# Science Questions

## Why are there two high tides a day?

Tue, 7th Jan 2014

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### Question

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

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.

#### Make a comment

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. Pmb, Sat, 7th Dec 2013

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. David Cooper, Sat, 7th Dec 2013

pmb,
How does gravity push water away from the moon?
Bored chemist, Sat, 7th Dec 2013

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. A Davis, Sun, 8th Dec 2013

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.

However, the point on Earth's surface closest to the Moon is closer than the center of the Earth, and feels a stronger pull than the center of the Earth, so water piles up higher there (a high tide).

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.

Now, if you add a second mass (the Sun), which does not always line up with the Moon:

When the Sun & Moon line up (Full Moon & New Moon), their tides add, leading to a very high tide.

When the Sun & Moon are at 90 degrees apart in the sky (Half Moon), their tidal influences almost cancel, leading to a much reduced tidal range.

At certain times of the year (near eclipses), the Sun, Moon & Earth line up in the same plane, and also when the Moon's elliptical orbit brings it particularly close to the Earth, the high tides are slightly higher than the monthly high tides - a "king tide".
evan_au, Sun, 8th Dec 2013

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. David Cooper, Sun, 8th Dec 2013

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.
Pmb, Sun, 8th Dec 2013

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...  bizerl, Mon, 9th Dec 2013

I was lead to believe that "centrifugal" was a dirty word in physics? bizerl, Mon, 9th Dec 2013

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 Pmb, Mon, 9th Dec 2013

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. A Davis, Tue, 10th Dec 2013

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. David Cooper, Tue, 10th Dec 2013

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. A Davis, Sun, 15th Dec 2013

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 - 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. David Cooper, Thu, 26th Dec 2013

While force of gravity follows an inverse square law with distance, the tidal force follows an inverse cube law 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. evan_au, Fri, 27th Dec 2013

Thanks - that's a good link. It also agrees that the simplest explanation in this case is indeed the right one. David Cooper, Sat, 28th Dec 2013

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. Colmik, Fri, 10th Jan 2014

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 Bill, Mon, 4th May 2015

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.
rmolnav, Fri, 29th May 2015

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 PmbPhy, Fri, 29th May 2015

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? rmolnav, Fri, 29th May 2015

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 PmbPhy, Fri, 29th May 2015

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.
rmolnav, Fri, 29th May 2015

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?

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. PmbPhy, Fri, 29th May 2015

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. PmbPhy, Fri, 29th May 2015

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. PmbPhy, Fri, 29th May 2015

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 ...
rmolnav, Sun, 31st May 2015

Sorry. An "n" is missing at "But the object does**´t suffer that reaction force". It is clear, isn´t it? rmolnav, Sun, 31st May 2015

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).
rmolnav, Mon, 1st Jun 2015

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.  rmolnav, Mon, 1st Jun 2015

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
PmbPhy, Tue, 2nd Jun 2015

"... 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.

rmolnav, Tue, 2nd Jun 2015

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. rmolnav, Tue, 2nd Jun 2015

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.
rmolnav, Thu, 4th Jun 2015

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. PmbPhy, Thu, 4th Jun 2015

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 ...

rmolnav, Fri, 5th Jun 2015

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.

rmolnav, Fri, 5th Jun 2015

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 Niño 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?
rmolnav, Sun, 7th Jun 2015

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? PmbPhy, Sun, 7th Jun 2015

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.
rmolnav, Mon, 8th Jun 2015

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.

rmolnav, Sat, 13th Jun 2015

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 (360º/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.  rmolnav, Sun, 14th Jun 2015

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.  rmolnav, Wed, 17th Jun 2015

The 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. evan_au, Thu, 18th Jun 2015

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 360º 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.

rmolnav, Thu, 18th Jun 2015

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. rmolnav, Thu, 18th Jun 2015

Who "we"? Southampton has four high tides per day, some Mediterranean ports have almost no tidal range at all. alancalverd, Thu, 18th Jun 2015

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 90º 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 ...
rmolnav, Sat, 20th Jun 2015

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
rmolnav, Thu, 1st Oct 2015

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! ProjectSailor, Thu, 1st Oct 2015

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. rmolnav, Sat, 3rd Oct 2015

(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".

rmolnav, Tue, 13th Oct 2015

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)
rmolnav, Sun, 1st Nov 2015

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. Wiz, Thu, 28th Jul 2016

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. rmolnav, Mon, 3rd Oct 2016

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

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. evan_au, Mon, 3rd Oct 2016

#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) nº1 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 "nº1" 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).
rmolnav, Tue, 4th Oct 2016

This implies that the nº1 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. evan_au, Tue, 4th Oct 2016

What?

A centripetal force acts towards the centre of rotation... chris, Tue, 4th Oct 2016

#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. rmolnav, Wed, 5th Oct 2016

#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.
rmolnav, Wed, 5th Oct 2016

#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) (!!!) rmolnav, Thu, 6th Oct 2016

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 ...
rmolnav, Thu, 1st Dec 2016

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. Janus, Fri, 2nd Dec 2016

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), 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. evan_au, Sat, 3rd Dec 2016

#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. rmolnav, Tue, 6th Dec 2016

#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 !!
rmolnav, Tue, 6th Dec 2016

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. evan_au, Tue, 6th Dec 2016

#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 !!
rmolnav, Wed, 7th Dec 2016

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. Bill S, Wed, 7th Dec 2016

#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. rmolnav, Wed, 7th Dec 2016

#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 90º 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. rmolnav, Thu, 8th Dec 2016

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