[Jaaanosik (OP)]

...    I'm not certain they're correct.
...

They are correct, the textbook:




The initial values are posted.
B rotates around O.
The flywheel spins around B.

...
That's it... that's all there is to it.  The acceleration of an arbitrary point A on the body   (which could be  A1 , A2, A3 or A4 in your very first post,   is the same in every frame which is just an offset of the frame K.  By an "offset" I mean it moves with constant velocity relative to the frame K.   As you may know, using Galilean relativity, all inertial frames can be described as these sorts of offsets from each other.   So what we are saying is that the acceleration of any point A on the body is the same when measured in ANY inertial reference frame  (using Galilean relativity and not Special relativity).   

The acceleration is a function of , just see the equations.
K'1 and K'2 observe different values because curvature radius changes, therefore the results are different.

[Jaaanosik (OP)]

Thanks Halc for your post. It is long, I'll address some parts so we can focus on the analysis.

...

The first 3 lines seem plausible. The next 4 A values are position coordinates only, not velocities or accelerations.

These are just initial values of Omega, omega and positions B and A points with their respective unit vectors.

Velocities are given here:

Looking at A1 column. V_B is moving at -1I due to the precession. That's wrong since it's the motion of points A3,A4, and B, but not A1 and A2 which are off to the side, moving faster and not quite perpendicular.
V_rel is the 3rd line, listed as -0.5k, meaning the distance between A and B is 0.25 (not elsewhere specified).
Ω x r_A/B seems to be their attempt to correct the discrepancy in the first line, but is done wrong. To get 0.25j, B would need to be right at the origin O and not where the picture puts it. So this middle line is wrong.
...

There is no precession. It is straight calculation .

A positions are specified , (x,y,z) origin is in B.
 
The calculation is correct A1, A2 points are offset to the side they are further away from the Omega axis of rotation compared to A3, A4. are at right angle to . are parallel to  , the results are correct. There is no attempt for any corrections. The calculation is straightforward.

is correct as well.

There is no gravity in this thought experiment.

The fact that you're talking about cycloids and cusps and such seems to indicate analysis of a rolling wheel in freefall and not a precessing gyroscope sort of setup depicted.
It's fine to do this, but then ditch the gyro picture and do an analysis of a wheel rolling with linear motion. I don't see any calculations being done by you, but if they're not coming out with identical acceleration values, it's a big clue that the calculations are being done incorrectly. And again, you seem to be attempting a calculation of different scenario than the one depicted in the OP.

There are just three inertial frames, with their respective observers, K, K'1, K'2, nothing else.
The cycloids are the consequence of the relative motion between the rest frame K and K'1, K'2.
This it t=0, O, O'1, O'2 are aligned and that's the reason why K'1 and K'2 see different positions on their respective cycloids.

[Eternal Student]

Hi.

    I've checked again but there's nothing wrong with the derivation given in post #16,  which would be about 3 lines if you wrote it out neatly.  There are also multiple references for the final result.   Here's just two:

https://galileo.phys.virginia.edu/classes/252/lecture1.htm
    Which is a short piece of Mathematics, that document was prepared by the University of Virginia

https://plato.stanford.edu/entries/spacetime-iframes/
    Which discusses the philosophy along with the mathematics,   if you prefer a Philosophical slant.   That's a document prepared by Stanford University.

   So, I'm really sorry but something does seem to have gone wrong with either the formulae you have used or the arithmetic that was done while applying them.
 -  - - - - - - - - - - -

I'll address some parts so we can focus on the analysis

    I appreciate that it can be helpful to identify exactly where you have gone wrong in your calculations.  In my own experiences this can be very informative and educational, much more so than just looking at some other method or solution to a problem.   
    However, I don't have enough time available at the moment.  I can't do much more than provide the references given above, sorry.

Best Wishes.

[Jaaanosik (OP)]

Hi.

    I've checked again but there's nothing wrong with the derivation given in post #16,  which would be about 3 lines if you wrote it out neatly.  There are also multiple references for the final result.   Here's just two:

https://galileo.phys.virginia.edu/classes/252/lecture1.htm
    Which is a short piece of Mathematics, that document was prepared by the University of Virginia

https://plato.stanford.edu/entries/spacetime-iframes/
    Which discusses the philosophy along with the mathematics,   if you prefer a Philosophical slant.   That's a document prepared by Stanford University.
...

There is more going on with your statement (bold) than one might think. Let us investigate.
We will discuss only the rest frame K at the moment.
The point B acceleration is (0I,-1J,0K), can we say the point B rotates around origin O1=(0I,-99J,0K) with Omega=(0I,0J,0.1K)?

