[David Cooper]

The problem is not intelligence, thus conscious thinking, but unconscious one.

It's very rarely an intelligence issue. It's usually more about running the wrong algorithm. If you put too much trust in a rule with a fault in it, that sabotages any thinking you do that is dependent on that rule, regardless of how faultless your thinking is when applying that rule. A lot of the rules are deeply hidden though, and that makes them hard to test and to fix if they're faulty.

Our ideas change with time, because it takes time for chance to change anything durably.

There are some people who change position rapidly though - the slightest hint of a possible fault and they check everything they're doing, mend the broken part and then jump straight to an improved position. What is it that allows some people to fly while others can only drag themselves through the dust on their stomach?

[Le Repteux]

There are some people who change position rapidly though

I've been about ten years on the forums questioning Relativity. We were probably thousands out there doing that, but the only guy I saw changing his mind is me, and it still took ten years. :0) People that change their mind fast enough for the phenomenon to be readily observable are probably the exception. When I saw the photon traveling sideways in the laser of your simulation, I considered myself as a very lucky exception even if knew you helped a lot with your phlegmatic attitude. You had to repeat a few times before I saw the light, exactly like what you do here with Rmolnav, but considering that chance isn't part of the game as far as understanding ideas is concerned is no good for another kind of understanding: the one about feelings. We can easily get upset repeating things that we know right if we consider that anything else than chance is involved. On the other hand, if we know it's also a chance issue, we know it's a time one, and we can more easily let it work all by itself for a while. After all, ideas are not about immediate issues like being threatened by a firearm for instance.

[rmolnav]

Here's the point: if the centripetal force is always equal to the centrifugal force for the moon as a whole (or Earth, or anything else), you have no mechanism for keeping it in an elliptical orbit because it will either spiral out and never come back or spiral in collide with the other body.
Why don't you respond to that issue and explain how the position of equality of centripetal and centrifugal force remains at the middle of the body instead of varying in the manner it would need to to support an elliptical orbit.

However well anybody could explain that to you, you wouldn´t get it at all !!.
I´ve already said the question is far more complex than what you think:
- In most of earth´s locations, the center of curvature of its elliptical orbit is not exactly in the direction of the barycenter.
- At locations where the ellipse is more/less curved than if it were a circle, the radius of curvature is
smaller/bigger respectively.
- The required centripetal force (per unit of mass) is ω²r, so is associated inertial centrifugal force.
It is absolutely impossible for you to understand all that stuff, due to utter ignorance on basic Physics.
You yourself said a few days ago:
"(if the angular speed means the component of speed perpendicular to the centripetal force)"
and following day you said:

(I looked up angular speed to check it's meaning and it appeared to mean what I said.)

[/b]
You utterly ignore what a quite basic concept, angular speed, is. Educated adults learnt that as teenagers.
But you, even after looking up its meaning (something I can´t remember if I ever had to do at all), you keep being wrong !! Do you have an own flawed dictionary, matching those unbelievably erroneous  ideas of you ??
Because,
"Angular velocity, (is the) time rate at which an object rotates, or revolves, about an axis, or at which the angular displacement between two bodies changes.
In ... physics, angles are usually expressed in radians and angular velocities in radians per second. These measures are related through the following conversion factors: 1 degree equals π/180 (about 0.0175) radian; 1 rpm equals π/30 (about 0.105) radian per second.
In many situations, an angular velocity—usually symbolized by the Greek letter omega (ω) …”
(Encyclopaedia Britannica).
And you dare say:

You are pushing fake physics and that needs to be challenged. I may be getting some things wrong here and there, and I expect people to point that out so that I can improve my understanding of things. I actually learn from that and gain, correcting any mistakes that I've been making. You don't though ...

Your quite wrong basic ideas, deeply rooted in your mind, actually prevent you any correct learning.
Its not my problem if you have to say:

If you think I'm not understanding the symbols you're throwing at me, why not try illustrating their usage with actual numbers and clear descriptions.

The problem is completely yours ... Why would you understand me better than definitions on dictionaries ??
By the way, you also said you had recently looked up the definition of "centripetal force", your "grey area" (what I even had previously included for you on one of my posts ...). Surely you also got it wrong, as you ignore what angular speed is !!

