diminished.
3) But the acceleration of the three-part system would be the same (F and m have decreased same proportion).
4) But the "weight" of the balls would decrease.
5) Inertial effects on the ball would be the same (same mass and same real acceleration).
 It is not that simple ...Too many factors for you to draw correct conclusions !!
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 11/09/2018 10:49:04
No. That page is incorrect. I used to make the same mistake myself, even in my first post in this thread. I'll post the ones used at MIT.
Thanks Pete, I've come across quite a few like this that don't really cut the mustard, will be interested in what you have.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 11/09/2018 21:01:58
If you don´t agree with what quoted by me from the "super-expert" Dr. Parker, you had better refuting his paragraphs, rather than proposing new scenarios difficult for you to grasp, due to your difficulties as far as understanding basic Physics laws is concerned ...

I've given you a set of clear scenarios which show you a way to think through the issue, proving along the way that there is only one force acting here, and that is gravity. (You're trying to divide gravity into gravity and centripetal force, and then you're making an arbitrary split between where one of those ends and the other begins, leading to misleading ideas about the causes of something really simple.) Your response, as always, is to fail to learn and recognise where you've been wrong, but instead to attack trivial issues which can easily be fixed and which have no bearing on the principles being explored. If someone pointed at a group of trees to warn you about a tiger lurking in amongst them and if one of them was technically a bush, you'd complain that one of the "trees" is a bush, and declare that the warning about there being a tiger lurking there is clearly nonsense as a result. In this case though, the tree in question might actually be a tree rather than a bush.

Quote
E.g.:
 
(C) Keep the two end balls, but replace the rest with one big, lightweight sphere the size of a moon, but with very little mass so that we can ignore its gravitational pull on the two balls. If the line of balls was the same length as the length of this new setup, the forces in the tethers to the end balls are the same as they were in the long-string-of-balls version of the experiment.
Why are you so sure?

Because in the unstated details as to how I designed the sphere, I made sure that the mass distribution is in effect the same.

Quote
Many things change:

Many things could change, but they could effectively be kept the same, and anyone setting out to make a real version of the experiment would make sure they were kept the same.

Quote
It is not that simple ...Too many factors for you to draw correct conclusions !!

You've deliberately trying to engineer in differences that a good engineer would avoid, which means you're a saboteur, dishonestly setting out to hide the truth that the experiment (when done properly) reveals. Why don't you just try to recognise that truth instead and then use it to improve your understanding?

Correction: My use of the word "arbitrary" near the top of this post was incorrect, as there clearly is a section through the moon where the material is following the same orbit it would still follow for a moment even if the rest of the moon disappeared. Interestingly, that section does not include the centre of mass, and the location of this section changes as the moon moves closer to or further away from the planet (due to the differential gravity having a greater difference in strength across the near half of the moon then over the far half). Anyway, this means that it's a sensible place to make an artificial distinction between the part of gravity that you're counting as gravity and the other part that you're counting as centripetal force, so the error is only in making an artificial distinction (and not in doing so in an arbitrary way).
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 11/09/2018 21:25:50
I'm trying to properly describe a system where the earth and the moon are united by a long arm, and where their orbital speed is slowed down progressively. Both of them will then progressively be supported by the arm instead of falling towards one another, and the two opposed tidal bulges of each body will progressively become a single inertial (or equatorial) outward one because the two inward ones will progressively be supported by the arm.
I need to better understand how the tidal bulges would transform into inertial ones while we would progressively slow down the earth, so here is my second thought. If the actual orbital speed of the earth/moon system would suddenly slow down a bit, the two bodies would begin falling towards one another, the gravitational force would increase, and the tides would also begin to increase. At that moment, a supporting arm would not completely prevent the earth from falling, it would only prevent its inward tidal bulge from increasing, but without preventing the outward one from doing so. The earth would thus go on falling until the inward bulge would be completely supported, and its outward bulge would go on increasing during that time. One half of the earth would then stop falling, but not necessarily the other half. If that half was still going faster than orbital speed, then it would still be in free fall and it would stay a tidal bulge. For that bulge to become an inertial one, the speed would have to progressively get down again until the last bit of the earth would not be free falling anymore, and the resultant inertial bulge would then automatically be smaller than the previous tidal one.

I said in my previous analysis that the two kinds of bulges seemed equivalent, but are they really? If they were, there should be no way to differentiate them, and there is: one is in free fall and the other is not, and at almost the same rotational speed the inertial bulge is substantially smaller than the tidal one. Am I going to change my mind again? I feel like my last bit of earth not knowing if it is tidal or inertial. :0)
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 11/09/2018 23:01:26
Thanks Pete, I've come across quite a few like this that don't really cut the mustard, will be interested in what you have.
You're most welcome. See:
http://www.newenglandphysics.org/other/Taylor_Tides.pdf

It's from Chapter 14 of Classical Mechanics by John R. Taylor

See also:
http://www.newenglandphysics.org/other/French_Tides.pdf
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 12/09/2018 14:01:57
For that bulge to become an inertial one, the speed would have to progressively get down again until the last bit of the earth would not be free falling anymore, and the resultant inertial bulge would then automatically be smaller than the previous tidal one.
I made a mistake: for the outward bulge to become an inertial one, the earth must simply go on falling until the whole earth is supported by the arm, and whenever we slow the earth/moon system down a bit, the two bodies will fall towards one another until they get to their perigee, so for the whole earth to be supported by the arm, there is no need to slow it down again as I was suggesting, only wait till it completely rests on the support, what would not prevent the inertial bulge to reveal itself while the tidal one would collapse. It looks as if the inertial bulge was contained in the tidal one, or as if the tidal one would grow over the inertial one. If we would stop the rotation once the whole earth is supported though, the inertial bulge would collapse, and if we would then take off the arm, the tidal ones would build up, so this way at least, the two kinds of bulges don't appear to be linked. What about the NOAA explanation then? Can we show that their calculation or their drawings are wrong for instance? Unfortunately, there is no calculation to support their drawings, and the drawings themselves are only qualitative.
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 12/09/2018 21:09:47
You're most welcome. See:
............
Thanks Pete, I’m not in good WiFi area at moment, when I am I will read through.
If I have time I’ll read through this thread and see what these guys are arguing about.

PS French Tides sounds a bit dodgy, didn’t think you were into that  ;)
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 13/09/2018 08:51:50
It looks as if the inertial bulge was contained in the tidal one, or as if the tidal one would grow over the inertial one
Sorry I haven´t replied you before ... but I have to insist that it is difficult to draw correct conclusions from those imaginary cases, unless we quite clearly have in our mind the right meaning of basic Physics laws, and the terms we are mentioning ...
Are you sure you fully understand the term "inertial", and its different ways it can manifests itself ??
To me to talk about "the tidal bulge", and the "inertial bulge", is misleading, to say the least:
1) "Tidal bulge" is a kind of "redundancy" ... The "bulges" we are referring to are always one of the manifestation of tides, therefore they are always "tidal".
2) Inertia is the "tendency" of massive stuff to maintain its vector speed constant. And it can manifest itself in different ways ...
In our case, as all earth stuff is being "forced" to revolve around the barycenter, whatever force that is causing that revolving is called (by definition) "centripetal force" ... That massive stuff is kind of "stubborn", and "insists" in not to change the direction of its speed vector (inertia) ... That turs up as a centrifugal force, or, to avoid that controversial term, as an "outward" force (always in the sense opposite to moon´s location).
In both bulges "inertia" intervenes:
1) On sublunar area, inertia tries to diminish sea water level ... But, as moon´s pull there is bigger, the result is an increase of sea water level, a "proper" bulge ...
2) On the antipodes, as moon´s pull there is smaller, inertial forces prevail, and we also have a "proper" bulge.
Though I would´t call that last bulge "inertial bulge", because it is due not only to inertia ...
And don´t forget: similar opposite effects occur all over closer and further hemispheres, and water from all areas moves towards sublunar and antipodal areas, contributing to the increase of water level on both areas !!

 
   
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 13/09/2018 15:41:31
Are you sure you fully understand the term "inertial", and its different ways it can manifests itself ??
Yes, I do know what inertia means, and I'm even studying a mechanism that might very well be the cause of both mass and inertia. (https://www.thenakedscientists.com/forum/index.php?topic=71122.0). The inertial inertial bulge I'm talking about is similar to the equatorial one, it is also due to rotation, that I call inertial to distinguish it from an orbital one. The problem is that it is the same gravitational pulling that produces the two kinds of rotation, inertial or gravitational. At first, David's explanation seemed logical, so I thought it was right, then since you looked so convinced, I decided to really study your proposition, and I chose to take a closer look at the transition between an equatorial and an orbital bulge.

1) "Tidal bulge" is a kind of "redundancy" ... The "bulges" we are referring to are always one of the manifestation of tides, therefore they are always "tidal".
An equatorial bulge is not a tide, so maybe you don't really understand what inertial means after all! ( just joking!) Your analysis (which seems to be the mainstream one) is perfectly logical, but David's one is also perfectly logical, so to be able to find the truth, I think we have to do what I did, which is to look for a way to link the two explanations. That's what I'm doing in my last posts, but it doesn't seem to interest either of you.

I know at least another way to analyze the tides: tangential speed. I used it before in this thread but it didn't seemed to interest anybody either. I said that while the earth's C.G. was going at the right orbital speed around the moon, neither of its near or far sides were going at the right speed: due to the daily rotation of the earth, the near side is going too slow and the far side is going too fast. If we accelerate a satellite which is already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon. By chance, that single explanation seems to contain your two explanations: if we stop the tangential speed of the earth/moon system, it collapses but the tides don't disappear as in David's explanation, and as in your explanation, tangential speed is also what produces the tides, except that if we reversed the daily rotation of the earth, the tangential orbital speed of the two halves of the earth would change, and it would necessarily affect the tides. If we could manage so that the two halves would be going at the right orbital speed all the time for instance, then there would be no tides, but if we stopped the orbital speed of the earth/moon system, it would not prevent them to rebuild.

Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 13/09/2018 20:02:59
I think we have to do what I did, which is to look for a way to link the two explanations. That's what I'm doing in my last posts, but it doesn't seem to interest either of you.

I couldn't follow what you were describing, but we already have an answer which makes the linkage between the two explanations fully clear. My explanation treats the gravitational force as the only force involved, but rmolnav's explanation introduces an artificial complication by dividing the gravitational force into two components of the same force, one of which is labelled as centripetal force while the other simply remains as gravitational force. The component that he calls centripetal force is the component that makes parts of a body move along the orbital path that they naturally "want" to follow (if no other force was applied to them), and that material is able to follow that path because the material further out (which wants to pull away outwards) and the material further in (which wants to pull inwards) cancel each other out, pulling equally on the material near the middle (which is therefore not pulled off the path it wants to follow). Having made this artificial distinction between gravity (non-centripetal) and gravity (centripetal), he then uses the former component to complete the fake explanation of events on the inside (the stronger gravity there pulling up a "bulge"), and more of the latter component to complete the fake explanation of events on the outside (with the extra centrifugal force throwing out a "bulge"). It is a wonderful example of an exercise in self-deception where people (including good scientists) lose sight of the basic physics and create imaginary mechanisms that are far more complex than the simple reality behind what's going on. In this case, straight-line differential gravity accounts for all the action.

Quote
I said that while the earth's C.G. was going at the right orbital speed around the moon

It isn't - the part that is is near the centre of gravity, but it's slightly nearer to the other body due to differential gravity varying more over the near side (of the Earth to the moon) than the far side.

Quote
if we stop the tangential speed of the earth/moon system, it collapses but the tides don't disappear as in David's explanation, and as in your explanation, tangential speed is also what produces the tides...

And with rmolnav's explanation, if you stop the Earth and moon, the component of gravity that is labelled as centripetal force loses that label and becomes reunited with the rest of the gravitational force as gravitational force. That's the game he's playing: there is only one force involved throughout, but he takes part of it and rebrands it as something else and then makes an artificial mixture of the two which, unsurprisingly adds up to the same amount of force as in my explanation because there is only one force involved.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 13/09/2018 20:37:10
And with rmolnav's explanation, if you stop the Earth and moon, the component of gravity that is labelled as centripetal force loses that label and becomes reunited with the rest of the gravitational force as gravitational force.
It is so with rmolnav's explanation, but not with mine. After you began to answer, I added a few lines on the part you quoted, so please read them and tell me if it changes anything.
Quote from: Le Repteux
By chance, that single explanation seems to contain your two explanations: if we stop the tangential speed of the earth/moon system, it collapses but the tides don't disappear as in David's explanation, and as in your explanation, tangential speed is also what produces the tides, except that if we reversed the daily rotation of the earth, the tangential orbital speed of the two halves of the earth would change, and it would necessarily affect the tides. If we could manage so that the two halves would be going at the right orbital speed all the time for instance, then there would be no tides, but if we stopped the tangential speed of the earth/moon system, it would not prevent them to rebuild.

It isn't - the part that is is near the centre of gravity, but it's slightly nearer to the other body due to differential gravity varying more over the near side (of the Earth to the moon) than the far side.
I don't get it, can you change your wording please?
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 13/09/2018 21:03:18
It is so with rmolnav's explanation, but not with mine. After you began to answer, I added a few lines on the part you quoted, so please read them and tell me if it changes anything.

Your last edit to it was made hours before I read it, so it doesn't change anything, but I haven't managed to follow your explanation. I was hoping that my latest post would help you understand what's going on well enough to apply that to your own thought experiment and fix whatever issues you have there.

Quote
It isn't - the part that is is near the centre of gravity, but it's slightly nearer to the other body due to differential gravity varying more over the near side (of the Earth to the moon) than the far side.
I don't get it, can you change your wording please?

The gravitational pull of the moon on the near side of the Earth is strongest, while the pull on the far side is weakest, but the difference in its strength between the near side and centre is greater than the difference in strength between the centre and far side, so the material that follows its natural path is not at the centre of the Earth, but a small distance away from there towards the near side, and it's further away from the centre the closer together the Earth and moon are. This is a very minor point, but it's worth considering if you want to put the artificial divide (between gravitational force labelled as gravitational force and gravitational force labelled as centripetal force) in the right place, and indeed it's crucial if you want to get the explanation of the tides wrong with high precision.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 14/09/2018 07:39:37
Quote from: David Cooper on Yesterday at 20:02:59
And with rmolnav's explanation, if you stop the Earth and moon, the component of gravity that is labelled as centripetal force loses that label and becomes reunited with the rest of the gravitational force as gravitational force.
It is so with rmolnav's explanation, but not with mine.
I really can´t understand D.C. "blind" stand ...
" ... the component of gravity that is labelled as centripetal force" (AS A FUNCTION, NOT AS ITS ESSENCE), logically " ...if you stop the Earth and moon ... "losses that label", because if no revolving, NO CENTRIPETAL FORCE FUNCTION IS NECESSARY any more ...
But it doesn´t "become reunited with the rest of the gravitational force as gravitational force".... because it NEVER LOST ITS GRAVITATIONAL FORCE ESSENCE ...
On a couple of previous replies to D.C. I referred to that kind of philosophical BASIC point, to no avail !!
Similarly to what I already said, when Donald Trump acts as U.S. President (FUNCTION), he doesn´t loose his ESSENCE of man !!
Am I exaggerating when I say D.C. has problems relative to not only his "sight" and basic Physics, but also to basic Logics and Philosophy (or even to his own Language) ... ??
The FUNCTION of centripetal force can be assumed by ESSENTIALLY different forces: gravity, the tension of a string, the inward push of a locomotive wheel rim, the friction between road and a car tires ... But none of them looses its essence when bending the trajectory of the moving object !!
No wonder first thing he said about pretty clear concept of "centripetal force" was that it is a "grey" area, and that many people out there say things not quite rational about it  ... !!
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 14/09/2018 15:11:18
Am I exaggerating when I say D.C. has problems relative to not only his "sight" and basic Physics, but also to basic Logics and Philosophy (or even to his own Language) ... ??
From my viewpoint, you are effectively exaggerating, but most people would think you're not, because most of us don't realize that we are all blind to other's arguments. What we usually do is chose our party and stick to it. It took a while before I decide to study correctly your proposition, and I could only do that while looking for a way to unite the two thinking. It's chance that permits us to change our mind, nothing else than chance, and it takes time for chance to do its duty, a lot more time than we think, so it is useless to think that it is others that don't understand, it doesn't lead to understanding, and saying it just aggravates the problem.

There might be something important to discover if we try to unite your two propositions, something we don't yet know about gravitation, so I'll go on trying to find the link. David is right and you are too, but there seems to be no link between the two and it is impossible, so let's find that link instead of pulling constantly on our own side like orbiting lonesome cowboys. :0) The pulling is normal, it's how things work, but let's not aggravate it.

I made a proposition that seemed to reconcile the two thinking, but neither of you commented it. Who will be the first to comment? You're the only ones that can tell if it seems to dovetail yours or not. Here is that proposition again:
Quote from: Le Repteux
I know at least another way to analyze the tides: tangential speed. I used it before in this thread but it didn't seemed to interest anybody either. I said that while the earth's C.G. was going at the right orbital speed around the moon, neither of its near or far sides were going at the right speed: due to the daily rotation of the earth, the near side is going too slow and the far side is going too fast. If we accelerate a satellite which is already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon. By chance, that single explanation seems to contain your two explanations: if we stop the tangential speed of the earth/moon system, it collapses but the tides don't disappear as in David's explanation, and as in your explanation, tangential speed is also what produces the tides, except that if we reversed the daily rotation of the earth, the tangential orbital speed of the two halves of the earth would change, and it would necessarily affect the tides. If we could manage so that the two halves would be going at the right orbital speed all the time for instance, then there would be no tides, but if we stopped the orbital speed of the earth/moon system, it would not prevent them to rebuild.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 14/09/2018 17:18:30
My explanation treats the gravitational force as the only force involved, but rmolnav's explanation introduces an artificial complication by dividing the gravitational force into two components of the same force, one of which is labelled as centripetal force while the other simply remains as gravitational force. The component that he calls centripetal force is the component that makes parts of a body move along the orbital path that they naturally "want" to follow (if no other force was applied to them)
The orbital path is due to two different kinds of motion, direct and tangential, so both of you need those two components to explain it. Using only the direct one like you do doesn't explain the path, and using a combination of the two motions the way rmolnav uses it doesn't explain why the tides don't need tangential motion to build up. The only way that seems to work is using orbital motion as I did, which is while considering that the two halves of the earth are not going at the right speed during their transit. It produces the same bulges during the transit, which is what rmolnav defends, and the bulges stay there if we stop the tangential motion, which is what you defend.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 14/09/2018 21:26:06
The gravitational pull of the moon on the near side of the Earth is strongest, while the pull on the far side is weakest, but the difference in its strength between the near side and centre is greater than the difference in strength between the centre and far side, so the material that follows its natural path is not at the centre of the Earth, but a small distance away from there towards the near side, and it's further away from the centre the closer together the Earth and moon are. This is a very minor point, but it's worth considering if you want to put the artificial divide (between gravitational force labelled as gravitational force and gravitational force labelled as centripetal force) in the right place, and indeed it's crucial if you want to get the explanation of the tides wrong with high precision.

I've missed a factor there though, because I divided the globe into two equal halves around the centre of gravity instead of making the section between them curved such that every point on the join is the same distance away from the moon's centre of gravity, and that should cancel out some of the previously mentioned effect, and I can't rule out the possibility that it cancels out 100% of it - there's some fun maths to do on that to find out, though I can't justify putting the time into it myself at the moment.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 14/09/2018 22:14:24
But it doesn´t "become reunited with the rest of the gravitational force as gravitational force".... because it NEVER LOST ITS GRAVITATIONAL FORCE ESSENCE ...

There's only one gravitational pull being applied by the moon to the Earth (or by the Earth to the moon). It pulls most strongly on the near side and least strongly on the far side. That's differential gravity - the force diminishing over distance. What you're doing is rebranding the amount of that force applied at the centre of the Earth as centripetal force, and then at the near side where the gravitational force is stronger, you're attributing the excess to differential gravity, while at the far side where the gravitational force is weaker, you're attributing the bulge there to centrifugal force flinging the material outwards.

In a case where there's no orbit (because the two bodies are falling directly towards each other, you have no centripetal force (or nothing that can currently be described as such by current definitions of the word), so all the action has to be attributed to differential gravity.

This means that when the Earth and moon are at a certain distance apart, such as 250,000 miles, there is an amount of gravity being applied by the moon to the centre of the Earth which we can call n. If we were to stop the system to remove the orbit, the amount of gravity applying there would still be n (for a moment). It's identical in both cases,  but in the former case, you're artificially reducing that value from n to 0 and you're then rebranding the gravitational force as centripetal force. On the near side where the gravitational force is n+p, you continue to treat the n as 0 (because you say the centripetal force is n instead), and you count p as gravitational force. On the far side where the gravitational force is n-q, you continue to treat the n as zero (because you say the centripetal force is n instead), and the q becomes the centrifugal force that tries to throw material outwards (or it becomes a centripetal force of -q). But all you're actually doing is chopping up gravitational force into artificial chunks with silly labels attached to them which pretend there's something more complicated going on than the actual reality of it. You say it never lost its gravitational force essence, but in reality it never stopped being gravitational force in any way at all.

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On a couple of previous replies to D.C. I referred to that kind of philosophical BASIC point, to no avail !!
Similarly to what I already said, when Donald Trump acts as U.S. President (FUNCTION), he doesn´t loose his ESSENCE of man !!

Why are you telling me that when you're the one making the mistake of treating him as if he's no longer a man? With this tide business, we're looking for an explanation to the tidal forces, and what you're doing is splitting off a massive chunk of gravitational force and presenting it as centripetal force while you keep some of it back and call it gravitational force, though only applying that at the near side, and then you introduce a centrifugal force at the far side which is also a deficit of centripetal force (not enough of it to hold that material in as strongly as it holds material at the centre) and which is also a deficit of gravitational force due to gravity decreasing over distance.

Quote
Am I exaggerating when I say D.C. has problems relative to not only his "sight" and basic Physics, but also to basic Logics and Philosophy (or even to his own Language) ... ??

The problems are all at your end, and all the problems you mention certainly do apply there.

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The FUNCTION of centripetal force can be assumed by ESSENTIALLY different forces: gravity, the tension of a string, the inward push of a locomotive wheel rim, the friction between road and a car tires ... But none of them looses its essence when bending the trajectory of the moving object !!
No wonder first thing he said about pretty clear concept of "centripetal force" was that it is a "grey" area, and that many people out there say things not quite rational about it  ... !!

The reason I don't like gravity being described as centripetal force is precisely that people like you are so misled by that usage that you end up treating the centripetal force as a different mechanism from gravitational force. The only mechanism in play here is straight-line differential gravity - the other mechanism is a ludicrous fabrication which comes out of people like you confusing themselves with words and dodgy labels.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 14/09/2018 22:55:49
...but there seems to be no link between the two...