The B acceleration results are equal for O, O1 but there is a difference between r and Omega values.



Question: Can an observer at B determine what is true origin O or O1 from inside the box without any signal from the outside, assuming the B observer has A points accelerations known to him?
Even more important question: Is information lost/destroyed to the B observer inside of the box?

[Eternal Student]

Hi.

   I didn't get around to making a reply for @geordief ,  sorry.

If an object is completely rigid and uniformly dense and the acceleration was evenly distributed (this reminds me a bit of gravity) would acceleration affect that body in a fundamentally different way than it would a body where  the acceleration would cause internal stresses?

Can we have an idealized  acceleration that is indistinguishable  from gravity?

    That is going to take a while to discuss.   I'm tempted to suggest creating another thread.
Very briefly:    There are some differences between a gravitational field that might be produced by a mass and a uniform acceleration.  There are articles discussing this and they are based on the idea that a gravitational source should create a force with a gradient (an increase in the magnitude of the force) as you move toward the object.  As such there are tidal forces on any object of non-zero width in the region or, to paraphrase it, there are ways to tell if you are in an elevator (without windows) that is being uniformly accelerated or if you were genuinely in a gravitational field.   This doesn't cause much problem for General Relativity or "the equivalence principle".  It's only necessary to consider local frames and they can be as local, or as small as required.

   As regards your question about a body with varying density and stress:   All bodies would exhibit stress from tidal forces.  You may be considering a related term which is "strain",  some objects are stretchy and they will deform under this stress.

Best Wishes.

[geordief]

Hi.

   I didn't get around to making a reply for @geordief ,  sorry.

If an object is completely rigid and uniformly dense and the acceleration was evenly distributed (this reminds me a bit of gravity) would acceleration affect that body in a fundamentally different way than it would a body where  the acceleration would cause internal stresses?

Can we have an idealized  acceleration that is indistinguishable  from gravity?

    That is going to take a while to discuss.   I'm tempted to suggest creating another thread.
Very briefly:    There are some differences between a gravitational field that might be produced by a mass and a uniform acceleration.  There are articles discussing this and they are based on the idea that a gravitational source should create a force with a gradient (an increase in the magnitude of the force) as you move toward the object.  As such there are tidal forces on any object of non-zero width in the region or, to paraphrase it, there are ways to tell if you are in an elevator (without windows) that is being uniformly accelerated or if you were genuinely in a gravitational field.   This doesn't cause much problem for General Relativity or "the equivalence principle".  It's only necessary to consider local frames and they can be as local, or as small as required.

   As regards your question about a body with varying density and stress:   All bodies would exhibit stress from tidal forces.  You may be considering a related term which is "strain",  some objects are stretchy and they will deform under this stress.

Best Wishes.

Thanks(I know you are busy so  there is no urgency to reply)

I have often heard  it said that we measure acceleration  in  a frame of reference by installing an accelerometer)

What if we dispensed with the accelerometer,?

If we observed  any change in relative motion  between any two parts of the system  in the frame of reference that would indicate acceleration.,I suppose

I think my idea about an infinitely  rigid body must be a non starter unless, perhaps we are thinking of point particles (are there point particles with mass?the neutrino?)

[Jaaanosik (OP)]

Hi.

   I didn't get around to making a reply for @geordief ,  sorry.

If an object is completely rigid and uniformly dense and the acceleration was evenly distributed (this reminds me a bit of gravity) would acceleration affect that body in a fundamentally different way than it would a body where  the acceleration would cause internal stresses?

Can we have an idealized  acceleration that is indistinguishable  from gravity?

    That is going to take a while to discuss.   I'm tempted to suggest creating another thread.
Very briefly:    There are some differences between a gravitational field that might be produced by a mass and a uniform acceleration.  There are articles discussing this and they are based on the idea that a gravitational source should create a force with a gradient (an increase in the magnitude of the force) as you move toward the object.  As such there are tidal forces on any object of non-zero width in the region or, to paraphrase it, there are ways to tell if you are in an elevator (without windows) that is being uniformly accelerated or if you were genuinely in a gravitational field.   This doesn't cause much problem for General Relativity or "the equivalence principle".  It's only necessary to consider local frames and they can be as local, or as small as required.

   As regards your question about a body with varying density and stress:   All bodies would exhibit stress from tidal forces.  You may be considering a related term which is "strain",  some objects are stretchy and they will deform under this stress.

Best Wishes.

