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Old 3rd August 2018, 01:23 AM   #121
Dave Rogers
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Originally Posted by Kid Eager View Post
More grist for the mill, cutting out the (alleged) middle man, as it were:
http://theelectromagneticnatureofthings.com

Mind you, it might be complete bollocks, but I leave it to y'all with the mad physics skills to interpret all the squiggly symbols.
That's a completely different assertion to the ones SDG's making, which suggests that SDG doesn't even understand the material he's trying to reproduce. Basically the argument is that, because the kinetic energy is time dependent in one frame of reference and time independent in another frame of reference, there must be measurable work done on the ball in one frame and not in the other. However, even that's wrong; work is measured as the product of force times distance, and the measurement of distance travelled in a given iterval is frame dependent.

The website claims that:

Quote:
The kinetic energy change predicts some work to be done on the ball in the ground reference frame as a function of time changing velocity, in other words we expect force magnitude changes based on the work-energy principle but we do not expect any force magnitude changes in the moving reference frame since the steel ball moves in a uniform circular motion therefore there is no work done on the ball from the moving reference frame point of view.
Everything after "in other words" is incorrect. In fact, it's not the force term but the distance term that varies in the moving frame of reference. In the moving frame, the vector product of the force and the velocity - that is, the rate at which work is done on the ball - is always zero, because the direction of movement is always perpendicular to the force vector. In the static frame, this is no longer the case, because there is a constant vector in the direction of motion added to the velocity; the resultant velocity is no longer perpendicular to the force vector, so work is done on the ball. This means that, in the static frame, the kinetic energy of the ball has a periodic variation. This requires no change in the magnitude of the force.

So the origin of the misunderstanding in the website is simply a failure to understand simple vector arithmetic. A first year undergraduate should be able to spot this pretty quickly.

Dave
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Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right?

Tony Szamboti: That is right
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Old 3rd August 2018, 01:30 AM   #122
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Originally Posted by Dave Rogers View Post
No. The above is either irrelevant or incorrect.

Your definition of the problem gives a funtional form for the angular velocity of the wheel, and the angular velocity of the balls attached to its perimeter must follow the same functional form. Your mathematical analysis gives a different functional form for the angular velocity of the ball to the one you put in the definition of the problem. No further analysis is needed to conclude that your mathematical analysis is incorrect.

Dave
Dave,
The points BA, LA, TA where the elastic rods are attached to the rigid wheel will rotate with the wheel. The balls will not follow the same rotation though.
The ball L is being accelerated by the string in a straight line in the ground reference frame, no energy goes to the rotation of the ball.
The balls B and T are being accelerated in a curved trajectory by the string in the ground reference frame, some energy goes to the rotation of the balls.

It was already established the motion and acceleration in a straight line is not the same as the motion and acceleration on a curved trajectory energy wise.
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Old 3rd August 2018, 02:54 AM   #123
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Originally Posted by SDG View Post
Dave,
The points BA, LA, TA where the elastic rods are attached to the rigid wheel will rotate with the wheel. The balls will not follow the same rotation though.
The ball L is being accelerated by the string in a straight line in the ground reference frame, no energy goes to the rotation of the ball.
The balls B and T are being accelerated in a curved trajectory by the string in the ground reference frame, some energy goes to the rotation of the balls.

It was already established the motion and acceleration in a straight line is not the same as the motion and acceleration on a curved trajectory energy wise.
SDG
If the balls are not rigidly attached then they won't follow a cycloid; cycloidal motion assumes that the balls are rigidly attached to the wheel. So either your results are wrong, or your model is inapplicable.

Dave
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Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right?

Tony Szamboti: That is right
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Old 3rd August 2018, 12:38 PM   #124
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Originally Posted by SDG View Post
Dave,
The points BA, LA, TA where the elastic rods are attached to the rigid wheel will rotate with the wheel. The balls will not follow the same rotation though.
The ball L is being accelerated by the string in a straight line in the ground reference frame, no energy goes to the rotation of the ball.
The balls B and T are being accelerated in a curved trajectory by the string in the ground reference frame, some energy goes to the rotation of the balls.

It was already established the motion and acceleration in a straight line is not the same as the motion and acceleration on a curved trajectory energy wise.
SDG
It is only different because you arbitrarily decide to view the ball as a unit.

Each point is accelerated and moves in the same way as if it was a linear motion, however, the distance to the center of the circle, in the circular case, determines the velocity, and hence the energy that point receives. So, when you contemplate a group of points that comprise a ball, they will be moving not only at different speeds, but in different directions: The points farthest from the wheel center will move faster, the ones closest will move slower. Also the points heading the middle of the ball will move at a different angle than those trailing the middle of the ball. So, when you contemplate the ball as a unit, it will seem that it gains a rotary energy, but it all boils down to different energies applied to different parts of the ball.

In contrast, a similar cluster of points, comprising a ball, being accelerated linearly, will all receive the same amount of energy, in the same direction.

That is all there is to it.

So linear acceleration or circular acceleration, the amount of energy is the same, but the distribution in an object with some extent is different.

Hans
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Old 8th August 2018, 11:58 AM   #125
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Mmm, where did he go?

Hans
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Old 8th August 2018, 12:02 PM   #126
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Old 8th August 2018, 01:14 PM   #127
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Old 8th August 2018, 01:52 PM   #128
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Old 10th August 2018, 09:04 AM   #129
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Originally Posted by Dave Rogers View Post
If the balls are not rigidly attached then they won't follow a cycloid; cycloidal motion assumes that the balls are rigidly attached to the wheel. So either your results are wrong, or your model is inapplicable.

