[PmbPhy]

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?

[rmolnav]

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]

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]

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]

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. 

[evan_au]

In the universe there have been satellites which even broke/exploded due to really high internal, tidal stresses.

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.

[rmolnav]

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]

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.

[alancalverd]

Who "we"? Southampton has four high tides per day, some Mediterranean ports have almost no tidal range at all.

[rmolnav]

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]

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
 

[ProjectSailor]

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!

[rmolnav]

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]

(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

[rmolnav]

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)
 

[Wiz]

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.

[rmolnav]

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.

[evan_au]

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

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.

[rmolnav]

#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).

[evan_au]

nº1 bulge, due to lunar attraction, logically happens below the Moon

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.