I've shown you the link between the two - they produce the same end result because they are doing the same thing, my approach getting there directly by recognising a single thing as itself (so that there is only one mechanism needed), and rmolnav's approach where he splits that same single thing into two things which he then treats as separate mechanisms, bringing in a whole lot of unnecessary complexity which is only needed because of his error in splitting something up that should be considered as the single thing that it actually is.

Quote
I made a proposition that seemed to reconcile the two thinking, but neither of you commented it. Who will be the first to comment?

Quote from: Le Repteux
If we accelerate a satellite which is already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon.

A satellite is moving relative to the Earth, so slowing it down relative to the Earth makes it fall. Let's suppose we have two satellites crossing over the near point (nearest to the moon) in opposite directions. If we slow them down, they both fall, even though one is going faster than the Earth along the Earth's orbit and the other satellite is going slower than the Earth along that path. You are merely adding confusion to things by providing an illustration that doesn't shed light on the issue under discussion here.

Quote
By chance, that single explanation seems to contain your two explanations:

It doesn't relate to either of them. You need imagine your satellites as stationary relative to a non-rotating Earth and then think about how they'll fall. The one on the near side will fall down more slowly than it would if it was 90 degrees further round the planet at the side because the moon's gravitational pull on it will be slightly higher than it is on the Earth as a whole, whereas the one on the far side will fall down more slowly too because the moon's gravity is pulling the Earth away from it slightly more strongly than it is pulling the satellite.

Quote
...except that if we reversed the daily rotation of the earth, the tangential orbital speed of the two halves of the earth would change, and it would necessarily affect the tides. If we could manage so that the two halves would be going at the right orbital speed all the time for instance, then there would be no tides,

That is the bit I can't make sense of. What is this right orbital speed that will eliminate the tides?

Quote
but if we stopped the orbital speed of the earth/moon system, it would not prevent them to rebuild.

And I don't know how you're getting there either, which is why I didn't comment before.

The orbital path is due to two different kinds of motion, direct and tangential, so both of you need those two components to explain it.

I don't need it at all, and relativity tells you that - that's why I was so sure that I was right when I began questioning rmolnav about the involvement of centripetal force - the perpendicular movement is very clearly a complete irrelevance to the mechanism. Whatever perpendicular movement there is, it simply applies to all the components equally, just like when you throw a ball to and fro across a train carriage (while sitting inside it at opposite sides from a friend) and you can explain the entire mechanism behind each throw without considering the movement of the train even though the train may be doing 200mph. If it's a round carriage (which, while impractical, could certainly be made), we don't need to factor in the tangential movement when the ball reaches either side because that tangential movement is in the same direction as the one the train is moving in. There is no tangential force outwards away from the direction the train is moving in, so why would you want to ram it into the calculations?
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 15/09/2018 08:18:07
What about the NOAA explanation then? Can we show that their calculation or their drawings are wrong for instance? Unfortunately, there is no calculation to support their drawings, and the drawings themselves are only qualitative.
I have to further "work" on last posts, before replying them. But one thing now: you and D.C. talk about rmolnav theory ... It is not just "mine":
It is true that since my very first post here (#20, 29/5/2015 !!), I´ve had quite clear the importance of earth circular movement "dancing" with the moon, and inherent centripetal (function exerted by moon´s gravity and internal stresses) and centrifugal forces. Then I said:
"1) “Centrifugal” force is NOT a forbidden word whatsoever: it is just a poorly understood and poorly explained force. I´m not going to deliver now any further explanation about why I say so, perhaps in another post if found convenient" ...
But a couple of aspects of my stand did change, especially in relation with my discussion with the NOAA scientist I already mentioned, and I even included part of our discussion.
Initially I didn´t agree 100% with them , but from our discussion, and further ruminating of facts, I got to fully understand and agree with THEIR stand  ...
Regarding with what quoted, I imagine you refer to NOAA work:
 https://tidesandcurrents.noaa.gov/restles3.html
But we (Colin2B and I) have also mentioned here:
https://tidesandcurrents.noaa.gov/publications/Tidal_Analysis_and_Predictions.pdf
a more than 300 page thorough work on tides, as referred to on #268.
Neither on what quoted "there is (any) calculation to support their (drawings) theory", as you say in relation to the shorter work.
But the long one goes through every imaginable detail affecting tides, using all available tools such as astronomical calculations (based on those forces they consider are actually acting on earth material stuff), real measurements in situ and from satellites, etc.
E.g., just on page 11 of the "Introduction" Dr. Parker says:
"The details of how these overtides (higher harmonics of the astronomical constituents) and compound tides (new tidal constituents resulting from the interaction of two or more astronomical constituents) are actually generated have often been ignored. In some cases the mechanisms that produced them were not even fully understood, and the shallow-water constituent frequencies were merely determined by adding or subtracting astronomical frequencies. Not understanding the nonlinear hydrodynamics that generate these constituents can cause problems in tidal prediction accuracy …"
Despite all that thorough work, and all the prizes he got (see #268), could Mr. Parker be wrong on the very core of tidal generating forces ... ??
Well, theoretically he could ... But the odds would be, I guess, less than 1:1000 !! 

 
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 15/09/2018 12:11:54

Quote from: rmolnav on Yesterday at 07:39:37
Am I exaggerating when I say D.C. has problems relative to not only his "sight" and basic Physics, but also to basic Logics and Philosophy (or even to his own Language) ... ??
From my viewpoint, you are effectively exaggerating, but most people would think you're not, because most of us don't realize that we are all blind to other's arguments
O.K. Naturally you are quite free to being so understanding D.C. ...
As I´ve posted here many times, I´ve discussed the issue on many sites, and even directly by email with many scientists, always giving logically MY arguments, but I´ve changed some details as said on my last post a few hours ago.
But D.C. time and again has loftly despised:
A) What said by eminent scientists, such what NOAA´s one I interchanged emails with ... That scientist had told me:
""The publication you are referring to is "Our Restless Tides", a 10-page pamphlet developed in the 1950's to provide a basic description of the forces which create the tides.  It's intended audience were the grade school children and adults of that time.  It used terminology of science and forces which were common in the 1950s.  Such as centrifugal force.  Centrifugal force was always an "imaginary force" (not a real / measurable force).  But that type of description made the concepts easier to understand and explain.  That  description and use of centrifugal force continued to be common practice until the 1970-80's.  At that point, the terminology shifted and the textbooks used in grade schools were changed to use a more modern terminology and description of this "effect" being a result of inertia rather than an "imaginary force”",
I quoted that here (#155), and D.C. replied:
"Here, we see that the scientist may not have a firm understanding of the tides himself though - all he's doing here is trying to get rid of the idea of centrifugal force by referring to the real mechanism that is behind centrifugal force, but he hasn't realized that this isn't the mechanism behind the tides”  ...
and, as I´ve also said many times, he is the author (or, at least one of them) of
https://tidesandcurrents.noaa.gov/restles3.html ...
...where they say:
"1. The Effect of Centrifugal Force. It is this little known aspect of the moon's orbital motion which is responsible for one of the two force components creating the tides. As the earth and moon whirl around this common center-of-mass, the centrifugal force produced is always directed away from the center of revolution. All points in or on the surface of the earth acting as a coherent body acquire this component of centrifugal force. And, since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon. This fact is indicated by the common direction of the arrows (representing the centrifugal force Fc) at points A, C, and B in Fig. 1, and the thin arrows at these same points in Fig. 2.
It is important to note that the centrifugal force produced by the daily rotation of the earth on it axis must be completely disregarded in tidal theory. This element plays no part in the establishment of the differential tide-producing forces"
B) He also despise first line dictionary definitions of "centripetal force" ... (Again: NOT as a "new" force, independent from others, but as a FUNCTION, a kind of "job", essentially real forces (such as gravity) can exert when acting "transversely" (not with null component perpendicular to the trajectory) to an object moving with a "not null" speed ...)
C) When I asked D.C.:
"Could you please send some link, from a University or similarly reliable source, where one could see that is not another absurd idea of yours ??"
he replied:
"I am always more interested in actual science than error-ridden authorities. I haven't seen anyone in science support my position (primarily because I haven't looked for that) - what I'm saying is based 100% on what I see when I look directly at the physics involved in this. If a force is generated by rotation (??)..."
and he continued with more absurdities ...
We both may be kind of stubborn, but he is also impervious to properly learning basic Physics he utterly misunderstand !!
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 15/09/2018 14:34:33
There's only one gravitational pull being applied by the moon to the Earth (or by the Earth to the moon). It pulls most strongly on the near side and least strongly on the far side. That's differential gravity - the force diminishing over distance. What you're doing is rebranding the amount of that force applied at the centre of the Earth as centripetal force, and then at the near side where the gravitational force is stronger, you're attributing the excess to differential gravity, while at the far side where the gravitational force is weaker, you're attributing the bulge there to centrifugal force flinging the material outwards.
A) "What you're doing is rebranding the amount of that force applied at the centre of the Earth as centripetal force" ...
NO REBRANDING WHATSOEVER ! That gravitational force there matches exactly with the centripetal force necessary for the revolving, and does that "job" without any force excess or deficit.
Or, as on NOAA scientist stand, centrifugal force there is equal but opposite to moon´s pull, and adding up both vectors there is no spare force to do anything else than the very revolving.
B) "... and then at the near side where the gravitational force is stronger, you're attributing the excess to differential gravity..."
 NOT AL ALL !! Don´t you also say:
"There's only one gravitational pull being applied by the moon to the Earth" ...?
At the near side ONLY part of that REAL pull ("differential" is just a mathematical tool of our minds) is necessary for the revolving of stuff there, and the rest remains kind of "free", and "tries" to accelerate ("straight line" acceleration) > > > near hemisphere tides.
Or, as on NOAA scientist stand, adding up centrifugal force and moon´s pull there, last one prevails > > > tides on hemisphere closer to moon.

C) " ... while at the far side where the gravitational force is weaker, you're attributing the bulge there to centrifugal force flinging the material outwards" ...
Well, just "trying" to fling the material outwards !!
It couldn´t be otherwise: moon´s pull there is insufficient for the FUNCTION of centripetal force required to produce on material stuff the centripetal acceleration they are being "forced" to get by the rest of the planet ... And INERTIA reveals itself that way > > > further hemisphere tides !!


Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 15/09/2018 15:39:54
We both may be kind of stubborn
To me, it's not a voluntary behavior, so there is no name for it in our actual language. First, I think we are blind to other's ideas because we automatically resist to a change, but more specifically, because our mind doesn't necessarily have the background ideas needed to understand some of them. Not because some ideas are more difficult to learn than others, but because we simply don't all learn the same things. To change our mind about something we believe right is then a chance issue: It may happen that one of our ideas suffer what I call a mutation, or it may happen that an argument is so close to what we had in mind that we suddenly understand it. It's as if the two ideas would then suddenly interfere. That's what happened to me when I understood the beaming phenomenon while I was studying David's simulation of the Michelson/Morley experiment (http://www.magicschoolbook.com/science/relativity.html): all of a sudden, relativity was clear. Thanks to David's phlegm repeating the same thing over until I understood it. That's what he does here too, he doesn't seem able to get involved with emotions, and that's precisely the right thing to do even if we think we are right otherwise we wouldn't ever be able to change our mind on anything and there would be no intellectual evolution. You seem quite phlegmatic too otherwise the discussion wouldn't have lasted that long. I still can't chose between you two though so I'll go on trying to find the link.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 15/09/2018 16:54:38
I don't need it at all, and relativity tells you that - that's why I was so sure that I was right when I began questioning rmolnav about the involvement of centripetal force - the perpendicular movement is very clearly a complete irrelevance to the mechanism. Whatever perpendicular movement there is, it simply applies to all the components equally, just like when you throw a ball to and fro across a train carriage (while sitting inside it at opposite sides from a friend) and you can explain the entire mechanism behind each throw without considering the movement of the train even though the train may be doing 200mph. If it's a round carriage (which, while impractical, could certainly be made), we don't need to factor in the tangential movement when the ball reaches either side because that tangential movement is in the same direction as the one the train is moving in. There is no tangential force outwards away from the direction the train is moving in, so why would you want to ram it into the calculations?
Orbital motions are not considered as rest frames, so I think that, in your example, the 200mph speed should be considered as an orbital speed. This way, the rotation speed of your round carriage would add or subtract to the orbital speed of the train when they coincide, what should affect the trajectory of the ball the same way accelerating or slowing down a satellite affects its trajectory. The ball is free to move radially, but the carriage is not, so it won't, but it could move tangentially if the ball does, and I think it should because a satellite on a circular path does when we change its speed: if we slow it down, it accelerates tangentially until it gets at its perigee, and if we accelerate it, it slows down tangentially until it gets at its apogee. Notice that this motion would slow down the earth's daily rotation because it opposes it, which gives us a true physical mechanism to explain that observation. Friction due to tides cannot slow down the earth's daily rotation unless it develops a force that opposes to the direction of that rotation, and there is no way for such a friction to produce such a force, which is probably why there is nothing about force in that standard explanation.
Title: Re: Why do we have two high tides a day?
Post by: PmbPhy on 15/09/2018 18:41:01
French Tides
You're most welcome. See:
............
Thanks Pete, I’m not in good WiFi area at moment, when I am I will read through.
If I have time I’ll read through this thread and see what these guys are arguing about.
You're welcome. Notice the emphasis on inertial forces, in particular the centrifugal force of the earth orbiting the center of mass of the earth-moon system.
Quote
PS French Tides sounds a bit dodgy, didn’t think you were into that  ;)
Call me naive, but I don't get it this joke. ????
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 16/09/2018 00:26:34
Orbital motions are not considered as rest frames,

Think about my train experiment again, but let's put it in space and turn it into a spaceship designed to resemble a train. We can put a very strong magnet at one side of the carriage and then play with an iron ball at the other side. We have a zero-G environment (technically microgravity - we can effectively ignore all gravity, and we can also remove all the air to avoid air currents messing things up). If I push the iron ball very gently towards the window at my side of the carriage, the magnet at the far side of the carriage will slow it down, then begin to pull it away from the window again. The carriage may be stationary in space, or it may be moving at close to the speed of light, but the behaviour of the iron ball that I see is the same in both cases. It may be that the iron ball is simply moving along a straight line and back, or it may be moving through space along a curved path (which is nearly perpendicular to the straight line that it would follow in a stationary system), but the perpendicular component of the movement is completely irrelevant to the cause of the ball being slowed by the magnet and then accelerated back towards it.

Now repeat the above with a magnet that gradually moves along its side of the carriage to vary the direction that it pulls on the ball. What difference does this make? Does it break relativity? No - the behaviour that I see as I move with the carriage is still the same regardless of the speed of the carriage, and if the magnet is directly opposite the ball at the point when the ball stops moving towards the window and starts to move away from it again, the force being applied to it is identical to the force being applied in the previous case without the magnet moving, as is the differential nature of the force, becoming less strong at greater distance. Relativity shows those of us who understand the relevant physics that there is no causal role for any perpendicular component of the movement - it all comes down to a straight-line pull.

Here's another idea that might help you understand this. Imagine that we have a moon that can ping in and out of existence at the press of a button. What happens to the tidal forces then? We make the moon appear, the Earth starts to accelerate a little towards it, the nearside sea is pulled more strongly towards it and the farside water is pulled less strongly, so we have tidal forces. Then we make the moon disappear again, and the forces disappear because the acceleration has been removed. If we start with a stationary Earth, we can make the moon appear several times in the same place (some distance away from the Earth), and each time we do so, tidal forces appear on the Earth due to differential gravity. Each time we remove the moon, the tidal forces go away. The Earth's now moving towards the place where the moon kept appearing, so let's now make the moon appear in a new location to the side instead. Tidal forces are again generated - they are caused by differential gravity as before, and the movement of the Earth doesn't do anything weird to this at all because we simply get "bulges" on opposite sides aligned with where the moon is. If we keep changing the places where we make the moon appear, we can repeatedly change the direction the Earth's moving in, and we can make it go round in something that approximates to a circle: a polygon with thousands of straight sides. The straight sides are made when the Earth isn't being accelerated, and the corners are made at moments when the moon exists to apply a force. At any of those moments when the moon exists, that force is straight-line gravity, diminishing in strength over distance. We can make these moments shorter and shorter, and we can also use more and more of them to make the Earth follow a path closer and closer to a circle or ellipse, but all we ever have at any moment is straight-line acceleration, and as soon as the moon isn't there, that acceleration is gone and there are no tidal forces. If the moon is there all the time, and constantly moving rather than appearing repeatedly in slightly different places, nothing has changed - it's still applying a series of straight-line accelerations towards its current position.

If we use a black hole instead of a moon, we can make a planet follow a square "orbit" by making the black hole appear for a moment four times per orbit in different places. Each time there will be tidal forces generated for a moment, but again we just have momentary straight-line accelerations. Every time the planet reaches a corner, a momentary acceleration force is applied which changes its course, while the perpendicular component of its movement is completely unchanged. That perpendicular component is shared with the water in the planet's ocean, so it retains that component of its motion too. Here we have a planet making a 90 degree turn, but the perpendicular (to the force) component of its motion is not changed at all. The water on the near side is pulled by the acceleration with greater force, so it tries to lift slightly, and the water at the far side is pulled with less force, so it tries to hold back slightly, but all this does is lead to pressure differences and slight leads and lags in reaction as those pressure differences are established, and then the whole caboodle (Earth plus water) accelerates and makes the 90 degree turn together. Switch frame of reference and you can view this as a 180 degree change of direction, completely ruling out any role for the perpendicular component of the planet's movement.

Now, I have no doubt that some minds are incapable of processing such ideas, and even if I was to go to the trouble of writing a program to illustrate how this square orbit works and how the tidal forces play out within it, such minds would not understand it and would simply regard the whole thing as nonsense (due to its exotic nature) and assert that I don't understand basic physics, but those with more flexible minds should be able to see that what I have just set out is entirely correct and that it shows more than adequately why the perpendicular movement has no role.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 16/09/2018 01:06:28
NO REBRANDING WHATSOEVER ! That gravitational force there matches exactly with the centripetal force necessary for the revolving, and does that "job" without any force excess or deficit.
Or, as on NOAA scientist stand, centrifugal force there is equal but opposite to moon´s pull, and adding up both vectors there is no spare force to do anything else than the very revolving.

And in a case where two bodies are moving directly towards each other? What then? You stop calling it centripetal force and use a more fundamental description of it instead, at which point you should recognise that straight-line differential gravity is the better explanation.

Quote
B) "... and then at the near side where the gravitational force is stronger, you're attributing the excess to differential gravity..."
 NOT AL ALL !! Don´t you also say:
"There's only one gravitational pull being applied by the moon to the Earth" ...?
At the near side ONLY part of that REAL pull ("differential" is just a mathematical tool of our minds) is necessary for the revolving of stuff there, and the rest remains kind of "free", and "tries" to accelerate ("straight line" acceleration) > > > near hemisphere tides.

If you're disagreeing with me, then that can only mean that the excess there isn't just down to differential gravity because you're also labelling a smaller amount of the gravity applying there as centripetal force than you do at the centre.

Quote
C) " ... while at the far side where the gravitational force is weaker, you're attributing the bulge there to centrifugal force flinging the material outwards" ...
Well, just "trying" to fling the material outwards !!
It couldn´t be otherwise: moon´s pull there is insufficient for the FUNCTION of centripetal force required to produce on material stuff the centripetal acceleration they are being "forced" to get by the rest of the planet ... And INERTIA reveals itself that way > > > further hemisphere tides !!

żAnd you still can't see that your approach is just an unnecessarily complex way of explaining something that is more directly accounted for by straight-line differential gravity (which works in every case, including the one where your explanation breaks) where there's absolutely no need to make an artificial distinction between components of gravity that are labelled as centripetal force and those that aren't?

It would be interesting to see if your NOAA scientist is capable of understanding this.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 16/09/2018 17:12:12
Think about my train experiment again, but let's put it in space and turn it into a spaceship designed to resemble a train. We can put a very strong magnet at one side of the carriage and then play with an iron ball at the other side.
You didn't comment my example and I'm not satisfied with your new explanation, so let me go on explaining my point while using your own example in case it would help. If we add to your orbiting train a circular carriage rotating in the same direction than its orbital trajectory, the train will be orbiting at the right speed all the time, but not the two halves of the carriage. Let's now replace the two rotating halves by two balls rotating at the two ends of an elastic. In principle, while the balls will transit on the line between the earth and the train, the one that is going too fast on its orbital trajectory around the earth will decelerate and get away from the earth a bit, and the one that is going too slow will accelerate and get closer to the earth a bit, what will slow down the rotation speed of the balls with time since those two accelerations are both opposed to the rotational direction of the balls. Of course, the elastic will prevent the balls from moving away from one another as much as if they were orbiting around one another because the more they pull on the elastic, the more its force increases, which is the inverse of how gravitation works, but the elastic should nevertheless stretch a bit each time the balls would transit, and shrink a bit each time they would cross over the train. If you think I make a mistake, tell me where so that I can change my mind, and if you think I'm right, then tell me why you think I cannot apply that principle to the tides.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 16/09/2018 20:35:44
You didn't comment my example...

I can't follow it well enough to be able to comment on it in any useful way, and the same applies to the new stuff. You seem to be introducing complications which obscure things rather than simplifying things to isolate the different factors.

Quote
...and I'm not satisfied with your new explanation,

All it is is a simple proof that relativity does apply and that it shows the perpendicular component to be irrelevant. Think about how my program handles the orbits. It applies forces from moon to planet and planet to moon many times a second and uses those forces to adjust the directions and speeds of the moon and planet. Both bodies move in a series of short, straight lines, and their orbital paths are polygons with many thousands of sides. The shorter you make those straight lines (which you can do by applying smaller forces at higher frequency), the closer the orbits get to being smooth, curved paths rather than polygons, and the simulation becomes more accurate if you do this. Ideally you would apply a near-infinite number of tiny forces, but whatever size they are, they always behave just like the bigger, less frequent applications of forces.