Dave
Dave,
let us assume the same wheel with the balls on the elastic rods on the ground, the wheel axle is attached to a concrete block.
When we start to accelerate the wheel the balls inertia will resist the acceleration and it will load 'the springs' of the elastic rods.
The balls will accelerate slower than the point on the wheel below the balls but the balls will follow a circular motion within the ground frame as well.
The same happens to the cycloid. The balls cycloids will have a different trajectory compared to the point on the wheel during the time the elastic rod 'springs' are loaded but the balls will definitely follow a cycloid trajectory.
SDG
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Old 10th August 2018, 10:22 AM   #130
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Originally Posted by MRC_Hans View Post
It is only different because you arbitrarily decide to view the ball as a unit.

Each point is accelerated and moves in the same way as if it was a linear motion, however, the distance to the center of the circle, in the circular case, determines the velocity, and hence the energy that point receives. So, when you contemplate a group of points that comprise a ball, they will be moving not only at different speeds, but in different directions: The points farthest from the wheel center will move faster, the ones closest will move slower. Also the points heading the middle of the ball will move at a different angle than those trailing the middle of the ball. So, when you contemplate the ball as a unit, it will seem that it gains a rotary energy, but it all boils down to different energies applied to different parts of the ball.

In contrast, a similar cluster of points, comprising a ball, being accelerated linearly, will all receive the same amount of energy, in the same direction.

That is all there is to it.

So linear acceleration or circular acceleration, the amount of energy is the same, but the distribution in an object with some extent is different.

Hans
Hans,
imagine this rotation:


\omega = 1rad/s
The radius r to the center of the red ball r=1m
The radius r to the center of the blue ball r=1.1m
The radius r to the center of the yellow ball r=0.9m
The tangential velocities:
The red ball v=1m/s
The blue ball v=1.1m/s
The yellow ball v=0.9m/s

Let us assume all balls have m=2kg
The translational kinetic energy = 1/2 m_ball v^2

The red ball E = 1J
The blue ball E = 1.21J
The yellow ball E = 0.81J

This is to demonstrate the energy DELTA.
The outside half of a rigid ball will have more translational kinetic energy than the inside part of the same ball when compared to the center of the ball. It is 0.21J between blue - red and 0.19J between red - yellow.
The energy total is 3.02J


This is not going to happen in the case of the linear acceleration.
The energy total is 3J if we accelerate the balls in the linear acceleration to 1m/s velocity.
The calculation includes only translational kinetic energy. The rotational kinetic energy adds up to the translational kinetic energy in the rotational acceleration case.

SDG

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Old 10th August 2018, 01:07 PM   #131
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Originally Posted by SDG View Post
Hans,
imagine this rotation:
https://i.imgur.com/Og5VbNn.png?1

\omega = 1rad/s
The radius r to the center of the red ball r=1m
The radius r to the center of the blue ball r=1.1m
The radius r to the center of the yellow ball r=0.9m
The tangential velocities:
The red ball v=1m/s
The blue ball v=1.1m/s
The yellow ball v=0.9m/s

Let us assume all balls have m=2kg
The translational kinetic energy = 1/2 m_ball v^2

The red ball E = 1J
The blue ball E = 1.21J
The yellow ball E = 0.81J

This is to demonstrate the energy DELTA.
The outside half of a rigid ball will have more translational kinetic energy than the inside part of the same ball when compared to the center of the ball. It is 0.21J between blue - red and 0.19J between red - yellow.
The energy total is 3.02J


This is not going to happen in the case of the linear acceleration.
The energy total is 3J if we accelerate the balls in the linear acceleration to 1m/s velocity.
The calculation includes only translational kinetic energy. The rotational kinetic energy adds up to the translational kinetic energy in the rotational acceleration case.

SDG
Mmmm, did you read my post, or what? I wrote that the difference, which exists only because you arbitrarily opt to view the ball as a unit, is that in the circular case, various parts of the ball are accelerated in different ways.

And, of course, the median energy is higher than for the center of the ball.

So yes, there is a difference, because the movement is different.

And your point is?

Hans
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Old 10th August 2018, 01:21 PM   #132
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Originally Posted by MRC_Hans View Post
Mmmm, did you read my post, or what? I wrote that the difference, which exists only because you arbitrarily opt to view the ball as a unit, is that in the circular case, various parts of the ball are accelerated in different ways.

And, of course, the median energy is higher than for the center of the ball.

So yes, there is a difference, because the movement is different.

And your point is?

Hans
Hans,
what did you mean by this?
Quote:
So linear acceleration or circular acceleration, the amount of energy is the same, but the distribution in an object with some extent is different.
What energy is the same?

How we should view the ball if not as a rigid body?
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Old 10th August 2018, 01:25 PM   #133
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Originally Posted by SDG View Post
Hans,
what did you mean by this?

What energy is the same?

How we should view the ball if not as a rigid body?
SDG
The energy for a given point is the same.

We can view it as an arbitrary cluster of points.

But, I ask again: WHAT IS YOUR POINT? All this is simple geometric nitpicked to some arbitrary level of detail. Why?

Hans
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Old 10th August 2018, 01:41 PM   #134
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Originally Posted by MRC_Hans View Post
The energy for a given point is the same.

We can view it as an arbitrary cluster of points.

But, I ask again: WHAT IS YOUR POINT? All this is simple geometric nitpicked to some arbitrary level of detail. Why?

