# Thread: Relativistic Rolling Wheel II

1. cinci: What I’d like you to do is show what a chord looks like at some other point of the wheel using the method that you now agree works for the roll out.

JT: I do agree that using your methodology gives us the correct roll-out. However, I do not agree that your method gives you the right shape of the chord. Your method says that if the chord is a straight horizontal line in the axle frames, then it is also a straight horizontal line in the road frame. Your method also says that the chord fit inside the circular fender in the axle frame, but that same chord is too long to fit inside the elliptical fender in the road frame. That is just plain wrong. Your methodology does not tell you that the chord curves in the road frame so that it does fit inside the fender in both frames.

I only used your method to get you to see that your own model is telling you that the wheel rolls out gamma*2piR in the road frame. That is something you were missing all this time because you thought you could build the rim with the wheel at rest and then pivot it around a point on the road. Once you starting with the marble contraption, you were effectively starting in the axle frame, and that is where you saw for yourself that the the rim should be built at speed, not at rest.

JT: So the real question should be, "Why did cincirob claim for years that the wheel would roll out 2piR in the road frame?" The answer is because you were modeling the wheel as pivoting at the contact point, and the rim on that type of "wheel" would not have to be built at speed.

cinci: I never claimed that the Gron analysis didn’t roll out 2*2piR. I do claim that a realistic model wouldn’t roll out 2*2piR because there is no relative velocity at the contact point.

JT: We are talking about your marble contraption, not Gron's model. You said your marble contraption was a good model for the wheel, because after you connected the marbles with links, and removed the hula hoop, it correctly modeled a rolling wheel. That is the wheel that you now agree rolls out gamma*2piR. So, I suggest you discard your edge-pinned-disk model now, since it contradicts your newer marble-contraption model.

cinci: No, my friend, after you link up the marbles you have an effectively solid wheel so we are talking about Gron’s model; if I do that analysis for the contact chord in the Gron wheel, I get the answer gamma*2piR and so will you.

And I could comment on all the stuff above and will if you desire. But first, let’s settle the issue at hand. You just agreed that that my analytical approach to Gron’s wheel gives the same roll out Gron gets by looking at it in the wheel frame, gamma*2piR. That means that the Gron analysis approach to the road frame will not give the gamma*2piR rollout when you look at it in the road frame. In fact it gives about one fourth of it. So this is a clear paradox for Gron’s model.
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2. Originally Posted by cincirob
cinci: What I’d like you to do is show what a chord looks like at some other point of the wheel using the method that you now agree works for the roll out.

JT: I do agree that using your methodology gives us the correct roll-out. However, I do not agree that your method gives you the right shape of the chord. Your method says that if the chord is a straight horizontal line in the axle frames, then it is also a straight horizontal line in the road frame. Your method also says that the chord fit inside the circular fender in the axle frame, but that same chord is too long to fit inside the elliptical fender in the road frame. That is just plain wrong. Your methodology does not tell you that the chord curves in the road frame so that it does fit inside the fender in both frames.

I only used your method to get you to see that your own model is telling you that the wheel rolls out gamma*2piR in the road frame. That is something you were missing all this time because you thought you could build the rim with the wheel at rest and then pivot it around a point on the road. Once you starting with the marble contraption, you were effectively starting in the axle frame, and that is where you saw for yourself that the the rim should be built at speed, not at rest.

JT: So the real question should be, "Why did cincirob claim for years that the wheel would roll out 2piR in the road frame?" The answer is because you were modeling the wheel as pivoting at the contact point, and the rim on that type of "wheel" would not have to be built at speed.

cinci: I never claimed that the Gron analysis didn’t roll out 2*2piR. I do claim that a realistic model wouldn’t roll out 2*2piR because there is no relative velocity at the contact point.

JT: We are talking about your marble contraption, not Gron's model. You said your marble contraption was a good model for the wheel, because after you connected the marbles with links, and removed the hula hoop, it correctly modeled a rolling wheel. That is the wheel that you now agree rolls out gamma*2piR. So, I suggest you discard your edge-pinned-disk model now, since it contradicts your newer marble-contraption model.

cinci: No, my friend, after you link up the marbles you have an effectively solid wheel so we are talking about Gron’s model; if I do that analysis for the contact chord in the Gron wheel, I get the answer gamma*2piR and so will you.