If you go the other way and make the straight lines longer by applying forces less often but at higher strength, you reduce the number of sides of the polygons, and they can go right down to 5, 4, 3 or even just 2 sides - you lose more and more accuracy compared to real planets and moons as you do this, but the principle as to how the forces are applied remains identical. Each momentary force applied leads to a change in direction of travel of the planet/moon and sets a new speed for it. Every time this happens, that change in direction which may be anything from a tiny fraction of a degree to a substantial turn can be converted into a precise 180 degree turn simply by viewing it with the right reference frame. When you do this, you see that relativity absolutely does apply to this issue and that there can be no role for the perpendicular component of movement in affecting the tidal forces. Once you understand that, you know that any thought experiment that attempts to show a role for the perpendicular component of motion to affect the forces along the line which the gravitational force acts along is doomed to fail unless it can break relativity (including both SR and LET) and thereby earn a Nobel prize.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 17/09/2018 18:13:37
How could D.C. draw correct conclusions from just his “vision” (?) of dynamical phenomena, if:
A) No matter how many times he is told not rectilinear movements ALWAYS require some force perpendicular to the trajectory to intervene, with the FUNCTION of centripetal force, he despises that basic Physics fact …
B) He insists in calling a revolving movement around moon-earth barycenter just a movement “perpendicular” to straight line gravitational forces, even having been told that it is a two-dimension movement, and that the acceleration inherent in that type of movements (centripetal) is perpendicular to the speed of the revolving (or rotating) object, therefore with same direction of gravitational forces in our case.  And that, logically, also inherent inertial force (centrifugal) has same direction...
C) He seems to ignore the concept of “inertia" … or at least never mentions it, or “inertial” as an adjective applied to concepts such as “force”, “effect” …
D) ... ("impossible" to list all his misconceptions ...)           
BY THE WAY, on a link posted by PmbPhy some days ago it is said:
"… the production of ocean tides is basically the CONSEQUENCE of the gravitational action of the moon- and, to a lesser extent, the sun …The analysis of the phenomenon is, however, considerably helped by introducing the concept of INERTIAL FORCES as developed in the present chapter.
… With respect to the CM of the earth-moon system … the earth´s center of mass has an acceleration of magnitude Ac … every point in the earth receives this same acceleration from the moon´s attraction.
… for a particle at the earth´s center, the CENTRIFUGAL FORCE and the moon´s gravitational attraction are equal and opposite. If, however, we consider a particle on the earth´s surface at the nearest point to the moon …, the gravitational force on it is greater than the CENTRIFUGAL FORCE by an amount that we shall call Fo …
By an exactly similar calculation, we find that the tide-producing force on a particle of mass m at the farthest point from the moon … is equal to -Fo …”

What I´ve been saying here time and again !!
LOGICALLY, the authors, as the NOAA scientists D.C. loftly despises (!!), neither say “the production of ocean tides is ONLY due to differential gravitational action" (or something with that meaning), nor “forget” an acting inertial force such as centrifugal force, as he stubbornly keeps doing !!
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 17/09/2018 19:01:54
Show your numbers then and claim your Nobel Prize if they vary depending on the direction and speeds of planet and moon at moments when the two are the same distance apart. The perpendicular component of motion does not affect the tidal forces. All you've done is make an artificial divide by splitting the gravitational force into centripetal and non-centripetal, and once you've added everything up, you'll be left with the same forces that are worked out much more simply just by treating the gravitational force as a single force (which is exactly what it is).

As for not mentioning inertia, it should be obvious to anyone that there is such a thing, but with gravitational pull, the speed of acceleration of an object being accelerated by gravity is the same regardless of the mass involved, so it has no impact on how much the Earth and moon accelerate towards each other (one ton of rock is accelerated just as much as a billion tons of rock ). The further parts are pulled less strongly and the near parts more strongly, so forces are generated as those parts try to lift. That's all there is to it and you need to stop asserting otherwise.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 18/09/2018 11:48:27
What an "interesting" case for the study Neil Comins did:
Sources of Misconceptions in Astronomy, by Neil F. Comins
University of Maine

"...In pursuing them and their origins, I use the definition of misconception as a resilient concept held to be correct, but that is actually at variance with accepted scientific knowledge. Resilient here means that the concept is incorporated in a person's conceptual framework and therefore resistant to change. Because they are woven into conceptual fabrics, many misconceptions help prevent further correct understanding of related topics. The description of "conceptual barriers to learning," by Hawkins, Apelman, and Colton & Flexner, cited in Confrey (1987) helped me focus on an operating definition.
For the past eighteen months I have been working with students taking the above- mentioned introductory college astronomy course in an effort to understand the origins of their misconceptions about astronomy. I will describe the methodology I am using shortly. In the process, the students have shared with me over five hundred, fifty misconceptions they held prior to taking the course. In this paper I report on a set of origins of astronomical misconceptions that derives from their lists and from interviews I held with small groups of students. The origins, listed below, can be used to explain virtually all the stated misconceptions. This set is certainly not unique; other categories of internal and external error leading to misconceptions can certainly be identified or constructed. Nor is this set necessarily complete, although it has been satisfactory in explaining the misconceptions on hand, a list that is continually growing. It is also important to point out that the list of origins is heterogeneous, some deriving from sources external to the holder of the misconception, some internal, and some with both internal and external characteristics. 
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 18/09/2018 16:53:51
I can't follow it well enough to be able to comment on it in any useful way, and the same applies to the new stuff. You seem to be introducing complications which obscure things rather than simplifying things to isolate the different factors.
I guess that what you can't follow is not the way we can change the trajectory of satellites. It is well established that if we accelerate a satellite that is already on a circular orbit, it will start decelerating and getting away from the earth until it gets at is apogee, and vice-versa if we decelerate it. Thus what you can't do is probably applying this principle to the tides, or to any example where different parts of an object are not traveling at the same orbital speed as its center of gravity. Instead, you prefer making your point with your own examples. I think you don't need to do that with me, because I already understand your point. Moreover, my principle doesn't seem to contradict yours: whenever we stop the orbital speed in my example, the tidal bulges stay the same, because they are not affected by the rotational inertial force as in rmolnav's principle. Nevertheless, inertia is certainly accounted for when we accelerate or decelerate a satellite, and the change in speed certainly changes the parameters of its orbital trajectory with time, so applying that principle to tides doesn't seem to contradict rmolnav's principle either.

Quote
You seem to be introducing complications which obscure things rather than simplifying things to isolate the different factors.
Your principle is simple for bodies with no orbital speed, but it is certainly not that simple for those with speed otherwise I guess rmolnav would have accepted it. I do accept it, but I had to be able to figure out how inertial motion would affect it. In your simulation, I assume that the whole earth (with its bulges) is first moved tangentially away from the moon a bit, and that you then apply the gravitational force at that distance to bring it back on its trajectory. The first motion is thus inertial, and the second one is gravitational. This way, there is visibly no need to use any inertial force to take the earth away, it moves all by itself and the bulges too, and when the gravitational force begins to pull, it pulls differently on its different parts, so the bulges change if the distance to the moon has changed. It works fine, and my explanation also works fine since the orbital speed I'm using to move the bulges tangentially away would be simulated the same way. What would be different is that the tangential orbital speed of the earth surface would change constantly due to its daily rotation, and as I explained, that should decelerate its rotation with time.

Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 18/09/2018 17:05:38
Resilient here means that the concept is incorporated in a person's conceptual framework and therefore resistant to change.
We all resist to change, and every existing thing does, so it is useless to try to bypass the problem. The only way is to deal with it, and as far as our ideas are concerned, it is to go on pushing them until others agree with them, but there is still a bug, because people can visibly adopt false ideas.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 18/09/2018 21:03:36
Nevertheless, inertia is certainly accounted for when we accelerate or decelerate a satellite, and the change in speed certainly changes the parameters of its orbital trajectory with time, so applying that principle to tides doesn't seem to contradict rmolnav's principle either.

If you look at the perpendicular component of movement, there is no acceleration (or deceleration) - the inertia of the bodies just keeps them moving that way at a constant speed (until there's a change in the direction the force is being applied in). The only change occurs along the direction in which the force is acting, and on that line you get the same size of tidal forces being generated regardless of whether the bodies are moving apart or together at that time. The perpendicular component is irrelevant.

As I said before. rmolnav needs to show his numbers if he thinks otherwise. I wrote a program which shows my numbers, and I invited him to produce his maths to put into the program to replace mine. He hasn't done so, and he probably isn't even capable of doing so. If he ever does though, I will be more than happy to take his code of this (pseudo-code will do - the algorithm is all I need from him because it isn't my job to build that for him) and put it into the program for him, and it can then run both methods at the same time while displaying the numbers for each side by side. If his maths is right, the numbers will necessarily match up with mine, but it will take a lot more processing to work them out each time because he is splitting the task up into artificial components. There is only one force driving tidal forces, and that's gravitational pull acting in a straight line and reducing in strength over distance.

Think about what happens when the Earth is directly between the sun and moon. What is the mechanism now for the tidal forces? Let's move the Earth and moon to a different orbit to make the two lots of pull equally strong. Be aware that this means moving them further away from the sun rather than nearer to it - the sun's input to the tidal forces is only smaller because its pull varies less across the Earth than the moon's pull does due to the greater distance to the sun. The Sun is pulling more strongly on F, the far side from the moon, and less strongly on N, the near side to the moon, while the moon is pulling more strongly on N and less strongly on F. These pulls are in opposite directions, so the result is enhanced tidal forces. That is the correct explanation. But what happens if we start chopping up the forces in the way rmolnav seeks to do and attributing extra mechanisms to the system?

If we're playing rmolnav's game, we take each of these forces at the centre of the Earth and label them as centripetal force. At point F (the far side of the Earth from the moon), we attribute the material's trying to lift there mainly to centrifugal force, plus a little bit to the sun's greater pull there. At point N (the near side of the Earth to the moon), we attribute the material's trying to lift there mainly to the moon's greater pull, plus a little due to centrifugal force. We supposedly have centrifugal force trying to fling material out at opposite sides in opposite directions. The Earth is neither accelerating towards the sun or moon in this situation (because we've put the Earth at a distance from the sun where the two forces are equally strong), so the Earth is moving in a straight line (and is doing so in every frame of reference). Let me spell that out more clearly for rmolnav: the Earth is moving in a straight line and you think you have material being flung out sideways by centrifugal force. No - what's actually happening is that you're mauling the physics. Everything that's actually happening is down to straight-line differential gravity.

Let's do another version which makes things even more stark. Let's give the Earth two moons which always stay on opposite sides. The Earth isn't moved by them at all now, but we still have tidal forces, nearly twice as strong as we have with just the one moon. How does rmolnav's explanation work here? The gravitational pull from each at the Earth's centre is called centripetal force, and the two lots cancel out. Each bulge is driven partly by centrifugal force from the Earth's "movement" acting on the material furthest from the moon that's applying the centripetal force to the centre that drives this "movement", and partly by the greater gravitational pull applied there by the other moon. Take away the sun and the Earth's orbit so that all we have is the Earth and its pair of moons, and what do we have this time? We have a stationary Earth with material being flung out at opposite sides by, well, I suppose it should be called centrifugal farce.

Ask yourself now, who is the person here who repeatedly fails to understand the basic physics that's at work here? Is it the one who can't supply his numbers and who keeps arguing against the correct explanation while pushing a much more complex and contrived one where a force is artificially split into different components and where that explanation breaks down in multiple situations, or is it the one who supplies actual numbers, properly reasoned arguments backed by a computer simulation which runs his model and demonstrably works, and whose explanation covers all cases?
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 18/09/2018 22:28:57
David, you did not comment a part of my previous answer that I found interesting, so let me present it again: 
In your simulation, I assume that the whole earth (with its bulges) is first moved tangentially away from the moon a bit, and that you then apply the gravitational force at that distance to bring it back on its trajectory. The first motion is thus inertial, and the second one is gravitational. This way, there is visibly no need to use any inertial force to take the earth away, it moves all by itself and the bulges too, and when the gravitational force begins to pull, it pulls differently on its different parts, so the bulges change if the distance to the moon has changed.
If I am right about the way your simulation works, it is so simple that I think Rmolnav should be able to understand it if we describe it properly. It contains the two principles, inertial and gravitational, and no inertial force is needed. I think the misunderstanding comes from using force instead of motion to describe the tangential motion. if we throw a stone with a sling, it is not a force that moves the stone away from us, only the tangential motion. The stone doesn't move radially away from us, it moves tangentially away from the point where it was released by the rotating sling, it transforms the rotational speed of the sling into inertial motion. Of course that the pulling from the moon exerts a force on the earth's mass so that it doesn't get away, but we can easily see that it is not a force that moves the earth away from the moon either, only a tangential motion. That straight tangential motion could also be curved by the pulling of a sling if there was no gravitation, but in this case, only the pulling of gravitation is at work.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 20/09/2018 00:28:38
David, you did not comment a part of my previous answer that I found interesting

I didn't have anything to say about it and still don't. If something's moving perpendicular to the force that's being applied, it simply maintains that perpendicular component of the movement while the acceleration adds or reduces movement in the direction of the force. After that change has been made, the object will naturally be moving in a slightly different direction from the previous one (unless the perpendicular component was zero). The result is simply what comes from that, and there's no other factor, except that any non-zero perpendicular component of movement leads to the next application of the force being applied in a slightly different direction the next time. Where material goes or tries to go will fit in perfectly with the idea of it being directed there by "centrifugal force" in some cases, but it's actually just going where it goes (or trying to) based on the amount of gravitational force acting on it and trying to alter its current movement accordingly.

We can see now a clear case in my previous post involving a planet with a moon going round it where rmolnav's explanation breaks horribly, and I want to expand that a bit further.

I did a few in-the-air calculations last night and reckon that if you put the Earth and moon about ten times as far from the sun as they actually are (i.e. to somewhere near Saturn's orbit), the strength of the Sun and moon's gravitational pull on the Earth would be equal. [If someone else wants to attempt that calculation, it would be interesting to see if we agree, but the principle involved here is right regardless of the actual distance required.] Imagine the Earth between the moon and the sun there. The pull is equal (on average) from each, so the Earth must be moving in a straight line at that moment. The strength of the moon's gravitational pull on the near and far sides of the Earth varies more than the Sun's because the moon and Earth are close together with the moon's gravity falling off in strength quickly over distance due to the proximity of the source. The sun also drives much smaller tidal forces because the strength of its gravity falls less slowly over distance by the time it reaches the Earth, and as we're putting the Earth out near Saturn, that fall off in strength over distance will be ten times smaller still, meaning that it won't add much to the tides at all.

Rmolnav's explanation says that the the bump on the far side from the moon is caused by centrifugal force, but in this case it clearly can't be because the Earth's moving in a straight line. The bump is still there though in full (and with a tiny bit extra added by the sun). So in this situation, what does he have to do to account for that bulge? He has no option other than to switch to the straight-line differential gravity explanation. He can justify this though on the basis that the sun and moon's pull cancel out all the centripetal force and leave the gravitational force simply being counted as one thing instead of being (artificially) divided. Take away the sun, and then he says the same tidal force (minus the tiny bit that was added by the sun) is now suddenly caused by centrifugal force instead.

The Earth obviously isn't as far out as Saturn, so let's now look at it where it actually is. We have situations once a month here where the Earth is between the moon and the sun, and it isn't following a straight line, but a curve that curves the opposite way from the one rmolnav normally thinks of it doing. If he wants to have centrifugal force involved here, it has to be throwing a "bulge" towards the moon rather than away from it, and yet it isn't the sun's gravity that's driving most of that "bulge" - it's the moon's stronger pull there that's responsible - the only component of the tidal force there that could be attributed to centrifugal force is the part caused there by the sun. Think about that carefully - this is a more severe curve that the Earth's following than the one in a case where the sun doesn't exist, and it curves the opposite way too. That should mean much greater centrifugal force, but at the same time, this is balanced out by the sun's stronger gravitational pull on the material that's trying to lift, and by the fact that the strength of the sun's pull falls off less over distance at this range. The end result is that the part of the "bulge" here (on the side of the Earth nearest the moon) which is attributed to centrifugal force would indeed only be the part caused by the sun, while the larger component of the "bulge" is attributed to the moon's differential gravity even though the moon is on the outside of the curve of the path that the Earth is following, and that component of the "bulge" is not driven by centrifugal force in any way. This is quite amusing - we have two "bulges" in the same place (and adding together into a bigger "bulge"), one caused by centrifugal force and the other not caused by centrifugal force at all. Here it becomes manifest just how ludicrous rmolnav's explanation actually is.

(None of the bulge on the side nearest the sun is being driven by centrifugal force because it's at the inside of the curve, and most of it isn't being lifted by the sun's differential gravity, so what is lifting it? It is actually being pulled up by the sun's gravity, but it's primarily the moon's differential gravity holding it back less strongly there than the rest of the Earth that allows it to try to lift.)

The numbers do actually match up if you crunch them all, but this only happens as a necessary accident. The reality is that this business of dividing gravity up into centripetal and non-centripetal force is just a silly abstraction which necessarily fits the numbers, but has no mechanistic role in what's happening whatsoever. The real mechanism is simply straight-line differential gravity in all cases. The real mechanism is the simple addition of all the gravitational forces acting on each particle - you sum the vectors and end up with one single result per particle which dictates how it tries to accelerate. The physics of this is ridiculously simple, but when people learn too much through abstractions, they can be misled by their education and end up tripping up over issues of this kind. Even good scientists can make such errors, but there is no shame attached to them from doing so - to err is human.

If we consider a case where the Earth is at the 90 degree corner of a right-angled triangle with the moon and sun at the other corners. What's going on in this case? There are four "bulges", but the smaller ones appear as "dips". Well, we know these bulges aren't real - they merely drive the generation of real tidal bulges. We actually have four sides with tidal forces acting to lift material there. If we attribute one of them to centrifugal force, it has to be the one on the far side of the Earth from the sun. The other three, including the pair caused by the moon, are driven by differential gravity and not by centrifugal force. The more you look at this, the more rmolnav's explanation falls apart. The only question remaining to be answered now is whether he recognises that.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 20/09/2018 12:00:23
Sorry, but rubbish and more rubbish ... !!
Another day, if I manage to keep patient, I´ll comment on last posts. But now, something about another of a few days ago.
On #211 I said:
"...IN A QUITE SIMILAR WAY, we could reason as follows:
If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)
Therefore, if with the differential moon´s gravity maintained, outer positive bulge doesn´t exist, the moon differential gravity cannot be the cause of outer bulge !!

Same way you´ve “reasoned” lot of times … Either both are right, or both are wrong …
Any guess ??"
Apart from what you replied then, a few days ago you said:
 
Let's give the Earth two moons which always stay on opposite sides. The Earth isn't moved by them at all now, but we still have tidal forces, nearly twice as strong as we have with just the one moon. How does rmolnav's explanation work here? The gravitational pull from each at the Earth's centre is called centripetal force, and the two lots cancel out. Each bulge is driven partly by centrifugal force from the Earth's "movement" acting on the material furthest from the moon that's applying the centripetal force to the centre that drives this "movement", and partly by the greater gravitational pull applied there by the other moon. Take away the sun and the Earth's orbit so that all we have is the Earth and its pair of moons, and what do we have this time? We have a stationary Earth with material being flung out at opposite sides by, well, I suppose it should be called centrifugal farce.
What a lot of absurdities ... !!
I´ve said here time and again that "centripetal force" is neither an "essentially" new force, nor just a  label we can give any force "at will" !! It is just a FUNCTION, which many essentially different types of forces can EXERT, if they are CAUSING the bending of the trajectory of a moving object ...
Your "second" moon, together with real one, are getting what I said on #211: the earth does´t move at all, without the necessity of any "megaman" ...
And what happens is what I said then, BUT TWO OPPOSITE-WAY TIMES:
"... outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)"
There is no "prevailing" centrifugal force whatsoever, because the earth doesn´t move ...
And neither are there bulges caused by the differential pull from either moon ...
What happens is that on each earth side the increase of sea level due to the pull from closer moon, is bigger than the DECREASE of sea level caused by the farther moon !!
And you even dare say:
The physics of this is ridiculously simple, but when people learn too much through abstractions, they can be misled by their education and end up tripping up over issues of this kind. Even good scientists can make such errors, but there is no shame attached to them from doing so - to err is human.
Again, what a record-braking case Neil Comins missed for his course:
Sources of Misconceptions in Astronomy, by Neil F. Comins
University of Maine

"...In pursuing them and their origins, I use the definition of misconception as a resilient concept held to be correct, but that is actually at variance with accepted scientific knowledge. Resilient here means that the concept is incorporated in a person's conceptual framework and therefore resistant to change. Because they are woven into conceptual fabrics, many misconceptions help prevent further correct understanding of related topics
Title: Re: Why do we have two high tides a day?
Post by: Halc on 20/09/2018 13:21:04
      BY THE WAY, on a link posted by PmbPhy some days ago it is said:
...
… for a particle at the earth´s center, the CENTRIFUGAL FORCE and the moon´s gravitational attraction are equal and opposite.
This illustrates the danger of considering centrifugal force to be a force.  If the two forces where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not.  It is accelerating due to the one and only force acting upon it here: gravity.

Yes, the tides can sometimes be explained in these term, but by eliminating the curved trajectory of Earth by putting it further out, or by putting it between two moons as David Cooper has done, the centrifugal explanation falls apart.  There is still gravity, but no centrifugal force, and yet the tide remains.

Let's do another version which makes things even more stark. Let's give the Earth two moons which always stay on opposite sides.
Or Earth centered between equal binary stars, unstable, but it gets rid of any possible acceleration altogether.  Tides totally without anything that can be considered centrifugal.

If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)
Therefore, if with the differential moon´s gravity maintained, outer positive bulge doesn´t exist, the moon differential gravity cannot be the cause of outer bulge !!
David’s examples very much did remove that rectilinear accelerated movement, and no compression of the atmosphere on the far side of Earth results.

Quote from: rmolnav
Your "second" moon, together with real one, are getting what I said on #211: the earth does´t move at all, without the necessity of any "megaman" ...
And what happens is what I said then, BUT TWO OPPOSITE-WAY TIMES:
"... outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)"
There is no compressive effect on the far side due to either moon.  The attraction of the far side is less than the attraction on the Earth below it.  That equates to tensile stress, and these stresses from each moon add to a double magnitude tide, not cancel out to something lower than what we see with one moon.

Quote
What happens is that on each earth side the increase of sea level due to the pull from closer moon, is bigger than the DECREASE of sea level caused by the farther moon !!
No, they’re both increases.  This seems to predict a partially cancellation of tides.

I see several references to post 211:
Quote
If, somehow, we "remove" the revolving, both earth and moon will accelerate towards each other … Though only a transient movement, bulges would remain (and increase) until the collision.
You seem to be envisioning removal of orbital speed, not the revolving of Earth and the moon.  If you remove the revolving, you get two tides per month, and we would get to see what is currently the far side of the moon.
Quote
OK. Therefore, revolving was not causing them[/i] (YOUR deduction).
IN A QUITE SIMILAR WAY, we could reason as follows:
If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)
On this I agree.  If a mysterious force grabbed only the rocky part of Earth and pushed it, but not its water away from the moon, then yes, the water would naturally slosh to the side away from the unnatural acceleration caused by this selective force.  But no such megaman exists.  The sun does it, but it does not decline to act on the water as well.  The megaman example does not reflect what might actually keep Earth from accelerating in the various examples above.