Very good post, the 'V' shape of the gravity field lines differs from the straight line uniform acceleration.
The curvature can be detected from within the box.
The same applies to my post above, the curvature radius can be detected.
The flywheel cannot rotate around O1 because the A points accelerations would be different.
The curvature radius is the main factor in the acceleration analysis and this is true even within K frame itself.

Having said that K'1 and K'2 see different flywheel trajectories, they do not agree on the curvature radius with the K frame. Every moving K' sees different curvature trajectory. That's the reason they do not agree on the A points acceleration analysis.

[alancalverd]

What if we dispensed with the accelerometer,?

Something of a problem! In the absence of an external frame we can only measure acceleration as the force per unit mass on a test mass that is free to move inside our accelerating body - i.e. an accelerometer.

Aside from spagettification in a steep gravitational gradient, if an entire rigid object is accelerating uniformly (which can only occur in the case of a flat object perpendicular to a uniform electric, magnetic or gravitational gradient) there is no way any element of it can be aware of its acceleration. The question then arises as to whether you could measure the change in length of a spaghettifying object since all your measuring methods would themselves either be spaghettifying, or constitute an accelerometer!.

[Jaaanosik (OP)]

Is there anything wrong with the analysis?
Did somebody find an error or mistake?

I agree, the claim the acceleration analysis done here breaks the relativity as we know it, that's a new hypothesis.
The relativity cannot work unless it is done through a preferred reference frame.
Is there a physicist who can show the analysis is not good?
The world of physics has some explanation to do.
This does not go away with just moving it to a different sub forum.

[Jaaanosik (OP)]

Is there anything wrong with the analysis?

I saw no analysis, just conclusions.
...

Reply #3 shows the rest frame calculations that are part of the analysis.
Do you agree with those calculations?
There is no point in checking other frames if we have no agreement in the rest frame.
Reply #23 is continuation of the analysis. It's very important post with implications.
Here are two posts that we can discuss, if interested.

[Jaaanosik (OP)]

Reply #3 shows the rest frame calculations that are part of the analysis.
Do you agree with those calculations?

Well, I said that the acceleration values could not be known without knowing the external forces being applied to the system.  But given the values in that table, one can extrapolate backwards to get the external forces.

No external forces specified, there are no external forces.
This is a closed system and for the sake of the argument we can say in the intergalactic space faraway from any gravity source.

I agree with the velocity table. Thing is, relative to any other inertial frame, the only values that change on either table is the first line, the IJK one, where the relative velocity of the frame in question is subtracted from those figures. None of the other lines in either table gets effected, so if you're getting different values, there must be a mistake in the calculations.
So for instance, in the K'₁ frame, the top line would be (-2I, 0J 0K) for each column.

I accept the acceleration table as well. I have comments on those lines
The first line is the acceleration of B at -1J. A nonzero total cannot be without an external force being applied to B large enough to effect that acceleration. I'm not disagreeing with the value, but just wanting to make you aware of this implication of external force being applied.
I don't know what the 2nd line is. It comes to zero for everything. (Ω x r_A/B) is the 2nd line of the velocity table, not zero, so that omega with a dot on it must be zero, but I don't know what that symbol means.

The 2nd line is the second expression from here:

It is ... an angular acceleration, but there is none.

3rd line is the disk spinning about the k axis and the values sum up to zero, so no external forces required.
5th line is the spinning of the disk about B, and again the values sum up to zero, so no external forces required.
The 4th line is unnatural and represents a torque in the direction of -i on the system. I agree with the line, but like line 1, can be used to extrapolate an external torque being applied to the system somewhere. This is expected since the setup implies a continuous change of angular momentum which would violate conservation of angular momentum laws without that application of external torque. And yes, this torque causes precession despite your refusal to name it thus.

Bottom line is that the post 3 values are consistent with the scenario.

Reply #23 is continuation of the analysis. It's very important post with implications.

That was a reply to ES, but I'll try to comment on it.

We will discuss only the rest frame K at the moment.
The point B acceleration is (0I,-1J,0K)
Can we say the point B rotates around origin O1=(0I,-99J,0K) with Omega=(0I,0J,0.1K)?

No, you can't. If acceleration was fixed at -1J, it would require a velocity relative to O1 of -10I, a contradiction with a velocity a tenth that.

Is that what you were looking for?

Yes, that's the point, the velocity can be determined by the A points accelerations.

Question: Can an observer at B determine what is true origin O or O1 from inside the box without any signal from the outside, assuming the B observer has A points accelerations known to him?

He knows his acceleration from his accelerometer, so he can point in the direction of O. He can sense the rate of change of that acceleration, so that gives him the ability to triangulate. Yes, he can determine the distance to O. For instance, B goes around O every 6.28 (seconds? units were not clearly given). But if the origin was at a distance of 100, that time would be over a minute.