And I could comment on all the stuff above and will if you desire. But first, let’s settle the issue at hand. You just agreed that that my analytical approach to Gron’s wheel gives the same roll out Gron gets by looking at it in the wheel frame, gamma*2piR. That means that the Gron analysis approach to the road frame will not give the gamma*2piR rollout when you look at it in the road frame. In fact it gives about one fourth of it. So this is a clear paradox for Gron’s model.
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Yes, let's settle the issue at hand. I say that your marble contraption is a good model for a rolling wheel. You explained how the rim of the marble-contraption-wheel could be built at speed, and I agree with that. I showed that the marble-contraption-wheel would roll out gamma*2piR in the road frame, and you agreed with that. However, that contradicts your pivoting wheel model, which only rolls out 2piR in the road frame. Which model do you think is better? Please tell me because I can't tell from your ambiguous statements. I say the marble-contraption-wheel is better.

3. OK, so let's nail the coffin tighter then. The following Minkowski diagrams depict a single spoke whose tip is aligned with the 90 degree fender benchmark, per ground. Figure 1 depicting a rolling wheel. Figure 2 depicting a translating wheel.

3-space diagram of the rolling wheel (curved spoke per ground) ...

3-space drawing of a purely translating wheel (linear spoke per ground) ...

Note that the wheel occupies the same axle space at any time t', no matter if the wheel rolls (figure 1) or translates (figure 2). The wheel is depicted at 5 equally separated events in the axle system. The wheel has a size and shape at any instant in any spacetime system. It's always round in the axle system, the axle always being stationary. It's shape in the ground system, at (say) time t=0, is dictated by the intersection of the slanted x,y plane at time t=0 with the 4 dimensional worldtube of the wheel in the axle system (represented by only x',y',t').

It's very clear (at a glance) from both figures, that the wheel has the same perimeter shape and size in either case per ground ... a moving 1/gamma contracted ellipse. It has to, because in either case, the wheel atoms occupy the very same volume of axle 3-space over duration. The requirement to maintain Born rigidity is the reason why, and given the 2 specific types of wheel motions involved. It's that simple. As such, and as KJW already pointed out prior, the rotation does not affect the wheel's perimeter shape and size.

One can extrapolate from these figures, that rotation affects the wheel's internal configuration of atoms, eg the velocity, shape & size, and location of individual wheel (and spoke) atoms. Rotation also affects the density of wheel atoms in any internal area of the wheel. Some of these features are depicted directly or indirectly by figure 1, while some other features would require more moments (than just 1) of the ground system to be considered. Nonetheless, all the features stated in this paragraph are determinable given multiple transformations of the rolling wheel "for different ground times t".

And again, a spoke (generally) is curved per ground if the wheel rolls, as Figure 1 depicts. The spoke is (always) linear per ground if the wheel only translates, as Figure 2 depicts. The depicted spoke is sky-blue, a solid line if above the slanted x,y plane at t=0 (representing ground times t > 0), and a dashed line if below (behind) the slanted x,y plane at t=0 (representing ground times t < 0). The highlighted DOTs represent the particular atom of the spoke that exists in the ground system at time t=0, and that moving atom is then momentarily co-located with some stationary axle frame coordinate being transformed to the ground system via the inverse LTs ... using only the translation velocity v between the axle and ground systems.

Thank You,

4. JT: Yes, let's settle the issue at hand. I say that your marble contraption is a good model for a rolling wheel. You explained how the rim of the marble-contraption-wheel could be built at speed, and I agree with that. I showed how that the marble-contraption-wheel would roll out gamma*2piR in the road frame. This contradicts your pivoting wheel model, which only rolls out 2piR in the road frame. Which model do you think is better? Please tell me because I can't tell from your ambiguous statements. I say the marble-contraption-wheel is better.

cinci: Stop changing the subject. This is not about the pivot model; it’s about Gron’s analysis. My point has always been that Gron transforms into the road frame incorrectly when he gets an ellipse. His wheel frame says the roll out is gamma*2piR and we all agree that in the wheel frame that’s what he gets.

In the road frame his model should say the same thing but it doesn’t. When you transform it the way I say he should, you get the same roll out calculated in both frames. That means his transformation is wrong because he doesn’t get the same roll out in both frames. After you agree that this is a flaw in his model then we can talk about better models.
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5. cinci: Right now JT has proved that my suggested approach to Gron’s analysis is consistent with the Gamma*2piR result for roll out by calculating it in the road frame and getting the same answer in the wheel frame. Gron’s doesn’t.

SYA: So the big mystery is as to what cincirob really means in the highlight above?

Starting from the axle frame, Gron's analysis predicts the ground observer must record the roll-out as gamma*2piR, even though a stationary tape measure at rest with the axle records the rotating rim at 2piR. So when you say "Gron's doesn't", what on earth do you mean?

cinci: It means that when you perform the analysis JT did using my suggested transformation you get the right answer in the road, but if you use Gron’s transformation and calculate roll out in the road you will not.

