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Old 19th August 2018, 09:42 PM   #201
SDG
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Originally Posted by Little 10 Toes View Post
#2 is contradictory. B, L, and T are "obviously" different in these points, but the rods are in a fixed position. Your picture that you selected do not show different rod lengths.
We want the rods and the balls to have the same attributes, to behave the same way when they are next to each other in the same conditions. When this is confirmed then we want to see if there is a delta in their behavior if they are at different points on the cycloid.
As I described in the post #197:
Quote:
For example when rods are next to each other, not moving, loaded with the same energy in the balls, the balls will oscillate in unison till they stop at the same time.
That's the reason the rods are supposed to have the same length.
Does this answer the question?
SDG
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Old 19th August 2018, 10:03 PM   #202
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Originally Posted by Reality Check View Post
This is a statement of the Einstein equivalence principle that Feynman is describing. So:
20 August 2018 SDG: Ignorance of an Feynman describing the Einstein equivalence principle in reply to a question about this thread.

20 August 2018 SDG: Where is the experiment in a freely falling laboratory in this thread?
Does the Feynman link the comparison of the standing rocket on the ground to the rocket accelerating at 1g to the Equivalence principle in the quoted text?
Yes, he does.
What ignorance are you talking about?
Do you understand how my questions in the post #192 relate to GR?
SDG
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Old 20th August 2018, 09:45 AM   #203
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Originally Posted by SDG View Post
The claim:
The proposed experiment leads to detection of straight line constant linear motion from within a moving reference frame without any signal from the outside. The conservation of energy analysis leads to different acceleration predictions between the ground and moving reference frames. Only one acceleration prediction can be true.

I know it is a loose definition but I hope it is good enough for now.
Well, ... it's a claim. I have no idea how you will manage to support it, but hey, have at it!

Quote:
a) The damping is the same in all rods when rods are exposed to the same experimental conditions. For example when rods are next to each other, not moving, loaded with the same energy in the balls, the balls will oscillate in unison till they stop at the same time.
We'll see if it can affect the outcome.

Quote:
b) The ground and the moving/train reference frames.
The ground being Earth, right? Of course, that's also a moving reference frame, but .... do press on.

Hans
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Old 20th August 2018, 09:50 AM   #204
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Originally Posted by MRC_Hans View Post
Well, ... it's a claim. I have no idea how you will manage to support it, but hey, have at it!







We'll see if it can affect the outcome.







The ground being Earth, right? Of course, that's also a moving reference frame, but .... do press on.



Hans


Thanks Hans,
I need an answer to my question regarding the centripetal forces in my post #192.
SDG


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Old 20th August 2018, 09:55 AM   #205
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Originally Posted by SDG View Post
Now I would like to discuss the rotation, angular velocity \omega, acceleration and force on a circle trajectory at different radii.
https://i.imgur.com/lYtMT9w.png
The lines across the ball represent radius, equipotential line.
Black - radius infinity, \omega = 0rad/s, straight line acceleration: 1m/s^2
Green - radius 100m, \omega = 0.1rad/s, acceleration on a circle: 1m/s^2
Orange - radius 1m, \omega = 1rad/s, acceleration on a circle: 1m/s^2

The balls undergo the same acceleration.
Are the F_cp forces equal in all instances or F_cp_1m > F_cp_100m > F_cp_infinity?
This has to be settled before we continue.
SDG
OK, stop right there. This is a game, of course, but not that game. You don't get to demand partial answers in steps. As I said before, YOU have the burden of proof (that is one rule of this game), so YOU do the work. If you want to know the centripetal forces in those instances, go calculate them.

... But I can tell you they won't be equal.

Hans
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Old 20th August 2018, 09:55 AM   #206
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F_cp_1m > F_cp_100m > F_cp_infinity
That's how it appears to me.
Anybody disagrees?
SDG


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Old 20th August 2018, 09:56 AM   #207
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Originally Posted by SDG View Post
I need an answer to my question regarding the centripetal forces in my post #192.
I suggest you stop playing the stupid Gotcha! games and state what the forces are, if it's relevant to your claim.