[Le Repteux]

Shake hands and say goodby, the game is over and nobody scored, I mean nobody succeeded to convince anybody that wasn't already convinced. I even tried to change viewpoints once in case I would understand better what the other camp was understanding, and it didn't work. As far as understanding ideas is concerned, good will doesn't even help. It does as far as feelings are concerned though, so smile, you're on candid camera!  :0)

[jimbobghost]

I was interested in this topic, as an old sailor; altho I have little knowledge in the science as discussed.
i would like to thank all those who participated in a robust discussion, in which i learned very much (altho the depth of the science remains over my head.

i was impressed by the vigor of the discussion, without anyone coming to blows ...and the tolerance of the moderator for not arbitrarily deleting comments or in any way moving them. (i have been in other forums in which the moderators enjoyed displaying their power)

i look forward to following other topics with as much to offer.

[David Cooper]

Here's the point: if the centripetal force is always equal to the centrifugal force for the moon as a whole (or Earth, or anything else), you have no mechanism for keeping it in an elliptical orbit because it will either spiral out and never come back or spiral in collide with the other body.
Why don't you respond to that issue and explain how the position of equality of centripetal and centrifugal force remains at the middle of the body instead of varying in the manner it would need to to support an elliptical orbit.

However well anybody could explain that to you, you wouldn´t get it at all !!.

I'm not convinced that you are able to do the maths. I produced a computer program which generated the right numbers for an orbiting planet and moon and the tidal forces generated. I asked you to produce your maths routine for me to fit into the program in place of mine. You still haven't taken up that invitation, which is a pity, because producing a program would force you to produce your maths, and that would help everyone understand your position properly. We could write code to show diagrams of how the point of balance between centripetal and centrifugal force moves relative to the centre of the moon and planet throughout each orbit.

I´ve already said the question is far more complex than what you think:

...which is presumably why you can't do the maths. And it never occurs to you that the simplest explanation which doesn't produce all that complexity (and which fits all the situations where your explanation breaks down) might be superior (i.e. correct rather than wrong).

- In most of earth´s locations, the center of curvature of its elliptical orbit is not exactly in the direction of the barycenter.

It sounds as if there's some interesting maths involved there which would be worth exploring, so why hide it? And if it's hard, I'm sure there are some experts here who can help. It sounds as if that bit's just a matter of calculating the gradient of a line perpendicular to the tangent to the ellipse at a specific point.

- At locations where the ellipse is more/less curved than if it were a circle, the radius of curvature is smaller/bigger respectively.

More maths help needed for that then.

- The required centripetal force (per unit of mass) is ω²r, so is associated inertial centrifugal force.

So, are you applying this from the centre of curvature then instead of from where the actual centripetal force is coming from?

It is absolutely impossible for you to understand all that stuff, due to utter ignorance on basic Physics.

Not at all - I can understand things fine once you put numbers to them, or spell out your methods. I'd never have guessed that you were cheating on the direction in which the centripetal and centrifugal force are being applied, but it's beginning to look as if you might be. If so, then your physics is even more fake than I thought. That's why you need to lay all your cards on the table so that everyone can see exactly what kind of game you're playing here.

You yourself said a few days ago:
"(if the angular speed means the component of speed perpendicular to the centripetal force)"
and following day you said:

(I looked up angular speed to check it's meaning and it appeared to mean what I said.)

[/b]

And at any single moment in time, what direction is that speed measured in? Can it be anything other than perpendicular to the line in which the centripetal force is acting? Clearly if you're measuring it over 5 degrees, it's more complex, but we're dealing here with a moment in time in which two forces (centripetal and centrifugal) are to be calculated for a moon or planet - we don't need to wait for it to move even a millionth of a degree to measure that because all we need's a value for that instant, and the resulting speed is equal to the perpendicular component of movement.

You utterly ignore what a quite basic concept, angular speed, is. Educated adults learnt that as teenagers.
But you, even after looking up its meaning (something I can´t remember if I ever had to do at all), you keep being wrong !! Do you have an own flawed dictionary, matching those unbelievably erroneous  ideas of you ??

It was Wikipedia, but it doesn't discuss the issue of measuring the angular speed for an instant as opposed to over a distance. There must be an angular speed for zero degrees though, and so far as I can see, that should be the perpendicular component of the movement. Is that wrong? If so, tell me what the correct answer is so that I can learn something.