Hans
Hans,
if the first post from the other forum
https://www.thenakedscientists.com/f...?topic=73606.0
has a valid point then all these solutions presented here are wrong because they do not account for the rotational kinetic energy:
https://en.wikipedia.org/wiki/Tautochrone_curve

If we take physical bob then this pendulum is not going to work as presented here:
https://en.wikipedia.org/wiki/File:I...al_pendula.gif

This all leads to difference between the linear and rotational acceleration.
This difference appears to break distribution of energy in the presented experiment as it is expected by the relativity. That is the point.

When you say: "The energy for a given point is the same."
For what point? The point representing what? A rigid body? A part of a rigid body?
SDG
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Old 11th August 2018, 04:26 AM   #135
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Originally Posted by SDG View Post
This all leads to difference between the linear and rotational acceleration.
This difference appears to break distribution of energy in the presented experiment as it is expected by the relativity. That is the point.
That is, obviously, nonsense. First of all, think again: If a simple problem of geometric could falsify the theory of relativity, that would have been discovered within the first decade after it's presentation. In fact, Einstein would most likely have discovered it himself, before he presented the theory.

All we have here is a construction of arbitrary definitions that appear to become a mystery if you ignore some of their implications, and that has nothing to do with relativity.

Quote:
When you say: "The energy for a given point is the same."
For what point? The point representing what? A rigid body? A part of a rigid body?
A part of the body sufficiently small that the differential in acceleration can be considered negligible. An atom, if you will.

The whole point is that various parts of a rotating body move at different speeds and in different directions. So if you choose to consider a part of some extent (for instance, one of the balls in your example), then the calculation of the kinetic energy possessed by that part as a whole will be rather complex.

And then, one way to do it is to calculate it's tangential energy and the remaining rotational energy as two different parts. But it's no mystery.

Hans
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Old 12th August 2018, 12:30 PM   #136
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Originally Posted by SDG View Post
Hans,
if the first post from the other forum
https://www.thenakedscientists.com/f...?topic=73606.0
has a valid point then all these solutions presented here are wrong because they do not account for the rotational kinetic energy:
https://en.wikipedia.org/wiki/Tautochrone_curve

If we take physical bob then this pendulum is not going to work as presented here:
https://en.wikipedia.org/wiki/File:I...al_pendula.gif
If the pendulum or the balls in the examples have a considerable rotational energy, like if the balls are large relative to the curve (perhaps even disks), and we continue to ignore that rotational energy, yes, then we are going to have a discrepancy.

Perhaps this is the reason the balls in those examples are all depicted quite small, and perhaps that is why pendulums are usually designed with a bob of negligible mass.

As for pendulums, have you considered clocks that have a balance (instead of a pendulum)? Here, it is all rotational energy. But it works nicely, if calculated in the proper way.

Hans
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Old 12th August 2018, 05:30 PM   #137
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Originally Posted by MRC_Hans View Post
That is, obviously, nonsense. First of all, think again: If a simple problem of geometric could falsify the theory of relativity, that would have been discovered within the first decade after it's presentation. In fact, Einstein would most likely have discovered it himself, before he presented the theory.

All we have here is a construction of arbitrary definitions that appear to become a mystery if you ignore some of their implications, and that has nothing to do with relativity.



A part of the body sufficiently small that the differential in acceleration can be considered negligible. An atom, if you will.

The whole point is that various parts of a rotating body move at different speeds and in different directions. So if you choose to consider a part of some extent (for instance, one of the balls in your example), then the calculation of the kinetic energy possessed by that part as a whole will be rather complex.

And then, one way to do it is to calculate it's tangential energy and the remaining rotational energy as two different parts. But it's no mystery.

Hans
What does this sentence mean? "a simple problem of geometric"

Einstein knew that linear acceleration is not equal to rotational acceleration. The curvature will cause the difference.
That's the reason the equivalence principle requires the linear acceleration to be compared to the 'uniform gravitational field'.

The curtate cycloid trajectory has the linear motion/acceleration and curved rotational motion/acceleration combined in one motion.

There is no mystery. The argument is clear.
The Lagrangian taught at the undergraduate level is wrong, as shown in the other forum. I am not sure if there is a book with the correct 'simple pendulum' Lagrangian at all. Can somebody point me to one?
SDG

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Old 12th August 2018, 05:37 PM   #138
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Originally Posted by MRC_Hans View Post
If the pendulum or the balls in the examples have a considerable rotational energy, like if the balls are large relative to the curve (perhaps even disks), and we continue to ignore that rotational energy, yes, then we are going to have a discrepancy.

Perhaps this is the reason the balls in those examples are all depicted quite small, and perhaps that is why pendulums are usually designed with a bob of negligible mass.

As for pendulums, have you considered clocks that have a balance (instead of a pendulum)? Here, it is all rotational energy. But it works nicely, if calculated in the proper way.

Hans
The example from the other forum showed that the rotational kinetic energy is 0.1% from the translational kinetic energy in the case of pendulum string length l=1m, bob mass m=1kg, r=0.05m.
This is significant. The relativity needs to account for 'every photon' of energy for the relativity to be true. If not than the relativity is 'just' an approximation.
SDG

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Old 12th August 2018, 06:30 PM   #139
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Originally Posted by SDG View Post
The example from the other forum showed that the rotational kinetic energy is 0.1% from the translational kinetic energy in the case of pendulum string length l=1m, bob mass m=1kg, r=0.05m.
This is significant. The relativity needs to account for 'every photon' of energy for the relativity to be true. If not than the relativity is 'just' an approximation.
SDG
Even if this were true, I believe you have it back to front: the pendulum calculations involve approximation, not the other way around.