Dave

ETA: I also suggest you either start treating forces properly as vectors or at least state when you're discussing their magnitudes only.
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Old 20th August 2018, 10:21 AM   #208
MRC_Hans
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Originally Posted by SDG View Post
F_cp_1m > F_cp_100m > F_cp_infinity
That's how it appears to me.
Anybody disagrees?
SDG


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Look, this is not a late nite discussion in the pub (although that might have been preferable), this is supposed to be some kind of scientific discussion. So since the calculation of these forces should be a fairly simple matter, do go ahead and calculate them. Don't ask US. Do it yourself.

Hans
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Old 20th August 2018, 11:38 AM   #209
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The following plot shows cycloid trajectory radius (blue), cycloid omega - \omega_ground (red), cycloid velocity (black) for the cycloid when the train moves at 1m/s and the wheel \omega = 0.5rad/s. The wheel radius is 1m:

The blue line, radius, represents equipotential line in the ground reference frame.

The resulting free body diagram shows the forces for the different radii on the cycloid (ground reference frame) and the resulting force combined with the loading force F_l during the rotational acceleration.


The magnitudes of the vectors are arbitrary, to demonstrate the delta between the forces. The magnitudes are not calculated.
F_cp_infinty is the ground reference point for the ball L - at the inflection point where the radius is infinity.
F_cp_9m is the ground reference point for the ball T - at the top of the cycloid where the cycloid trajectory radius is 9m.
F_cp_1m is the ground reference point for the ball B - at the bottom of the cycloid where the cycloid trajectory radius is 1m.

The equipotential line for the moving reference frame is 1m radius circle.
The moving reference frame predicts the same force for B, L, T balls around the circle.
There is a delta between forces at B, L, T points in the ground reference frame in contradiction to the moving reference frame that does not predict any delta between the forces at B, L, T points.
The balls cross different potential lines in the ground frame. The balls convert potential energy to kinetic energy and kinetic energy back to potential energy along the cycloid. The energy has to be conserved.
This issue does not arise for the moving reference frame.
This analysis supports the claim.

Discussion...
Comments...
SDG
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Old 20th August 2018, 01:11 PM   #210
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Originally Posted by SDG View Post
The following plot shows cycloid trajectory radius (blue), cycloid omega - \omega_ground (red), cycloid velocity (black) for the cycloid when the train moves at 1m/s and the wheel \omega = 0.5rad/s. The wheel radius is 1m:
https://i.imgur.com/b2wj0pG.png?1
The blue line, radius, represents equipotential line in the ground reference frame.

The resulting free body diagram shows the forces for the different radii on the cycloid (ground reference frame) and the resulting force combined with the loading force F_l during the rotational acceleration.
https://i.imgur.com/Q3jma22.png

The magnitudes of the vectors are arbitrary, to demonstrate the delta between the forces. The magnitudes are not calculated.
F_cp_infinty is the ground reference point for the ball L - at the inflection point where the radius is infinity.
F_cp_9m is the ground reference point for the ball T - at the top of the cycloid where the cycloid trajectory radius is 9m.
F_cp_1m is the ground reference point for the ball B - at the bottom of the cycloid where the cycloid trajectory radius is 1m.

The equipotential line for the moving reference frame is 1m radius circle.
The moving reference frame predicts the same force for B, L, T balls around the circle.
There is a delta between forces at B, L, T points in the ground reference frame in contradiction to the moving reference frame that does not predict any delta between the forces at B, L, T points.
The balls cross different potential lines in the ground frame. The balls convert potential energy to kinetic energy and kinetic energy back to potential energy along the cycloid. The energy has to be conserved.
This issue does not arise for the moving reference frame.
This analysis supports the claim.