Because,
"Angular velocity, (is the) time rate at which an object rotates, or revolves, about an axis, or at which the angular displacement between two bodies changes.
In ... physics, angles are usually expressed in radians and angular velocities in radians per second. These measures are related through the following conversion factors: 1 degree equals π/180 (about 0.0175) radian; 1 rpm equals π/30 (about 0.105) radian per second.
In many situations, an angular velocity—usually symbolized by the Greek letter omega (ω) …”
(Encyclopaedia Britannica).

Ah, I get it now - I had misunderstood that in a big way. So it would be the same angular speed for two objects with one twice as far away and moving twice as fast. A simple misunderstanding which can be corrected with ease. Thanks for your help. But what does this do to the balance between centripetal and centrifugal force? Does it put them out of balance at some points of the orbit, and if so, how far out of balance can it go? This is what I want to see, and it's why it would really help if you'd provide some actual numbers to use as illustrations.

And you dare say:

You are pushing fake physics and that needs to be challenged. I may be getting some things wrong here and there, and I expect people to point that out so that I can improve my understanding of things. I actually learn from that and gain, correcting any mistakes that I've been making. You don't though ...

Your quite wrong basic ideas, deeply rooted in your mind, actually prevent you any correct learning.

I make a few mistakes here and there, and like I said, I correct them as soon as they become clear to me. This rotating frame stuff isn't something I've explored before as it's warped physics, so I expect to get things wrong here and there, and there may be places where I'm mixing rotating and non-rotating frames too, leading to further errors. I'll keep fixing those errors whenever they show up. None of that will alter the fact that this is warped physics though - centrifugal force is not real, and using it as part of a mechanism in an explanation of tidal forces is a very bad idea indeed. I'm still interested in exploring it though just because it's an interesting abstraction, and that's why I'd like to put it into the computer program. I can't do that though until I understand the algorithm you're running.

Its not my problem if you have to say:

If you think I'm not understanding the symbols you're throwing at me, why not try illustrating their usage with actual numbers and clear descriptions.

The problem is completely yours ... Why would you understand me better than definitions on dictionaries ??

You're the one who has the algorithm that I'm trying to extract. I assume that you understand it, and I want to see how the centrifugal and centripetal forces do battle for the moon and planet to make them follow their elliptical paths. Up until now, I took from your words that they are always equal at the centre of the body in question, but that would lead to a spiral orbit. I'd have thought you'd be keen to show how that is avoided. Is there some reason why you don't want to provide any illustrations through numbers, or are you just unable to calculate them due to complications? If it's the latter, how do you know that they aren't so unequal that the point of balance is sometimes outside of the moon or planet entirely (as I suggested earlier)?

By the way, you also said you had recently looked up the definition of "centripetal force", your "grey area" (what I even had previously included for you on one of my posts ...). Surely you also got it wrong, as you ignore what angular speed is !!

What has understanding angular speed got to do with the fundamental difference between a centripetal force that's balanced by reactive centrifugal force and a centripetal force that isn't balanced by anything?

[rmolnav]

I need more time to go slowly through your last post, and reply properly ...
But, in relation to the definition of "angular speed", I wonder how you can say:

It was Wikipedia, but it doesn't discuss the issue of measuring the angular speed for an instant as opposed to over a distance. There must be an angular speed for zero degrees though, and so far as I can see, that should be the perpendicular component of the movement

Their article is long. I had a look over it minutes ago, and haven´t seen what you say ...
Its beginning is:
"In physics, the angular velocity of a particle is the rate at which it rotates around a chosen center point: that is, the time rate of change of its angular displacement relative to the origin (i.e. in layman's terms: how quickly an object goes around something over a period of time - e.g. how fast the earth orbits the sun). It is measured in angle per unit time, radians per second in SI units, and is usually represented by the symbol omega (ω, sometimes Ω).
By the way, "ω", one of the symbols you said I´m "throwing at you" ... when what I had said was:
"- ω²r (angular speed ω and distance to barycenter r)".
Do you need a kind of safety helmet for such "dangerous" cases?