And another thing - why are you here attempting to have a debate by proxy?
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Old 13th August 2018, 05:47 AM   #140
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Originally Posted by SDG View Post
The example from the other forum showed that the rotational kinetic energy is 0.1% from the translational kinetic energy in the case of pendulum string length l=1m, bob mass m=1kg, r=0.05m.
This is significant. The relativity needs to account for 'every photon' of energy for the relativity to be true. If not than the relativity is 'just' an approximation.
SDG
Actually, I believe it is not so much relativity as thermodynamics, but never mind. Since you seem to be able to easily calculate the figure, it seems that science can account for it.

SO again, it seems any "problems" lie in the approximations used.

Hans
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Old 13th August 2018, 05:54 AM   #141
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Originally Posted by SDG View Post
Einstein knew that linear acceleration is not equal to rotational acceleration. The curvature will cause the difference.
Fine. Glad we sorted that out.

Quote:
That's the reason the equivalence principle requires the linear acceleration to be compared to the 'uniform gravitational field'.
Yes, because a uniform gravitational field causes linear acceleration.

Quote:
The curtate cycloid trajectory has the linear motion/acceleration and curved rotational motion/acceleration combined in one motion.
Fine. So some added complexity. But obviously well understood.

Quote:
There is no mystery. The argument is clear.
The Lagrangian taught at the undergraduate level is wrong, as shown in the other forum.
Look, practically all physics taught at the undergraduate level is "wrong" if you nitpick it closely enough. It nearly always involves approximations. After all, the idea is to teach physics, not complex math.

Hans
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Old 13th August 2018, 06:08 AM   #142
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Originally Posted by SDG View Post
Dave,
let us assume the same wheel with the balls on the elastic rods on the ground, the wheel axle is attached to a concrete block.
When we start to accelerate the wheel the balls inertia will resist the acceleration and it will load 'the springs' of the elastic rods.
The balls will accelerate slower than the point on the wheel below the balls but the balls will follow a circular motion within the ground frame as well.
The same happens to the cycloid. The balls cycloids will have a different trajectory compared to the point on the wheel during the time the elastic rod 'springs' are loaded but the balls will definitely follow a cycloid trajectory.
SDG
No, they won't. As the loadings of the springs change, then the lengths of the springs will change, so the positions of the balls will be different to those they would occupy if they were following a cycloidal path.

You're trying to hav it both ways; you're trying to make up a scenario in which the balls aren't rigidly attached to the wheel but still follow a cycloidal path. Those two conditions are by definition contradictory.

And at the heart of it are two completely erroneous claims, so far as I can see (and it's pretty damned difficult to see them because you're failing dismally to express them clearly):

(1) The rate of rotation of a ball following a cycloid is not constant, because it's given by the product of the frame velocity and the rate of curvature of the cycloid. This is wrong, because the rate of rotation isn't given by that, it's actually given by the product of the rate of curvature of the cycloid and the speed of the ball along the cycloidal path, which is not constant; the two variations of the former and the latter cancel out to give a constant rate of rotation.

(2) The kinetic energy of the ball is constant in the moving frame of reference, but is not constant in the fixed frame of reference; and this is impossible because the magnitude of the force on the ball does not change, therefore the work done on it can have no time dependence. This is also wrong, because the work done on the ball is the scalar product of the force vector and the motion vector of the ball; in the moving frame the force vector is always perpendicular to the motion vector, resulting in zero work done, but in the fixed frame there is an additional constant vector added to the motion vector, resulting in variations in both the magnitude of the motion vector and its angle to the force vector. This inevitably means that work is done on the ball in the stationary frame.

Those are your error and Jano's error, described in detail. Until you stop repeating them there's nothing more to be said.

Dave
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Old 14th August 2018, 07:21 AM   #143
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Originally Posted by MRC_Hans View Post
Fine. Glad we sorted that out.



Yes, because a uniform gravitational field causes linear acceleration.



Fine. So some added complexity. But obviously well understood.



Look, practically all physics taught at the undergraduate level is "wrong" if you nitpick it closely enough. It nearly always involves approximations. After all, the idea is to teach physics, not complex math.

Hans
The first post of the other forum thread is short and straight to the point.
It explains the issue of the rotation very well.
It is much better approach to teach physics. To explain that the 'simple pendulum' is an approximation. If it is not said then the students might think it is precise solution.

We are striving to be precise. Here is one example:
General Relativity and Space Geodesy. Ludwig Combrinck.
From the link:


f_Other - includes satellite rotation effects
f_Emp - empirical corrections

f_Emp tells a lot, even with all the knowledge and hard work we are not able to make a precise calculation.

SDG

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Old 14th August 2018, 08:50 AM   #144
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Originally Posted by Dave Rogers View Post
No, they won't. As the loadings of the springs change, then the lengths of the springs will change, so the positions of the balls will be different to those they would occupy if they were following a cycloidal path.

You're trying to hav it both ways; you're trying to make up a scenario in which the balls aren't rigidly attached to the wheel but still follow a cycloidal path. Those two conditions are by definition contradictory.

And at the heart of it are two completely erroneous claims, so far as I can see (and it's pretty damned difficult to see them because you're failing dismally to express them clearly):

(1) The rate of rotation of a ball following a cycloid is not constant, because it's given by the product of the frame velocity and the rate of curvature of the cycloid. This is wrong, because the rate of rotation isn't given by that, it's actually given by the product of the rate of curvature of the cycloid and the speed of the ball along the cycloidal path, which is not constant; the two variations of the former and the latter cancel out to give a constant rate of rotation.