Quote
Therefore, if with the differential moon´s gravity maintained, outer positive bulge doesn´t exist, the moon differential gravity cannot be the cause of outer bulge !!
I’m obviously in the camp that says the differential gravity very much does cause the far bulge.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 20/09/2018 14:26:11
Quote from: rmolnav on 18/09/2018 06:48:27]
    "...In pursuing them and their origins, I use the definition of misconception as a resilient concept held to be correct, but that is actually at variance with accepted scientific knowledge. Resilient here means that the concept is incorporated in a person's conceptual framework and therefore resistant to change. Because they are woven into conceptual fabrics, many misconceptions help prevent further correct understanding of related topics
Again, no need to accuse others to be resistant, just to consider that resistance to change is mass, that it is universal, and that it applies to anything that exists, even to our ideas. On these nice words, I think I'll leave the discussion. Sometimes it is no use to insist. Time also has its word to tell.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 20/09/2018 21:10:17
The biggest winner in any argument of this kind is the one who is most wrong, realises it, and shifts to a new position in order to be right. Such a winner may emerge from here some day, but there are still no signs of movement - instead we see the usual human response which is to dig a deeper trench.

I've learned a lot from this thread though by being pushed into thinking about things I wouldn't have explored otherwise, so I'm grateful to rmolnav for that - he has been useful. I'm sure he hasn't wasted any of his time here though because his superior intellect will doubtless have converted everything into a gain: as Neil Comins might say, a fool finds it very hard to learn anything from a genius, but a genius can always find something to learn from a fool.

I want to share another thing that I've learned from this. I'd never stopped to think carefully before about how sun's pull damps down the tides when the sun and moon are 90 degrees apart in the sky - I've always imagined the sun to be subtracting from the tidal forces in some way, but no: they add tidal force, causing material to try to lift out just as much as they do when the sun and moon pull from the same direction or from 180 degrees apart.

Imagine the Earth with two moons 90 degrees apart, both pulling with the same strength from the same distance away. Each moon is trying to generate two "bulges", but they are not bulges. When you combine these four bulges, you get no tides, but there are strong tidal forces trying to lift material up all round the Earth. Imagine the Earth in the middle of the screen with one moon out to the left and the other down below the Earth. The combined acceleration will move the Earth down at 45 degrees towards the bottom left. The point on the Earth close to the bottom moon is called BN (bottom near). Opposite that (on the other side of the Earth) we have the point BF (bottom far). [Maybe I should have gone for top, but no time to change it all - just press on.] The point on the Earth nearest the moon out to the side is SN (side near) and the point opposite it is SF (side far).

At BN, the material is being pulled down by the bottom moon more strongly than the Earth as a whole, but it's being pulled towards the side moon with the same strength as the Earth as a whole.

At BF, the material is being pulled down by the bottom moon less strongly than the Earth as a whole, but it's being pulled towards the side moon with the same strength as the Earth as a whole.

In both of these cases, we have tidal forces trying to lift material, and the equivalent happens at points SN and SF too, so are there four "bulges" or are there none?

Let's look at a point on the Earth's surface half way round from BN to SN, this being the point that leads the Earth towards the bottom left. The force from both moons on this point is greater than on the Earth as a whole, so again the material will lift, but this point is further away from the bottom moon than BN, and further away from the side moon than SN, so it isn't going to get double lift, and the two forces are being applied at 90 degrees, so they add as vectors, reducing the magnitude of the lift further. How much lift is it? Cos 45 = 0.707, so that's roughly the strength of each component of the lift (compared with the strength at BN or SN, and as they're acting at 90 degrees to each other, we have 0.707^2 + 0.707^2 = 1, and the square root's the same: we have the same amount of lifting force there as we do at BN and SN.

We should do the same thing at two more points to check that the tides cancel out completely (ignoring the slightly smaller tidal forces on the far side due to the strength of gravity falling off less over distance on the way there from the centre because of greater distance from the source). At the point on the surface half way between BF and SF, we have a lesser pull there with the same numbers working through to a value of 1, so it's the same lift there. One more point will complete the picture: at the point half way round from SF to BN, we have a stronger pull than average downwards, and a lesser pull than average to the left, and yes - it's the same numbers again and the same lift. So, we have no tides, but we have tidal forces all round the Earth lifting material regardless.

It the case of the real Earth, moon and sun, you can imagine this working the same way, but with the sun's input being less than the moon's (it's a greater force from the sun, but with less change over distance). When they are 90 degrees apart, all the material round the equator is being lifted a microscopic amount; more strongly in line with the moon, but still lifting all the way round. When the moon and sun are pulling from the same or opposite sides, they're then lifting material more strongly if it's in line with them, but the points on the equator which see the moon and sun down at the horizon are not being lifted at all, and it is only at times like this that the tidal force at such locations goes down to zero.

(And I apologise to rmolnav for posting more absolute cobblers here, but I kinda like correct "nonsense".)
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 21/09/2018 15:06:09
#309 Halc
Thank you and welcome. You seem to argue “reasonably”, and I do hope we can discuss fruitfully …
You say several things I can´t agree with.
But today, I´ll only tackle what I consider ESSENTIAL for any possible understanding: NOT to have completely opposite ideas regarding INERTIA, that can manifest itself in different ways, one of them as “centrifugal force", real force (or, at least, causing quite real effects ...)
This illustrates the danger of considering centrifugal force to be a force.  If the two forces where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not.  It is accelerating due to the one and only force acting upon it here: gravity.
When I first entered here (more than three years ago !!), I used to approach this issue considering earth´s CM revolving “uses” 100% moon´s gravity pull there to cause the required centripetal acceleration for that movement. No “spare” pull (which, if existing, could produce other effects) there …
But on closer to moon hemisphere there is an excess of moon´s pull. Any given material particle there can´t “use” that excess of pull to get a bigger acceleration, because the rest of the planet “forces” them to move together, with same acceleration … That imbalance generates internal stresses in both senses (between contiguous parts opposite forces are exerted on each other, due to Newton´s 3rd Motion Law).
The opposite occurs on farther hemisphere …
THAT FIELD of interaction forces, acting either “moonwards”, or outwards (arguably “centrifugally”), would be what actually causes tides !!
But I discussed the issue (directly through emails) with NOAA scientist author (at least one of them) of:
 https://tidesandcurrents.noaa.gov/restles3.html
and among other things he told me:
"The publication you are referring to is "Our Restless Tides", a 10-page pamphlet developed in the 1950's to provide a basic description of the forces which create the tides.  It's intended audience were the grade school children and adults of that time.  It used terminology of science and forces which were common in the 1950s.  Such as centrifugal force.  Centrifugal force was always an "imaginary force" (not a real / measurable force).  But that type of description made the concepts easier to understand and explain.  That  description and use of centrifugal force continued to be common practice until the 1970-80's.  At that point, the terminology shifted and the textbooks used in grade schools were changed to use a more modern terminology and description of this "effect" being a result of inertia rather than an "imaginary force”.
Initially I didn´t agree, and insisted: only previously mentioned outward internal forces could be considered “centrifugal”, to fully satisfy Newton´s 2nd Motion Law
Later, after father ruminating on the issue, certainly helped by the effort to refute many people´s propositions (here and on other sites), I got pretty sure NOAA mentioned scientists (and the authors of “French Tides” site referred to on # 276  ) are quite correct in their approach (though certainly many current books, even well known dictionaries, keep the “old” definition as an “imaginary” or “fictitious force" ...)
In any case, both ways are equivalent, because deducting required centripetal force vector from moon´s pull on a given earth particle (the force the particle “interchanges” with the rest of the planet), gives same result than adding the “arguably” called centrifugal force vector to moon´s pull …
But it is necessary to tackle the “apparently” insolvable fact that earth´s CM is actually accelerating …
It is rather tough to explain through posts here. I did have in mind a couple of ways. With your post, yet another. In chronological order:
A) One in relation to the erroneously (to me) used concept of “non-ínertial reference system" …
B) Another way (though kind of equivalent) in relation to the scenarios used by D.C., imagining earth not revolving …
C) Last one, in relation to what you say: “… If the two forces (moon´s pull and centrifugal force) where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not".
ANOTHER DAY I´ll try and do it.
Anyhow I´ll copy and paste something I´ve already said a couple of times (way A), needing to elaborate on).
PmbPhy had said:
"In particular, the centrifugal force is an inertial force which is observed in rotating (non-inertial) frames”.
I replied:
"I´m afraid that is rather confusing ... What do you mean with "... "which is observed in …"?
Does it mean it is a real force "observed" by us? Or, in that "particular" case, is it rather a kind of tricky tool WE apply to bring a "fictitious" situation (an accelerating frame of reference) back to reality (with all REAL effects which have disappeared in that frame of reference)?"
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 21/09/2018 17:14:37
Yes, the tides can sometimes be explained in these term, but by eliminating the curved trajectory of Earth by putting it further out, or by putting it between two moons as David Cooper has done, the centrifugal explanation falls apart.  There is still gravity, but no centrifugal force, and yet the tide remains.
Come on !!. You yourself are saying "by eliminating the curved trajectory" ... !!
BASIC LOGICS: The fact that there can be imagined (or even real) scenarios without the intervention of any centrifugal force (logically, without any curved trajectory) where the two bulges do build, DOESN´T MEAN in other real scenarios (with curved trajectories) CENTRIFUGAL FORCE can´t be an acting REAL force …
There are different ways INERTIA can manifest itself !! 

   
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 21/09/2018 18:46:19
BASIC LOGICS: The fact that there can be imagined (or even real) scenarios without the intervention of any centrifugal force (logically, without any curved trajectory) where the two bulges do build, DOESN´T MEAN in other real scenarios (with curved trajectories) CENTRIFUGAL FORCE can´t be an acting REAL force …

The actual case gives us situations where the curve that the Earth is following curves the opposite way to the path you claim for it. The Earth would fall 85 miles towards the moon in a day if the perpendicular component of its movement was removed, but it would also fall 14,000 miles towards the sun in a day if the perpendicular movement of its orbit round the sun was removed. Now think about the two curves added together (with the perpendicular movement not removed) and how the 85 miles one way is out-gunned by the 14,000 the other way. In this case, any centrifugal force involved would need to be trying to lift material towards the moon and not towards the sun, so the majority of the tidal force making material stick out towards the sun must come from the moon's differential gravity operating on that far side of the Earth (far side from the moon) and not from centrifugal force, while the minority of the tidal force acting to lift material there is caused by the sun's differential gravity. On the near side of the Earth (to the moon), the minority of the tidal force trying to lift material there is caused by the sun's differential gravity, and this is the only part that you could claim is caused by centrifugal force. The majority of the tidal force trying to lift material there is caused by the moon's differential gravity there, and you cannot justify calling that centrifugal force as you only have enough mathematical justification to cover the minority part of the lift there. Then when you see that the minority part of it there (the sun's contribution) is supposedly driven by centrifugal force while the majority part of it (the moon's contribution) cannot be driven by centrifugal force in any way, the mistake in your analysis becomes clear - you are making artificial distinctions. This is a real case that happens once every month.

Another real case which destroys your explanation happens twice a month, and that's the one where the sun and moon are nearly 90 degrees apart in the sky with the sun dictating most of the curved path that the Earth is following and with the moon's pull acting at 90 degrees to the direction of the only component of force that could be labelled as centrifugal. The two tidal inputs from the moon in this case can only be explained through differential gravity with no possible role for centrifugal force to aid them. These are real cases where your explanation is shown to be plain wrong, but the differential gravity explanation works in 100% of cases. An explanation which works some of the time but is wrong the rest of the time is simply an incorrect explanation. I've shown you specific cases where it's clearly 100% wrong, but there are lots of other cases with different angles where it's clearly partly wrong because the angle in which the part of the force that could be labelled as centrifugal force is not aligned with the moon. The only case where it is aligned correctly is where the moon is directly between the Earth and the sun, so there is only one moment per month when your numbers stack up 100%. The rest of the time your explanation doesn't fit the facts - it only looked as if it might fit in cases where the sun was ignored, so it turns out that you've been dealing with imaginary scenarios all this time yourself!
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 21/09/2018 21:23:18
There is no compressive effect on the far side due to either moon.  The attraction of the far side is less than the attraction on the Earth below it.  That equates to tensile stress, and these stresses from each moon add to a double magnitude tide, not cancel out to something lower than what we see with one moon.
Sorry, but that is erroneous …
If we want not to mix causes and effects, we must analyze the dynamics of the two-moon case very, very carefully.
As a triple object and symetric system, the two moon can be rotating around the earth CM, logically at same distance and with same angular speed, and the earth stand still (for the sake of simplification, I suppose the system without any orbital movement around the sun).
There are “infinite” pulls from each moon on earth particles …
But the total force exerted on the earth is null …
The total force exerted by each moon is equal, but opposite, to the one exerted by the other …
The distribution of each moon´s pull is logically the closer, the bigger (inversely proportional to the square of the distance), logically ALWAYS towards the respective moon.
Being earth not moving at all, there are no inertial effects. If now we consider SEPARATELY the effects of each moon´s pull, whatever the reason why the earth doesn´t move, each moon pulls ocean water toward itself, same way you admitted yesterday:
Quote
OK. Therefore, revolving was not causing them[/i] (YOUR deduction).
IN A QUITE SIMILAR WAY, we could reason as follows:
If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)
On this I agree.  If a mysterious force grabbed only the rocky part of Earth and pushed it, but not its water away from the moon, then yes, the water would naturally slosh to the side away from the unnatural acceleration caused by this selective force.  But no such megaman exists.  The sun does it, but it does not decline to act on the water as well.  The megaman example does not reflect what might actually keep Earth from accelerating in the various examples above.
The specular feature of the whole system makes what in my analogy the "megaman" did: not to let the earth move.
Further-to-each-moon hemisphere is pulled by that moon towards earth´s middle section: solid earth would be compressed (again: considering ONLY each moon effect separately), and water would move towards considered moon, causing there a sea level decrease …
But closer-to-each-moon hemisphere is ALSO pulled by the moon at that side, more strongly than previously mentioned opposite pulls, causing bigger similar effects but in the opposite sense.
Adding up those opposite effects (due to forces actually acting on each hemisphere particle), we have the two bulges …
BUT THEY HAVE NOTHING TO DO with the so called “differential gravity” exerted separately by either moon
In all D.C. imaginary cases, in order to have a second proper bulge we must let the earth accelerate … because INERTIAL effects are absolutely necessary.
So called "differential gravity”, without those inertial effects, can´t cause the second bulge
By the way, in the two-moon case the size of the bulges doesn´t increase to double whatsoever. …
The second moon pull kind of substitutes real case inertial effects (centrifugal forces).
To double the size of the bulges it´d be necessary to have the earth revolving around each moon-earth pair center of mass, and that is not possible simultaneously !!





Title: Re: Why do we have two high tides a day?
Post by: Halc on 21/09/2018 21:27:26
#309 Halc
Thank you and welcome.
And thank you for the welcome. I’ll be glad when I become human and I get more than a postage-stamp size window in which to compose my responses.  As it is, I am forced to do it in an offline wordprocessor.

Quote
But today, I´ll only tackle what I consider ESSENTIAL for any possible understanding: NOT to have completely opposite ideas regarding INERTIA, that can manifest itself in different ways, one of them as “centrifugal force", real force (or, at least, causing quite real effects ...)
I suppose it is a matter of interpretation if it is real or not.  I kind of go with force being the F=ma in Newton’s 2nd law, whereas inertial effects (centrifugal being among them) are the ‘m’ in that equation, not the F.  It isn’t causing any real acceleration of anything in the direction of the supposed force.

Quote
When I first entered here (more than three years ago !!), I used to approach this issue considering earth´s CM revolving “uses” 100% moon´s gravity pull there to cause the required centripetal acceleration for that movement. No “spare” pull (which, if existing, could produce other effects) there …
But on closer to moon hemisphere there is an excess of moon´s pull. Any given material particle there can´t “use” that excess of pull to get a bigger acceleration, because the rest of the planet “forces” them to move together, with same acceleration … That imbalance generates internal stresses in both senses (between contiguous parts opposite forces are exerted on each other, due to Newton´s 3rd Motion Law).
You mean the CM can be used to track Earth’s motion as if it were a point mass.  Indeed, it doesn’t work when the discussion is about more local effects like tidal forces.  It also doesn’t work in general.  I can have two pairs of all identical mass objects with identical difference in separation by CM, but one is a pair of spheres and the other is a pair of long barbells arranged perpendicularly.  The spheres will have more gravitational attraction than the crossed barbells despite the same mass and distanace between their respective centers of mass, because most of the mass is nowhere near that center like it is with the spheres.  So point taken.  You can’t treat objects as point masses when the difference matters.
You describe tidal forces nicely there.  Yes, they put stresses on things, enough to tear them apart if they get within the Roche limit.

Quote
Initially I didn´t agree, and insisted: only previously mentioned outward internal forces could be considered “centrifugal”, to fully satisfy Newton´s 2nd Motion Law
And since nothing is accelerating outward, this would be consistent with an outward centrifugal force not being real.  Sorry, just thinking out loud when you haven’t got to the point yet.


Quote
In any case, both ways are equivalent, because deducting required centripetal force vector from moon´s pull on a given earth particle (the force the particle “interchanges” with the rest of the planet), gives same result than adding the “arguably” called centrifugal force vector to moon´s pull …
Still agree…  As I said, I suspect the difference is interpretational.


Quote
But it is necessary to tackle the “apparently” insolvable fact that earth´s CM is actually accelerating …
It is rather tough to explain through posts here. I did have in mind a couple of ways. With your post, yet another. In chronological order:
A) One in relation to the erroneously (to me) used concept of “non-ínertial reference system" …
That seems unnecessary if you get rid of the sun and everything else and just have Earth and moon in the inertial frame of their mutual CM.  Keeping the sun and its frame lets you see that the ‘centrifugal’ tide is actually accelerating away from the moon at full moon.


Quote
B) Another way (though kind of equivalent) in relation to the scenarios used by D.C., imagining earth not revolving …
???  So they wouldn’t be called tides anymore because the distortion would be permanent and not noticed.  This is assuming Earth rotating in sync with the month, not stopped altogether, which would have the tides occurring twice a month.  The day is lengthening, but it will never be longer than the lunar month.

Quote
C) Last one, in relation to what you say: “… If the two forces (moon´s pull and centrifugal force) where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not".
ANOTHER DAY I´ll try and do it.
Sorry, but I never figured out from these three ways what this insolvable fact is concerning earth’s CM accelerating.  My comment quoted in C there was just another way of pointing out why I don’t consider inertia to be a force.


Quote
Anyhow I´ll copy and paste something I´ve already said a couple of times (way A), needing to elaborate on).
PmbPhy had said:
"In particular, the centrifugal force is an inertial force which is observed in rotating (non-inertial) frames”.
I replied:
"I´m afraid that is rather confusing ... What do you mean with "... "which is observed in …"?
Does it mean it is a real force "observed" by us? Or, in that "particular" case, is it rather a kind of tricky tool WE apply to bring a "fictitious" situation (an accelerating frame of reference) back to reality (with all REAL effects which have disappeared in that frame of reference)?"
Centrifugal force seems to be the force that holds the water in the pail being swung around, similar to how gravity holds water in the same pail hanging from my arm.  But both are wrong.  It is the pull of my arm that holds the water in the pail in both cases.  Without that, the water would drift out of the pail, even in the gravity field.  Water isn’t held with pails on the space station, despite no lack of gravity up there.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 22/09/2018 08:39:39
#316 Halc
When I read carefully your post, I´ll try and refute what found necessary ... But just one thing now.
Months ago I discussed the issue with a physicist on the email. I even posted here part of the "story".
He is in line of your and D.C.´s stand, as far as moon (and sun) related tides are concerned.
But, on the site linked below, he explained why we weight less on the equator, and why the so called "equatorial bulge" (the circular and permanent one due to daily earth spinning) happens ...
I commented on that here (#141... for some reason I started referring to tidal effects on the moon, due to similar reasons to the ones on our planet):
"Moon is rotating around the barycenter, and centripetal (mainly earth pull) and centrifugal forces do happen at different parts of it. And also own moon gravitational forces.
At closer moon half, earth attraction prevails. At further moon half, earth pull is insufficient to supply the centripetal force necessary for the rotation ... Own moon gravity has to supply the deficit, and all those further parts kind of lighten, due to same reasons explained on:
https://www.lockhaven.edu/~dsimanek/scenario/centrip.htm
"The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v2/R (the centripetal force). Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation"
Please kindly analyze his explanation, and say if you agree or doesn´t ...
He certainly doesn´t use the term "centrifugal force", but the root of his argument is on the REAL effects of INERTIA ...
If that happens as an inertial reaction of the water (and the man´s mass) when "forced" to follow the equatorial curved path, why wouldn´t happen something similar when the water has to revolve, following during some 28 days a circumference (considering only earth-moon "dance") of a radius equal to the distance between earth CM and earth-moon barycenter ??
 
Title: Re: Why do we have two high tides a day?
Post by: Halc on 22/09/2018 13:42:37
#316 Halc
When I read carefully your post, I´ll try and refute what found necessary ...
You should perhaps quote the parts of mine that you are refuting, because I pretty much agree with your whole post except the last bit.
I looked up centrifugal force and realized it is not a fictional force. In an inertial frame, there is no such force.  But it is a term legitimately used in rotating frames, where none of Newton's 3 laws apply.  An object at rest tends to accelerate, and the force that causes this acceleration is centrifugal.  No reaction stands opposite this action.  Moving objects do not follow straight lines.  Not sure if all such accelerated motion can be considered centrifugal since the acceleration is more typically inward instead of outward, and 'centrifugal' is typically used only for the tendency towards outward.  If not, I don't know the name of the reactionless force that appears to accelerate objects inward in such rotating frames.

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"This is the reason for the equatorial bulge of the Earth due to its own axial rotation[/b]"
Please kindly analyze his explanation, and say if you agree or doesn´t ...
This part I is fine.  He doesn't really discuss tides, but it is mentioned once in the 'stress' section.  Yes, stress causes tides.  Such stresses can kill you if it is strong enough.

I took issue with his blocky diagram above in the 'stress' section, which shows two stacked blocks in a gravitational field and a weight force preventing the whole setup from accelerating from the gravity.  It labels the spring as tension, despite the arrows depicting the spring pushing on the two blocks, which would compress the spring.  The arrows are correct, but text incorrectly labels this as tension.  If the arrows represent tension T, they need to point the other way, and the tension would be negative.  So while I understand the concept of stress and strain (I took civil engineering courses), the description of it there needs a bit of review.