[rmolnav]

I was interested in this topic, as an old sailor; altho I have little knowledge in the science as discussed.
I would like to thank all those who participated in a robust discussion, in which i learned very much (altho the depth of the science remains over my head.
I was impressed by the vigor of the discussion, without anyone coming to blows ...and the tolerance of the moderator for not arbitrarily deleting comments or in any way moving them. (i have been in other forums in which the moderators enjoyed displaying their power)
I look forward to following other topics with as much to offer.

You are welcome!
I´m usually not that "vigorous" ... The bottom of the question is rather tricky, and I do understand other side stand. But I can´t understand why people with serious misconceptions in basic Physics can boldly despise, not my stand (that obviously could be erroneous) but basic Physics principles and works of eminent scientists (see #421).
I´m not going to "invite" you to join the discussion, because if you have little knowledge in what discussed, most probably you would say erroneous things.
But had you any question about basic details, don´t hesitate and ask. I´ll try and do my best.
When I started here (more than three years ago !!) I sent posts trying to put clear rather simple things, a necessary kind of foundation for any further discussion (you can see, e.g., #35 and 38).
And I had to repeat things ... Nobody could now read all that, but I can help telling people on which posts some particular question was dealt with.
By the way, we have not been continuously discussing all that time. There were a couple of "stand by" periods of several months. And most posts were sent in last few months (discussing with people who had not intervened previously, apart from a few cases) ... 

[David Cooper]

@ rmolnav

I´ve already said the question is far more complex than what you think:
- In most of earth´s locations, the center of curvature of its elliptical orbit is not exactly in the direction of the barycenter.
- At locations where the ellipse is more/less curved than if it were a circle, the radius of curvature is smaller/bigger respectively.

Because of the first line of that, I misunderstood where you were going with those comments last time. In reality, you were actually just stating the really obvious: these "complications" aren't complications at all and should cause us no difficulty whatsoever. The barycentre is exactly where the barycentre is expected to be and the moon and planet are exactly where they should be. If we're going to put your method into a simulation, we'll always know where these three things are, how they're moving and their direction of travel - we don't need to make any calculations about the shape of the orbit as it will come out automatically if the rules are right. What we need to do is run the simulation using your rules, but I need clarification from you as to how they apply.

- ω²r (angular speed ω and distance to barycenter r)
are in balance with each other.

This can be applied by working out the moon/Earth's component of speed perpendicular to the direction of the centripetal force, then working out the angle the object would move in a given length of time as viewed from the barycentre. Do you agree with that? (We always have coordinates for these three things and vectors for their movements, so we simply work from those.)

What I most want to know though is how you would calculate the centripetal force. Is it done by applying it from the barycentre too? If you apply the Earth's pull on the moon from the Earth in a conventional way, it will be too weak once you've subtracted the centrifugal force from it, but centrifugal force is an imaginary phenomenon which is only relevant to rotating frames, so do you move this (the source of the centripetal force) to the barycentre to compensate? If so, then that means you'd also have to move the origin of the moon's pull on the Earth to the barycentre as well.

What I want to do is add this on top of my existing simulation to see if your method makes the planet and moon there follow the same orbit as is generated by the existing code. Once that works, we can then continue to apply your method to calculate the tidal forces and see if they too match up to the ones calculated by differential gravity, and we can also look to see how the centripetal and centrifugal forces compare from one side of each body to the other. I'm sure that the numbers will match up when we've got it right, but this is something that needs to be done as it will clarify exactly what you're doing.

[rmolnav]

We could write code to show diagrams of how the point of balance between centripetal and centrifugal force moves relative to the centre of the moon and planet throughout each orbit.

You haven´t got it right yet … No wonder, as centripetal force is your “grey area” … (one of them, actually).
For any considered centripetal force, causing the turning of an object linear speed, there is an opposite centrifugal force equal in size, as a manifestation of inertia. There is never any imbalance between them.
You must be calling centripetal force what is not exactly (=mathematically) that.

Quote
- In most of earth´s locations, the center of curvature of its elliptical orbit is not exactly in the direction of the barycenter.
It sounds as if there's some interesting maths involved there which would be worth exploring

Let us suppose, e.g., the orbiting object is at one of the extremes of the minor axis of its elliptical orbit. It is obvious that the perpendicular to the orbit is mentioned axis, and the center of curvature has to be a point of that axis.
But the other object which causes the gravitational field is on one of the focusses of the ellipse, separated from minor axis (unless it were the singular case of a circle, where the two focusses join to be its “center”).