(2) The kinetic energy of the ball is constant in the moving frame of reference, but is not constant in the fixed frame of reference; and this is impossible because the magnitude of the force on the ball does not change, therefore the work done on it can have no time dependence. This is also wrong, because the work done on the ball is the scalar product of the force vector and the motion vector of the ball; in the moving frame the force vector is always perpendicular to the motion vector, resulting in zero work done, but in the fixed frame there is an additional constant vector added to the motion vector, resulting in variations in both the magnitude of the motion vector and its angle to the force vector. This inevitably means that work is done on the ball in the stationary frame.

Those are your error and Jano's error, described in detail. Until you stop repeating them there's nothing more to be said.

Dave
Dave,
I do understand what you are saying and I agree with you, but there is more to this than meets the eye. I'll explain.

Here is a proper application of the conservation of energy analysis that includes the rotation issues and what to do with it. When it is OK to ignore it and when it is not OK to ignore it.
We take this pendulum:
https://en.wikipedia.org/wiki/File:I...al_pendula.gif
The violet pendulum represents an ideal cycloid, rotating wheel energy analysis.
Now we take our experiment with the well balance horizontal wheel and the balls on the rods.
The radius r=1m, that is the distance from the axle to the center of the ball.
The train moves at 1m/s.
We have a capability to make the rods stiff, the balls, the rods, the wheel are one rigid body.
We will accelerate the wheel to 1rad/s. Now we have an ideal cycloid in the ground reference frame. The centripetal acceleration is 1m/s^2.
We can make the rods elastic again and the balls will not change position in the relation to the wheel.

The energy analysis to determine velocity v at the bottom/top of the ideal cycloid:
1/2m v^2 = mgh
m is eliminated
1/2 v^2 = gh
g is 1m/s^2 in our case, that's the centripetal acceleration.
h = 2m
v = sqrt(2*2) = 2m/s - this is the velocity at the top/bottom of the cycloid.

This is true and it is only true when we ignore the rotational inertia 1/2 I_ball \omega^2. I am fine with ignoring the rotational energy in this case.
Why? Because it is important to understand the system and the boundaries of the system. The boundaries include the initial state of the system.
We are starting the analysis after the acceleration is done.
The wheel and the balls already gained the rotational kinetic energy and the rotational energy is not changing anymore.

The proposed experiment is different though. It is the analysis of the acceleration itself. The boundaries of the system and its analysis are different.
The questions are related to how the rotational kinetic energy 'will enter' the balls.
The moving frame says all the balls will be picking up the rotational kinetic energy at the same rate.
The ground frame says the \omega_ground is not constant through out the cycloid trajectory so the balls should pick up different rotational energy at different points on the cycloid.
The consequence of this should be a positional change between the balls on the wheel. They would not maintain the same distance between themselves as it is in the steady state.

Having said that I am really skeptical the balls will be acquiring the same rotational kinetic energy along the cycloid during the acceleration. I just do not see how this can be possible.
For me personally, this can be settled only by an experiment.
SDG
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Old 14th August 2018, 09:33 AM   #145
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Originally Posted by SDG View Post
We take this pendulum:
https://en.wikipedia.org/wiki/File:I...al_pendula.gif
The violet pendulum represents an ideal cycloid, rotating wheel energy analysis.
No, it doesn't. It's a completely different construction of a completely different experiment, and is pretty much irrelevant other than that it includes a picture of a cycloid.

Originally Posted by SDG View Post
Now we take our experiment with the well balance horizontal wheel and the balls on the rods.
The radius r=1m, that is the distance from the axle to the center of the ball.
The train moves at 1m/s.
We have a capability to make the rods stiff, the balls, the rods, the wheel are one rigid body.
We will accelerate the wheel to 1rad/s. Now we have an ideal cycloid in the ground reference frame. The centripetal acceleration is 1m/s^2.
We can make the rods elastic again and the balls will not change position in the relation to the wheel.
This is an unnecessary complication. Let's simply imagine a scenario where the wheels are rotating in equilibrium and the various elastic components are also in equilibrium.

Originally Posted by SDG View Post
The energy analysis to determine velocity v at the bottom/top of the ideal cycloid:
1/2m v^2 = mgh
m is eliminated
1/2 v^2 = gh
g is 1m/s^2 in our case, that's the centripetal acceleration.
h = 2m
v = sqrt(2*2) = 2m/s - this is the velocity at the top/bottom of the cycloid.
No, it isn't. It's the difference in velocities between the velocity at the top and the velocity at the bottom of the cycloid. But the energy analysis, in this instance, is irrelevant, because you've specified - from the fact that the tyrain is moving at 1m/s, the wheel is rotating at 1rad/s and the radius is 1m - that the ball is, in the stationary reference frame, moving at 2m/s at the top and 0m/s at the bottom of the cycloid.

Originally Posted by SDG View Post
This is true and it is only true when we ignore the rotational inertia 1/2 I_ball \omega^2.
No, it's true whether you include rotational inertia or not, because it's specified in the definition of the problem.

Originally Posted by SDG View Post
The proposed experiment is different though. It is the analysis of the acceleration itself. The boundaries of the system and its analysis are different.
The questions are related to how the rotational kinetic energy 'will enter' the balls.
The moving frame says all the balls will be picking up the rotational kinetic energy at the same rate.
The ground frame says the \omega_ground is not constant through out the cycloid trajectory so the balls should pick up different rotational energy at different points on the cycloid.
Would you like me to explain why the part in bold is wrong for the sixth time?

The rate of rotation of the ball may be expressed as omega, the rate of curvature of the cycloid, multiplied by ds/dt, the rate at which the ball traverses the cycloid. You've already seen that at the bottom of the cycloid, where the rate of curvature is infinite, the speed of the ball is zero, and at the top of the cycloid, where the rate of curvature is a minimum, the speed of the ball is at its maximum.