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He certainly doesn´t use the term "centrifugal force", but the root of his argument is on the REAL effects of INERTIA ...
Yes.  In an inertial frame, the equator bulge occurs from a combination of at least 3 effects: You weigh less (and so does the water) there because you're further from the typical particles that make up the planet, and because part of the gravitational force is used for acceleration and need not be resisted by the force of weight from below.  The third part is due to the fact that the typical particle is concentrated more below and less to the sides, and this component makes you weigh a bit more, but nowhere near enough to overcome the negative effects of the other two.
The second component I listed is the inertial one.

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If that happens as an inertial reaction of the water (and the man´s mass) when "forced" to follow the equatorial curved path, why wouldn´t happen something similar when the water has to revolve, following during some 28 days a circumference (considering only earth-moon "dance") of a radius equal to the distance between earth CM and earth-moon barycenter ??
Inertial effects are indeed involved, but not directly like that.  The far tide has far more acceleration about the barycenter (which represents the inertial frame of a two-body model), but both tides are the same magnitude, which they would be even if the moon was further away putting the barycenter above the surface.  The direct inertial explanation of the tide would push the water down, not upward, since it is now rotating on the same side of the barycenter.  Inertial of the water on the sides would tend to carry it to the far side.

So no, I cannot agree with this wording.  In a uniform gravitational field (say at a sufficient distance from the galactic black hole to have a gravitational force on earth similar to that of our moon (2e20N), you have the same inertial forces at work, but undetectable tides.  It is the non-uniformity of the gravitational field, not inertia, that causes the stress, and  tides which are strain resulting from that stress.
Title: Re: Why do we have two high tides a day?
Post by: Bored chemist on 22/09/2018 13:58:07
Hello
What is tides how it work
Why not Google it and find out?
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 22/09/2018 14:34:01
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But today, I´ll only tackle what I consider ESSENTIAL for any possible understanding: NOT to have completely opposite ideas regarding INERTIA, that can manifest itself in different ways, one of them as “centrifugal force", real force (or, at least, causing quite real effects ...)
I suppose it is a matter of interpretation if it is real or not.  I kind of go with force being the F=ma in Newton’s 2nd law, whereas inertial effects (centrifugal being among them) are the ‘m’ in that equation, not the F.  It isn’t causing any real acceleration of anything in the direction of the supposed force.
Come on !! Many forces can be real without "causing any real acceleration of anything in the direction of the supposed force" ... if the affected object is not free to move ! In that case what they cause is deformations, of the object and/or contiguous objects (or parts of an object) !!
That´s the essence of internal stresses and strains, and even the actual manifestation of tides on solid parts of earth. And on our oceans they can produce water pressure changes, and subsequently the transportation of water towards where the bulges build !!
You mean the CM can be used to track Earth’s motion as if it were a point mass.  Indeed, it doesn’t work when the discussion is about more local effects like tidal forces.  It also doesn’t work in general.  I can have two pairs of all identical mass objects with identical difference in separation by CM, but one is a pair of spheres and the other is a pair of long barbells arranged perpendicularly.  The spheres will have more gravitational attraction than the crossed barbells despite the same mass and distanace between their respective centers of mass, because most of the mass is nowhere near that center like it is with the spheres.  So point taken (??).  You can’t treat objects as point masses when the difference matters.
Sorry, but that is utterly absurd ...
A not uniforme distribution of masses, even two o more separated objects, can be treated (for certain purposes) as a whole, with a common center of mass, useful for its dynamic analysis ...
E.g., the very stuff we are talking about: earth-moon system.

It has a common CM (the barycenter), which is what follows the elliptical orbit around the sun. It´s quite reasonable to analyze the complex dynamic of moon-related tide formation in different stages: mentioned orbiting of common CM, earth-moon revolving/rotation around the barycenter, gravitational and inertial forces on the earth as a whole (where earth CM has an important roll), and then the detail of forces acting on each area of our planet (internal force changes due to earth-moon dynamic included) ...
Those last ones are mainly in the moon´s CM-barycenter-earth´s CM direction, both "moonwards" and outwards (or "moonfugal" or "centrifugal"...)
And they are the direct cause of tides ... certainly generated by moon varying pull, but also the inertial effects on all and each particle of the earth.
As I´ve said many times, earth particles only can "feel" gravity at their respective location, and pushes and/or pulls from contiguous matter. And they always have their tendency not to change their velocity vector (inertia), what causes inertial effects ...
What they can´t "feel" is either the so called "differential gravity" or the gravity on locations many km away, let alone they can "subtract" different gravitational pulls to react to that difference !!
Title: Re: Why do we have two high tides a day?
Post by: Halc on 22/09/2018 15:16:10
That´s the essence of internal stresses and strains, and even the actual manifestation of tides on solid parts of earth. And on our oceans they can produce water pressure changes, and subsequently the transportation of water towards where the bulges build !!
Agree with this.  Its about stress and strain.

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Quote from: Halc
You mean the CM can be used to track Earth’s motion as if it were a point mass.  Indeed, it doesn’t work when the discussion is about more local effects like tidal forces.  It also doesn’t work in general.  I can have two pairs of all identical mass objects with identical difference in separation by CM, but one is a pair of spheres and the other is a pair of long barbells arranged perpendicularly.  The spheres will have more gravitational attraction than the crossed barbells despite the same mass and distanace between their respective centers of mass, because most of the mass is nowhere near that center like it is with the spheres.  So point taken (??).  You can’t treat objects as point masses when the difference matters.
Sorry, but that is utterly absurd ...
A not uniforme distribution of masses, even two o more separated objects, can be treated (for certain purposes) as a whole, with a common center of mass, useful for its dynamic analysis ...
E.g., the very stuff we are talking about: earth-moon system.
Earth and moon are hardly shaped like barbells.  If all objects can be treated as point masses, then I should weigh more and more as I dig myself into a hole.

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It has a common CM (the barycenter), which is what follows the elliptical orbit around the sun.
Well, to be fully anal about it, and in an ideal situation with no other planets, we would orbit the CM of the solar system, not the sun itself.  The two points are almost the same thing if we discard the other planets.
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It´s quite reasonable to analyze the complex dynamic of moon-related tide formation in different stages: mentioned orbiting of common CM, earth-moon revolving/rotation around the barycenter, gravitational and inertial forces on the earth as a whole (where earth CM has an important roll), and then the detail of forces acting on each area of our planet (internal force changes due to earth-moon dynamic included) ...
Those last ones are mainly in the moon´s CM-barycenter-earth´s CM direction, both "moonwards" and outwards (or "moonfugal" or "centrifugal"...)
Complicated but reasonable to analyze the dynamics this way, sure.  Some of them assume a rotating frame of reference, and it is reasonable to analyze the dynamics in that frame.

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And they are the direct cause of tides ... certainly generated by moon varying pull, but also the inertial effects on all and each particle of the earth.
The tides are directly proportional to the varying pull, and not directly proportional to the inertial effects.  So the latter is a related cause, but not a direct cause.  As I said, it is reasonable to analyze it in such terms, but it is more complicated due to this non-proportional relationship.  The inertial effects from the sun are far greater (about 175x) than those from the moon, yet produces a significantly smaller tidal effect.

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As I´ve said many times, earth particles only can "feel" gravity at their respective location, and pushes and/or pulls from contiguous matter. And they always have their tendency not to change their velocity vector (inertia), what causes inertial effects ...
Yep
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What they can´t "feel" is either the so called "differential gravity" or the gravity on locations many km away, let alone they can "subtract" different gravitational pulls to react to that difference !![/b]
Right.  No individual particle can detect a gravity gradient.  It requires pairs of separated particles.  What the particles might 'feel' is the changing forces put on them by their immediate neighbors, which in turn is effectively them 'feeling' the differential gravity on locations many km away.  This is similar to the way you feel a train go by 100m away.  The stresses are communicated between particles at different locations.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 22/09/2018 19:00:22
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If that happens as an inertial reaction of the water (and the man´s mass) when "forced" to follow the equatorial curved path, why wouldn´t happen something similar when the water has to revolve, following during some 28 days a circumference (considering only earth-moon "dance") of a radius equal to the distance between earth CM and earth-moon barycenter ??
Inertial effects are indeed involved, but not directly like that.  The far tide has far more acceleration about the barycenter (which represents the inertial frame of a two-body model), but both tides are the same magnitude, which they would be even if the moon was further away putting the barycenter above the surface.  The direct inertial explanation of the tide would push the water down, not upward, since it is now rotating on the same side of the barycenter.  Inertial of the water on the sides would tend to carry it to the far side.
So no, I cannot agree with this wording.  In a uniform gravitational field (say at a sufficient distance from the galactic black hole to have a gravitational force on earth similar to that of our moon (2e20N), you have the same inertial forces at work, but undetectable tides.  It is the non-uniformity of the gravitational field, not inertia, that causes the stress, and  tides which are strain resulting from that stress.
I´m afraid you haven´t fully grasp the difference between rotating and revolving (what earth does around the barycenter), not in their layman acceptation but in physicist´s …
Disregarding earth daily spinning (which has nothing to do with moon-earth dynamics), only earth´s CM follows a circular path around the barycenter. Any other earth point follows identical circumferences, but with their centers at its actual distance from the CM, and keeping its orientation relative to far distant stars. Perhaps best analogy is the movement of a child  waist making a hulla-hoop rotate …
As said on one of the NOAA previously linked sites:
" The Effect of Centrifugal Force. It is this little known aspect of the moon's orbital motion which is responsible for one of the two force components creating the tides. As the earth and moon whirl around this common center-of-mass, the centrifugal force produced is always directed away from the center of revolution. All points in or on the surface of the earth acting as a coherent body acquire this component of centrifugal force. And, since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”
At sublunar area, direct moon´s pull is bigger than above mentioned centrifugal force, it prevails, and we have there the more obvious bulge ...
And at antipodal area, being moon´s pull there smaller, centrifugal force (or “outward” inertial force if you like) prevails, and we also have a bulge ...
Not sure if all such accelerated motion can be considered centrifugal since the acceleration is more typically inward instead of outward, and 'centrifugal' is typically used only for the tendency towards outward.
I still have to post here my ideas about the generally badly used issue of inertial and non-inertial reference frames ...
I´m procrastinating in that realm, because I know it is tough to convey.
But please note that the term "centrifugal acceleration" doesn´t exist ... "Centripetal force" is the name we give to any force that is bending a moving object trajectory. And "centripetal acceleration" the one that object gets due to mentioned force.
Centrifugal force is a controversial term, but it is always an inertial reaction of the affected body to the fact that it is being obliged to change the direction of its speed ...
But if the object moved outward, that would actually mean that it´s got somehow free, because both centripetal and centrifugal forces had ceased to exist, at least partially.
And regarding:
"...In a uniform gravitational field (say at a sufficient distance from the galactic black hole ... you have the same inertial forces at work, but undetectable tides ..."
 please kindly note that in that case centrifugal forces would also be very, very tiny, proportional to v˛/r, being "r" enormous !! 
Title: Re: Why do we have two high tides a day?
Post by: Halc on 23/09/2018 01:47:58
Oh, hey, I am a human now, instead of a modal robot.  The edit window didn't get larger.  Still have to do it offline :(

I´m afraid you haven´t fully grasp the difference between rotating and revolving (what earth does around the barycenter), not in their layman acceptation but in physicist´s …
Thanks for that correction.  Rotate is what something does around its own axis, and revolve is around a different one, even if off-center within the thing revolving.  Orbital motion is a form of revolving.

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As said on one of the NOAA previously linked sites:
Which admitted to being for children and laymen, and just wrong after the 50’s….   Hardly an argument from authority.
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since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”
This contradicts with the actual situation since the barycenter is within Earth and thus points on the lunar side of that get centrifugal force directed towards the moon.  They're being sloppy, something they'd not get away with in a publication meant for a higher audience.
Yes, I understand this reasoning.  You are now just repeating it instead of addressing my objections to it in prior posts.  It isn’t completely wrong (the centrifugal force needs to be described in the frame in which it exists), just not proportional to the tides, and most importantly, it doesn’t explain tides in situations where there are no centrifugal forces, such as all these hypothetical situations with multiple moons and such.  That lack of a general explanation makes it wrong I think

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At sublunar area, direct moon´s pull is bigger than above mentioned centrifugal force, it prevails, and we have there the more obvious bulge ...
This is the actual cause, and it works without computing the difference with centrifugal forces to see which is prevailing.  It is a general solution that works even in non-revolving situations.

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Quote from: Halc
Not sure if all such accelerated motion can be considered centrifugal since the acceleration is more typically inward instead of outward, and 'centrifugal' is typically used only for the tendency towards outward.
I still have to post here my ideas about the generally badly used issue of inertial and non-inertial reference frames ...
Other frames are valid, but the laws of physics are not the same as in inertial frames.  Centrifugal force is a valid force in rotating frames, which is contrary to what I first said.  Your use of it in descriptions of the scenario in inertial frames seems like bad use of frames.
The moon and Earth are stationary in the rotating frame of the barycenter, and gravity prevents the natural tendency for both objects to accelerate away due to centrifugal force.

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I´m procrastinating in that realm, because I know it is tough to convey.
But please note that the term "centrifugal acceleration" doesn´t exist
It does.  An object at rest begins to move if not impeded.  Sounds like acceleration to me.  The acceleration is initially outward, but quick bends until it is mostly inward, but always accelerating.
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... "Centripetal force" is the name we give to any force that is bending a moving object trajectory.
In an inertial frame yes, and then only sometimes. The thrust of a rocket turning left isn't usually described as centripetal. In a rotating frame, that centripetal force might be holding an object stationary against its unimpeded tendency to start moving.


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Centrifugal force is a controversial term, but it is always an inertial reaction of the affected body to the fact that it is being obliged to change the direction of its speed ...
But if the object moved outward, that would actually mean that it´s got somehow free, because both centripetal and centrifugal forces had ceased to exist, at least partially.
That bolded part I didn’t understand.  Maybe if you specify a frame it would make sense, but it sounds like you’re just describing an object moving forever in a line in an inertial frame, which yes, it does if no forces act upon it, per Newton’s first law.  I hadn’t seen that law described as something “getting free”.
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And regarding:
"...In a uniform gravitational field (say at a sufficient distance from the galactic black hole ... you have the same inertial forces at work, but undetectable tides ..."
 please kindly note that in that case centrifugal forces would also be very, very tiny, proportional to v˛/r, being "r" enormous !! 
No, the centrifugal forces would be identical to what the moon generates since they have to counter the gravitational pull of that black hole at the exact distance where the force was the same.  The acceleration of Earth in such a situation would be the same as in a pure Earth/moon system, countered by the same inertial resistance that you’ve been labeling centrifugal force.  I chose that scenario specifically for its identical gravitational forces.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 23/09/2018 08:15:27
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What they can´t "feel" is either the so called "differential gravity" or the gravity on locations many km away, let alone they can "subtract" different gravitational pulls to react to that difference !![/b]
Right.  No individual particle can detect a gravity gradient.  It requires pairs of separated particles.  What the particles might 'feel' is the changing forces put on them by their immediate neighbors, which in turn is effectively them 'feeling' the differential gravity on locations many km away ... The stresses are communicated between particles at different locations.
Right ... But that "chain of transmission" of information what actually transmits is not "differential gravity", but the part of total forces acting directly on each particle (moon´s pull, gravitational pull from the rest of the planet, and pushes or pulls from contiguous particles, in your own words "the changing forces put on them by their immediate neighbors"), not used to cause the centripetal acceleration required for the revolving of the considered particle !!.
Those "communicated" stresses, according to 3rd Newton´s Motion Law, happen in pairs of opposite forces exerted on each other (between two considered "neighbors") ... Most of them are parallel to global moon´s pull, half of them centripetal ("moonwards") and half of them outward, or in the broad sense of the term, centrifugal ("moonfugal").
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 23/09/2018 08:35:06
By the way, what you said:
I took issue with his blocky diagram above in the 'stress' section, which shows two stacked blocks in a gravitational field and a weight force preventing the whole setup from accelerating from the gravity.  It labels the spring as tension, despite the arrows depicting the spring pushing on the two blocks, which would compress the spring.  The arrows are correct, but text incorrectly labels this as tension.  If the arrows represent tension T, they need to point the other way, and the tension would be negative.  So while I understand the concept of stress and strain (I took civil engineering courses), the description of it there needs a bit of review.
I have to check the details of that figure and explanation, and I´ll come back to that. But please keep in mind that, as said on recent post in relation to earth internal stresses, the concept "tension" (of a spring in this case) always includes the inherent two opposite forces ... Just an "arrow" where two contiguous objects A and B should always include the information if it is the force exerted by A on B, or by B on A !!   
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 23/09/2018 14:51:12
Orbital motions are not considered as rest frames
I think you mean ‘inertial frames’

I haven’t had time to follow this thread and so am not sure what the point of contention is about, so best I keep out particularly as there is a lot of antagonism going on. However, skimming through I did notice your comment below and as it doesn’t appear to have been picked up I will comment.

I know at least another way to analyze the tides: tangential speed. I used it before in this thread but it didn't seemed to interest anybody either. I said that while the earth's C.G. was going at the right orbital speed around the moon, neither of its near or far sides were going at the right speed: due to the daily rotation of the earth, the near side is going too slow and the far side is going too fast. If we accelerate a satellite which is ,  already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon.

This is a valid analysis. I thought about it and decided it’s easier to think of the moon where you don’t have the complication of the barycentre being inside the earth.
The moon will orbit with its centre of mass at an equilibrium position based on the balance between centrifugal force and gravitational force (rotating, noninertial frame). As you say, the inner part will want to move to a lower orbit and the outer to a higher orbit. Because the moon keeps its same face to the earth it will be elongated along the earth-moon line by the difference in force.
When looking at the earth, the barycentre is within the earth and, as you say, the far side is going to fast and the near side too slow. You comment on the earth’s daily rotation, but this is an added complication we can remove. Despite what some books or websites show, the moon does not orbit around the equator so the earth’s spin doesn’t align with it. Fortunately, as with most motion/force scenarios, we can deal with these as separate issues and say that the spinning creates a symmetrical equatorial bulge and take that out of the equation. In this case we are left with the earth’s centre of mass orbiting the barycentre at it’s equilibrium position and, as you correctly say, the far side will go too fast and the nearside too slow, hence each will try to seek its own equilibrium point.

     BY THE WAY, on a link posted by PmbPhy some days ago it is said:
..the CENTRIFUGAL FORCE and the moon´s gravitational attraction are equal and opposite.
If the two forces where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not. 
@PmbPhy wouldn’t make that mistake. I think by now you will have realised that he was talking about non inertial, rotating frames.
I think the whole subject of centrifugal and what are called fictitious forces is badly taught. NASA tends to refer to all 3 as inertial forces and includes G force in linear acceleration as a 4th. I think it’s better terminology.
Certainly these inertial forces are only seen in a non-inertial, rotating frame (or a close approximation), so they aren’t prime movers, but they do have real, measurable effects.

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As said on one of the NOAA previously linked sites:
Which admitted to being for children and laymen, and just wrong after the 50’s….   Hardly an argument from authority.
Quote
since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”
This contradicts with the actual situation since the barycenter is within Earth and thus points on the lunar side of that get centrifugal force directed towards the moon.  They're being sloppy, something they'd not get away with in a publication meant for a higher audience.
I think they are considering the earth to be a rigid body, so any force at the centre of mass will be transferred to near and far sides. Although the water isn’t rigid it is incompressible and would tend to find it’s own level so might be considered to act as the earth does. Whether it’s a reasonable assumption is up for debate, but I don’t think the article was intended to be completely rigorous.
Title: Re: Why do we have two high tides a day?
Post by: Halc on 23/09/2018 16:41:59
Quote from: Halc
If the two forces where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not.
@PmbPhy wouldn’t make that mistake. I think by now you will have realised that he was talking about non inertial, rotating frames.
I have since corrected my statement above.  Yes, in a rotating frame reference, the two forces exist and can cancel, holding the object stationary.

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I think the whole subject of centrifugal and what are called fictitious forces is badly taught. NASA tends to refer to all 3 as inertial forces and includes G force in linear acceleration as a 4th.
What are the other two?  Centrifugal is an inertial force in a rotating frame.  G force is an inertial force in an accelerating reference frame.  There are inertial effects involved anytime some mass is accelerated in an inertial frame.  That doesn't mean that inertial effects are all centrifugal.

Tides happen or not in both cases, so tides are not a function of either.  That was my point.  Tides are a function of the stresses and strain on an object due to non-uniform gravitational field, and are a function of only that.

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Quote from: NOAA
since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”
Quote from: Halc
This contradicts with the actual situation since the barycenter is within Earth and thus points on the lunar side of that get centrifugal force directed towards the moon.  They're being sloppy, something they'd not get away with in a publication meant for a higher audience.
I think they are considering the earth to be a rigid body, so any force at the centre of mass will be transferred to near and far sides.
They seem to be making an explicit reference to forces on specific points, not about where those forces get transferred due to the rigidity of the body. It was a nit, not really detracting from the point being made by NOAA.

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Although the water isn’t rigid it is incompressible and would tend to find it’s own level so might be considered to act as the earth does. Whether it’s a reasonable assumption is up for debate, but I don’t think the article was intended to be completely rigorous.
Sure.  For one, the tides would be higher if the day were longer since it would give the water more time to move to where gravity is shoving it.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 23/09/2018 18:52:07
Thanks for that correction.  Rotate is what something does around its own axis, and revolve is around a different one, even if off-center within the thing revolving.  Orbital motion is a form of revolving.
There also can be “rotations” around other axis … E.g., the “hammer" to be thrown by the athlete: it simultaneously follows an “orbit”, and "spin" around its own axis (hammer hook always points towards the athlete ...).
Curiously it is the same case of our moon, because it is tidal locked to the earth.
But earth, in its dance with the moon, keeps its orientation relative to distant stars, and that is "revolving" (orbiting without rotation).
That´s why I prefer to reserve the term “spinning” for when around own axis, “revolving” for orbiting without any spinning, and “rotation” for orbiting with simultaneous spinning (when due to the dynamics of same pair of objects “dance”, such as moon´s rotation around barycenter) ...
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”since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”
This contradicts with the actual situation since the barycenter is within Earth and thus points on the lunar side of that get centrifugal force directed towards the moon.
Sorry, but that is erroneous. You seem not to have fully grasped important nuances of actual earth revolving …
I suggest you put your open hand horizontally on a table. Spot the table point app. under the middle of the first bone of your middle finger (B, from barycenter). And try slowly to make your hand “revolve”, with the first knuckle of same finger (C, app. at hand center) following a circle around B .
If initially you had imagined a “moon” a couple of feet away in line with your middle finger, and that “moon” rotated around B simultaneously to hand revolving, ALWAYS in line with B and C and at side opposite to C, you would have a scenario quite similar to actual earth-moon dance.
If you carefully observe the end of any of your fingers (or any other point of your hand), you will find that THEY ALL follow equal circumferences, ALWAYS keeping their location farthest from the moon (logically, within each own path).
THAT´S, I think, A GOOD WAY to see what quoted from NOAA´s scientits work IS QUITE CORRECT:
" since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 23/09/2018 22:25:28
What are the other two?  Centrifugal is an inertial force in a rotating frame.  G force is an inertial force in an accelerating reference frame.  There are inertial effects involved anytime some mass is accelerated in an inertial frame.  That doesn't mean that inertial effects are all centrifugal.
They are Coriolis force, which we all recognise, and Euler force -  tangential force caused by angular acceleration/deceleration - which happens during any pitch, roll, yaw manoeuvre.