Discussion...
Comments...
SDG
Quote:
The claim:
The proposed experiment leads to detection of straight line constant linear motion from within a moving reference frame without any signal from the outside.
According to your claim, you can detect the linear motion without reference to the outside (ground frame). You have not shown this. In fact you have explicitly compared the ground reference with the moving reference. Any idiot can detect speed when comparing those frames.

Hans
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Old 20th August 2018, 01:12 PM   #211
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Do you want me to point out for a ninth time that your analysis incorrectly assumes that the rate of motion of the ball along the cycloid is constant, when in fact it varies? Because if you haven't been able to get that fact into your head yet, I really can't see the point.

Dave
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Old 20th August 2018, 02:04 PM   #212
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I'm also wondering how the guy in the train will be able to perceive a cycloid unless the platform with which the wheel is fixed moves within the confines of the train car.
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Old 20th August 2018, 02:15 PM   #213
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Thumbs down A fantasy that any of the questions in post 192 relate to GR

Originally Posted by SDG View Post
Does the Feynman link the comparison of the standing rocket on the ground to the rocket accelerating at 1g to the Equivalence principle in the quoted text?
Yes, he does.
What ignorance are you talking about?
Do you understand how my questions in the post #192 relate to GR?
SDG
ETA: Adding more displays of ignorance is bad, SDG.

21 August 2018 SDG: A fantasy that any of the questions in post 192 relate to GR.
Post 192 has questions about a Fcp and rotating balls of various radii.
Fcp is centripetal force in which case the questions are ignorant. The centripetal force for a pendulum does not depend on the size of the pendulum bob. It depends on the bob mass m, velocity v, and string length r and also the surface gravity g but we can assume this s a terrestrial pendulum.

21 August 2018 SDG: Cannot understand a Feynman quote ending "That is the basis of the principle of equivalence" is about the principle of equivalence!

21 August 2018 SDG: Cannot understand a simple "where is the experiment" question about this thread

20 August 2018 SDG: Ignorance of an Feynman describing the Einstein equivalence principle in reply to a question about this thread.

20 August 2018 SDG: Where is the experiment in a freely falling laboratory in this thread?

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Old 20th August 2018, 02:31 PM   #214
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Originally Posted by MRC_Hans View Post
According to your claim, you can detect the linear motion without reference to the outside (ground frame). You have not shown this. In fact you have explicitly compared the ground reference with the moving reference. Any idiot can detect speed when comparing those frames.

Hans
Hans,
Let us imagine we are in the train car and we do not see outside.
The train is not moving and the wheel is rotating at 0.5rad/s.
We start rotational acceleration.

The top ball will start to move in this direction in relation to the wheel:


The bottom ball will start to move in this direction in relation to the wheel:


The initial balls trajectories are parallel.

Now the train moves at 1m/s, the wheel rotation is 0.5rad/s and we start rotation.

The top ball will start to move in this direction in relation to the wheel:


The bottom ball will start to move in the same direction as in the previous instance because the cycloid trajectory radius at the bottom of the wheel is 1m.
The initial balls trajectories are not parallel in this case.
This is how we know the train car is moving without any signal from the outside. The initial balls trajectories are observed from within. We do not need any signal from the outside.
There is no explicit comparison between the frames. Where do you see that?
SDG
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Old 20th August 2018, 02:35 PM   #215
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The balls will move exactly the same in both cases. Why do you think they wont ? What force will be different ?
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Old 20th August 2018, 02:39 PM   #216
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Originally Posted by Dave Rogers View Post
Do you want me to point out for a ninth time that your analysis incorrectly assumes that the rate of motion of the ball along the cycloid is constant, when in fact it varies? Because if you haven't been able to get that fact into your head yet, I really can't see the point.