So observer B with help of A points accelerations can determine the velocity, the curvature radius.
B observer does not need any signal from the outside to do so, I hope it makes sense.

The question is can the motion satisfy analysis of K frame with curvature radius 1m and at the same time K'1 analysis with curvature radius 4m?
What are the accelerometers at A points going to measure? 1m or 4m radius, because they cannot both.

[Halc]

No external forces specified, there are no external forces.
This is a closed system and for the sake of the argument we can say in the intergalactic space faraway from any gravity source.

OK, so you're in denial of Newtonian physics since your assertion violates the first and third laws of motion as well as conservation of angular momentum. Somehow you'll probably blame that on Einstein as well.
The close system seems to have its center of mass at B, and that center of mass is accelerating, impossible for a closed system.

Yes, that's the point, the velocity can be determined by the A points accelerations.

That can't be true else there's be an absolute motion detector. So for circular motion about some axis, one can determine velocity and distance to that axis relative to the frame in which that axis is stationary. That's different than the absolute statement you gave. So our guy on the rolling wheel is always going to measure v=0.5 relative to B, and tangential to the direction of B (r=0.25 away), and no abstract frame change is going to alter that physical measurement.

Question: Can an observer at B determine what is true origin O or O1 from inside the box without any signal from the outside, assuming the B observer has A points accelerations known to him?

OK, I misread this question the first time. The A-point masses define where B is (their mutual center of gravity). If he knows the accelerations of those points, that's the same as the observer at B knowing his own acceleration. All he has to do is average them. That reduces the answer to the one I gave: that B can determine the distance to O by looking at his own instruments instead of those at the A points.

So observer B with help of A points accelerations can determine the velocity, the curvature radius.

He needs no help from the A points to do that, and again, only the velocity relative to O.  'The velocity' and 'the curvature radius' without a frame reference are a meaningless phrases. You said you're only considering frame K at this point, the one in which the axis of Ω is stationary, so in light of that reference, yes the observer at B can determine those things, and without help.

The question is can the motion satisfy analysis of K frame with curvature radius 1m and at the same time K'1 analysis with curvature radius 4m?

It can't determine velocity relative to an unspecified frame. If B is told to add 1I to all his velocity computations, then sure, he can say his velocity now ranges from 2 to zero relative to this new frame in which the Ω axis is moving at speed -1I. But that's giving him the answer. He doesn't measure that.

What are the accelerometers at A points going to measure?

Those are objective facts, physical measurements. They read what your chart says in post 3, minus the axis labels. They cannot read one thing and also read a different one. That would be a logical contradiction.
So for instance, in the OP scenario at t=0, the accelerometer at A4 reads 2.236, and thataway, a direction of a point that you happen to assign the coordinate (X=0 Y=0, Z=0.25), but the meter is unaware of those coordinates. It just displays an arrow pointing that way, essentially what a plumb bob does (except it points the opposite way). That, plus the magnitude, is your physical data, and it must be the same regardless of abstract frame choice.

1m or 4m radius, because they cannot both.

Accelerometers read a vector acceleration, not a scalar length. Meters is the wrong unit for an accelerometer to read. You might want to put one additional instrument next to it reading the derivative of the accelerometer. For the rolling wheel, it (the ones at A points) would read zero for the magnitude change and ω for the rate of direction change. For the complicated setup in the OP, the meters would both be bouncing all over the place and there would be no simple way to determine the center of anything from such unstable readings, but it gets easy if you combine (average) the readings at all 4 A points.

[Jaaanosik (OP)]

No external forces specified, there are no external forces.
This is a closed system and for the sake of the argument we can say in the intergalactic space faraway from any gravity source.

OK, so you're in denial of Newtonian physics since your assertion violates the first and third laws of motion as well as conservation of angular momentum. Somehow you'll probably blame that on Einstein as well.
The close system seems to have its center of mass at B, and that center of mass is accelerating, impossible for a closed system.

The origin O is the barycenter of the closed system, that's where is the center of mass, not B.
O is the origin of the inertial system.
There is no denial of Newtonian physics or conservation of angular momentum.

Yes, that's the point, the velocity can be determined by the A points accelerations.

That can't be true else there's be an absolute motion detector. So for circular motion about some axis, one can determine velocity and distance to that axis relative to the frame in which that axis is stationary. That's different than the absolute statement you gave. So our guy on the rolling wheel is always going to measure v=0.5 relative to B, and tangential to the direction of B (r=0.25 away), and no abstract frame change is going to alter that physical measurement.