Title: Re: Why do we have two high tides a day?
Post by: Halc on 24/09/2018 02:07:21
They are Coriolis force, which we all recognise, and Euler force -  tangential force caused by angular acceleration/deceleration - which happens during any pitch, roll, yaw manoeuvre.
OK.
I thought of Coriolis force after asking the question.  I never heard of Euler force, which seems to be the rotating equivalent of G-force in an accelerating rotating frame.  I wasn't considering frames with variable rotation.

My guess was a different one: In a rotating frame, a stationary object accelerates to move away due to centrifugal force.  Coriolis curves it to the side once it picks up speed and the acceleration vector swings in a direction against the rotation.  After a bit more time, that acceleration vector points nearly back to the axis.  What force accelerates objects back to the origin like that?

The example is stars in the sky, which, in the rotating frame of Earth's surface, go around once a day, all faster than light.  Their acceleration vector is very high, proportional to their distance, and directed straight at us.  What force is responsible for this apparent acceleration in the opposite direction of the centrifugal force?
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 24/09/2018 10:44:29
Quote from: Le Repteux on 14/09/2018 15:11:18
I know at least another way to analyze the tides: tangential speed. I used it before in this thread but it didn't seemed to interest anybody either. I said that while the earth's C.G. was going at the right orbital speed around the moon, neither of its near or far sides were going at the right speed: due to the daily rotation of the earth, the near side is going too slow and the far side is going too fast . If we accelerate a satellite which is ,  already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon. [/quote]
As “Colin2B” correctly says, it is much more logical to disregard the daily spinning of earth when analyzing moon (or sun) related tides … It causes the circular and permanent so called “equatorial bulge”, which has nothing to do with earth-moon dynamics …
Quite another thing is that due to that spinning, we usually perceive the tide cycle as if it had two high and two low tides a day, when the actual cycle has two high and two low tides in a little less than a month.
Apart from that, what you say:
"... while the earth's C.G. was going at the right orbital speed around the moon...
is erroneous: the earth´s C.G. doesn´t orbit around the moon whatsoever !!
The moon pull certainly “supplies” the forces necessary for the circular movement of the earth, which is a revolving around earth-moon barycenter, actually very, very far from the moon !!
That´s why I consider applying proper orbiting dynamics features to our case is an erroneous approach …
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 24/09/2018 10:56:18
This is a valid analysis (NOAA scientist´s). I thought about it and decided it’s easier to think of the moon where you don’t have the complication of the barycentre being inside the earth.
The moon will orbit with its centre of mass at an equilibrium position based on the balance between centrifugal force and gravitational force (rotating, noninertial frame). As you say, the inner part will want to move to a lower orbit and the outer to a higher orbit.
Though you are quite right on your reference to NOAA´s analysis, you aren´t on what in bold ...
If you read carefully second half of my post « on: Yesterday at 18:52:07 » (and recent one) I do hope you´ll see why I say so ...
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 24/09/2018 11:08:59
... the tides would be higher if the day were longer since it would give the water more time to move to where gravity is shoving it.
Again: you continue to mix daily earth spinning with moon related tides … Gravity and inertial effects (centrifugal forces) don´t "need" longer days to build full-size tides, because they have plenty of time to deform ocean surface: a little more than a week from low to high tide !!
If the day were longer, what would imply a smaller angular and tangential speed, what we would have is a smaller gap between sublunar meridian and actual bulge location (the same at antipodes), apart from some more time between our perception of low and following high tide ...
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 25/09/2018 17:12:04
Quote from: Le Repteux on 14/09/2018 10:11:18
Quote
I know at least another way to analyze the tides: tangential speed. I used it before in this thread but it didn't seemed to interest anybody either. I said that while the earth's C.G. was going at the right orbital speed around the moon, neither of its near or far sides were going at the right speed: due to the daily rotation of the earth, the near side is going too slow and the far side is going too fast. If we accelerate a satellite which is already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon.
This is a valid analysis. I thought about it and decided it’s easier to think of the moon where you don’t have the complication of the barycentre being inside the earth.
The moon will orbit with its centre of mass at an equilibrium position based on the balance between centrifugal force and gravitational force (rotating, noninertial frame). As you say, the inner part will want to move to a lower orbit and the outer to a higher orbit. Because the moon keeps its same face to the earth it will be elongated along the earth-moon line by the difference in force.
When looking at the earth, the barycentre is within the earth and, as you say, the far side is going to fast and the near side too slow. You comment on the earth’s daily rotation, but this is an added complication we can remove. Despite what some books or websites show, the moon does not orbit around the equator so the earth’s spin doesn’t align with it. Fortunately, as with most motion/force scenarios, we can deal with these as separate issues and say that the spinning creates a symmetrical equatorial bulge and take that out of the equation. In this case we are left with the earth’s centre of mass orbiting the barycentre at it’s equilibrium position and, as you correctly say, the far side will go too fast and the nearside too slow, hence each will try to seek its own equilibrium point.
Hi Collin, and thanks for noticing that my idea looks right.

As you said, it might be easier to use the moon instead of the earth to analyze it, so let's do that. I said that the proper rotation of the planet had something to do with the fact that the far and the near sides are not going at the right orbital speed, and I must defend that feature otherwise my idea wouldn't work. Of course, that kind of rotation also produces an equatorial bulge, but even if the rotation is as tiny as the moon's one, that's what makes its far and near sides not going at the same orbital speed as its center of gravity. These sides could very well be going at the right speed if they weren't tidally locked for instance, and in this case, if my explanation is right, there would be no tides. What would be producing the tides if the speed was wrong would also produce another kind of motion that we attribute to tidal friction: the slow deceleration of rotations. In the earth's case, when its far side starts going too fast around the moon, it also starts decelerating and getting away from it as if it was at its perigee, while its near side does the inverse, and they both go on doing that all along their transit. This way, no need for friction to explain the observation, the principle of orbital motion does it all, and it can also be applied directly to orbital rotations.

For instance we can consider that only the barycenter of the earth/moon system is going at the right orbital speed around the sun, and it explains why the two bodies are constantly moving away from one another while their orbital speed is constantly decelerating without needing to use the less evident tidal friction explanation, which is so unclear to me that I decided to start a thread about it (https://www.thenakedscientists.com/forum/index.php?topic=74894.new#new). Naturally, it also applies to the rotation of the solar system around the galaxy core, which should also be recessing. The only data that I found about that is the slow deceleration of the probes that we send throughout the solar system, for which the NASA is still looking for an explanation. If the orbital motion of the probes is constantly recessing, and if the calculations don't account for that, they are certainly not at the place where they are expected. They should actually be farther from the sun than expected, and moving at a slower orbital speed around it than expected. No need to change the gravitational theory to explain the phenomenon though, just the tidal one.

Let me now try to test my principle while applying it to the orbital rotation of the stars belonging to two different orbiting galaxies. If my proposition is right, while two galaxies would be orbiting around one another, the stars belonging to one of the galaxies would not be orbiting at the right orbital speed with regard to the other galaxy during their transit, so they should also be recessing slowly from the galaxy core with time, and the outer ones would necessarily be recessing faster than the inner ones for all their orbital distances to stay proportional. Now, the more the galaxy we are observing is far from us, the less it would have time to recess, so we see it as it was when the orbits of its stars were much closer to one another, thus also when their respective orbital speeds were much faster than they actually are, but without the shape of the galaxy looking any different from the one of the closer galaxies that we observe. Now the big question: if our estimate of the distant galaxies' dimension is wrong, could that mechanism replace the dark matter one?



Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 25/09/2018 19:00:45
I said that the proper rotation of the planet had something to do with the fact that the far and the near sides are not going at the right orbital speed, and I must defend that feature otherwise my idea wouldn't work. Of course, that kind of rotation also produces an equatorial bulge, but even if the rotation is as tiny as the moon's one, that's what makes its far and near sides not going at the same orbital speed as its center of gravity. These sides could very well be going at the right speed if they weren't tidally locked for instance, and in this case, if my explanation is right, there would be no tides.
I have to go more slowly through your post, but for now I have to say, as I´ve already done many times, that the two cases are quite different ...
As all earth particles follow identical circular paths (of a radius equal to the distance between earth CM and barycenter), "the fact that the far and the near sides are not going at the right orbital speed" hasn´t actually any reasonable meaning ... There is no a "proper" orbiting around the moon !
But being the moon tidal locked to the earth, and orbiting around the barycenter (to say around the earth is sufficiently right too...), that would allow you to apply your idea.
With my approach (and NOAA´s scientists, "French Tides" ...), further not visible side, apart from having a tangential speed bigger than the moon CM and still bigger than closer side, centrifugal forces are also the further the bigger (ω˛r per unit of mass). That increases tidal effects, because not only the so called "differential gravity" counts for tides.
That rotation could be considered as the addition of a revolving without rotation (as earth moves), and a spinning around its own axis over same some 28 days of the orbiting.
But the effect of that "spinning" shouldn´t be considered like an "equatorial bulge" as what happens on earth ... Moon´s one is part of earth-moon global dynamics, but earths´s one (with much, much higher angular speed) is not whatsoever ...   
Title: Re: Why do we have two high tides a day?
Post by: Halc on 25/09/2018 19:10:33
... the tides would be higher if the day were longer since it would give the water more time to move to where gravity is shoving it.
Again: you continue to mix daily earth spinning with moon related tides … Gravity and inertial effects (centrifugal forces) don´t "need" longer days to build full-size tides, because they have plenty of time to deform ocean surface: a little more than a week from low to high tide !!
A little more than 6 hours actually.  A week between spring tide and whatever they call the low between them.  The shoving I spoke of happens every 6 hours.  The river by me runs backwards twice a day due to that shoving.
Title: Re: Why do we have two high tides a day?
Post by: Halc on 25/09/2018 19:46:45
I know at least another way to analyze the tides: tangential speed. I used it before in this thread but it didn't seemed to interest anybody either. I said that while the earth's C.G. was going at the right orbital speed around the moon, neither of its near or far sides were going at the right speed: due to the daily rotation of the earth, the near side is going too slow and the far side is going too fast. If we accelerate a satellite which is ,  already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon.

This is a valid analysis. I thought about it and decided it’s easier to think of the moon where you don’t have the complication of the barycentre being inside the earth.
The moon will orbit with its centre of mass at an equilibrium position based on the balance between centrifugal force and gravitational force (rotating, noninertial frame). As you say, the inner part will want to move to a lower orbit and the outer to a higher orbit.
My bold.  I disagree.  This explanation predicts that tides would be negative if the Earth spun the other way since the inner part has the greatest tangential speed and wants to move outward, and the outer part has the least tangential speed and wants to move inward.  Venus is such a case (it rotates backwards), at it has normal solar tides towards and away from the sun just like Earth.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 26/09/2018 07:07:46
... the tides would be higher if the day were longer since it would give the water more time to move to where gravity is shoving it.
Again: you continue to mix daily earth spinning with moon related tides Gravity and inertial effects (centrifugal forces) don´t "need" longer days to build full-size tides, because they have plenty of time to deform ocean surface: a little more than a week from low to high tide !!
A little more than 6 hours actually.  A week between spring tide and whatever they call the low between them.  The shoving I spoke of happens every 6 hours.  The river by me runs backwards twice a day due to that shoving.
Your basic error comes from same misconception as if somebody said moon (or sun) rotates around earth once every 24 hours ... That is only an apparent movement !
"Sublunar" bulge is always almost in line with the moon, and its actual cycle is, as moon´s, 28/29 days ... A little more than 7 (rather than 6) days  ahead of next low tide !!
Solid earth below (and partially ocean deep water) experiences a daily spin (which has nothing to do with earth-moon dynamics), and  as we also move with solid earth,  what you say apparently happens !!
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 26/09/2018 07:26:16
Quote from: Colin2B on 23/09/2018 14:51:12
This is a valid analysis. I thought about it and decided it’s easier to think of the moon where you don’t have the complication of the barycentre being inside the earth.
The moon will orbit with its centre of mass at an equilibrium position based on the balance between centrifugal force and gravitational force (rotating, noninertial frame). As you say, the inner part will want to move to a lower orbit and the outer to a higher orbit.
My bold.  I disagree.  This explanation predicts that tides would be negative if the Earth spun the other way since the inner part has the greatest tangential speed and wants to move outward, and the outer part has the least tangential speed and wants to move inward. 
You keep insisting on your error, probably because you haven´t read (or disagree with) some of my recent posts, one of them #328:
... what quoted from NOAA´s scientits work IS QUITE CORRECT:
" since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon
ALL material earth points have the same tangential speed (considering only moon-earth "dance"), because their paths are identical circles, logically with same cycle period (28/29 days) ...
Title: Re: Why do we have two high tides a day?
Post by: Halc on 26/09/2018 12:29:16
Your basic error comes from same misconception as if somebody said moon (or sun) rotates around earth once every 24 hours ... That is only an apparent movement !
It is apparent movement that I'm talking about.  The tides go up and down twice a day, and that requires movement relative to Earth's surface of thousands of cubic km of water each 6 hours.  In most places, this is not a problem, since the necessary movement is only about 1km or less, and 6 hours is plenty of time for gravity to shove water that far.  In fact, the variance in the Atlantic comes more from resonance than it does from the gravity boost, so a longer day might actually make the tides lower.

So my example of a place that would get much higher tides were the day to be longer is the Mediteranean.  No resonance there, and all the water needed to raise the tides need to fit through Gibraltar, and the gap simply isn't large enough to allow the tide to come in before it starts running out a few hours later.

Quote
"Sublunar" bulge is always almost in line with the moon, and its actual cycle is, as moon´s, 28/29 days ... A little more than 7 (rather than 6) days  ahead of next low tide !!
I'm well aware of that, but no water is stationary relative to this bulge.  If it was, it would kill us: New York City and Shanghai wiped out twice daily by a wave of water moving west at over 1000 km/hr.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 26/09/2018 16:21:49
This explanation predicts that tides would be negative if the Earth spun the other way since the inner part has the greatest tangential speed and wants to move outward, and the outer part has the least tangential speed and wants to move inward.  Venus is such a case (it rotates backwards), at it has normal solar tides towards and away from the sun just like Earth.
This argument looks solid, but I didn't find any data on the Venus tides to check it out. Do you have a link to them? A similar argument can also be raised about how tidal locking could happen to the moon's surface since it is still rotating a bit, thus still going at a different orbital speed around the earth than its own CG. The answer may reside in the way the tidal bulges affect the proper speed. In my explanation, it is the orbital speed that produces the tides, and the bulges are not only moving radially, but they also move tangentially a bit in a direction that opposes the direction of rotation. In other words, it is the bulges that produce the deceleration of the proper rotation, and it happens where the orbital speed counts, not in between, so in reality, the rotation might stop decelerating just because the bulges are already going at the right speed on their orbital trajectory while the whole planet is still rotating faster than that a bit, and we can apply the same reasoning to Venus since its proper rotation is also very slow.

In fact though, the bulges suffer a deceleration followed by an acceleration, they raise and fall in the same day, so with my explanation, they should appear and disappear without decelerating the proper rotation of the planet when it's slow, while still decelerating it when it's fast, which looks difficult to explain. After all, maybe the bulges don't need orbital rotation to raise, but still need it to decelerate the proper rotation. I don't like the friction explanation, I can't understand it, so I'm looking for another one, and the orbital speed one seems to apply so well to orbital receding.

 
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 26/09/2018 17:07:09
That rotation could be considered as the addition of a revolving without rotation (as earth moves), and a spinning around its own axis over same some 28 days of the orbiting.
But the effect of that "spinning" shouldn´t be considered like an "equatorial bulge" as what happens on earth ...
All the rotations depend on inertial motion, whether they are held by gravitation or by a rope, and inertial motion isn't a radial motion, it is tangential, so the outward motion when the force stops pulling is only incidental. It certainly takes a force to keep the inertial motion circular, but it is the same force whether it is a rope or gravitation. It can't be both though, either it is a rope or it is gravitation. When the earth spins, it is inertia that produces its outward motion (not force), but it is gravitation that produces the inward force (not motion), so the reason why the equatorial bulge doesn't spread away from the earth is  gravitation, and the reason why a body on orbital motion doesn't fly away too. It is the same kind of inertial motion, and the same kind of force, so to me, the two kinds of rotation are equivalent.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 26/09/2018 19:13:00
So my example of a place that would get much higher tides were the day to be longer is the Mediteranean.  No resonance there, and all the water needed to raise the tides need to fit through Gibraltar, and the gap simply isn't large enough to allow the tide to come in before it starts running out a few hours later.
On the one hand, at some areas of the Mediterranean resonance actually causes some tides.
On the other, the reason why global moon and sun related tidal effects are only very slightly perceived there is not just a "lack of time" for that Gibraltar "bottle neck" to allow sufficient water in. Open ocean bulges build with not only water from other meridian areas (not too far from the equator), but also from other N and S higher latitude areas ... Otherwise tides almost wouldn´t happen at such high latitudes, where they can actually be very strong.
 
Quote
"Sublunar" bulge is always almost in line with the moon, and its actual cycle is, as moon´s, 28/29 days ... A little more than 7 (rather than 6) days  ahead of next low tide !!
I'm well aware of that, but no water is stationary relative to this bulge.  If it was, it would kill us: New York City and Shanghai wiped out twice daily by a wave of water moving west at over 1000 km/hr.
I can´t understand what you mean! Precisely because of daily earth spinning, that moon-related bulge (per se moving much slowly, 40,000 km/app. 28 days) relatively to earth solid parts it does move at 40,000 km/app. 24h = 1,667 km/h ... But no serious problems because those tidal waves are relatively very, very small (not more than a few meters, if any, of difference between high and low tide, some 10,000 km apart (at the equator) !!)
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 29/09/2018 22:21:37
My bold.  I disagree.  This explanation predicts that tides would be negative if the Earth spun the other way since the inner part has the greatest tangential speed and wants to move outward, and the outer part has the least tangential speed and wants to move inward.  Venus is such a case (it rotates backwards), at it has normal solar tides towards and away from the sun just like Earth.
My fault. I was in a hurry and didn’t give a full explanation, thinking @Le Repteux  would work out what I was saying.
Let’s start with the moon which was the example I suggested. It orbits around the barycentre with it’s same face to the earth, so it is not spinning relative to the earth/moon line. The distance from the barycentre to the moon’s centre of mass can be determined by the balance between the centripetal force (due to earth’s gravitational pull) and the centrifugal force (due to speed of orbit of the centre of mass). If we consider points on the earth/moon line on the near and far sides of the moon, it is reasonable to suggest that although these points have the same angular velocity as the CG they are moving at a faster or slower orbital speed than the CG, hence they could be considered to be seeking a different orbit. The nearside moving towards earth, the farside away.
The next question is an interesting one. You introduce the spin of a planet eg Venus that changes the planet’s surface speed. However, I would argue, as I suggested to @Le Repteux, that the spin has no effect on the orvbit. If you consider the locus of points on the surface the spin does not cause any net increase or decrease in the orbital speed and it is orbital speed which is the prime cause of any centrifugal forces. At any instant in time the forces on any mass on the earth/moon line are the gravitational pull and the centrifugal force and we can treat the spin separately because, it has no effect on the net orbital speed of any part of the mass. Also, as I mentioned to @Le Repteux  we can treat the spin as a separate component of the overall system and it's only effect is to create centrifugal force causing an equal bulge on the equator of the spin, it does not cause a net motion towards or away from the earth.
Title: Re: Why do we have two high tides a day?
Post by: Halc on 30/09/2018 01:11:09
The next question is an interesting one. You introduce the spin of a planet eg Venus that changes the planet’s surface speed. However, I would argue, as I suggested to Le Repteux, that the spin has no effect on the orbit.
I didn’t suggest that the spin has any (short term) effect on the orbit.  Yes, Le Repteux has a thread open about tidal forces slowly pushing orbiting things away, and that is true.  The spin of the sun has a higher angular velocity than any of the planets, so that tide slowly pushes each of them away.  Most planets also spin faster than their orbits, putting thrust on the sun that also contributes to higher orbits.  Venus is an exception there, and its negative spin actually degrades its orbit, but not as much as the tide on the sun expands that orbit.  All this is relevant to the other thread, but your response above seems to concern this point.

The discussion was about tides, with a suggestion that the tides might be partially caused by higher linear speeds at points furthest out, but I pointed out that points on Venus furthest from the sun have the lowest linear speed and should be ‘seeking lower orbit’, and the points closest to the sun have the greatest linear speed and thus should be ‘seeking higher orbit’.  If that were so, the tides on Venus would be to the sides, not towards and away from the sun as all solar tides are.

Quote
If you consider the locus of points on the surface the spin does not cause any net increase or decrease in the orbital speed and it is orbital speed which is the prime cause of any centrifugal forces. At any instant in time the forces on any mass on the earth/moon line are the gravitational pull and the centrifugal force and we can treat the spin separately because, it has no effect on the net orbital speed of any part of the mass. Also, as I mentioned to Le Repteux we can treat the spin as a separate component of the overall system and it's only effect is to create centrifugal force causing an equal bulge on the equator of the spin, it does not cause a net motion towards or away from the earth.
All this seems to be about orbital speed, something on which I was not commenting in this topic, until what I wrote above in this post now.

The comment of mine that you quoted in the immediately preceding post concerned centrifugal explanations for the tides.  Instead, I agree with your last line there that we can treat the spin as a separate component of the overall system and yes, all it does is that standing bulge everywhere at the equator.
Tides are partially an effect of spin.  The spin rate coupled with the resonant frequency of oceans and shorelines causes higher tides in some places than others.  Move the continents or alter the spin and the tides will be higher in different places.  But high tide is always towards and away from the gravity gradient and the bulge will form regardless of the axis or intensity of spin.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 30/09/2018 18:48:20
The spin of the sun has a higher angular velocity than any of the planets, so that tide slowly pushes each of them away.  Most planets also spin faster than their orbits, putting thrust on the sun that also contributes to higher orbits
Sorry, but I can´t understand what you mean ...
The discussion was about tides, with a suggestion that the tides might be partially caused by higher linear speeds at points furthest out
Colin2B and me have said several times that idea (about actual linear speeds and orbits), considering as a whole the spinning added to the revolving, is rather a "bad" approach ... Especially when, as in our case of earth tides, linear speeds caused by those different movements are so different ...
Keep in mind that, on the one hand, v (linear speed) = ω r (r: radius of curvature, app. the distance from center of each path to the earth considered point).
The angular speed ω of the spinning is some 28 times bigger than the one of the revolving.
And the radius some 3/2 times bigger too.
And on the other hand, related centrifugal forces are proportional to ω˛r ... Enormously higher in the spinning !!
"Normal" bulges are actually quite insignificant compared to the so called equatorial bulge !!
And to tackle earth tides as in what quoted is utterly absurd !!