Dave
Dave,
yes, the rate of motion of the ball along the cycloid is not constant, it varies.
The velocity is shown in the plot. I have no issue with that.
What are the consequences of this varying velocity?
The changing radius/curvature of the cycloid motion.
What is the consequence of the varying radius/curvature of the cycloid motion?
This:


SDG
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Old 20th August 2018, 02:41 PM   #217
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Originally Posted by Dr.Sid View Post
The balls will move exactly the same in both cases. Why do you think they wont ? What force will be different ?
Check this free body diagram:


SDG
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Old 20th August 2018, 02:46 PM   #218
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I suggest for everybody to cool down. Read it multiple times, sleep on it multiple times.
... then we can discuss,
SDG
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Old 20th August 2018, 02:47 PM   #219
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Hans,
some kids might be reading this. It's not nice to use words like 'idiot' here.
SDG
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Old 20th August 2018, 03:52 PM   #220
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Originally Posted by SDG View Post
Check this free body diagram:
https://i.imgur.com/Q3jma22.png

SDG
Fcp only depends on radius and angular speed. It does not depend on train speed. Since radius and angular speed are the same in both frames of reference, Fcp is also the same in both frames of reference.
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Old 20th August 2018, 04:15 PM   #221
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Originally Posted by Dr.Sid View Post
Fcp only depends on radius and angular speed. It does not depend on train speed. Since radius and angular speed are the same in both frames of reference, Fcp is also the same in both frames of reference.
This is very important to understand:
Originally Posted by SDG View Post
Now I would like to discuss the rotation, angular velocity \omega, acceleration and force on a circle trajectory at different radii.

The lines across the ball represent radius, equipotential line.
Black - radius infinity, \omega = 0rad/s, straight line acceleration: 1m/s^2
Green - radius 100m, \omega = 0.1rad/s, acceleration on a circle: 1m/s^2
Orange - radius 1m, \omega = 1rad/s, acceleration on a circle: 1m/s^2

The balls undergo the same acceleration.
Are the F_cp forces equal in all instances or F_cp_1m > F_cp_100m > F_cp_infinity?
This has to be settled before we continue.
SDG
... and this:
Originally Posted by SDG View Post
Right, there is no d\omega/dt in this case.
What I have in mind is constant \omega and centripetal acceleration. What centripetal acceleration there would be on a circle based on radius, \omega parameters?
Notice how the radius/circle/equipotential line splits the ball into to parts that are not equal for 1m and 100m radii. How is this related to the centripetal force?
SDG
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Old 20th August 2018, 04:42 PM   #222
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Originally Posted by Dr.Sid View Post
Fcp only depends on radius and angular speed. It does not depend on train speed. Since radius and angular speed are the same in both frames of reference, Fcp is also the same in both frames of reference.
The train speed is a factor contributing to varying velocity along the cycloid trajectory.
Then this post is important:
Originally Posted by SDG View Post
Dave,
yes, the rate of motion of the ball along the cycloid is not constant, it varies.
The velocity is shown in the plot. I have no issue with that.
What are the consequences of this varying velocity?
The changing radius/curvature of the cycloid motion.
What is the consequence of the varying radius/curvature of the cycloid motion?
This:


SDG
This is how the F_cp depends on the train speed.
SDG
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Old 20th August 2018, 05:04 PM   #223
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Originally Posted by SDG View Post
...The initial balls trajectories are not parallel in this case.
This is how we know the train car is moving without any signal from the outside. The initial balls trajectories are observed from within. We do not need any signal from the outside.
There is no explicit comparison between the frames. Where do you see that?
SDG


From your post 209 above.


Originally Posted by SDG View Post
The following plot shows cycloid trajectory radius (blue), cycloid omega - \omega_ground (red), cycloid velocity (black) for the cycloid when the train moves at 1m/s and the wheel \omega = 0.5rad/s. The wheel radius is 1m:

The blue line, radius, represents equipotential line in the ground reference frame...
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Old 20th August 2018, 05:30 PM   #224
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Cycloid trajectory has nothing to do with the fact that Fcp only depends on omega and radius.
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Old 20th August 2018, 05:38 PM   #225
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Originally Posted by SDG View Post
Dave,
yes, the rate of motion of the ball along the cycloid is not constant, it varies.
The velocity is shown in the plot. I have no issue with that.
What are the consequences of this varying velocity?
The consequences are that they cancel out with the variation in curvature of the cycloid such that the motion of the balls is the same in both frames of reference. I suggest you read your own ******* reference, http://theelectromagneticnatureofthings.com/cwt/, again, in particular equations 9 and 10, which actually do the calculation and conclude that the accelerations are identical in both frames. As, of course, I've already pointed out to you.