It’s perfectly clear what I meant if you include the underlined part of the sentence. Things will be a lot clearer to you if you read whole sentences instead of “cherry picking”. But assuming you just don’t read that well, I’ll lay it out for you table style:

……………………………………….....……………..My Trans. Approach………….Gron’s Trans Approach

Calculated rollout in Axle Frame……………….gamma*2piR………………….gamma*2piR

Calculated rollout in Road Frame……………..gamma*2piR…………………not gamma*2piR

Calc. Roll out at v = .866……………………………….4*piR…………………………….25*piR

Mystery solved?
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6. Originally Posted by SYA
Starting from the axle frame, Gron's analysis predicts the ground observer must record the roll-out as gamma*2piR, even though a stationary tape measure at rest with the axle records the rotating rim at 2piR. So when you say "Gron's doesn't", what on earth do you mean?

Originally Posted by cincirob
It means that when you perform the analysis JT did using my suggested transformation you get the right answer in the road, but if you use Gron’s transformation and calculate roll out in the road you will not.

It’s perfectly clear what I meant if you include the underlined part of the sentence. Things will be a lot clearer to you if you read whole sentences instead of “cherry picking”. But assuming you just don’t read that well, I’ll lay it out for you table style:

……………………………………….....……………..My Trans. Approach………….Gron’s Trans Approach

Calculated rollout in Axle Frame……………….gamma*2piR………………….gamma*2piR

Calculated rollout in Road Frame……………..gamma*2piR…………………not gamma*2piR

Calc. Roll out at v = .866……………………………… 4*piR…………………………… 0.25*piR

Mystery solved?
Your posts are of the worst written posts I've ever read, and I've read all the same one's you have for many years. The reason they are so bad, comes from your confusion regarding how relativity works. It's akin to someone trying to define a circle mathematically, who does not know math. One can just imagine how poorly worded the questions or descriptions will be. Now, in regards to your roll out comments ...

Wow, you're even more worse off than I thought. OK then, since you are the only fellow who believes the above highlight is actually true, surely you can explain why. Yes? Let us all see how logical your response is, and properly define the mystery here.

Thank You,

7. Originally Posted by cincirob
Stop changing the subject. This is not about the pivot model; it’s about Gron’s analysis. My point has always been that Gron transforms into the road frame incorrectly when he gets an ellipse. His wheel frame says the roll out is gamma*2piR and we all agree that in the wheel frame that’s what he gets.

In the road frame his model should say the same thing but it doesn’t. When you transform it the way I say he should, you get the same roll out calculated in both frames. That means his transformation is wrong because he doesn’t get the same roll out in both frames. After you agree that this is a flaw in his model then we can talk about better models.
So far, this is your only explanation of the failed Gron roll out, per axle. That is ...

1) cincirob has his own concoction to determine roll out.
2) cincirob's roll out does not match Gron's required roll out, in cincirob's belief system.
3) Ergo, Gron's analysis is wrong.

And lets not forget, Gron's rolling disk is assumed to exist in the Born rigid state while rotating at a steady relativistic rate.

Thank You,

8. This --> http://www.thephysicsforum.com/speci...html#post12056

which has been edited slightly since its original posting.

Thank You,

9. Originally Posted by cincirob
JT: Yes, let's settle the issue at hand. I say that your marble contraption is a good model for a rolling wheel. You explained how the rim of the marble-contraption-wheel could be built at speed, and I agree with that. I showed how that the marble-contraption-wheel would roll out gamma*2piR in the road frame. This contradicts your pivoting wheel model, which only rolls out 2piR in the road frame. Which model do you think is better? Please tell me because I can't tell from your ambiguous statements. I say the marble-contraption-wheel is better.

cinci: Stop changing the subject. This is not about the pivot model; it’s about Gron’s analysis.
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What about your marble contraption? You said that was a good model for a rolling wheel. Can't we talk about that anymore?

Originally Posted by cincirob
cinci: My point has always been that Gron transforms into the road frame incorrectly when he gets an ellipse. His wheel frame says the roll out is gamma*2piR and we all agree that in the wheel frame that’s what he gets.

In the road frame his model should say the same thing but it doesn’t. When you transform it the way I say he should, you get the same roll out calculated in both frames. That means his transformation is wrong because he doesn’t get the same roll out in both frames. After you agree that this is a flaw in his model then we can talk about better models.
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Sorry but I don't agree that Gron got a wrong answer for the roll out. Gron's model says the wheel rolls out a distance of gamma*2piR in the road frame. The reason you think his model gets the wrong answer is because all you are doing is looking at how narrow the ellipse is, and deciding that a straight horizontal chord near the bottom would be much shorter than its rest length. You always forget to apply ROS! The chord on the ellipse is not a straight horizontal line!