This calculation is for 100m curvature radius:


This is the calculation for 1m curvature radius:


This is the calculation for 4m curvature radius:


A3, A4 points acceleration values determine B, A points velocity and the curvature radius, nothing else is required.
The acceleration is absolute, only one set of values can be on A3 accelerometer and A4 accelerometer.

Question: Can an observer at B determine what is true origin O or O1 from inside the box without any signal from the outside, assuming the B observer has A points accelerations known to him?

OK, I misread this question the first time. The A-point masses define where B is (their mutual center of gravity). If he knows the accelerations of those points, that's the same as the observer at B knowing his own acceleration. All he has to do is average them. That reduces the answer to the one I gave: that B can determine the distance to O by looking at his own instruments instead of those at the A points.

So observer B with help of A points accelerations can determine the velocity, the curvature radius.

He needs no help from the A points to do that, and again, only the velocity relative to O. 'The velocity' and 'the curvature radius' without a frame reference are a meaningless phrases. You said you're only considering frame K at this point, the one in which the axis of Ω is stationary, so in light of that reference, yes the observer at B can determine those things, and without help.

The bold part, please, see the calculation above.
The acceleration is absolute therefore the acceleration determines the velocity and the curvature radius.

The question is can the motion satisfy analysis of K frame with curvature radius 1m and at the same time K'1 analysis with curvature radius 4m?

It can't determine velocity relative to an unspecified frame. If B is told to add 1I to all his velocity computations, then sure, he can say his velocity now ranges from 2 to zero relative to this new frame in which the Ω axis is moving at speed -1I. But that's giving him the answer. He doesn't measure that.

What are the accelerometers at A points going to measure?

Those are objective facts, physical measurements. They read what your chart says in post 3, minus the axis labels. They cannot read one thing and also read a different one. That would be a logical contradiction.
So for instance, in the OP scenario at t=0, the accelerometer at A4 reads 2.236, and thataway, a direction of a point that you happen to assign the coordinate (X=0 Y=0, Z=0.25), but the meter is unaware of those coordinates. It just displays an arrow pointing that way, essentially what a plumb bob does (except it points the opposite way). That, plus the magnitude, is your physical data, and it must be the same regardless of abstract frame choice.

If this board is placed on a table it will measure the gravity, if rotated, different axis will measure the gravity.
Accelerometers measure absolute acceleration.
This board is calibrated for m/s^2.

1m or 4m radius, because they cannot both.

Accelerometers read a vector acceleration, not a scalar length. Meters is the wrong unit for an accelerometer to read. You might want to put one additional instrument next to it reading the derivative of the accelerometer. For the rolling wheel, it (the ones at A points) would read zero for the magnitude change and ω for the rate of direction change. For the complicated setup in the OP, the meters would both be bouncing all over the place and there would be no simple way to determine the center of anything from such unstable readings, but it gets easy if you combine (average) the readings at all 4 A points.

[Jaaanosik (OP)]

...

This calculation is for 100m curvature radius

These numbers are fiction. For instance, top line lists V_B at -10I where in fact at r=100m and Ω=1, V_B would be -100I.
So you're trying to compute stuff possibly from a curvature, but the curvature doesn't give any indication of how fast somethings is traversing that curve. It could be anything. You made up some numbers. Ω wasn't held constant, and neither was any of the proper accelerations, so it wasn't based on those.

I pointed out in my Reply #23
and it follows so the flywheel maintains its absolute rotation around K, k axis because 0.1 + 0.9 = 1Ω.

So to expand, the analysis goes backwards.

The A3, A4 accelerations are measured and the values are above.
What is the B velocity if B acceleration is known (above)? The first line in the velocity table because that's the only possible calculation.
It cannot be 1m/s (or anything else) because it would require to have different A3, A4 points acceleration, see the table bellow.


The direction of the acceleration in B points towards the origin and the radius is how far is the origin.
Once we know velocity and the acceleration we get the curvature radius from

It is crucial to resolve this part of analysis in order to discuss anything else further.

[GertrudeFranklin]

What a detailed explanation with calculus. Thanks! I came on the thread and were ready to post a few questions, but your last post dismissed them all. Thank you another time! And what I would to add - https://essays.studymoose.com/do-my-homework [nofollow] is a source that saves me when I feel I don't have sufficient time and need StudyMoose experts to do my homework for me. I find clear explanations and solutions with detailed steps, not only for physics, but for any academic domains.

[Jaaanosik (OP)]

What a detailed explanation with calculus. Thanks!
I came on the thread and were ready to post a few questions, but your last post dismissed them all.
Thank you another time!