Dave
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Old 21st August 2018, 01:32 AM   #226
MRC_Hans
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Originally Posted by SDG View Post
Hans,
Let us imagine we are in the train car and we do not see outside.
The train is not moving and the wheel is rotating at 0.5rad/s.
We start rotational acceleration.

The top ball will start to move in this direction in relation to the wheel:
https://i.imgur.com/rxTBNAg.png

The bottom ball will start to move in this direction in relation to the wheel:
https://i.imgur.com/gXVOp7I.png
Top? Bottom? The wheel is horizontal.
Anyway, how do you get different movements on the two balls?

Quote:
The initial balls trajectories are parallel.
Parallel? They are moving round.

Quote:
Now the train moves at 1m/s, the wheel rotation is 0.5rad/s and we start rotation.

The top ball will start to move in this direction in relation to the wheel:
https://i.imgur.com/Qf3ie7M.png

The bottom ball will start to move in the same direction as in the previous instance because the cycloid trajectory radius at the bottom of the wheel is 1m.
The initial balls trajectories are not parallel in this case.
They are "parallel" (I suppose you mean identical), when observed from within the train.

Quote:
This is how we know the train car is moving without any signal from the outside. The initial balls trajectories are observed from within. We do not need any signal from the outside.
There is no explicit comparison between the frames. Where do you see that?
SDG
You use the apparent cycloid movement movement that is not observable from within the train.

Hans
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Old 21st August 2018, 01:36 AM   #227
MRC_Hans
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Originally Posted by SDG View Post
The train speed is a factor contributing to varying velocity along the cycloid trajectory.
Then this post is important:


This is how the F_cp depends on the train speed.
SDG
No. You are simply mistaken. Look, even when the train is stopped, Earth is rotating, and orbiting the sun, and moving through the galaxy, all at rather high speeds. Why do you assume that this won't affect your wheel, if the train speed does?

Hans
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Old 21st August 2018, 01:40 AM   #228
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Originally Posted by SDG View Post
This is how the F_cp depends on the train speed.
SDG
No it does not
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Old 21st August 2018, 08:26 AM   #229
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Originally Posted by Dr.Sid View Post
Cycloid trajectory has nothing to do with the fact that Fcp only depends on omega and radius.
Cycloid trajectory has own \omega and radius.
If you have in mind the circle \omega and radius then is it OK to use mix reference frames in the analysis?
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Old 21st August 2018, 08:36 AM   #230
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Originally Posted by SDG View Post
Now I would like to discuss the rotation, angular velocity \omega, acceleration and force on a circle trajectory at different radii.
https://i.imgur.com/lYtMT9w.png
The lines across the ball represent radius, equipotential line.
Black - radius infinity, \omega = 0rad/s, straight line acceleration: 1m/s^2
Green - radius 100m, \omega = 0.1rad/s, acceleration on a circle: 1m/s^2
Orange - radius 1m, \omega = 1rad/s, acceleration on a circle: 1m/s^2

The balls undergo the same acceleration.
Are the F_cp forces equal in all instances or F_cp_1m > F_cp_100m > F_cp_infinity?
This has to be settled before we continue.
SDG

The balls are all equal mass.

The balls undergo the same centripetal acceleration.

F = ma

The forces are equal.

It's true that for the first two balls, the force and centripetal acceleration is not the same on every point within the ball. But it still adds up (integrates) to F = ma = mω2rcm where cm refers to the center of mass, for the whole ball.

