[Colin2B]

I agree with Rmolnav about rotation being involved in the tides.

Yes, you might like to look at the NOAA site https://tidesandcurrents.noaa.gov/restles1.html
They take input from survey vessels and buoys all over the world and the barycentre model is the one that most closely matches the observations. It takes account of both differential gravity and rotation.
You have to be careful comparing different frames because although centrifugal force can appear to disappear in some, a rotating earth is not an inertial frame.

[rmolnav]

For clarity purposes, I need to specify that I changed my mind. At first, I agreed with David's differential pulling on free falling bodies, and now, I agree with Rmolnav about orbital rotation being involved in the tides. Please, reread that post and tell me if you understand it differently.

Thank you ... I´m going to reread that post carefully, and reply tomorrow.

[David Cooper]

Yes, you might like to look at the NOAA site https://tidesandcurrents.noaa.gov/restles1.html
They take input from survey vessels and buoys all over the world and the barycentre model is the one that most closely matches the observations. It takes account of both differential gravity and rotation.

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

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

At any given moment, the Earth and moon are effectively under straight-line differential gravity, and that produces all the right numbers. These forces apply in every billionth of a second, and every quintillionth too - the orbit can be completely ignored. Furthermore, the clincher is that the forces are the same at any given moon-Earth distance regardless of whether the directions they are moving in at the time.

[David Cooper]

For clarity purposes, I need to specify that I changed my mind. At first, I agreed with David's differential pulling on free falling bodies, and now, I agree with Rmolnav about orbital rotation being involved in the tides. Please, reread that post and tell me if you understand it differently.

I see the point you were trying to make there now. The reason people are tripping up on this though is precisely that the two explanations produce the same numbers, so it's easy to become confused. If you want to use the centrifugal explanation, you should use it for both sides and recognise that it doesn't have a more significant role on one side than the other. When you get to a case though where two bodies are moving directly towards each other, there is no centrifugal force, but the tidal forces continue to act in full. This, as I've pointed out before, could be considered to be a case of centrifugal force in which no revolution is involved, but if this was accepted as centrifugal force, that then means that when a car breaks in a straight line, you would be able to say that you are thrown forwards by centrifugal force. The word is not normally used that way, but logically it should be because the cause is the same, so the word would have to be given an improved definition to include that case. (The same applies to any alternative wordings involving centripetal force.) That would provide a greater excuse for attributing the tides to centrifugal force, but it is still highly misleading because considering the perpendicular component is entirely superfluous.

[rmolnav]

I agree with Rmolnav about rotation being involved in the tides.

Yes, you might like to look at the NOAA site https://tidesandcurrents.noaa.gov/restles1.html
They take input from survey vessels and buoys all over the world and the barycentre model is the one that most closely matches the observations. It takes account of both differential gravity and rotation.
You have to be careful comparing different frames because although centrifugal force can appear to disappear in some, a rotating earth is not an inertial frame.

Thank you.
As you know, some time ago we commented that NOAA work, which had been kindly mailed to me by one of the authors, with whom I had been discussing the issue ... And months ago I (and you too) copied here not only the link, but also some of what we interchanged ...
But recently I have not posted the link again, preferring to discuss myself basic Physics with D. C. ...  In any case in #155 I copied one of the things that NOAA scientist told me:
""The publication you are referring to is "Our Restless Tides", a 10-page pamphlet developed in the 1950's to provide a basic description of the forces which create the tides.  It's intended audience were the grade school children and adults of that time.  It used terminology of science and forces which were common in the 1950s.  Such as centrifugal force.  Centrifugal force was always an "imaginary force" (not a real / measurable force).  But that type of description made the concepts easier to understand and explain.  That  description and use of centrifugal force continued to be common practice until the 1970-80's.  At that point, the terminology shifted and the textbooks used in grade schools were changed to use a more modern terminology and description of this "effect" being a result of inertia rather than an "imaginary force”...
but D.C., loftly deposing mentioned NOAA scientist, replied:
"Here, we see that the scientist may not have a firm understanding of the tides himself though - all he's doing here is trying to get rid of the idea of centrifugal force by referring to the real mechanism that is behind centrifugal force, but he hasn't realized that this isn't the mechanism behind the tides”
I still have to post some ideas regarding the issue of reference systems, which I think is also misunderstood by many ... But they are not easy to convey !!
They are in line with what I already replied to PmbPhy (?), in the specific thread "What is centrifugal force?" ... You can find it easily, because I think it is the last one, though it was months ago ...
 

[rmolnav]

I'm comparing two kinds of rotation to see if they are equivalent, so when I say orbital speed, I'm talking of the orbital rotation, and when I say rotational speed, I'm talking of the non orbital one.

