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

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

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....

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.

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

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 Virginiahttps://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....

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?

Hi. I didn't get around to making a reply for @geordief , sorry.Quote from: geordief on 10/04/2023 00:42:12If 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.

What if we dispensed with the accelerometer,?

Quote from: Jaaanosik on 12/04/2023 12:00:30Is there anything wrong with the analysis?I saw no analysis, just conclusions....

Is there anything wrong with the analysis?

Quote from: Jaaanosik on 13/04/2023 03:12:34Reply #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.

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

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 linesThe 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.

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.QuoteReply #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.Quote from: Jaaanosik on 10/04/2023 19:11:19We 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?

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

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)?

QuoteQuestion: 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.

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?

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.

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

So observer B with help of A points accelerations can determine the velocity, 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?

What are the accelerometers at A points going to measure?

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

Quote from: Jaaanosik on 13/04/2023 18:42:19No 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.

Quote from: Jaaanosik on 10/04/2023 19:11:19Yes, 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.

QuoteQuestion: 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.QuoteSo 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.

QuoteThe 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.QuoteWhat 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.

Quote1m 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.

...QuoteThis calculation is for 100m curvature radiusThese 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.

This calculation is for 100m curvature radius

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

Quote from: Jaaanosik on 14/04/2023 16:49:28 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 areΩ 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 ωa_{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...

Can we define Ω 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.

B spin angular velocity Ω=1, right?

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

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.

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

and it is negated by centrifugal force

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

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.

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