The chord near the bottom is very nearly its rest length. At the contact point it is its rest length. You claim you understand the Gron model, but you always get this wrong, even after I've pointed this out to you many times over the years. Now can we agree once and for all that the Gron model does not have a straight horizontal chord near the bottom of the ellipse?

10. See the difference?

See the difference?

11. Remember now, from post #95 ... link -> Relativistic Rolling Wheel II (shown in part below) ...

Originally Posted by cincirob
You mentioned my 30 (actually 40+) year career as an engineer. In that time I dealt with many design ideas from other people and in many cases had to dissuade them for various reasons. I know how that is done; one has to give valid reasons for ones arguments. And I know you haven’t come close to that. Simply saying my ideas are bad or claiming I don’t understand relativity doesn’t cut it. Those are only your opinions. Claiming rotation makes Gron’s analysis correct with no real explanation of how doesn’t cut it either. You wouldn’t last 2 weeks in the jobs I’ve held with your attitude and approach. You are a phony.
Most your questions regarding the rolling wheel are answered here ... link -> Relativistic Rolling Wheel II

Note that Minkowski spacetime diagrams model OEMB, verbatum. A picture paints a 1000 words.

Here are some supporting rolling wheel figures ... link ->Relativistically Rolling Wheel

Now, I could easily repeat the relativistic analysis numerically that I did here ... link -> Relativistically Rolling Wheel ... for the spoke depicted in the referenced figures above. However, KJW realized that you had difficulty with the analysis and its related numbers, and so he asked that JT and I "show restraint" with the posting numerical analyses, per this ... link -> Relativistically Rolling Wheel ... and which we have honored. Since numerical analyses do not help you, diagrams and figures are all that's left given you cannot envision how the LTs themselves apply, theoretically. Think carefully about figures JTyesthatJT just posted for you, before responding in haste.

Thank You,

12. JT: Sorry but I don't agree that Gron got a wrong answer for the roll out. Gron's model says the wheel rolls out a distance of gamma*2piR in the road frame. The reason you think his model gets the wrong answer is because all you are doing is looking at how narrow the ellipse is, and deciding that a straight horizontal chord near the bottom would be much shorter than its rest length. You always forget to apply ROS! The chord on the ellipse is not a straight horizontal line!

cinci: You have to apply ROS to the chord when you do it as I suggest. That doesn’t change just because the shape is different. Since you would start with a larger chord in the road frame using my method than Gron’s, you get a still bigger curved chord than Gron.

Think about it. In the wheel frame the rack is contracted and the pinion is full size. When you transform then to the rack frame, the rack gets larger and the pinion gets smaller. It’s not logical to think this would produce the same roll out.
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JT: Sorry but I don't agree that Gron got a wrong answer for the roll out. Gron's model says the wheel rolls out a distance of gamma*2piR in the road frame. The reason you think his model gets the wrong answer is because all you are doing is looking at how narrow the ellipse is, and deciding that a straight horizontal chord near the bottom would be much shorter than its rest length. You always forget to apply ROS! The chord on the ellipse is not a straight horizontal line!

cinci: You used my suggested transformation procedure to conclude that it will predict the correct roll out when you analyze roll out in the road frame. If you can do that with my procedure, surely you can do it with Gron's. Just do it and end the arguing.
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13. To SYA: If you can't understand what I'm asking about roll out, then don't bother posting to this thread; my explanation would satisfy any reasonable person. I did it in two sentences in my message to JT above. Now don't you feel silly? If you still don't get it, you're just posting to confuse the issues without trying to understand it. Most would call that spamming.
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14. Originally Posted by cincirob
To SYA: If you can't understand what I'm asking about roll out, then don't bother posting to this thread; my explanation would satisfy any reasonable person. I did it in two sentences in my message to JT above. Now don't you feel silly? If you still don't get it, you're just posting to confuse the issues without trying to understand it. Most would call that spamming.
Very sorry, but this is a public forum. And, your posts are not consistent with mainstream relativity. As such, I'll be hanging around a bit longer to reduce confusion for innocent readers. For example, you posted this ...

Originally Posted by cincirob
I’ll lay it out for you table style:

……………………………………….....……………..My Trans. Approach………….Gron’s Trans. Approach

Calculated rollout in Axle Frame……………….gamma*2piR………………….gamma*2piR

Calculated rollout in Road Frame……………..gamma*2piR…………………not gamma*2piR

Calc. Roll out at v = .866……………………………… 4*piR…………………………… 0.25*piR

Mystery solved?
As I mentioned already, you are flat wrong regarding the highlights. Add, your approach uses the length contraction formula (LCF), not the Lorentz transformations. As such, you should not label your column "My Trans. Approach", and should be labeled something like "My Approach" or "My LCF approach".