Quote
- At locations where the ellipse is more/less curved than if it were a circle, the radius of curvature is smaller/bigger respectively.
More maths help needed for that then.

I already referred to those mathematical concepts (#432):
"Curvature, in mathematics, (is) the rate of change of direction of a curve with respect to distance along the curve. At every point on a circle, the curvature is the reciprocal of the radius; for other curves (and straight lines, which can be regarded as circles of infinite radius), the curvature is the reciprocal of the radius of the circle that most closely conforms to the curve at the given point (see figure)”. (Encyclopedia Britannica).

Quote
- The required centripetal force (per unit of mass) is ω²r, so is associated inertial centrifugal force.
So, are you applying this from the centre of curvature then instead of from where the actual centripetal force is coming from?

Where do you think the "actual centripetal force" is pointing at?

But what does this do to the balance between centripetal and centrifugal force? Does it put them out of balance at some points of the orbit, and if so, how far out of balance can it go?

See what above ...

I want to see how the centrifugal and centripetal forces do battle for the moon and planet to make them follow their elliptical paths.

Once more: you got it wrong. Those forces don´t “battle for the moon and planet to make them follow their elliptical paths" whatsoever ...

I took from your words that they (centripetal and centrifugal f.) are always equal at the centre of the body in question, but that would lead to a spiral orbit

Please kindly tell us where “my words” you took that from are … I´ve never said they are equal ONLY at the center of the body: They are ALWAYS equal (each considered pair) !!
And, is that “spiral orbit” the result of one of your so good computer programs??

how do you know that they (centripetal and centrifugal f.) aren't so unequal that the point of balance is sometimes outside of the moon or planet entirely (as I suggested earlier)?

Come on ! How deeply rooted are your misconceptions !!

Quote
By the way, you also said you had recently looked up the definition of "centripetal force", your "grey area" (what I even had previously included for you on one of my posts ...). Surely you also got it wrong, as you ignore what angular speed is !!
What has understanding angular speed got to do with the fundamental difference between a centripetal force that's balanced by reactive centrifugal force and a centripetal force that isn't balanced by anything?

My last sentence applies here too ...

[rmolnav]

I was interested in this topic, as an old sailor; altho I have little knowledge in the science as discussed.

Youtube video of a conference by eminent astronomer mentioned by me on #421 titled:
What if the Moon Didn't Exist? — Neil F. Comins
could interest you.
You can see he is really an authority just googling "Neil F. Comins".
It´s rather long, because audience, mainly young students (by the way, not "too" knowledgable), is asked questions by NFC, and he answers them ...
In case you don´t want to be watching it for so much time, he refers to the issue we are discussing here (causes of tides) since app. time 15:40.

[David Cooper]

We could write code to show diagrams of how the point of balance between centripetal and centrifugal force moves relative to the centre of the moon and planet throughout each orbit.

You haven´t got it right yet … No wonder, as centripetal force is your “grey area” … (one of them, actually).
For any considered centripetal force, causing the turning of an object linear speed, there is an opposite centrifugal force equal in size, as a manifestation of inertia. There is never any imbalance between them.
You must be calling centripetal force what is not exactly (=mathematically) that.

If there is never any imbalance between them, how do you prevent spiral orbits? A circular orbit might work, but as soon as you have an elliptical one, your balanced forces are unable to accelerate or decelerate an object that's moving outwards or inwards, so it will keep moving outwards or inwards and will never switch from one to the other.

Once more: you got it wrong. Those forces don´t “battle for the moon and planet to make them follow their elliptical paths" whatsoever ...

In your model then, do they follow elliptical paths by magic? You're pushing a particular explanation which is based on the physics of an abstraction rather than on real physics, but if it deserves to be called physics at all it should still work mathematically, so it should be possible to simulate it. Do you not want to see it being simulated and functioning as claimed? Is there something about the way it works that you want to hide? You have a programmer here who is willing to build this for you for free - an opportunity not to be missed. All I need from you is a clear description of your mechanisms.

I took from your words that they (centripetal and centrifugal f.) are always equal at the centre of the body in question, but that would lead to a spiral orbit

Please kindly tell us where “my words” you took that from are … I´ve never said they are equal ONLY at the center of the body: They are ALWAYS equal (each considered pair) !!