This can all be analysed mathematically with no need for an experiment. If you correctly determine the expression for ds/dt, the rate at which the ball traverses the cycloid, you will find that it is the inverse of omega, the rate of curvature of the cycloid, and that therefore the product of the two is constant. Since you've already specified in the definition of the problem that the rate of rotation is constant, this shouldn't be a surprise to you.

Now, will you please go away and do that calculation? Until you post an expression for ds/dt, the rate of travel of the ball along the cycloid, I'm not going to bother paying you any more attention.

Dave
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Old 14th August 2018, 10:12 AM   #146
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Originally Posted by Dave Rogers View Post
No, it doesn't. It's a completely different construction of a completely different experiment, and is pretty much irrelevant other than that it includes a picture of a cycloid.



This is an unnecessary complication. Let's simply imagine a scenario where the wheels are rotating in equilibrium and the various elastic components are also in equilibrium.



No, it isn't. It's the difference in velocities between the velocity at the top and the velocity at the bottom of the cycloid. But the energy analysis, in this instance, is irrelevant, because you've specified - from the fact that the tyrain is moving at 1m/s, the wheel is rotating at 1rad/s and the radius is 1m - that the ball is, in the stationary reference frame, moving at 2m/s at the top and 0m/s at the bottom of the cycloid.



No, it's true whether you include rotational inertia or not, because it's specified in the definition of the problem.



Would you like me to explain why the part in bold is wrong for the sixth time?

The rate of rotation of the ball may be expressed as omega, the rate of curvature of the cycloid, multiplied by ds/dt, the rate at which the ball traverses the cycloid. You've already seen that at the bottom of the cycloid, where the rate of curvature is infinite, the speed of the ball is zero, and at the top of the cycloid, where the rate of curvature is a minimum, the speed of the ball is at its maximum.

This can all be analysed mathematically with no need for an experiment. If you correctly determine the expression for ds/dt, the rate at which the ball traverses the cycloid, you will find that it is the inverse of omega, the rate of curvature of the cycloid, and that therefore the product of the two is constant. Since you've already specified in the definition of the problem that the rate of rotation is constant, this shouldn't be a surprise to you.

Now, will you please go away and do that calculation? Until you post an expression for ds/dt, the rate of travel of the ball along the cycloid, I'm not going to bother paying you any more attention.

Dave
Dave,

This from you:
Quote:
"Since you've already specified in the definition of the problem that the rate of rotation is constant, this shouldn't be a surprise to you."
... tells me were are not on the same page. It appears to me you are still stuck in a steady state on an ideal cycloid.

I am not saying the rate of rotation is constant in the proposed experiment!
I am talking about an acceleration of the wheel when the train is moving at the constant velocity already.
The balls go from one curtate cycloid to another during the acceleration in the ground reference frame.
The wheel \omega in the moving frame is not constant as well. The wheel is accelerating its rotation.

This is different from a steady state cycloid.

SDG
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Old 14th August 2018, 11:57 AM   #147
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Originally Posted by Dave Rogers View Post
No, it doesn't. It's a completely different construction of a completely different experiment, and is pretty much irrelevant other than that it includes a picture of a cycloid.



This is an unnecessary complication. Let's simply imagine a scenario where the wheels are rotating in equilibrium and the various elastic components are also in equilibrium.



No, it isn't. It's the difference in velocities between the velocity at the top and the velocity at the bottom of the cycloid. But the energy analysis, in this instance, is irrelevant, because you've specified - from the fact that the tyrain is moving at 1m/s, the wheel is rotating at 1rad/s and the radius is 1m - that the ball is, in the stationary reference frame, moving at 2m/s at the top and 0m/s at the bottom of the cycloid.



No, it's true whether you include rotational inertia or not, because it's specified in the definition of the problem.



Would you like me to explain why the part in bold is wrong for the sixth time?

The rate of rotation of the ball may be expressed as omega, the rate of curvature of the cycloid, multiplied by ds/dt, the rate at which the ball traverses the cycloid. You've already seen that at the bottom of the cycloid, where the rate of curvature is infinite, the speed of the ball is zero, and at the top of the cycloid, where the rate of curvature is a minimum, the speed of the ball is at its maximum.

This can all be analysed mathematically with no need for an experiment. If you correctly determine the expression for ds/dt, the rate at which the ball traverses the cycloid, you will find that it is the inverse of omega, the rate of curvature of the cycloid, and that therefore the product of the two is constant. Since you've already specified in the definition of the problem that the rate of rotation is constant, this shouldn't be a surprise to you.

Now, will you please go away and do that calculation? Until you post an expression for ds/dt, the rate of travel of the ball along the cycloid, I'm not going to bother paying you any more attention.

Dave
Dave,
somebody already linked this site in this thread: http://theelectromagneticnatureofthings.com/cwt/
All the calculations are done there,
SDG
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Old 14th August 2018, 08:26 PM   #148
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Originally Posted by SDG View Post
Dave,
somebody already linked this site in this thread: http://theelectromagneticnatureofthings.com/cwt/
All the calculations are done there,
SDG
An obviously ignorant web page, SDG.
Quote:
This author points out that the Equivalence Principle does not hold in the real life conditions. This is confirmed by the shown analysis and if follows from Einstein/Feynman comments as well and this should be an indication that an issue is looming around the corner.
This is the Equivalence Principle
Quote:
In the theory of general relativity, the equivalence principle is any of several related concepts dealing with the equivalence of gravitational and inertial mass, and to Albert Einstein's observation that the gravitational "force" as experienced locally while standing on a massive body (such as the Earth) is the same as the pseudo-force experienced by an observer in a non-inertial (accelerated) frame of reference.
The textbooks point out that the Equivalence Principle is tested in real life conditions and has passed them!