It's also true that different and slightly unequal portions of the balls lie on opposite sides of the imaginary "radius line" (actually a cylindrical section) depending on the radius. Doesn't matter. Why not? The reason is different depending on whether or not the ball is free to rotate at an angular velocity other than ω (for instance, zero). At this point I've lost track of which version of the experiment we're describing, so I won't venture further.
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Old 21st August 2018, 08:37 AM   #231
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Reference frames don't even matter, as speed is simply not part of the Fcp equation. Only thing which differs in reference frames is the speed of the whole wheel. That alone does not introduce any force.
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Old 21st August 2018, 08:37 AM   #232
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Originally Posted by Dave Rogers View Post
The consequences are that they cancel out with the variation in curvature of the cycloid such that the motion of the balls is the same in both frames of reference. I suggest you read your own ******* reference, http://theelectromagneticnatureofthings.com/cwt/, again, in particular equations 9 and 10, which actually do the calculation and conclude that the accelerations are identical in both frames. As, of course, I've already pointed out to you.

Dave
Dave,
Yes, the accelerations are identical. I agree with that.
Nevertheless, even though the accelerations are equal the forces does not have to be equal.
F=ma
The rate of change in force does not mean the acceleration will change in the same rate as well. The inertial mass could be changing too.
Originally Posted by SDG View Post
Now I would like to discuss the rotation, angular velocity \omega, acceleration and force on a circle trajectory at different radii.

The lines across the ball represent radius, equipotential line.
Black - radius infinity, \omega = 0rad/s, straight line acceleration: 1m/s^2
Green - radius 100m, \omega = 0.1rad/s, acceleration on a circle: 1m/s^2
Orange - radius 1m, \omega = 1rad/s, acceleration on a circle: 1m/s^2

The balls undergo the same acceleration.
Are the F_cp forces equal in all instances or F_cp_1m > F_cp_100m > F_cp_infinity?
This has to be settled before we continue.
SDG
SDG
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Old 21st August 2018, 08:41 AM   #233
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Originally Posted by SDG View Post
Dave,
Yes, the accelerations are identical. I agree with that.
Nevertheless, even though the accelerations are equal the forces does not have to be equal.
F=ma
The rate of change in force does not mean the acceleration will change in the same rate as well. The inertial mass could be changing too.


SDG
so now you have magical mass changing balls?
anything goes apparently when you want to negate standard physics
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Old 21st August 2018, 08:48 AM   #234
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Originally Posted by SDG View Post
Dave,
Yes, the accelerations are identical. I agree with that.
Nevertheless, even though the accelerations are equal the forces does not have to be equal.
F=ma
The rate of change in force does not mean the acceleration will change in the same rate as well. The inertial mass could be changing too.


SDG
This is getting pathetic. You've just admitted that you've been wrong all along, though you don't even realise it, and now you're trying to invoke physical absurdities to make it look like you had a point all along.

I've been telling you all along that your claim violates, not just Einsteinian, but also Galilean, relativity; that you are, in effect, trying to revoke Newtonian mechanics. You're now suggesting that the inertial mass of a specific body is a variable in the limit of low velocities. Newton would have you sitting in the corner wearing a pointed hat.

Dave
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Old 21st August 2018, 08:50 AM   #235
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Originally Posted by SDG View Post
The rate of change in force does not mean the acceleration will change in the same rate as well. The inertial mass could be changing too.

What is causing the mass of the balls to change?

Or when you say "inertial mass" do you actually mean moment of inertia?
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Old 21st August 2018, 09:17 AM   #236
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Originally Posted by MRC_Hans View Post
Top? Bottom? The wheel is horizontal.
The cycloid from the vertical rolling wheel has bottom and top.
The top of the wheel has velocity 2v in the ground frame and the bottom has 0 velocity.
When we rotate the wheel to the to horizontal position then there is still a point on the cycloid that has velocity 2v. This is what I consider the top of the cycloid. Zero velocity is the bottom of the cycloid.

Quote:
Anyway, how do you get different movements on the two balls?

Parallel? They are moving round.