It's hard to work out what you mean when so many of the things you say seem to contradict other things elsewhere. If you have centrifugal force winning out on the far side and forming a tidal bulge, how can it be equal to the centripetal force there? If centrifugal force is losing out on the near side and extra gravitational pull is forming a tidal bulge there too, again how can the centrifugal force be equal to the centripetal force there? That's why I want to see actual numbers so that I can get a clear picture of what you're talking about, but better still would be clear rules of how they'd be calculated so that I can write a program to do all the hard work for you for a multitude of cases.

And, is that “spiral orbit” the result of one of your so good computer programs??

I'm predicting that your program will produce spiral orbits if you have the two forces in balance all the time. It will only fail to produce a spiral orbit if it's a circular orbit. Let's put that to the test by writing the actual program. All I need is your rules. You've given me enough to calculate the centrifugal force, but I need to know how you calculate the centripetal force and where it is applied from (meaning from the barycentre or directly from the moon/planet as in real physics).

how do you know that they (centripetal and centrifugal f.) aren't so unequal that the point of balance is sometimes outside of the moon or planet entirely (as I suggested earlier)?

Come on ! How deeply rooted are your misconceptions !!

Are you unable to explain how to put my misconceptions right? If the centripetal and centrifugal forces are in balance, there is no outward or inward acceleration, but an elliptical orbit requires times when there is an inward or outward acceleration. Can you not see the problem there? Does your mechanism only work for circular orbits?

What has understanding angular speed got to do with the fundamental difference between a centripetal force that's balanced by reactive centrifugal force and a centripetal force that isn't balanced by anything?

My last sentence applies here too ...

No it doesn't - you appear to be blinded by the fake physics of rotating frames, imagining that they provide a view of reality. In real physics, gravity is not balanced by any opposing force.

[rmolnav]

Sorry, but I´m running out of patience ...
After hundreds of posts discussing the issue, you keep altering my words and mixing up concepts time and again. This way we might reach post #1000 !
E.g.: I´ve directly referred to the cause of antipodal bulge tens and tens of times ... Please kindly tell us where "on earth" I said what in italics: in its supposed "battle" with centripetal force we...

... have centrifugal force winning out on the far side and forming a tidal bulge, how can it be equal to the centripetal force there?

I regret to have to say it again, but I´m afraid you´ve got a problem either with your vision or with your reason ... or both.

[rmolnav]

I was interested in this topic, as an old sailor; altho I have little knowledge in the science as discussed.

As a continuation of my reply of yesterday, just to tell you that before than the timing I gave, NFC starts to say interesting things about tides, app. at 06:00. I had the other figure on a file with the reference since months ago, but this morning I watched the video again, and learnt that.
In fact, he starts talking about another "show" or conference by defenders of the idea that differential gravity is what actually causes tides, and includes a clip of a video ... He clearly says that is wrong, and makes other comments worthy to hear.
E.g., regarding the so common idea that the moon orbits the earth, he also clearly says it is wrong ... And he explains the joint movement of moon and earth as something quite different than a kind of "mutual orbiting" (last wording is mine), not following their tangents thanks to mutual pull, but not falling directly onto each other thanks to inertial "outward" forces (as he prefers to call them instead of the controversial term "centrifugal force") ...
By the way, that´s an idea I brought up here several times, though not qualifying so strongly the other model as wrong ... Last time perhaps on #364, 2nd part of my series "MY ULTIMATE GO ?"

[David Cooper]

Sorry, but I´m running out of patience ...

Why don't you just show me how you apply the centripetal force so that I can finish writing the program to simulate your mechanism. How long would it take for you to do that? I need to know if you apply the moon's gravitational pull from the barycentre (within the Earth) instead of from the moon? Is that what you don't want people to see?

Please kindly tell us where "on earth" I said what in italics: in its supposed "battle" with centripetal force we...

You have spoken many times about centrifugal force winning out on the far side and causing a bulge there. The word "battle" is appropriate. Whether you ever used the word "winning" either I can't remember - I'm not a quoting machine, but go by the meaning (or the closest thing to a meaning that can be extracted from your mangled wordings).

... have centrifugal force winning out on the far side and forming a tidal bulge, how can it be equal to the centripetal force there?