The "Einstein/Feynman comments" are descriptions of the Equivalence Principle. General relativity is based on that principle and has passed tests of general relativity. That shows that GR works and so the principle is very probably correct.

ETA: A quick search for "Jaroslav Kopernicky" gave a "gravity is electromagnetism" crank as the first results. The title of the blog is "The Electromagnetic Nature Of Things", the article is talking about gravity, you come to your own conclusion.

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Old 14th August 2018, 08:44 PM   #149
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Originally Posted by Reality Check View Post
An obviously ignorant web page, SDG.



This is the Equivalence Principle





The textbooks point out that the Equivalence Principle is tested in real life conditions and has passed them!



The "Einstein/Feynman comments" are descriptions of the Equivalence Principle. General relativity is based on that principle and has passed tests of general relativity. That shows that GR works and so the principle is very probably correct.



ETA: A quick search for "Jaroslav Kopernicky" gave a "gravity is electromagnetism "crank as the first results. The title of the blog is "The Electromagnetic Nature Of Things", the article is talking about gravity, you come to your own conclusion.


I am 'skeptical' GR works.
Check the linked paper a few posts above.
Why the f_Emp exists in the satellite equation?



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Old 15th August 2018, 12:59 AM   #150
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Originally Posted by SDG View Post
I am 'skeptical' GR works.
Check the linked paper a few posts above.
Why the f_Emp exists in the satellite equation?



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Ah, now we get to the heart of the matter.

You don't believe GR works, so are willing to overlook dubious sources and flawed experiments if they validated your suspicion?

What event triggered this "skepticism"? You could have chosen something a little more obscure that wasn't supported by so much evidence...
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Old 15th August 2018, 01:41 AM   #151
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Originally Posted by SDG View Post
Dave,
somebody already linked this site in this thread: http://theelectromagneticnatureofthings.com/cwt/
All the calculations are done there,
SDG
No, they aren't; stop telling lies. At no point in that site are any calculations done on the rate of rotation of a ball on the edge of a wheel. You clearly don't understand the calculations you're trying to parrot.

As for the actual calculations on that site, they suffer from the precise problem you've been whingeing about all along; they take approximations. In particular, the harmonic motion of the oscillator is calculated for an inertial frame of reference in equations 11-14, and not adjusted for the fact that the oscillator in the thought experiment is in a rotating frame of reference. And, unsurprisingly, a small discrepancy is found.

There is nothing going on here except mathematical incompetence based in an inability to understand physics.

Dave
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Old 15th August 2018, 01:44 AM   #152
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Originally Posted by Kid Eager View Post
You don't believe GR works, so are willing to overlook dubious sources and flawed experiments if they validated your suspicion?
It's worse than that. He's raising objections which, if they were true, would overturn Galilean relativity as comprehensively as Einsteinian, and therefore refute Newtonian mechanics. They'd also refute simple logic, because he's trying to prove that the rate of rotation of a ball rigidly attached to the edge of a wheel is different to that of the wheel - though he doesn't seem to understand his own line of argument well enough to realise that that's what he's trying to prove.

Dave
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Old 15th August 2018, 03:36 AM   #153
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Originally Posted by Dave Rogers View Post
No, they aren't; stop telling lies. At no point in that site are any calculations done on the rate of rotation of a ball on the edge of a wheel. You clearly don't understand the calculations you're trying to parrot.

As for the actual calculations on that site, they suffer from the precise problem you've been whingeing about all along; they take approximations. In particular, the harmonic motion of the oscillator is calculated for an inertial frame of reference in equations 11-14, and not adjusted for the fact that the oscillator in the thought experiment is in a rotating frame of reference. And, unsurprisingly, a small discrepancy is found.

There is nothing going on here except mathematical incompetence based in an inability to understand physics.

Dave
Dave,
you were asking for cycloid ds/dt equations. Those are on the site and they are correct.
Good comments about the oscillator, agreed,
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Old 15th August 2018, 04:00 AM   #154
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Originally Posted by SDG View Post
Dave,
you were asking for cycloid ds/dt equations. Those are on the site and they are correct.
OK, now multiply that by the expression for omega for the cycloid, and you should see everything cancel out to leave a constant rate of rotation of the ball for a constant rate of rotation of the wheel. (If you don't get that, you've done it wrong, because your starting point is that they're both rotating at the same angular velocity.) In fact, you'll find that the angular velocities of the ball and the wheel are equal; again, they must be, because they're constrained to be by the definition of the problem. And that means that, for any rate of angular acceleration of the wheel, the angular acceleration of the ball is equal to that of the wheel. And, finally, that means that, for any constant angular acceleration of the wheel, the angular acceleration of the ball is constant, and therefore the rate at which the energy of rotation of the ball increases is also constant. Since that's true in both the fixed and the moving frames, there is no discrepancy.

And that's the seventh time I've explained it.

Dave
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Old 15th August 2018, 04:14 AM   #155
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Originally Posted by Dave Rogers View Post
It's worse than that. He's raising objections which, if they were true, would overturn Galilean relativity as comprehensively as Einsteinian, and therefore refute Newtonian mechanics. They'd also refute simple logic, because he's trying to prove that the rate of rotation of a ball rigidly attached to the edge of a wheel is different to that of the wheel - though he doesn't seem to understand his own line of argument well enough to realise that that's what he's trying to prove.

Dave
Dave,
Do you understand the proposed experiment?
Do you understand that there is energy transfer from the wheel rotational acceleration to the balls? The balls will gain the rotational kinetic energy.
I am going to repeat this is not a steady cycloid motion.