And regarding Venus special case, I´m afraid you say wrong things, but prefer not to refute them without having quite clear what on my first phrase in bold ... 
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 30/09/2018 20:08:25
I just had a flash. If the earth would stop spinning, the equatorial bulge would get down and the tidal bulges would follow it, but for my proposition to be true, they should also disappear, and eventually reappear if ever the orbital speed was also stopped because then, the two sides of the earth would not be accelerating at the same rate towards the moon. If only the spinning is stopped, then the only fleeing motion left is due to orbital speed, and all the different parts of the earth are fleeing almost the same distance away from the moon in the same time, but there is still a difference in the force from the moon on those different parts since they are not necessarily at the same distance from it: if the cg of the earth is pulled a certain distance towards the moon to absorb the fleeing motion, then the outer part is pulled less and the inner one more. This way, contrary to what I was proposing, the spinning would have effectively no effect on the tidal bulges, and it seems to match David's explanation, but I'm not sure rmolnav's will agree with it since he seems to need a centrifugal force where I only need a tangential motion.
Title: Re: Why do we have two high tides a day?
Post by: Halc on 01/10/2018 00:47:34
The spin of the sun has a higher angular velocity than any of the planets, so that tide slowly pushes each of them away.  Most planets also spin faster than their orbits, putting thrust on the sun that also contributes to higher orbits
Sorry, but I can´t understand what you mean ...
The sun spins every 25 days, which is faster than the orbit of any planet (88 days of Mercury being the fastest), so the tides on the sun push all of the planets outward, and correspondingly slow the sun’s spin.

Quote
And regarding Venus special case, I´m afraid you say wrong things, but prefer not to refute them without having quite clear what on my first phrase in bold ...
In my quote above your bolded phrase, I was speaking of the tides on the sun caused by any planet (including Venus), not of the tides on any particular planet.  Venus is not a special case in this regard. That part you quoted was related to how tides might have an effect on orbital radius, which is a discussion going on in another thread.  This thread is about two tidal bulges instead of just one.
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 01/10/2018 07:53:33
All this seems to be about orbital speed, something on which I was not commenting in this topic
In that case we are talking at cross purposes and misunderstanding each other; my reply to @Le Repteux  was only to do with orbital speed.

I just had a flash. If the earth would stop spinning, the equatorial bulge would get down and the tidal bulges would follow it,   ............. contrary to what I was proposing, the spinning would have effectively no effect on the tidal bulges,
The spinning has no effect on tides whatsoever. The centrifugal force causing the equatorial bulge is balanced by earth’s gravity so the ‘new’ level is the geoid ie sea level and any tide raising forces produce an increase relative to this. Not sure if that is what you were trying to say, if so my comments are only for clarification.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 01/10/2018 12:05:13
The sun spins every 25 days, which is faster than the orbit of any planet (88 days of Mercury being the fastest), so the tides on the sun push all of the planets outward, and correspondingly slow the sun’s spin.
Quote
And regarding Venus special case, I´m afraid you say wrong things, but prefer not to refute them without having quite clear what on my first phrase in bold ...
In my quote above your bolded phrase, I was speaking of the tides on the sun caused by any planet (including Venus), not of the tides on any particular planet.  Venus is not a special case in this regard.
But those "tides" on the sun ...
1) ... are almost negligible, even the ones caused by biggest planets ...
2) ... given the distances involved, tangential component of each "bulge" would also be quite negligible ...
2) ... those "individual" bulges would be scattered around sun´s surface kind of at random (and varying with time) ...
3) ... and therefore it is not like the earth-moon scenario whatsoever, where we continuously have a gap between a bulge and actual sublunar meridian (the bulge ALWAYS eastwards from the moon), which is what pulls forward the moon, not just earth eastward spinning ...
So, to extrapolate earth-moon scenario to sun-planets case is quite erroneous ...   
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 01/10/2018 14:38:42
The centrifugal force causing the equatorial bulge is balanced by earth’s gravity so the ‘new’ level is the geoid ie sea level and any tide raising forces produce an increase relative to this. Not sure if that is what you were trying to say, if so my comments are only for clarification.
Yes, that's what I was saying about the spinning, but as I said, I prefer to use the concept of centrifugal motion instead of centrifugal force, just in case @rmolnav would understand what @David Cooper was telling him. If we draw a tangential speed vector on a circle, the tip of the arrow is where the body would be if no force was applied on it, and the force produces the acceleration needed to bring it back on the circle in the same time it took to get away. That centripetal force can be called inertial if the body is held by a rope, or gravitational if it held by gravitation, but to me, we shouldn't oppose it to a centrifugal one because it is a bit misleading. We could if the body's speed was affected by it, but it is not the case since it is perpendicular to the tangential motion. The only opposition the force faces is from the body's mass, as if it was at rest at a certain distance away from the circle. Of course it is different if the trajectory is an ellipse, because then, the force is not often perpendicular to the tangential vector and part of it serves to accelerate or decelerate the body tangentially to its trajectory.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 01/10/2018 19:04:25
If we draw a tangential speed vector on a circle, the tip of the arrow is where the body would be if no force was applied on it, and the force produces the acceleration needed to bring it back on the circle in the same time it took to get away. That centripetal force can be called inertial if the body is held by a rope, or gravitational if it held by gravitation, but to me, we shouldn't oppose it to a centrifugal one because it is a bit misleading. We could if the body's speed was affected by it, but it is not the case since it is perpendicular to the tangential motion. The only opposition the force faces is from the body's mass, as if it was at rest at a certain distance away from the circle.
1) "If we draw a tangential speed vector on a circle, the tip of the arrow is where the body would be if no force was applied on it ...": "Almost" quite right: you should have added "after a second" (or any other used unit of time).
2) "... and the force produces the acceleration needed to bring it back on the circle in the same time it took to get away. That centripetal force can be called inertial if the body is held by a rope, or gravitational if it held by gravitation ...": quite right the first phrase, but not the second.
 The centripetal force that makes an object rotate (instead of following in the direction of a previously acquired "tangential" speed), is always an "active" force (not to be called "inertial"), both if gravitational or the tension of a rope ...
3) "...  but to me, we shouldn't oppose it to a centrifugal one because it is a bit misleading. We could if the body's speed was affected by it, but it is not the case since it is perpendicular to the tangential motion. The only opposition the force faces is from the body's mass..."
It´s not "we" who "oppose it (the centripetal f.) to a centrifugal one" ... Centrifugal f. is the way "inertia" (certainly due to the body´s mass, and its speed) can manifest itself, not always in the same way ...
Even being "perpendicular to the tangential motion", it could for instance cause deformations ...
Rather than "a bit misleading" I would say it is something difficult to grasp, let alone to successfully convey to others ... I´ll keep trying. As I´ve already said, I have some "fresh" ideas for it, but prefer not to post them until having them better elaborated, precisely not to "mislead" anybody (or the fewer the better ...)
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 01/10/2018 21:04:20
Centrifugal f. is the way "inertia" (certainly due to the body´s mass, and its speed) can manifest itself, not always in the same way ...
Even being "perpendicular to the tangential motion", it could for instance cause deformations ...
A force causes an acceleration in the direction it is applied, whether the body is moving or not. We can even consider that the body is at rest to measure the force. What opposes the force is the mass of the body, and that body has to move for the force to be measured. When the force pulls on the body, the body moves towards the force, not away, and this motion is not clear when studying rotation. We may still imagine that a rotation produces an outward force, thus centrifugal, but it is an outward motion that it produces, and that motion is then pulled back in the direction of the force. If that outward motion is due to the spinning earth, then it produces equatorial bulges, and if it is due to its orbital motion, then it produces the tides, but it is the same kind of motion and the same kind of force, so there is no need to separate them, and it is what you seem to be doing.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 02/10/2018 12:13:31
We may still imagine that a rotation produces an outward force, thus centrifugal, but it is an outward motion that it produces, and that motion is then pulled back in the direction of the force. If that outward motion is due to the spinning earth, then it produces equatorial bulges, and if it is due to its orbital motion, then it produces the tides, but it is the same kind of motion and the same kind of force, so there is no need to separate them, and it is what you seem to be doing.
Well, first of all, an "outward motion" is not a proper physical variable, let alone a force ...
But within layman language, I would tell you those water motions (when equatorial bulge and tides) happen in opposition to own water weight (the most important force there), and after the imagined tangential movement and centripetal force pulling back, water remains further from earth CG, actually weighing a little bit less.
I, and eminent physicist, call that diminishing of weight centrifugal force ... If you prefer, I also could call it an "outward" force, certainly one of the ways inertia can manifest itself ...
 
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 02/10/2018 18:53:22
Just some additional details about what said on my last post:
...those water motions (when equatorial bulge and tides) happen in opposition to own water weight (the most important force there), and after the imagined tangential movement and centripetal force pulling back, water remains further from earth CG, actually weighing a little bit less
What in bold refers only to antipodal bulge ...
As I´ve said many times, considering only the earth revolving around moon-earth barycenter, all earth particles follow identical circles, and their locations (within own circular path) are ALWAYS the farthest from the moon. That implies centrifugal force is always in the sense opposite to the moon.
Therefore, where sublunar bulge water (and solid elements) weights actually increase a tiny little bit ...(due to that inertial effect, the centrifugal force, in same sense as earth´s own pull).
But as moon´s pull there is bigger than that inertial force, sublunar bulge builds (towards the moon).
But at the antipodes moon´s pull is smaller than centrifugal force, the later prevails, and also a bulge (in the sense opposite to the moon) builds there.
In the case of the so called equatorial bulge all centrifugal forces are radially in the "fugal" sense from earth c.g., and the only existing gravitational pull is own weight of earth particles, uniform around the equator ... That´s why there the weights of all particles diminish, proportionally to ω˛r (r the distance to earth axis of spinning, the closer to the equator the bigger ...).
Solid earth get deformed (angular speed ω is some 28 times bigger than in the moon-earth dance), and water from both hemispheres piles up towards the equator.

 
 
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 02/10/2018 21:07:59
Well, first of all, an "outward motion" is not a proper physical variable, let alone a force ...
During rotation, the motion is not directly outwards, it is tangential, and the distance being increasing if we cut the force is thus only incidental. That motion is due to inertia, reason why we call it inertial. While saying "an outward motion is not a proper physical variable", what do you mean exactly? Do you mean that inertial motion is not physical?
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 03/10/2018 10:20:45
Well, first of all, an "outward motion" is not a proper physical variable, let alone a force ...
During rotation, the motion is not directly outwards, it is tangential, and the distance being increasing if we cut the force is thus only incidental. That motion is due to inertia, reason why we call it inertial. While saying "an outward motion is not a proper physical variable", what do you mean exactly? Do you mean that inertial motion is not physical?
An adjective such as "inertial" can be used scientifically, or in a broader layman sense ...
Therefore, in that last sense, you can use it for any noun with almost any relation to inertia ...
But in physics science, the "variables" can be speed, mass, acceleration, force, momentum ... all with their own units. Never just a "motion" ...

And that "motion" is not actually due to "inertia", it is due to an initial speed, whatever its cause.
Due to "inertia", objects "tend" to keep constant their velocity vector. If no force opposes that inertial "tendency", objects keep moving without any "problem". But if objects are "forced" to change their velocity vector, they somehow react (they are kind of stubborn: "no, no, I don´t want to change my velocity vector"...). And inertia manifests itself, in different ways ...
In our case, inertia manifests itself as a force, equal but opposite to the centripetal force, the "guilty" for the velocity vector change ... An outward force ... centrifugal (in its broad sense) because it is kind of "fugitive" (< > fugal) from the moon (the "center" of gravitational pull, source of acting centripetal forces ...)


Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 03/10/2018 15:38:25
In our case, inertia manifests itself as a force, equal but opposite to the centripetal force, the "guilty" for the velocity vector change ... An outward force ... centrifugal (in its broad sense) because it is kind of "fugitive" (< > fugal) from the moon (the "center" of gravitational pull, source of acting centripetal forces ...)
To me, the definition of inertia is a bit weird: a body goes straight ahead because it has inertia, and it resists to change direction or speed also because it has inertia. A force is not a motion, and inertia is about both, so it means that it is probably still misunderstood. When nothing forces a body to change direction or speed, it goes straight ahead, as if something inside it would know how to move this way, and whenever something is on the way, it resists to change direction or speed, but it still changes them after a while if it is free to move, and it is probably that time that we call inertia. If it would change instantly, there would be no resistance and no inertia, and there would be no motion either. We can't measure it at our scale, but if I'm right about inertia, the atoms are probably going straight ahead for a while during rotation for us to be able to measure a force, and they probably feel no force during that time otherwise they would not be free to move. That's what makes me think that the equatorial bulge and the tidal ones are driven by the same phenomenon, and I think that you could admit it on that basis, but you probably have another viewpoint on inertia that makes you think your way. We behave like atoms as far as inertia is concerned, we all resist to change, and we all finally change since we are free to move, but it takes time. The problem with us is that even if we cannot know if the direction we took is the right one, we still think it is if nothing forces us to change it, but why would an atom change its direction or speed when it is not forced to? :0)
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 04/10/2018 07:35:58
... a body goes straight ahead because it has inertia, and it resists to change direction or speed also because it has inertia
To me those two phrases of you are kind of the two sides of a coin: the body, either "feels" a null total force exerted on it, or it feels a not null force ... Inertia phenomenon makes it maintain its velocity vector (including the case of a null vector) in the first case (without any other "inertial" effect), or, in the second case, it experiences an acceleration vector (a=f/m) ... It is in this last case when other inertial effects may occur, not always the same way, depending on the individual forces acting on the object (the breakdown of mentioned total force vector).
 
When nothing forces a body to change direction or speed, it goes straight ahead, as if something inside it would know how to move this way, and whenever something is on the way, it resists to change direction or speed, but it still changes them after a while if it is free to move, and it is probably that time that we call inertia. If it would change instantly, there would be no resistance and no inertia, and there would be no motion either .
Sorry, but that´s utterly erroneous !!
Since the very first instant (an infinitesimal fraction of time), mentioned acceleration vector changes the object speed, though logically only an infinitesimal amount of change ... (dv=a*dt: infinitesimal speed vector change equals to acceleration vector multiplied by the infinitesimal fraction of time the total force has been acting).
And please, try not to make more difficult for people to understand the already "tricky" concept of "inertia" saying things such as "it is probably that time that we call inertia" ...!!
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 04/10/2018 14:56:21
Since the very first instant (an infinitesimal fraction of time), mentioned acceleration vector changes the object speed, though logically only an infinitesimal amount of change ...
If nothing is faster than light, then nothing changes instantly, thus when we apply a force on a body, it necessarily takes a while before it changes speed or direction. If bodies could move instantly, we wouldn't have the time to feel any force, and they would go from one point to the other without going through the transitional ones. Now, if bodies can absolutely not change speed or direction instantly, it is straightforward to deduce that this may be the cause for their resistance to change, and I don't feel I'm misleading people while suggesting that. You may not be tempted to discuss that point, but others might be interested. On the other hand, if you could just admit that possibility for a while, I think you may understand more easily what I'm saying about rotational motion.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 04/10/2018 16:00:19
Since the very first instant (an infinitesimal fraction of time), mentioned acceleration vector changes the object speed, though logically only an infinitesimal amount of change ...
If nothing is faster than light, then nothing changes instantly, thus when we apply a force on a body, it necessarily takes a while before it changes speed or direction. If bodies could move instantly, we wouldn't have the time to feel any force, and they would go from one point to the other without going through the transitional ones. Now, if bodies can absolutely not change speed or direction instantly, it is straightforward to deduce that this may be the cause for their resistance to change, and I don't feel I'm misleading people while suggesting that. You may not be tempted to discuss that point, but others might be interested. On the other hand, if you could just admit that possibility for a while, I think you may understand more easily what I'm saying about rotational motion.
Sorry, but if you have NO IDEA of what the maths of infinitesimal values of variable and functions are, as clearly your comments show, this is not the place for me to try and explain that too ...
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 04/10/2018 16:05:29
MY ULTIMATE GO ?
Once again I´m going to try and convey a core idea of my stand, now in a “fresh” way …
It will have three parts. I´ll post them neither on same post, nor on three different ones, but on two posts … (that way readers will have more time to “ruminate” what proposed).

A) Not to need to include a figure, please kindly IMAGINE one with two celestial objects on same “vertical” line (on your screen).
Let us imagine the lower one (M) on a fix location, and the upper one (E) moving “horizontally” from right to left, and let us consider M´s gravity starts to pull downwards E just when reaching M´s vertical (remember, M is considered somehow on a fix location).
If within certain game of relation between masses, distance and speed of E, E will start “orbiting” around M.
M´s gravity, acting as centripetal force, changes E´s speed vector direction, but it is not “able” to change its size. A “free fall” to many, but rather a “partially free fall” to me, because E´s inertia avoids a direct straight line fall.

B) Let us now imagine M, at the very moment E gets at M´s vertical, starts moving also from right to left, with same velocity as E.
Being now M´s pull on E always vertical, E will fall towards M, following a parabolic line.
Now M´s pull is able not only to bend E´s speed vector, but also to increase its vertical speed with an acceleration proportional to M´s pull … same way as if neither of them had an initial horizontal speed …
That´s what I consider is a “fully free” fall ...

And now we should not say E is orbiting “around” M, because it will fall onto it !!
 
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 04/10/2018 16:18:11
Sorry, but if you have NO IDEA of what the maths of infinitesimal values of variable and functions are, as clearly your comments show, this is not the place for me to try and explain that too ...
To study a relativistic phenomenon, we need Relativity math, but even then, math won't help us if our theory is wrong, so since I find my motion simulations interesting, for the moment, I prefer to think that our theory on motion is wrong. By the way, in my simulations, the particles do move while making very smal steps. They progress by steps whose length is .01 times the steps of the photons, which are only 1 pixel long on the screen. I could have used a lot smaller steps, but it would have slowed down the simulation too much. There is no way to move anything on a screen without using steps, and I can't see how nature could avoid that.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 05/10/2018 07:51:03
MY ULTIMATE GO ? (2nd part)

C) Let us now suppose that, when E reaches M´s vertical with its initial “horizontal" speed, apart from starting M´s gravitational pull, M somehow is given an initial speed parallel to E´s, not to the left as in case B but to the right … And not just as big as E´s (as in case B) but much, much bigger (see below how much).
If with the correct values for masses, M-E distance, and initial speeds, we would have our real case, E being the Earth and M the Moon …
We know the result is that E, instead of orbiting around M as in case A, and somehow opposite to what in case B (when E got “fully free” to fall directly onto M), now E is “forced” to bend its trajectory much more, because M´s location is changing in a sense opposite to E´s speed … E and M revolves/rotates around their common center of mass. E´s trajectory curvature is much, much bigger than when orbiting around M (case A).
For similar (but kind of opposite) reasons to case B, it is erroneous to consider E “orbits” around M, and to say that E is in a “free fall” … In cased A we had a “partially free" fall, and in B a “fully free” fall … But now I consider we have no free fall at all, because initial speeds and mutual pulls are causing a kind of dance of the couple, keeping their separation actually constant …
With this and last post I´ve already delivered what “promised”: a fresh explanation of one of the main roots of my stand …
But I´ll leave for another post the elaboration of main consequences of that, as far as I can understand ...   
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 05/10/2018 16:24:15
Let us now suppose that, when E reaches M´s vertical with its initial “horizontal" speed, apart from starting M´s gravitational pull, M somehow is given an initial speed parallel to E´s, not to the left as in case B but to the right …
Notice that when the pulling begins, the motion is straight. If it was already curved, no force would be needed to curve it. A force that curves a trajectory needs to be applied to a straight trajectory, so since we don't observe such a motion at our scale, atoms are probably going straight line for a while when we apply a force on them. They probably change their direction, and then go on straight line, and change their direction again, and go straight line, indefinitely. That's probably the main reason for their quantified energy. Their quantified light is probably linked to their quantified motion. They probably move by steps to accommodate individually each photon that strikes them, and it is probably while they are executing a step that they emit a photon. That would produce straight steps, and each of them would have a specific speed and a specific direction. In the case of a curved motion, the steps would have a constant length in the direction which is perpendicular to the force, and their length in the direction of the force would depend on its intensity. This way, there can be no outward force during an orbital motion, only inward. Nothing forces a body to go straight line, it moves all by itself.
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 05/10/2018 23:35:18
Notice that when the pulling begins, the motion is straight. If it was already curved, no force would be needed to curve it. A force that curves a trajectory needs to be applied to a straight trajectory, so since we don't observe such a motion at our scale, atoms are probably going straight line for a while when we apply a force on them.
The gravitational force is continuous, it doesn’t switch on and off so the object does not have the option of travelling in a straight line even for a fraction of time. In order to keep an object’s orbit (curved path) the gravitational force has to be continuous otherwise it would fly off at a tangent, it will not keep a curved path without that force.


That's probably the main reason for their quantified energy. Their quantified light is probably linked to their quantified motion. They probably move by steps to accommodate individually each photon that strikes them, and it is probably while they are executing a step that they emit a photon.
‘Quantified light’ has nothing to do with quantified motion. The energy of a photon is quantised because the electron, when bound to an atom, can only release or absorb energy in quanta. It’s worth noting that a free electron is not so constrained and can take on any energy value.