I regret to have to say it again, but I´m afraid you´ve got a problem either with your vision or with your reason ... or both.

If you don't have a difference, you don't have bulges forming. Are you playing games of avoidance? You have a mechanism which I want to put into a simulation to run your model, and I'd have thought you'd be keen to see it in action in a simulation too? Anyone who's had the misfortune to read through this entire thread deserves to be rewarded with the sight of a working simulation which puts all the numbers in front of them and makes it absolutely clear how your mechanism works. Once they've seen it, they'll be much better placed to judge whether it's real physics or a mere abstraction.

[rmolnav]

Quote
Please kindly tell us where "on earth" I said what in italics: in its supposed "battle" with centripetal force we...
You have spoken many times about centrifugal force winning out on the far side and causing a bulge there. The word "battle" is appropriate. Whether you ever used the word "winning" either I can't remember - I'm not a quoting machine, but go by the meaning (or the closest thing to a meaning that can be extracted from your mangled wordings).

My mangled wordings ?? Logically, my English doesn´t always let me be 100% clear. But I´ve often given you definitions quoted from dictionaries, and you, time and again, keep misusing them !
"Centrifugal force winning out" ? ... I have often said that indeed, or something with same meaning. But I´m afraid you wrongly consider the "enemy" is not the real one, the one I logically referred to most probably due to your deeply rooted misconceptions. 

I need to know if you apply the moon's gravitational pull from the barycentre (within the Earth) instead of from the moon?

Only the fact that you so deeply feel I´m wrong could make you ask me that !
It´s not a question of where I apply the moon´s pull from: the moon ONLY can exert its pull from each of its particles ... And, by the way, ONLY on each of earth´s particles, inversely proportionally to the square of individual distances (without nature "making" any vector subtraction, the so called "differential gravity").

[David Cooper]

I need to know if you apply the moon's gravitational pull from the barycentre (within the Earth) instead of from the moon?

Only the fact that you so deeply feel I´m wrong could make you ask me that !

It has nothing to do with whether your mechanism is real or the fake physics of an abstraction. What matters here is how your method should be applied. The centrifugal force is calculated from the barycentre. I want to know if the centripetal force is also calculated from there, because if you calculate it from the moon/planet, you don't need the centrifugal force as the correct orbit is already generated by the centripetal force alone.

It´s not a question of where I apply the moon´s pull from: the moon ONLY can exert its pull from each of its particles ... And, by the way, ONLY on each of earth´s particles, inversely proportionally to the square of individual distances (without nature "making" any vector subtraction, the so called "differential gravity").

You've just described differential gravity and you're giving centrifugal force no role. Nature adds the forces without doing any maths - we merely crunch numbers that we map to the events so that we can simulate the events.

Do you actually understand your method or are you just making it up as you go along?

[rmolnav]

What matters here is how your method should be applied. The centrifugal force is calculated from the barycentre