It appears to me your potential questions got answered, not dismissed.
It is good to have questions.

The calculations are based on textbooks, the main questions are what input parameters where used, that is important.
What parameters, why those parameters,... more things to uncover.
The calculations can be checked by anybody with the input parameters layed out.

If there is something new then I am going to say it is new, as in my post #28.
I am not claiming it is a theory, because I see it as a hypothesis. After a review a hypothesis can become a theory if agreed by reviewers.

An example, we know that information cannot be destroyed in QM:
https://en.wikipedia.org/wiki/No-hiding_theorem#Conservation_of_quantum_information

So here is a hypothesis:
The information about velocity and curvature radius cannot be hidden/destroyed in classical physics, mechanical systems.
It is conserved in the absolute proper acceleration.

Wouldn't it be nice to get a unification/agreement between QM and classical physics?

[Halc]

and it follows so the flywheel maintains its absolute rotation around K, k axis because 0.1 + 0.9 = 1Ω.

OK, I think I actually finally see what you're trying to do, and where your mistakes are. Rotation (as well as acceleration) being absolute means they cannot be frame dependent. So Ω and ω must remain 1 and 2 respectively. To suggest otherwise is to say that it takes a different amount of time to go all the way around relative to another frame.  Spin rates cannot change under Newtonian mechanics, so Ω and ω are fixed.

What you describe is a pseudo-rotation radius based on the momentary frame-dependent curvature of the trajectory of a point on a cycloid.  That's a valid (albeit pretty complicated) way to go about it, but assign it a new symbol. Don't overload Ω like you've done since Ω is fixed at 1 (and the instruments at B can measure it directly). So call it δ or something. Then the equations will work out and you get the correct answer for any frame.

The r=4 acceleration line matches the (wrong) values from post 3, so I'll use that one as my example with corrected values.
Values for motion of B are relative to the K'1 frame where the pseudo-rotation about the stationary axis at (0, 0, -3) at rate δ which is 0.5. Other values are relative to B and not the frame of choice.
V_B = (-2I, 0, 0)      ( 4δ )
All the other velocity lines still reference Ω or ω and not δ and are relative to B, not relative to any chosen frame, so they remain the same as the K values in post 3 which are
Ω x r_A/B    ¼j  -¼j  0  0   (precession rate, a function of Ω, not of δ)
v_rel  -½k ½k ½i  -½i     (spin velocity due to ω)

A similar adjustment to the acceleration lines. First one is based on δ, the rest on Ω or ω
a_B =  -1J           (centripetal due to δ)
Ω² x r_A/B = -¼i, ¼i, 0, 0   (centripetal accel due to Ω)
2Ω x v_rel =  0, 0, 1j, -1j   (from torque causing precession rate Ω)
a_rel = -i, i, -k, k    (centripetal acceleration due to ω)

The sum of these is identical in any frame. In fact, except for the V_B line, all lines are the same, and the accelerations are not in any way a function of V_B.

You should have realized that a mistake was made when the acceleration results came out frame dependent like that, different in both magnitude and direction.
For instance, A3 accelerates at 1 m/sec² but in the K2 frame it comes out to 1.096 m/sec² and pointing 14° in a different direction. That violates these values being objective. You seem to persist in asserting that the meter at B reads two different things, a logical contradiction. So consider my correction to your analysis above.

[Jaaanosik (OP)]

and it follows so the flywheel maintains its absolute rotation around K, k axis because 0.1 + 0.9 = 1Ω.

OK, I think I actually finally see what you're trying to do, and where your mistakes are. Rotation (as well as acceleration) being absolute means they cannot be frame dependent. So Ω and ω must remain 1 and 2 respectively. To suggest otherwise is to say that it takes a different amount of time to go all the way around relative to another frame.  Spin rates cannot change under Newtonian mechanics, so Ω and ω are fixed.

What you describe is a pseudo-rotation radius based on the momentary frame-dependent curvature of the trajectory of a point on a cycloid.  That's a valid (albeit pretty complicated) way to go about it, but assign it a new symbol. Don't overload Ω like you've done since Ω is fixed at 1 (and the instruments at B can measure it directly). So call it δ or something. Then the equations will work out and you get the correct answer for any frame.