Let’s try and keep this thread on real physics rather than made up ideas.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 06/10/2018 07:48:59
MY ULTIMATE GO ? (3rd part)

To diminish the risk of misinterpretations, before delivering the conclusions I referred to on 2nd part, I´ll repeat here the "breakdown" of earth complex movement into its simpler (and more easily graspable) components ...
Earth´s movement (as a whole) is basically the addition of three simple, but quite different, movements:
1) Earth-moon common center of mass orbits the sun (once a year). Both celestial objects, as a "couple", are kind of linked to that CM (called barycenter), and move with it.
 That movement, and sun´s pull, originates what I call sun-related tidal effects (sea tides included).
For the sake of simplicity, I usually disregard that movement, when analyzing main component of tides: moon-related tides.
2) The "couple", as I said yesterday on 2nd part, revolves/rotates around the barycenter (once every some 28 days).
That is the movement I always refer to (unless clearly said the contrary), discussing and analyzing it as if only that movement were happening ... Moon´s pull, and dynamical effects originated by that movement´s features, causes two opposite bulges that continuously change position, logically with same periodicity of some 28 days (relative to the rest of the universe).
Details of the singularity that earth revolves (instead of rotating like the moon) are explained on #328, with a "handy" analogy ...
3) Earth also has its daily spinning, which, apart from causing the permanent equatorial bulge (by the way, thanks to huge centrifugal forces inherent to such fast circular movement), makes us to perceive the period of the movement of the pair of bulges as if it were once a day, instead of once every some 28 days.
Some time ago I posted the link of a youtube video relative to that:
"I suggest anybody interested to have a look at :
where it´s clearly seen that daily movement of the bulges is only apparent, that they are almost still and it is the solid part of our planet (though also the bulk of ocean waters due to friction) what is actually spinning …
The formation of the bulges is a rather slow process (some 28 days the complete cycle) … Nothing to do with all those daily local whirlpools, due to the much faster Earth spin, and with any other local singularity"
 
   
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 06/10/2018 13:16:08
by the way, thanks to huge centrifugal forces inherent to such fast circular movement
There is no more centrifugal force in the case of the equatorial bulge than in the case of the tidal ones. Both forces are due to massive bodies refusing to change their speed or their direction. They resist whether we pull them tangentially or perpendicularly to the motion they already have. They resist as if they were at rest. There is no centrifugal force when a car accelerates, it only resists to accelerate. If it stops accelerating, it only moves in the direction it is already moving, not in the other. The same thing is happening in the case of bodies on circular motion: no centrifugal force is exerted by them.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 06/10/2018 14:43:14
The gravitational force is continuous, it doesn’t switch on and off so the object does not have the option of traveling in a straight line even for a fraction of time.
Gravitation looks continuous at our scale like any force, but considering that the atoms exhibit a randomness behavior when they are observed separately, it may very well be discontinuous at theirs. In other words, the constant forces or motions that we observe at our scale may very well be issued from a statistical phenomenon at theirs, and if it is the case, they could be free to move in whatever direction they chose for a while providing they chose the right one more often. In my simulations on motion, (http://www.motionsimulations.com/) I have no other option than to move my particles by steps, and I also move them in the direction of the force, but in reality, I could move them statistically and I would get the same result after a while. If we could nudge an individual atom for so short a time that it wouldn't have time to exhibit a statistical answer, I think it might very well not take the direction or the speed we expect it to take.

‘Quantified light’ has nothing to do with quantified motion. The energy of a photon is quantised because the electron, when bound to an atom, can only release or absorb energy in quanta. It’s worth noting that a free electron is not so constrained and can take on any energy value.
I wasn't talking about unbounded bodies either: to me, gravitation is a kind of bonding. To follow a curved trajectory, a body has to suffer acceleration, and it is precisely during acceleration that I suspect the particles to behave statistically.

Let’s try and keep this thread on real physics rather than made up ideas.
What I'm suggesting doesn't seem to contradict the observations, but tell me if you see any and I'll stop arguing. I'm not suggesting to change the physics, but to use it to make another step further. In fact, I use what I had to figure out to build my simulations to convince Rmolnav that the equatorial bulge and the tidal ones are equivalent. I'm not really expecting him to change his mind though, just to take a closer look at what the particles might be doing during motion.
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 06/10/2018 16:30:13
What I'm suggesting doesn't seem to contradict the observations, but tell me if you see any and I'll stop arguing. I'm not suggesting to change the physics, but to use it to make another step further.
It's OK to take the physics a (specualtive) step further, but not in this section.
It is worth setting up a new theory to discuss.
While we see quantum effects for energy, there is no evidence that that energy varies as the particle moves ie no stop start or quantised motion.
If you want to discuss it, let's do it elsewhere, here it just causes confusion amongst the readers.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 06/10/2018 17:27:28
It's OK to take the physics a (speculative) step further, but not in this section.
What about the tides then? Aren't we speculating about the composition of the force? The way Rmolnav explains them, a centrifugal force is needed, whereas no centrifugal force is present when no orbital rotation is involved and the tidal bulges are still present.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 06/10/2018 22:01:10
MY ULTIMATE GO ? (4th part)

As said on 2nd part, I consider that rather than talking about “free fall” within a given gravitational field, it is more realistic to consider earth-moon interactions as if they were parts of an unique extended object, with no massive parts apart from where both celestial objects actually are, and rotating/revolving about their common center of mass, maintaining the distance between them due to their dynamic equilibrium.
An ice skating couple can spin like a single dancer, because they are linked usually through their hands … If one of them had a pony-tail, when spinning, instead of hanging vertically, the pony-tail would move upwards, against its own weight … Clearly due to centrifugal force.
The same would happen whatever the way they were linked to each other. If it were a kind of mutual “gravitational” pull sufficient not to need their hands (as earth-moon case), and they had suitable initial "tangential" speeds, the pony-tail would also “feel” that inertial force (why wouldn´t it?) … It would react to the TOTAL force acting DIRECTLY ON IT (WHATEVER the pull on other parts of the dancer, what the pony-tail “ignores" …), and raise where centrifugal force prevailed.
Earth particles react to those inertial forces in a similar way to what quite clearly explained for the diminishing of our weight on the equator, and causes the equatorial bulge (though the author prefers not to mention the adjective “centrifugal” … ):
“"The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v2/R (the centripetal force). Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force (man´s weight) is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation"
https://www.lockhaven.edu/~dsimanek/scenario/centrip.htm
That´s why the NASA scientists I´ve referred to many times say:
"At the center of the Earth there is a balance between gravitational attraction (trying to pull the Earth and moon together) and centrifugal force (trying to push the Earth and moon apart as they revolve around that common point).
At a location on the Earth’s surface closest to the moon, the gravitational attraction of the moon is greater than the centrifugal force of the Earth (moving around the center of the revolving Earth-moon system).
On the opposite side of the Earth, facing away from the moon, the centrifugal force is greater than the moon’s gravitational attraction.
In a hypothetical ocean covering the whole Earth with no continents there will be two tidal bulges resulting from these imbalances of gravitational and centrifugal forces, one facing the moon (where the gravitational force is greater than the centrifugal force) and one facing away from the moon (where the centrifugal force is greater than the gravitational force)”
I DO KNOW many people (physicists included) will say: "But you are using a non-inertial frame of reference, where centrifugal forces are “added” by us, in order to keep Newton´s Motion Laws valid. But those forces are “fictitious”, they don´t actually exist" … (or something similar).
This post is already rather long. I´ll leave my refuting of what usually is deduced from that question of frames of reference for another post.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 07/10/2018 10:50:39
There is no centrifugal force when a car accelerates, it only resists to accelerate. If it stops accelerating, it only moves in the direction it is already moving, not in the other. The same thing is happening in the case of bodies on circular motion: no centrifugal force is exerted by them.
Without sufficient imagination to think there can be manifestations of inertia different than what we are kind of used to, it´s impossible to properly grasp the concept of centrifugal force, and therefore tides ...
If a car accelerates in a straight line, LOGICALLY there cannot be any "centri***al" force, because there is no "center" at all !! Inertia manifests itself "trying" not to let the velocity of the passengers increase, and they somehow "feel" pushed backwards (relatively to the car). But if they had inflatable cushions between their backs and seat´s backs, and the acceleration were sufficiently big compared to the strength of the cushions, these could explode. And for them to blow up two opposite forces are required ... They backward one exerted by the passengers on the seat backs is clearly an inertial force (though not "centrifugal" whatsoever !!).
But if, e.g., one of the passenger shoulder were leaning on a side window glass not sufficiently strong, and the car were turning to the other side when going very fast, the window could get broken ...
Inertia certainly just tries to keep constant the passenger´s speed vector (similarly to the above mentioned straight-line case), but now the window is "forcing" the passenger to turn (centripetal force), and as an "inertial" reaction (3rd Newton´s Motion Law) the passenger pushes outwards on the window and brakes it (centrifugal force).   
I´ve said it many times: inertia manifests itself in different ways, depending on the type of individual forces exerted on the considered object, and especially on the type and degree of "freedom" to move the object actually has !!
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 07/10/2018 14:35:30
now the window is "forcing" the passenger to turn (centripetal force), and as an "inertial" reaction (3rd Newton´s Motion Law) the passenger pushes outwards on the window and brakes it (centrifugal force).
It is not a centrifugal force, but a relative or apparent centrifugal motion that results from the car changing direction and the passenger being forced to follow it. If the passenger was free to move, he would follow a straight line while the car would be moving away from him: in this case, motion is still relative but the driver knows it is the car that moves away because he suffers a force, and the passenger knows it too because he doesn't suffer a force while the car is still moving away.
Title: Re: Why do we have two high tides a day?
Post by: Colin2B on 07/10/2018 15:30:49
It is not a centrifugal force, but a relative or apparent centrifugal motion that results from the car changing direction and the passenger being forced to follow it. If the passenger was free to move, he would follow a straight line while the car would be moving away from him: in this case, motion is still relative but the driver knows it is the car that moves away because he suffers a force, and the passenger knows it too because he doesn't suffer a force while the car is still moving away.
You have to be careful not to mix frames, which is something many people get confused over.
From the inertial frame the car is being forced in a circle by friction with the road and the driver/passengers are trying to travel in a straight line, but again centripetal forces pull them into a curve.
From the rotating frame the driver and passengers feel centrifugal force throwing them radially outwards and they are restrained by a reactive centripetal  force - either friction from the seats or force from the side of the car.
Both views are valid, just don’t mix frames or everyone gets confused.

The radial motion experienced by the observer in the rotating frame can easily be seen by plotting the locus of the tangential motion relative to the observer on the rotating curve. If you’re not sure what I mean I could try and draw it for you.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 07/10/2018 16:54:57
Both views are valid, just don’t mix frames or everyone gets confused.
I didn't mix frames, I simply didn't use the wording. A reference frame is a viewpoint experimented by an observer, so comparing the viewpoints of the two observers like I did is like comparing the two reference frames. If there is no mistake, there is no need to use the reference frame wording for people to understand what is going on. Knowing that one of the two observers is accelerating is enough for them to know that it is the one that is moving away. When acceleration is involved, adding the reference frame principle to the explanations confuses them. There are tons of pages written about the twins paradox on the forums that try to use the reference frame wording without being able to stop the questioning, whereas the questioning immediately stops if we simply say that the accelerating twin is the one that moves away. There is no reason to change reference frames either if we know an observer is accelerating, and if no observer is accelerating, then there is simply no need to use the reference frame principle since there is no use to the observations made by either of the observers anyway, except if they need to meet, and then they can use doppler effect and aberration to know their relative speed and direction. Changing reference frames only helps us to illustrate up to what point inertial motion can be relative, not to know which one of the observers is getting younger or older.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 07/10/2018 17:32:59
From the rotating frame the driver and passengers feel centrifugal force throwing them radially outwards and they are restrained by a reactive centripetal  force - either friction from the seats or force from the side of the car.
Though I´m much "closer" to what you say than to what said by L.R., please kindly note that, if the window glass breaks, an outward quite real force (centrifugal force, in the broad sense of the term) has to exist, whatever the reference system !!
To me the whole issue of reference systems is generally misunderstood and wrongly used.
Locations, speeds, and accelerations can be easily handled using different reference systems: it is just a question of deducting vectors.
But real forces (apart from exclusively "inertial" forces), as far as I can understand, either exist or not, no matter which reference system we use, or from where we observe reality !!
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 07/10/2018 17:39:03
if the window glass breaks, an outward quite real force (centrifugal force, in the broad sense of the term) has to exist, whatever the reference system !!
If the window breaks, it is still the car that will be moving away from the passenger, not the inverse. The passenger will simply be moving in the same direction the car was moving when the window broke.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 07/10/2018 22:29:46
if the window glass breaks, an outward quite real force (centrifugal force, in the broad sense of the term) has to exist, whatever the reference system !!
If the window breaks, it is still the car that will be moving away from the passenger, not the inverse. The passenger will simply be moving in the same direction the car was moving when the window broke.
You insist in speaking in a kind of layman language, without taking into consideration what I already told you: in physics science, "motion" is not a variable ... What "motion" units would you use when calculating real variables of physics science such as momentum, energy, etc ?
And "motion" can´t be what directly causes the breaking of the window glass ... Inertial forces originated by movements, only if not completely "free", can break (or just deform) things !!
The passenger was TRYING to move "in the same direction the car was moving". But the solidity of the window "forced" him to turn with the car for some short time (centripetal force). As a reaction the passenger´s shoulder pushed outwards on the glass, with a real force equal but opposite to centripetal one, until it broke (the window frame was also pushing the glass inwards ...).
And that outwards force has a name: centrifugal force, as I said, in the broad sense of the term.
If you only have layman ideas about physical phenomena, you had better learning something of the science called "physics", especially "dynamics" ... I´m sure there are many sites on the internet where that won´t be difficult.
But let me suggest you something: keep an open mind if you do that ... 
 
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 08/10/2018 06:01:48
As a reaction the passenger´s shoulder pushed outwards on the glass, with a real force equal but opposite to centripetal one, until it broke
It's the glass that pushes inwards on the shoulder, so the shoulder resists to the force, and the glass breaks. Whether the force is direct or centripetal, bodies behave the same when the are accelerated: they resist, and they nevertheless accelerate in the direction of the force, not in the opposed direction. The word centrifugal contains the suffix fugal, that derives from the french word fugue, which means run away, and there is simply no run away when a body resists to a force. We're actually orbiting around one another so close and so fast that tidal bulges are beginning to grow on us. We should shut off the force before we get dislocated! :0)
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 08/10/2018 07:13:17
The word centrifugal contains the suffix fugal, that derives from the french word fugue, which means run away, and there is simply no run away when a body resists to a force.
Come on !
I´m Spanish, and know the origins of "fugal", actually from latin ...
But I said :
"... that outwards force has a name: centrifugal force, as I said, in the broad sense of the term" (mainly not to mislead people, because many dictionaries define "centrifugal force", in quite a restrictive sense, as a ficticious force, linked to the concept of non-inertial force).
But for the adjective "centrifugal" you can find, e.g. on Oxford English Dictionary:
"Physics
Moving or tending to move away from a centre.
The opposite of centripetal”
It is precisely when the circular movements are more kind of avoided when we actually have bigger centrifugal forces !!
By the way, as an “example sentence” they include:
"Einstein warmed to the idea that the gravitational field of the rest of the Universe might explain centrifugal and other inertial forces resulting from acceleration.’ ...
though I dare say that is a too far fetched idea ...
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 08/10/2018 11:28:45
MY ULTIMATE GO ? (5th part)

Before trying and convey my stand on non-inertial frames issue, let us go back to the drawing board, as if we didn´t have any education on dynamics …
Same image with Earth (E) on upper part, and vertically below the Moon (M).
Let us suppose we only know how they move, with kind of the mind of a “smart” child.
And then we learn that (as David Cooper said many times), if Earth and Moon tangential speeds somehow became null, they would begin to accelerate straightly towards each other, because of gravitational mutual attraction …
If just stopping their circular movements, they get “free” to accelerate towards each other … THOSE CIRCULAR MOVEMENTS ARE WHAT, WHEN EXISTING, WERE SOMEHOW COUNTERING THOSE STRAIGHT ACCELERATIONS NOW THEY HAVE GOT !!  No other logical possibility for a “clean” child mind …
How come? Let us see …
Curiously, with a rotating (so called "non-inertial”) frame of reference, same thing happens, but let us say “virtually” …
As the frame of reference rotates with E and M, E and M don´t "rotate" anymore
relatively to that frame of reference … Centripetal forces are not required whatsoever, and gravitational pull would make them fall onto each other … But that´s far from reality …
Then physicists go on: "How can we do “real” maths with that “artificial” (the adjective is mine ...) reference system ? … We have to apply a “fictitious” force, which we´ll call centrifugal force". That way WE GET BACK TO THE REAL SCENARIO … as far as dynamics is concerned.
Well, that force is certainly “fictitious”, because it has been introduced by us to work with that non-inertial frame of reference … But that means that the real circular movements, not actually needing the addition of any “artificial” force, somehow produce an inertial effect that keeps E and M without falling straightly onto each other !! To me (and not only me), that is a REAL centrifugal force !!
In case B (# 362), as E and M were moving horizontally at same speed, Moon´s pull was able not only to bend E´s trajectory, but also to accelerate E straightly towards M with its full strength, quite a “free” acceleration. That “freedom” was made possible by the movement of the moon towards the left, with same horizontal speed as E.
The real case is dynamically quite the opposite: the movement of M in opposition to E´s speed, and with speeds which make them to revolve/rotate around the barycenter, makes impossible the decrease of E-M distance, and even any proper orbiting of E around M. It gives us a “rotating” scenario, where all forces have basically same direction as the line between centers of mass, and where the angular position doesn´t matter much (as far as Earth-Moon dynamics is concerned), as long as we accept centrifugal forces keep the separation constant …
After all, if Earth and Moon were the only celestial objects in the universe, to talk about angular position of the system would have no sense at all …
But, still, they would have to keep rotating.  That way inherent inertial force, the centrifugal force, would keep them without following straightly onto each other !! 
Therefore, as I´ve said many times, I find quite correct what a NOAA scientist told me:
"... to provide a basic description of the forces which create the tides.  It's intended audience were the grade school children and adults of that time.  It used terminology of science and forces which were common in the 1950s.  Such as centrifugal force.  Centrifugal force was always an "imaginary force" (not a real / measurable force).  But that type of description made the concepts easier to understand and explain.  That  description and use of centrifugal force continued to be common practice until the 1970-80's.  At that point, the terminology shifted and the textbooks used in grade schools were changed to use a more modern terminology and description of this "effect" being a result of inertia rather than an "imaginary force”.
Initially I didn´t fully grasp his point ... But later I did.
The problem is that many books, dictionaries included, keep following models with ideas which are several decades out of date …

   
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 08/10/2018 16:48:54
It gives us a “rotating” scenario, where all forces have basically same direction as the line between centers of mass, and where the angular position doesn't matter much (as far as Earth-Moon dynamics is concerned), as long as we accept centrifugal forces keep the separation constant …
After all, if Earth and Moon were the only celestial objects in the universe, to talk about angular position of the system would have no sense at all …
Reference frames were invented to account for inertial motion, not rotation. In a merry go round, we know we are rotating because of the force, and we also know we are orbiting because of the tides. Rotation and orbiting motion are not relative motions since we know we are moving. We can use a Sagnac interferometer to measure our orbital speed, so even if the earth and the moon were alone in the universe, we could know we are moving. That's mainly why I decided to attribute mass and motion to light, and to make simulations based on the absolute direction and speed of light. If I had to make a simulation of orbital motion, I wouldn't use equations based on an instantaneous force like David did, I would use photons that take time to transfer the information, and doppler effect and aberration produced at the source and at the observer as the only available information, apart from the photon's force weakening with distance of course. This way, no need for an external reference to know that the system is moving, and it is the light exchanged between the particles that drives it.
Title: Re: Why do we have two high tides a day?
Post by: David Cooper on 08/10/2018 18:32:25
For any new readers of this thread who may only have read the start and end of it, the correct answer does not involve centrifugal or centripetal force.

The sun's pull on the Earth dictates the direction the Earth moves in much more strongly than the moon's pull, out-gunning it to the point that when the Earth is in between the Sun and moon, the moon is on the outside of the curved path that the Earth is following through space and not on the inside of it, so any imagined role for centrifugal force throwing water up on the sun-side of the Earth is revealed to be a fantasy, and the sun isn't responsible for lifting most of the water on that side either. If water was being thrown upwards by centrifugal force, it would be towards the moon at such times, but there is no such effect (other than from the Earth's rotation which raises the sea all the way round, most strongly at the equator).

The real answer is simply straight-line differential gravity, mostly from the moon. The moon's and sun's gravity decreases over distance, but it falls in strength most strongly over distance with greater proximity to the source, which is why the moon has a greater effect on the tides than the sun even though the sun's pull on the Earth is much stronger - the moon's effect on our tides is greater because the moon is so close to us, leading to its pull reducing more quickly over distance from one side of the Earth to the other, whereas the sun's so far away that its pull is much closer to an even pull across the planet.

The Earth simply moves where the combined pull of moon, sun, Juptier, etc. forces it to go, and the water of the sea has a slightly different amount of force acting on it due to differential gravity. The water nearest the moon is always pulled a bit more strongly towards the moon, and the water furthest from the moon is always pulled less strongly by it, so whenever the Earth is being pulled towards the moon (e.g. when the moon is between the sun and Earth), the water nearest the moon will try to move more quickly towards the moon than the Earth as a whole and the water furthest from the moon will try to move less quickly in that direction, whereas if the Earth is being pulled away from the moon (e.g. when the Earth is between the sun and moon), the water furthest from the moon will be pulled away most strongly (because it's held back the least by the moon's gravity) while the water nearest the moon will be held back most strongly by the moon.

The effect of these forces is that the pressure on the sea nearest to and furthest from the moon reduces a bit and the water tries to rise a tiny amount - it doesn't rise much directly because there isn't enough time for water to rush in from elsewhere to pile up there, so the actual tidal bulges that we see are a secondary effect built up by resonance responding to the input of the tidal forces. The tidal forces from the sun are smaller because the sun's gravity reduces less over distance where we are (it is spreading out less by the time it reaches us). When the tidal forces from the sun are aligned with the moon's, we get the biggest combined force and higher tides. When the tidal forces from the sun are acting at 90 degrees to those from the moon, we get lower tides, but note that the water is still being lifted slightly by the sun's tidal forces on the sides nearest and furthest from it, so it doesn't actually subtract from the moon's tidal forces, but merely masks their strength. This is why it's important to realise that the tides that we see are a secondary effect (through resonance) and not direct lifting of water into bulges.

Until the topic is locked, I'll repost this every page or two just to make sure that no one is misled by authoritative nonsense.
Title: Re: Why do we have two high tides a day?
Post by: Le Repteux on 08/10/2018 19:01:23
But, still, they would have to keep rotating.  That way inherent inertial force, the centrifugal force, would keep them without falling straightly onto each other !! 
It is their speed with regard to one another that keeps them from falling, and speed is not a force, only a motion.
Title: Re: Why do we have two high tides a day?
Post by: rmolnav on 08/10/2018 19:04:53
Reference frames were invented to account for inertial motion, not rotation
Well, I´m afraid your background can´t help me help understand what you actually mean ...
"Rotation" requires acceleration (centripetal) ... Therefore inertial phenomena appears !
I´m not expert on the history of the concept, but if e.g. you visit:
 NON-INERTIAL FRAMES - Oxford Physics - University of Oxfordhttps://users.physics.ox.ac.uk/~harnew/lectures/lecture12-mechanics-handout.pdf
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