What you call "my method" is not mine,´ it is what considered correct by scientists most knowledgable on tides, and I managed to grasp after some direct discussion with some of them ...  Such as the one who wrote:
https://tidesandcurrents.noaa.gov/publications/Tidal_Analysis_and_Predictions.pdf
As I previously said, Dr. Bruce Parker (loftly despised by you !!) is the author of that more than 300 pages book, 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).
Logically, I trust on him much more than on any of your (or mine, indeed) methods ...
Without a thorough understanding of the basic stuff of the subject, it is not possible to reach correct conclusions.
You keep mixing up basic concepts, and Mr. Parker´s book is full of details you could not even dream of understanding ... Try and have a go, e.g., with:
"... As mentioned above, because the ocean and connected bays are a forced oscillating system, the tide will oscillate with the same frequencies as the astronomical tide-producing forces, which are determined by the relative motions of the Earth, moon, and sun. There are many different tidal
frequencies because of the complex nature of the orbit of the moon around the Earth and of the orbit of the Earth around the sun. Astronomers have very precisely determined all of the required
astronomical frequencies. The fact that tidal energy will always be at known frequencies allows one to predict the tide at a specific location for any time in the future, as long as there are data to analyze to determine the amplitude and epoch (phase lag) for each important tidal constituent (the amplitude and epoch not being known ahead of time because they are determined by the hydrodynamics).
If the moon-Earth orbit and the Earth-sun orbit were both circular and were both in the plane of
the Earth's equator, there would only be two tidal frequencies. One could then predict the tide using only two semidiurnal tidal harmonic constituents (M2 and S2, which are defined in Section 2.2.2).
However, the orbital motions are much more complicated. Both orbits are elliptical, so the distance
between the moon and Earth changes throughout the month, and the distance between the Earth and sun changes throughout the year. Both orbital planes are also at angles relative to the Earth's
equatorial plane. Because of the angle between the moon’s orbit and the plane of the Earth’s
equator, the moon appears to an observer on Earth to move north of the equator and then south of
the equator and back north of the equator over roughly a month (actually 27.3 days). Similarly, the
sun appears north of the equator half of the year (summer in the Northern Hemisphere) and south
of the equator the other half of the year (winter in the Northern Hemisphere). To further complicate
matters, the angles between the orbital planes and the equator also slowly change periodically over the years. All these motions modulate (that is, periodically vary the strength of) the tidal forces, so that tidal energy is spread out among many more frequencies (in addition to M2 and S2).
The fact that the moon’s monthly orbit around the Earth is angled to the Earth’s equatorial plane means that the moon will be over the equator only twice a month (equatorial declination). The rest
of the month the moon will appear either north or south of the equator and the two tidal bulges (on
the opposite sides of the Earth) will be asymmetric with respect to the axis of rotation (see Figure
2.11; here as in Figure 2.8, the idealized equilibrium tide is being considered, i.e., an ocean covering the whole earth with no continents). As a result there will be a once per day inequality in the elevation of the two high waters (and in the elevation of the two low waters) formed by the tidal
bulge at any given latitude. This asymmetric declination effect puts energy into diurnal tidal
frequencies. The strength of the resulting diurnal tide will vary in strength from zero when the moon is over the equator, to a maximum value when the moon appears (to an observer on the Earth) farthest north of the equator (northern declination) or farthest south of it (southern declination).

[rmolnav]

You keep mixing up basic concepts

E.g.:

The centrifugal force is calculated from the barycentre. I want to know if the centripetal force is also calculated from there ...

Why do you say "The centrifugal force is calculated from the barycentre" ? It is not always so, and neither is the centripetal force ...
But in each case, being equal but opposite, both of them must be calculated from same point: the center of curvature of the trajectory of the object ...

You've just described differential gravity and you're giving centrifugal force no role. Nature adds the forces without doing any maths

Wrong! What nature actually adds is inertial forces (centrifugal forces in our case) to gravity, with their values at each location (no maths necessary). 
But to subtract gravities from distant locations our "intelligent" minds are required, and nature can´t do it !!

[David Cooper]

What you call "my method" is not mine,´ it is what considered correct by scientists most knowledgable on tides, and I managed to grasp after some direct discussion with some of them .

It is your method because it's the method that you're pushing here while claiming that it is superior to the differential gravity method. In claiming that superiority, you turn yourself into an authority and you should be able to back your method in a better way than waving at other authorities. You should understand how it works and be able to spell that out so that it can be programmed. We have a differential gravity program here that works, and we have that because I understand its mechanism and turned it into a program. I now want to write a program to illustrate your method. Is it really too much to ask that you provide the rules for it? If you don't have them, what right do you have to assert its superiority?

Such as the one who wrote:
https://tidesandcurrents.noaa.gov/publications/Tidal_Analysis_and_Predictions.pdf
As I previously said, Dr. Bruce Parker (loftly despised by you !!)...

Where is your evidence that I despise someone that I have no dislike of whatsoever?

...is the author of that more than 300 pages book, 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).

Lovely - he knows a lot about the tides, but misunderstands the tidal forces that drive them. Experts don't always get everything right - they are not gods.

Logically, I trust on him much more than on any of your (or mine, indeed) methods ...
Without a thorough understanding of the basic stuff of the subject, it is not possible to reach correct conclusions.

Logically, you should be trusting reason rather than authority.

You keep mixing up basic concepts, and Mr. Parker´s book is full of details you could not even dream of understanding ... Try and have a go, e.g., with:

You've quoted a chunk of very readable, instantly comprehensible information from his book which are very obvious to me. I could have written that myself. Unlike for you, these ideas come easily to me as I have a simulation of it all running in my head which matches up to the universe, so I can see all the complications playing out there.