The r=4 acceleration line matches the (wrong) values from post 3, so I'll use that one as my example with corrected values.
Values for motion of B are relative to the K'1 frame where the pseudo-rotation about the stationary axis at (0, 0, -3) at rate δ which is 0.5. Other values are relative to B and not the frame of choice.
V_B = (-2I, 0, 0)      ( 4δ )
All the other velocity lines still reference Ω or ω and not δ and are relative to B, not relative to any chosen frame, so they remain the same as the K values in post 3 which are
Ω x r_A/B    ¼j  -¼j  0  0   (precession rate, a function of Ω, not of δ)
v_rel  -½k ½k ½i  -½i     (spin velocity due to ω)

A similar adjustment to the acceleration lines. First one is based on δ, the rest on Ω or ω
a_B =  -1J           (centripetal due to δ)
Ω² x r_A/B = -¼i, ¼i, 0, 0   (centripetal accel due to Ω)
2Ω x v_rel =  0, 0, 1j, -1j   (from torque causing precession rate Ω)
a_rel = -i, i, -k, k    (centripetal acceleration due to ω)

The sum of these is identical in any frame. In fact, except for the V_B line, all lines are the same, and the accelerations are not in any way a function of V_B.

You should have realized that a mistake was made when the acceleration results came out frame dependent like that, different in both magnitude and direction.
For instance, A3 accelerates at 1 m/sec² but in the K2 frame it comes out to 1.096 m/sec² and pointing 14° in a different direction. That violates these values being objective. You seem to persist in asserting that the meter at B reads two different things, a logical contradiction. So consider my correction to your analysis above.

Thanks Halc, now we are getting into the inner workings...
First, just some thought exercise: Can we define Ω without a radius? Can we have radian without a radius?

If I understand correctly you are suggesting to name B orbital angular velocity δ and the value is 0.5 for 4m radius.
... and B spin angular velocity Ω=1, right?

Let us use your suggestion for the following calculation.
We take the flywheel alone and it falls in a 'circular' orbit around the Earth. The flywheel is flat.

We just replace O with the center of the Earth r=7000km, that is the distance between B and the center of the Earth.
 
The gravitational acceleration is 8.13475m/s^2 at that height.

If we go by the current GR there is no force acting on B, B observer is inertial by definition and the flywheel rotates around B.
We define Ω=(0,0,0) and to get some reasonable numbers let us increase ω=(0,200,0).
The current understanding of the physics says, there is only centripetal force from A points to B, nothing else.

Now let us examine an idea, what if gravity is a force and not curved space.
The calculation Ω=(0, 0, 0.001078011) and ω=(0, 200, -0.001078011) and r=7000000m, the flywheel dimensions are the same.


Your argument should be δ=(0, 0, 0.001078011) generates gravity centripetal force at B and it is negated by centrifugal force of the same magnitude, so there is no force acting on B and your Ω=(0,0,0); so there is only centripetal force on the flywheel to B, correct?

My calculation is different, it predicts a torque. This torque generates a precession, speed up of the flywheel orbit. If the flywheel rotates opposite way the calculation predicts recession, slow down of the flywheel orbit.
This has been settled by 'experiments' - satellite flyby anomalies: https://en.wikipedia.org/wiki/Flyby_anomaly
The satellites have reaction wheels on them for positioning/turning.

The GR cannot explain it, satellites speed ups and slow downs, my calculation can.
The hypothesis: The gravity is a real force.

[Halc]

Thanks Halc, now we are getting into the inner workings...

Well, no, now we're kind of done. The corrected calculation is shown, and it is also incomplete and doesn't work for frames with motion along the Y axis. You'd need another line to do it with your funny cycloid method because motion along a trochoid path isn't at a constant rate like it is for a circular path, so a component for the acceleration needed for those speed changes needs to be there. The component is zero if no Y motion is present, so the answer works.

Can we define Ω without a radius?

Of course. Ω here is just 2 rads/sec which makes no mention of radius, and is actually used with different radii in the numbers you put in post 3. It is a vector, so the direction of the axis is part of Ω, and is presumably north in the orbity thing below, but most satellites do not orbit about that axis.

If I understand correctly you are suggesting to name B orbital angular velocity δ and the value is 0.5 for 4m radius.

Yes for the 0.5 and 4m radius, but it isn't an orbital velocity since B isn't orbiting anything 4 meters away. B doesn't actually orbit anything, which is a word reserved for a sort of a gravity/satellite relationship. B traces a circular path around O (the origin of frame K), and since the disk with centroid at B has changing angular velocity, that circular motion is precession about O, not an orbit about O.
B never revolves, precesses, or orbits about the pseudo-point at K=-3 in K'₁, even if its motion maintains a constant distance from that point for a moment.

B spin angular velocity Ω=1, right?

The disk defining B spins at ω=2. B precession angular velocity is Ω=1. I'm agreeing with you, but wording it more correctly.

We take the flywheel alone and it falls in a 'circular' orbit around the Earth. The flywheel is flat.
We just replace O with the center of the Earth r=7000km, that is the distance between B and the center of the Earth.