But you had said:
"... an equatorial bulge would build up progressively on both systems until they reach their orbiting speed, and stopping the speed at that moment would not erase the bulge immediately, it would erase only when they would stop orbiting because they are getting supported again ..."
Sorry but I find it rather confusing:
1) "equatorial bulge ... on both systems" ? ... Moon related bulges happen on low latitude areas, but "equatorial" adj. should be reserved for the "circular" permanent bulge due only to daily spinning ...
2) "... their orbiting speed" ? : you use "orbiting" for the two systems ... ??
3) "...stopping the speed" ... Which speed ??
I suggest you to imagine the moon-earth pair as if it were a single object, but with null mass where there is no real material ...
That "single" object has a some 28 days period "spinning" around its c.g. (the barycenter).
That generates centrifugal forces on all material points, both on moon and on earth.
The ones affecting earth, although the barycenter is some 1/3 earth radius inside it, are all in the sense opposite to moon, because earth only revolves around the barycenter (it is not tidal locked to the moon).
ALL THAT, together with opposite sense gravity field due to the moon, results in what I said last time a couple of days ago:

Quote from: rmolnav on 08/09/2018 19:35:09
... at closer to moon hemisphere stronger moon´s pull prevails. And at further hemisphere, where moon´s pull is smaller, centrifugal force prevails.
... and water from each hemisphere "piles up" (though relatively very, very little) respectively at sublunar and antipodal areas !!

IF WE NOW ADDED earth daily spinnig (within that "imagined" single object) , we would have:

That is why "we have two high moon-related tides a month", and due to the daily earth spinning, we perceive them twice a day (on top of the permanent equatorial bulge, that has nothing to do with moon-earth dynamics).

If somehow what we had were ONLY the daily spinning, there are two possibilities:
1) If somehow earth is keeping at same actual distance from the moon: bulges would depend on the way we kept earth completely still ... In any "imaginable" case, further hemisphere would not deform "outwards" (unless we applied the so called "fictitious" forces on each material point, but It is better not to enter that question now ...)
Equatorial bulge would form independently ...
2) If letting the earth free to accelerate towards the moon:  then for some tens of hours, apart from the permanent equatorial bulge, we would certainly have tides similar to real ones (increasing in size) ...
But, as I already said, that would be due to both differential gravity, and the addition of a manifestation of inertia different to when the real case. Then it is when the explanation of D.C. could be "almost" completely correct:
 

Quote from: David Cooper on 07/09/2018 21:08:26
Differential gravity simply means that further-away parts of it have less gravity acting on them, so they accelerate less in the direction of that straight-line force

Though they actually would "try" to accelerate less, but are "forced" to accelerate the average due to the much bigger own earth gravity and compactness where solid earth ... That would produce internal stresses and changes in water pressure distribution > > deformation and tides ...

[Le Repteux]

These forces apply in every billionth of a second, and every quintillionth too - the orbit can be completely ignored.

If I use my small steps to analyze the orbital motion, then the orbital steps have two components: the inertial one brings the particle away from the orbital path a bit, and the gravitational one puts it back on track at the end. Of course, those two components are executed in the same time, but I see no way to get a curved trajectory on a screen without using the inertial one. I describe my mind experiment differently below in my answer to rmoldav, hoping it will be clearer.

But you had said:
"... an equatorial bulge would build up progressively on both systems until they reach their orbiting speed, and stopping the speed at that moment would not erase the bulge immediately, it would erase only when they would stop orbiting because they are getting supported again ..."
Sorry but I find it rather confusing:
1) "equatorial bulge ... on both systems" ? ... Moon related bulges happen on low latitude areas, but "equatorial" adj. should be reserved for the "circular" permanent bulge due only to daily spinning ...

I agree with your explanation and the NOAA one, so even if you don't seem to understand what I said, I think we simply say the same thing but with different words. I'm trying to properly describe a system where the earth and the moon are united by a long arm, and where their orbital speed is slowed down progressively. Both of them will then progressively be supported by the arm, and the two opposed tidal bulges of each body will progressively become a single inertial (or equatorial) outward one because the two inward ones will be supported by the arm. I think that this mind experiment shows that the two kinds of bulges (inertial and tidal) are equivalent, don't you agree?

[PmbPhy]

The correct explanation is found in texts on classical mechanics. I can probably find them online if anybody wants to read them. They explain it in detail. I.e. why the tidal force on a particle is away from the earth on the moon's side and  when its in the opposite side away from the earth as well, giving rise to the two ocean tides. The addition of the force due to the sun causes spring and neap tides.