In the Gron analysis, the axle holds atoms at the perimeter of the disk to be moving steadily at the relativistic rotation rate of v. As such, those atoms are length contracted along the circumferential axis of motion. Since Gron's scenario requires Born rigidity to be maintained during the steady rotation rate, there must exist "gamma times more atoms" lined up along that perimeter than it would possess for a non-rotating Born rigid disk. When the perimeter atoms are in contact with the ground, they are at a virtual rest and uncontracted wrt the ground. Sum the rest lengths of each individual perimeter atom when at rest with the ground, and one attains the proper-length of the disk's rotating perimeter, which also equals the roll out length of the disk over one full rotation per ground. A slowly rolling disk would possess no relativistic effects per axle or ground, and as such the atomic rest-lengths would sum to 2piR, and so the roll out is then 2piR. For a Born rigid disk that rolls at relativistic rate, the atomic rest-lengths would sum to gamma*2piR since there are gamma times more atoms along any circumference than a purely translating (non-rotating) disk. Now, how do they fit? Because the higher on the rolling disk you consider, the greater the atomic velocity and length contraction. More contraction, then more Born rigidly connected atoms can fit the perimeter as a whole.

From the ground POV, the atoms are at their rest length when at contact, each atom holding the other essentially non-length-contracted. One might imagine a pure carbon disk on a pure carbon road, all atoms identical. From the axle POV, the disk and road atoms "at contact" are both moving relativistically at the steady rate of v, and so both atoms length contracted equally (since co-moving), each of those atoms holding the other essentially at rest and uncontracted wrt itself. The axle measures the rotating disk perimeter at 2piR using a stationary tape measure at rest with the axle. Since the perimeter atoms are length contracted per axle, there are again "gamma times more atoms" aligned along the perimeter than a non-rotating disk. However, a ground ruler at the contact point is as length contracted as the ground atom contacting the disk's perimeter atom. As such, a moving length-contracted ruler (per axle) measuring a moving equally length-contracted perimeter atom (per axle) must record the rest length of the perimeter atom, since bodies are never contracted per themselves. Since the ground ruler must record the rest length of any co-moving disk atom at contact, the sum of all atoms over a full rotation must be gamma times the perimeter length per axle, so gamma*2piR, ie the roll out length for a full rotation of the relativistically rolling disk per ground.

Thank You,

15. SYA: From the ground POV, the atoms are at their rest length when at contact, each atom holding the other essentially non-length-contracted. One might imagine a pure carbon disk on a pure carbon road, all atoms identical. From the axle POV, the disk and road atoms "at contact" are both moving relativistically at the steady rate of v, and so both atoms length contracted equally (since co-moving), each of those atoms holding the other essentially at rest and uncontracted wrt itself. The axle measures the rotating disk perimeter at 2piR using a stationary tape measure at rest with the axle. Since the perimeter atoms are length contracted per axle, there are again "gamma times more atoms" aligned along the perimeter than a non-rotating disk. However, a ground ruler at the contract point is as length contracted as the ground atom contacting the disk's perimeter atom. As such, a moving length contracted ruler (per axle) measuring a moving equally length contracted perimeter atom (per axle) must record the rest length of the perimeter atom, since contracted rulers cannot record their own contraction. Since the ground ruler must record the rest length of the disk atom at contact, the sum of all atoms over a full rotation must be gamma times the perimeter length per axle, so gamma*2piR, the roll out length for a full rotation of the rolling disk per ground.

cinci: All this is exactly what I’ve been saying. What you are leaving out is the rest length of the rim in your transformation to the road. Remember the rack and pinion. You built it at speed and added teeth; at rest these teeth are smaller in the wheel frame because they are contracted. They will fit in the road frame, if you allow them to return to their rest length. I don’t believe your transformation of the shape of the wheel to the road frame permits this return to their rest length. The underlined sentence above says you need this rest length.