But there's no bar running from O this time applying torque to the disk, so the disk won't precess. It will just spin on the x axis permanently as it orbits. There's a lot of satellites that do this. None of them precess.

The gravitational acceleration is 8.13475m/s^2 at that height.
If we go by the current GR there is no force acting on B

You're mixing theories here. Under Newton there is gravitational acceleration since gravity is a force, and F=ma.  Under GR there is no force and no acceleration and the spinning disk follows a geodesic through curved spacetime. Either way, the proper acceleration of B is zero, but  the coordinate acceleration is around 8, assuming a sort of semi-inertial coordinate system that Newton might have used, or possibly a rotating reference frame that most people use from day to day.

B observer is inertial by definition

Locally inertial since he experiences negligible proper acceleration. In a larger coordinate system (one that includes Earth), it's not inertial because spacetime isn't Minkowskian there.

We define Ω=(0,0,0)

The flywheel is not in orbit, but just hovering over Earth? That makes no sense.

and to get some reasonable numbers let us increase ω=(0,200,0).

~1910 RPM. OK...

The current understanding of the physics says, there is only centripetal force from A points to B, nothing else.

It's a disk. The forces are from every direction, not just 4 points. There are differences, and ones that should be taking into account in your post 3 calculation but are not. The answer there is a good approximation, but since the disk has more angular moment about the k axis than does the collection of 4 point masses, some of the lines (notably the torque one) need a bit of adjustment.
I'm saying all this because I don't want you pointing back to my post saying that I agreed to something which is only approximately correct. So on that note, the centripetal force from all the mass of the disk adds up to zero, and so has no effect on B. B could be severed from the A points (becoming a small dot of matter surrounded by a detached spinning disk with a slightly larger hole in it) and the motions of everything would be essentially unaffected by this. The A points act on and cancel out each other and don't really have much if any effect on B. This is not true in the OP example where the A points applied lateral force and also torque to B which B had to transfer down the axle to O where the hidden parts of the close system could deal with them. That isn't happening with B in the sky.

Now let us examine an idea, what if gravity is a force and not curved space.
The calculation Ω=(0, 0, 0.001078011) and ω=(0, 200, -0.001078011)[/quote]Sorry, you changed ω. Pick one. And what's the point of tilting the disk slightly like that?
Also, 7 significant digits seems too many for a simple example.  How about 3.9 rads/hr for Ω?

 and r=7000000m, the flywheel dimensions are the same.

Your argument should be δ=(0, 0, 0.001078011) generates gravity centripetal force at B

Don't presume to tell me what my argument should be.  If we're talking the frame of the CoG of Earth, then δ and Ω are the same thing. So it's just Ω. Secondly, there's no such thing as gravity centripetal force. In Newtonian physics, there is no centripetal force on a satellite, only gravity. The accelerometer would be nonzero if there was centripetal force. Under GR, there is no force at all, centripetal, gravitational, or otherwise. So B has zero proper acceleration in both cases.

and it is negated by centrifugal force

There is no centrifugal force at all here. That's only meaningful in a rotating reference frame, and nobody has specified such a frame here.

My calculation is different, it predicts a torque. This torque generates a precession, speed up of the flywheel orbit.

Precession doesn't change the orbit altitude. It changes either the angular momentum direction of the spinning object or the direction of the orbital axis, or other variants like the precession of the major axis of an eccentric orbit, all quite normal things. I don't know what you mean by 'speed up'. Does the orbit gain or lose energy? Speed-up usually means a loss of orbital energy, in which case you need to show where that goes.

The link you give has to go with gravitational boosts of a flyby (like what the Voyager probes did with Jupiter and most of the other planets) to gain far more speed than could be provided by the limited propulsion system. It has nothing to do with orbits, which are not known to gain energy from nowhere.

How is this relevant to the topic here? You posted contradictory stuff. It takes 38 posts to figure out what you actually did, it was corrected, apparently to your dismay. So instead of learning, you go off and propose a new thing, this 'my calculation' which is strangely (but not unexpectedly) unspecified. Let me know how it works out.

[hamdani yusuf]

As the subject says, is proper acceleration absolute?
The answer is yes, because all inertial observers agree on the acceleration that would be measured by an accelerometer.

Accelerometers in ISS show near zero values, although it's accelerated toward the earth center significantly.

Gravity at the altitude of the ISS is approximately 90% as strong as at Earth's surface, but objects in orbit are in a continuous state of freefall, resulting in an apparent state of weightlessness.
https://en.wikipedia.org/wiki/International_Space_Station