They are "parallel" (I suppose you mean identical), when observed from within the train.

You use the apparent cycloid movement movement that is not observable from within the train.

Hans
Let's say we know that the train is either moving at 1m/s or is standing.
Can we find out if it is one or the other? How?

The balls are on the wheel in the opposite positions.
We do our experiment starting from an arbitrary point S.
We start the acceleration from \omega=0.5rad/s to 0.6rad/s and we will measure the initial trajectory of the balls.
Then we decelerate back to 0.5rad/s and we start acceleration to 0.6rad/s at the point S+pi/100 and we will measure the angle between the initial trajectories again.

We repeat this for the whole circle.
If we find some points along the circle that have initial trajectories not parallel then the train is moving.
If all the initial trajectories are parallel then the train is standing.

That would be bad to do it this way.
So how about 100 balls around the wheel and to do just one acceleration. That would be faster.

SDG
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Old 21st August 2018, 09:24 AM   #237
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Originally Posted by SDG View Post
The balls are on the wheel in the opposite positions.
We do our experiment starting from an arbitrary point S.
We start the acceleration from \omega=0.5rad/s to 0.6rad/s and we will measure the initial trajectory of the balls.
Then we decelerate back to 0.5rad/s and we start acceleration to 0.6rad/s at the point S+pi/100 and we will measure the angle between the initial trajectories again.

We repeat this for the whole circle.
If we find some points along the circle that have initial trajectories not parallel then the train is moving.
If all the initial trajectories are parallel then the train is standing.
You admitted five posts upthread that the accelerations are identical in both frames of reference. In other words, you just admitted that the above is not true. You're literally unable to follow your own line of argument.

Dave
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Old 21st August 2018, 09:26 AM   #238
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Originally Posted by SDG View Post
The cycloid from the vertical rolling wheel has bottom and top.
The top of the wheel has velocity 2v in the ground frame and the bottom has 0 velocity.
When we rotate the wheel to the to horizontal position then there is still a point on the cycloid that has velocity 2v. This is what I consider the top of the cycloid. Zero velocity is the bottom of the cycloid.


Let's say we know that the train is either moving at 1m/s or is standing.
Can we find out if it is one or the other? How?

The balls are on the wheel in the opposite positions.
We do our experiment starting from an arbitrary point S.
We start the acceleration from \omega=0.5rad/s to 0.6rad/s and we will measure the initial trajectory of the balls.
Then we decelerate back to 0.5rad/s and we start acceleration to 0.6rad/s at the point S+pi/100 and we will measure the angle between the initial trajectories again.

We repeat this for the whole circle.
If we find some points along the circle that have initial trajectories not parallel then the train is moving.
If all the initial trajectories are parallel then the train is standing.

That would be bad to do it this way.
So how about 100 balls around the wheel and to do just one acceleration. That would be faster.

SDG

If you're inside the train, the only trajectory you can measure for the balls is their rotational motion around the axis of the wheel. That won't tell you whether the train is moving or not.

If you're outside the train and you want to know if the train is moving or not, you could (if the setup permits) observe whether the balls on the wheel inside the train are moving in circles or in cycloids. Or you could just measure the trajectory of the damn train.
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Old 21st August 2018, 09:36 AM   #239
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Originally Posted by Myriad View Post
If you're inside the train, the only trajectory you can measure for the balls is their rotational motion around the axis of the wheel. That won't tell you whether the train is moving or not.

If you're outside the train and you want to know if the train is moving or not, you could (if the setup permits) observe whether the balls on the wheel inside the train are moving in circles or in cycloids. Or you could just measure the trajectory of the damn train.
If you were standing on the grd plane watching the train go by you might only see the wheel on its edge and not see a cycloid at all (just a ball moving back and forth in a straight line).
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Old 21st August 2018, 09:45 AM   #240
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Inertial mass can be changing too ? Well, that's outside my expertise. Guess I can't help you with your discovery. It's just way over my head. It would just .. change ? Wow ..

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