[rmolnav]

The correct explanation is found in texts on classical mechanics

Well, surely there must be "correct texts" on classical mechanics (but I´ve also seen some erroneous ...)
And also specific books on tides, as the one I´ve just now referred to on another "thenakedscientist" thread:
Dr. Bruce Parker is the author of the more than 300 pages book linked below, and he spent most of his career in NOAA and much of his time working on tide related problems as a specialty even while tackling jobs with a much broader scope. Positions he held at NOAA included: Chief Scientist of the National Ocean Service; Director of the Coast Survey Development Laboratory; Director of the World Data Center for Oceanography; Principal Investigator for the NOAA Global Sea Level Program; and head of the U.S. national tides and currents program (in a earlier organizational form before it became CO-OPS).
Among his awards are the NOAA Bronze Medal, the Department of Commerce Silver and Gold Medals, and the Commodore Cooper Medal from the International Hydrographic Organization.
Dr. Parker is presently a Visiting Professor at the Center for Maritime Systems at the Stevens Institute of Technology.
 Dr. Parker has written many papers on tidal subjects, some of which are included in the References section of this book, as well as many tidal analysis programs, some still being used in some form in CO-OPS.
He also had the privilege of organizing the program for the International Conference On Tidal Hydrodynamics in 1988 and editing the book that resulted.
Dr. Parker received his Ph.D. in physical oceanography from The Johns Hopkins University, and prior to that an M.S. in physical oceanography from the Massachusetts Institute of Technology, and a B.S./B.A.in biology and physics from Brown University ...
AND HE SAYS:
"At the center of the Earth there is a balance between gravitational attraction (trying to pull the Earth and moon together) and centrifugal force (trying to push the Earth and moon apart as they revolve around that common point).
At a location on the Earth’s surface closest to the moon, the gravitational attraction of the moon is greater than the centrifugal force of the Earth (moving around the center of the revolving Earth-moon system).
On the opposite side of the Earth, facing away from the moon, the centrifugal force is greater than the moon’s gravitational attraction. In a hypothetical ocean covering the whole Earth with no continents (see Figure 2. there will be two tidal bulges resulting from these imbalances of gravitational and centrifugal forces, one facing the moon (where the gravitational force is greater than the centrifugal force), and one facing away from the moon (where the centrifugal force is greater than the gravitational force).
Logically, he also analyses thoroughly local effects of tides all over the world ...
https://tidesandcurrents.noaa.gov/publications/Tidal_Analysis_and_Predictions.pdf

[David Cooper]

"...there will be two tidal bulges resulting from these imbalances of gravitational and centrifugal forces, one facing the moon (where the gravitational force is greater than the centrifugal force), and one facing away from the moon (where the centrifugal force is greater than the gravitational force)."

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

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

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

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

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

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

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

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

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

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

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

This explanation contains misleading, superfluous content. The tidal forces result from differential gravity alone as the perpendicular component of travel has no causal role. It's highly misleading because it encourages people to imagine that the perpendicular movement has an important role, but it has none - the perpendicular motion is incidental, and the "evidence" for the proposed mechanism described in (E) is actually a necessary consequence of differential gravity rather than a cause. The only force involved is gravitational - dividing it into "gravitational" and "centrifugal" is an artificial distinction which has no place in the explanation of tidal forces.

[Colin2B]

The correct explanation is found in texts on classical mechanics. I can probably find them online if anybody wants to read them. They explain it in detail. I.e. why the tidal force on a particle is away from the earth on the moon's side and  when its in the opposite side away from the earth as well, giving rise to the two ocean tides. The addition of the force due to the sun causes spring and neap tides.

I can’t immediately remember which text it’s in, but I assume you mean like this:
http://burro.astr.cwru.edu/Academics/Astr221/Gravity/tides.html

[PmbPhy]

The correct explanation is found in texts on classical mechanics. I can probably find them online if anybody wants to read them. They explain it in detail. I.e. why the tidal force on a particle is away from the earth on the moon's side and  when its in the opposite side away from the earth as well, giving rise to the two ocean tides. The addition of the force due to the sun causes spring and neap tides.

I can’t immediately remember which text it’s in, but I assume you mean like this:
http://burro.astr.cwru.edu/Academics/Astr221/Gravity/tides.html

No. That page is incorrect. I used to make the same mistake myself, even in my first post in this thread. I'll post the ones used at MIT.

[rmolnav]

If you don´t agree with what quoted by me from the "super-expert" Dr. Parker, you had better refuting his paragraphs, rather than proposing new scenarios difficult for you to grasp, due to your difficulties as far as understanding basic Physics laws is concerned ...
E.g.:
 

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

Why are you so sure?
Many things change:
1) Ball-big sphere-ball mass is smaller than the full string of balls.
2) The planet pull would diminish proportionally to mentioned mass has diminished.
3) But the acceleration of the three-part system would be the same (F and m have decreased same proportion).
4) But the "weight" of the balls would decrease.
5) Inertial effects on the ball would be the same (same mass and same real acceleration).
 It is not that simple ...Too many factors for you to draw correct conclusions !!