JT already proved that my transformation supports getting the proper length in the road. All you have to do now is prove that the ellipse supports it. At the moment, you’re just assuming it does. My 35 years of engineering tells me that one shouldn’t make too many assumptions.
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16. Originally Posted by cincirob
All this is exactly what I’ve been saying. What you are leaving out is the rest length of the rim in your transformation to the road.
Nope. The wheel's size and shape in the ground system is completely determinable from the information of the moving wheel in the axle system, as per the Gron analysis. And besides, it's precisely as pointed out by Erhenfest and Einstein. Erhenfest showed that the atoms along a circumference length-contract with greater rotation rate. As such, a Born rigid disk must eventually shatter because Born rigidity has its threshhold. Einstein pointed out that a disk riding observer co-rotating with the disk holds the geometry of the disk as non-euclidean, and its perimeter circumference as length gamma*2piR. That be its proper length at steady rotation rate. What's that mean for a rotating body? Well, an atom's proper length (Lp) is its moving-contracted-length (Lc) times the gamma factor, where gamma is defined by the relative velocity between the observer's frame and the atom's rest frame, so Lp = Lc*gamma. It means that if each atom measures itself (it holds itself as stationary), then the sum of all atomic proper-lengths along a circumference (in an ground instant) totals gamma*2piR ... 2piR being the rotating circumference's length as measured by a tape measure at rest with the axle. Since the rim atom of a rolling wheel is comoving with a ground atom when at contact, the ground cannot hold the rim atom as length contracted. Ie, it's at its rest length. As such, the rollout of the wheel over a full rotation must be the sum of the atomic rest lengths of atoms aligned along the perimeter. Again, that is gamma*2piR.

This is also consistent with the fact that if the axle records a full rotation over duration t', that the ground MUST hold that duration dilated at t = gamma*t'. There must be gamma times more atoms on the rolling perimeter than on the perimeter of a wheel of the same radius (per axle) that's non-rotating, for the wheel to roll over a full rotation without slippage ... because the atoms at contact are comoving and uncontracted per each other. Therefore, there is no need to transform from a non-inertial wheel frame to an inertial ground frame, because we already have all the required information without doing so. If one ever feels over zealous, they may try and do so if they so desire, even though it is not required. Gron's analysis didn't need to, and so it does not have the errors and shortcomings you have claimed it does. You just don't understand Gron's analysis, the result of a lack of (a complete) understanding of SR.

What you need to understand, is that the shortcomings of Gron's analysis that you pointed out regarding the roll-out length, are mistakes on your part.

Originally Posted by cincirob
Remember the rack and pinion. You built it at speed and added teeth; at rest these teeth are smaller in the wheel frame because they are contracted. They will fit in the road frame, if you allow them to return to their rest length. I don’t believe your transformation of the shape of the wheel to the road frame permits this return to their rest length. The underlined sentence above says you need this rest length.
No matter how many times you attempt to convert Gron's scenario into a differing version of your own, you will not succeed. Gron's scenario description assumed a Born rigid disk "rotating at steady rate". That is, it is simply assumed to exist as such, and we care not how it got there, or how it continues as such. There has never been a roll up process in Gron's scenario. As such, Gron added no extra atoms. He simply assumed "the Born rigid disk exists at speed".

When I considered a rack and pinion, I assumed what Gron did, that "the pinion simply exists at speed". It would be silly for anyone to assume their wonder-pinion to "not fit the rack at speed". What would be the purpose of that scenario? I assumed the pinion fits flush into the rack at speed while rolling at relativistic rate. For a radius of R' per axle, a specific number of teeth of a specific length are required, for otherwise the pinion would not fit the rack at speed. Then, the relativistic analysis of an impossible wonder-pinion can ensue, and the implications of the theory explored for relativistic rotation rates. So you should understand, that consideration of any roll-up process is not the Gron kinematic analysis, and as such cannot be used as an argument against it.

Now, if you would like to depart from Gron's scenario altogether, and consider a roll up process and its impact on rack and pinion, we can. However, that's not Gron's analysis, that's more akin to Erhenfest's analysis. And keep in mind that all these scenarios consider Born rigid wonder-bodies, because no real wheel, disk, or pinion could exist at relativistic rotation rates due to classical forces alone, let alone survive the roll up process. This has to be re-repeated in your case. As such, we can talk about what would be required of a Born rigid pinion during the roll up process, however that would be a departure from the Gron analysis under discussion, and switching the discussion as such does not change the fact that all your accusations of Gron's analysis are in error.

Originally Posted by cincirob
JT already proved that my transformation supports getting the proper length in the road. All you have to do now is prove that the ellipse supports it. At the moment, you’re just assuming it does. My 35 years of engineering tells me that one shouldn’t make too many assumptions.
What I stated above proves it. The only assumptions are that moving atoms length contract along their axis of motion per their uniform speed per axle, that Born rigidity dictates the number of required moving atoms aligned along its perimeter, and the ground system experiences a time dilation of t = gamma*t'. Everything is consistent, relativistically speaking.

The correct roll out length in the road per Gron, is gamma*2piR. The axle agrees that should happen, per Gron's analysis. You said Gron's analysis failed in that respect, per the axle. That's incorrect.