The moving reference frame predicts that all the balls will have the same ratio between the increase of the translational and rotational kinetic energies.

I am asking is this going to be the case for the ground reference frame \omega_ground is not constant in a sense it has different values along the cycloid.
The idea is to separate the balls from the rigid wheel through the elastic rods in order to give the balls a freedom of motion.
We can even seat the balls on a 'friction less' bearing at the top of the elastic rod so the rod will not torque the ball when the rod rotates beneath the ball.
The only forces applied on the ball is the rod pushing the ball and the string attaching the ball to the axle.

The question is what ratio between the translation and rotational kinetic energies is going to be applied to the balls at different points (B, L, T) along the cycloid in the ground reference frame?
Is there going to delta in the acceleration between B-BA points compared to L-LA points? B - ball at the bottom of the cycloid, BA - point on the wheel where B ball rod is attached to the wheel. The same naming applies to L, LA and T, TA points.

Is this a clear enough description of the experiment? Any more questions?
SDG
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Old 15th August 2018, 04:17 AM   #156
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Originally Posted by Kid Eager View Post
Ah, now we get to the heart of the matter.

You don't believe GR works, so are willing to overlook dubious sources and flawed experiments if they validated your suspicion?

What event triggered this "skepticism"? You could have chosen something a little more obscure that wasn't supported by so much evidence...
I am being realistic. How does GR work if we have to use empirical corrections along the GR?
Something is off,
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Old 15th August 2018, 04:22 AM   #157
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Originally Posted by Dave Rogers View Post
OK, now multiply that by the expression for omega for the cycloid, and you should see everything cancel out to leave a constant rate of rotation of the ball for a constant rate of rotation of the wheel. (If you don't get that, you've done it wrong, because your starting point is that they're both rotating at the same angular velocity.) In fact, you'll find that the angular velocities of the ball and the wheel are equal; again, they must be, because they're constrained to be by the definition of the problem. And that means that, for any rate of angular acceleration of the wheel, the angular acceleration of the ball is equal to that of the wheel. And, finally, that means that, for any constant angular acceleration of the wheel, the angular acceleration of the ball is constant, and therefore the rate at which the energy of rotation of the ball increases is also constant. Since that's true in both the fixed and the moving frames, there is no discrepancy.

And that's the seventh time I've explained it.

Dave
The balls are on the elastic rods. They have a freedom of motion compared to the wheel. They do not form a rigid body together.
Where does the bold part come from?
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Old 15th August 2018, 04:47 AM   #158
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Originally Posted by SDG View Post
The moving reference frame predicts that all the balls will have the same ratio between the increase of the translational and rotational kinetic energies.

I am asking is this going to be the case for the ground reference frame \omega_ground is not constant in a sense it has different values along the cycloid.
Yes, it is going to be the same for the ground reference frame, because variations in omega_ground are the inverse of variations in the tangential velocity along the cycloid, so the two cancel out precisely.

I give up. I've explained that to you eight times now. You simply aren't capable of understanding the explanation.

Dave
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Old 15th August 2018, 06:02 AM   #159
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Originally Posted by SDG View Post
Dave,
Do you understand the proposed experiment?
Do you understand that there is energy transfer from the wheel rotational acceleration to the balls? The balls will gain the rotational kinetic energy.
I am going to repeat this is not a steady cycloid motion.

The moving reference frame predicts that all the balls will have the same ratio between the increase of the translational and rotational kinetic energies.

I am asking is this going to be the case for the ground reference frame \omega_ground is not constant in a sense it has different values along the cycloid.
The idea is to separate the balls from the rigid wheel through the elastic rods in order to give the balls a freedom of motion.
We can even seat the balls on a 'friction less' bearing at the top of the elastic rod so the rod will not torque the ball when the rod rotates beneath the ball.
The only forces applied on the ball is the rod pushing the ball and the string attaching the ball to the axle.


The question is what ratio between the translation and rotational kinetic energies is going to be applied to the balls at different points (B, L, T) along the cycloid in the ground reference frame?
Is there going to delta in the acceleration between B-BA points compared to L-LA points? B - ball at the bottom of the cycloid, BA - point on the wheel where B ball rod is attached to the wheel. The same naming applies to L, LA and T, TA points.

Is this a clear enough description of the experiment? Any more questions?
SDG
It is a clear description of you constantly moving the goalposts.

Now, if the balls are mounted on frictionless bearings, they will not rotate with the wheel. They will still go through a circular motion, but they will not rotate themselves. And, of course, they will receive no rotational energy, because all parts of the ball have the same velocity and direction at any given moment.

However, that is NOT the situation you have described previously. There they were fixed to the wheel. That there is some elasticity in the attachment does not make them less fixed. It may delay their acceleration a little bit, and there may be some oscillations during the acceleration phase, but at the end of the day they are going to move just as if they were welded to steel rods.

You are just obfuscating, trying to make the setup unnecessarily complicated.

Hans
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Old 15th August 2018, 06:04 AM   #160
MRC_Hans
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Originally Posted by SDG View Post
I am being realistic. How does GR work if we have to use empirical corrections along the GR?
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SDG
We use such corrections when we don't know all parameters to sufficient precision. For example for satellites, which are subject to several unpredictable forces. (I'm sure the paper you quote lists examples)

ETA, I was wrong: It does not just list examples, it devotes several whole sections to describe such forces, in detail. But of course you ignored that, since it would destroy your argument.

Hans
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Last edited by MRC_Hans; 15th August 2018 at 06:15 AM.
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