[Colin2B]

No. That page is incorrect. I used to make the same mistake myself, even in my first post in this thread. I'll post the ones used at MIT.

Thanks Pete, I've come across quite a few like this that don't really cut the mustard, will be interested in what you have.

[David Cooper]

If you don´t agree with what quoted by me from the "super-expert" Dr. Parker, you had better refuting his paragraphs, rather than proposing new scenarios difficult for you to grasp, due to your difficulties as far as understanding basic Physics laws is concerned ...

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

E.g.:
 

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

Why are you so sure?

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

Many things change:

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

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

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

Correction: My use of the word "arbitrary" near the top of this post was incorrect, as there clearly is a section through the moon where the material is following the same orbit it would still follow for a moment even if the rest of the moon disappeared. Interestingly, that section does not include the centre of mass, and the location of this section changes as the moon moves closer to or further away from the planet (due to the differential gravity having a greater difference in strength across the near half of the moon then over the far half). Anyway, this means that it's a sensible place to make an artificial distinction between the part of gravity that you're counting as gravity and the other part that you're counting as centripetal force, so the error is only in making an artificial distinction (and not in doing so in an arbitrary way).

[Le Repteux]

I'm trying to properly describe a system where the earth and the moon are united by a long arm, and where their orbital speed is slowed down progressively. Both of them will then progressively be supported by the arm instead of falling towards one another, and the two opposed tidal bulges of each body will progressively become a single inertial (or equatorial) outward one because the two inward ones will progressively be supported by the arm.

I need to better understand how the tidal bulges would transform into inertial ones while we would progressively slow down the earth, so here is my second thought. If the actual orbital speed of the earth/moon system would suddenly slow down a bit, the two bodies would begin falling towards one another, the gravitational force would increase, and the tides would also begin to increase. At that moment, a supporting arm would not completely prevent the earth from falling, it would only prevent its inward tidal bulge from increasing, but without preventing the outward one from doing so. The earth would thus go on falling until the inward bulge would be completely supported, and its outward bulge would go on increasing during that time. One half of the earth would then stop falling, but not necessarily the other half. If that half was still going faster than orbital speed, then it would still be in free fall and it would stay a tidal bulge. For that bulge to become an inertial one, the speed would have to progressively get down again until the last bit of the earth would not be free falling anymore, and the resultant inertial bulge would then automatically be smaller than the previous tidal one.

I said in my previous analysis that the two kinds of bulges seemed equivalent, but are they really? If they were, there should be no way to differentiate them, and there is: one is in free fall and the other is not, and at almost the same rotational speed the inertial bulge is substantially smaller than the tidal one. Am I going to change my mind again? I feel like my last bit of earth not knowing if it is tidal or inertial. :0)

[PmbPhy]

Thanks Pete, I've come across quite a few like this that don't really cut the mustard, will be interested in what you have.

You're most welcome. See:
http://www.newenglandphysics.org/other/Taylor_Tides.pdf

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

See also:
http://www.newenglandphysics.org/other/French_Tides.pdf

[Le Repteux]

For that bulge to become an inertial one, the speed would have to progressively get down again until the last bit of the earth would not be free falling anymore, and the resultant inertial bulge would then automatically be smaller than the previous tidal one.

I made a mistake: for the outward bulge to become an inertial one, the earth must simply go on falling until the whole earth is supported by the arm, and whenever we slow the earth/moon system down a bit, the two bodies will fall towards one another until they get to their perigee, so for the whole earth to be supported by the arm, there is no need to slow it down again as I was suggesting, only wait till it completely rests on the support, what would not prevent the inertial bulge to reveal itself while the tidal one would collapse. It looks as if the inertial bulge was contained in the tidal one, or as if the tidal one would grow over the inertial one. If we would stop the rotation once the whole earth is supported though, the inertial bulge would collapse, and if we would then take off the arm, the tidal ones would build up, so this way at least, the two kinds of bulges don't appear to be linked. What about the NOAA explanation then? Can we show that their calculation or their drawings are wrong for instance? Unfortunately, there is no calculation to support their drawings, and the drawings themselves are only qualitative.

[Colin2B]

You're most welcome. See:
............

Thanks Pete, I’m not in good WiFi area at moment, when I am I will read through.
If I have time I’ll read through this thread and see what these guys are arguing about.

PS French Tides sounds a bit dodgy, didn’t think you were into that 

[rmolnav]

It looks as if the inertial bulge was contained in the tidal one, or as if the tidal one would grow over the inertial one

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