Thank You,

17. SYA: What I stated above proves it. The only assumptions are that moving atoms length contract along their axis of motion per their uniform speed per axle, that Born rigidity dictates the number of required moving atoms aligned along its perimeter, and the ground system experiences a time dilation of t = gamma*t'. Everything is consistent, relativistically speaking.

The correct roll out length in the road per Gron, is gamma*2piR. The axle agrees that should happen, per Gron's analysis. *You said* Gron's analysis failed in that respect, per the axle. *That's incorrect*.

cinci: Seriously, you aren’t really this dense, are you? You just repeated what everybody knows about calculating roll out in the axle frame. What I keep asking you to do is take your Gron transformed shape and show that you can make that same calculation in the road frame by looking at the contact area the ellipse produces.

You didn’t make a single calculation in the road frame or refer to the ellipse or your method of producing the ellipse.
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18. Originally Posted by cincirob
cinci: What I keep asking you to do is take your Gron transformed shape and show that you can make that same calculation in the road frame by looking at the contact area the ellipse produces.
*****************************
Alright, I will show you using an axle speed of v=0.866025c such that gamma=2. At time t'=0.000000 in the axle frame, these three points form a straight horizontal chord very close to the contact point:
LEFT__(x', y') = (-0.087156, -0.996195)
CENTER(x', y') = (0.000000, -0.996195)
RIGHT_(x', y') = (0.087156, -0.996195)

You can see that the length of that chord is 2*0.087156=0.174312. Since this chord is very close to the rim of the wheel, we can estimate gamma=2, and so the rest length of that chord would be approximately 2*0.174312=0.348624. So that is approximately the length I would expect to find in the road frame. Of course it is not going to be exact, since this chord is not actually located exactly at the contact point.

Let's see. At time t=0.000000 in the road frame, those three points form a v-shaped chord very close to the contact point:
LEFT__(x, y) = (-0.162507, -0.945706)
CENTER(x, y) = (0.000000, -0.996195)
RIGHT_(x, y') = (0.162507, -0.945706)

You can calculate the length of that v-shaped chord to be 2*sqrt(0.162507^2+.050489^2)=0.340339 which is pretty close to the predicted value of 0.348624. If I wanted it to be closer, I would choose a chord that was even closer to the contact point. Happy now?

19. JT: Alright, I will show you using an axle speed of v=0.866025c such that gamma=2. At time t'=0.000000 in the axle frame, these three points form a straight horizontal chord very close to the contact point:
LEFT__(x', y') = (-0.087156, -0.996195)
CENTER(x', y') = (0.000000, -0.996195)
RIGHT_(x', y') = (0.087156, -0.996195)

You can see that the length of that chord is 2*0.087156=0.174312. Since this chord is very close to the rim of the wheel, we can estimate gamma=2, and so the rest length of that chord would be approximately 2*0.174312=0.348624. So that is approximately the length I would expect to find in the road frame. Of course it is not going to be exact, since this chord is not actually located exactly at the contact point.

cinci: Well good going sport, but you don’t get to use the rest length for the Gron analysis. At least a hundred times on this forum either you or SYA have told me you transform the length that is measured in the wheel frame because the speed doesn’t make any difference.

What you get to do is contract 0.174312 by (1 - .866^2)^.5 and get .087156 which does’t come within a row of apples to match .348624.
You used my transformation approach and proved it works….again.
**********************

JT: Happy now?

cinci: Extremely!
****************************

20. Originally Posted by SYA
What I stated above proves it. The only assumptions are that moving atoms length contract along their axis of motion per their uniform speed per axle, that Born rigidity dictates the number of required moving atoms aligned along its perimeter, and the ground system experiences a time dilation of t = gamma*t'. Everything is consistent, relativistically speaking.

The correct roll out length in the road per Gron, is gamma*2piR. The axle agrees that should happen, per Gron's analysis. *You said* Gron's analysis failed in that respect, per the axle. *That's incorrect*.

Originally Posted by cincirob
Seriously, you aren’t really this dense, are you?
OK, so you yourself posted this ...

Originally Posted by cincirob
I’ll lay it out for you table style:

……………………………………….....…………….. My Trans. Approach ………….Gron’s Trans. Approach

Calculated rollout in Axle Frame……………….gamma*2piR………………….gamma*2piR

Calculated rollout in Road Frame……………..gamma*2piR…………………not gamma*2piR

Calc. Roll out at v = .866……………………………… 4*piR…………………………… 0.25*piR

Mystery solved?
As I mentioned already, you are flat wrong regarding the highlights. You remain wrong, still. You've been wrong about a great number of things, relativistically speaking.

All frames must calculate the same rollout of gamma*2piR upon the ground in the ground system.

The correct answers are these ...

……………………………………….....…………..................…...Gron ’s Trans. Approach

Ground's traversal in Axle Frame…………….....…………………. 2piR

Wheel's traversal in Ground Frame…………….....………………. gamma*2piR

Wheel's rollout upon ground, per Axle Frame……………….... gamma*2piR

Wheel's rollout upon ground, per Road Frame……………..... gamma*2piR

The rollout is upon the road, defined by where the rubber meets the road per the ground, over a full rotation.

Thank You,

21. Originally Posted by cincirob
cinci: Well good going sport, but you don’t get to use the rest length for the Gron analysis.
****************************
I didn't use the rest length for anything, I just mentioned what it was for your benefit. Here is a new version of my post without any reference to any lengths at all:

I will show you using an axle speed of v=0.866025c such that gamma=2. At time t'=0.000000 in the axle frame, these three points form a straight horizontal chord very close to the contact point:
LEFT__(x', y') = (-0.087156, -0.996195)
CENTER(x', y') = (0.000000, -0.996195)
RIGHT_(x', y') = (0.087156, -0.996195)

At time t=0.000000 in the road frame, those three points form a v-shaped chord very close to the contact point:
LEFT__(x, y) = (-0.162507, -0.945706)
CENTER(x, y) = (0.000000, -0.996195)
RIGHT_(x, y') = (0.162507, -0.945706)

All of the above was calculated using only the LT's and v. I'll leave it up to you to calculate the lengths involved. The ball is back in your court, hotshot.

22. SYA: As I mentioned already, you are _flat wrong_ regarding the *highlights*. You remain wrong, still. You've been wrong about a great number of things, relativistically speaking. Both the axle and ground frames must calculate (and record) the same gamma*2piR rollout in the ground system.

cinci: Yes, both frames should calculate the same roll out. Now prove that the ellipse analysis does instead of just assuming it.

JT just showed how to do it with my transformation; now do it with yours.
*******************************

23. cinci: Why should I do your calculations. You just wrote end results here; there's no way to tell what you did.
***************************

24. Originally Posted by cincirob
cinci: Why should I do your calculations. You just wrote end results here; there's no way to tell what you did.
***************************
The axle frame would say the points I chose, LEFT(x',y') and RIGHT(x',y'), are 10 degrees apart and centered on the bottom of the circular wheel at axle-time t'=0. The radius of the wheel is 1.000000. That is how I got these axle-frame coordinates:
LEFT__(x', y') = (-0.087156, -0.996195)
CENTER(x', y') = (0.000000, -0.996195)
RIGHT_(x', y') = (0.087156, -0.996195)

I thought you would recognize that as a straight horizontal chord, with endpoints on the circle. I also thought you knew how to transform those three points to the road frame as a v-shaped chord, with endpoints on the ellipse.

25. THREE WHEEL-POINTS, AS MEASURED BY THE AXLE FRAME, ALL AT ONE INSTANT OF TIME :
LEFT__(x', y', t') = (-0.087156, -0.996195, 0.000000)
CENTER(x', y', t') = (0.000000, -0.996195, 0.000000)
RIGHT_(x', y', t') = (0.087156, -0.996195, 0.000000)

THE SAME THREE WHEEL-POINTS, AS MEASURED BY THE AXLE FRAME, AT THREE DIFFERENT TIMES:
LEFT__(x', y' ,t') = (-0.325007, -0.945706, 0.281461)
CENTER(x', y' ,t') = (0.000000, -0.996195, 0.000000)
RIGHT_(x', y' ,t') = (0.325007, -0.945706, -0.281461)

Transforms to this:

THE SAME THREE WHEEL-POINTS, AS MEASURED BY THE ROAD FRAME, ALL AT ONE INSTANT OF TIME:
LEFT__(x, y, t) = (-0.162507, -0.945706, 0.000000)
CENTER(x, y, t) = (0.000000, -0.996195, 0.000000)
RIGHT_(x, y, t) = (0.162507, -0.945706, 0.000000)

26. Originally Posted by SYA
As I mentioned already, you are _flat wrong_ regarding the *highlights*. You remain wrong, still. You've been wrong about a great number of things, relativistically speaking. Both the axle and ground frames must calculate (and record) the same gamma*2piR rollout in the ground system.

Originally Posted by cincirob
Yes, both frames should calculate the same roll out. Now prove that the ellipse analysis does instead of just assuming it.

JT just showed how to do it with my transformation; now do it with yours.
You should have been able to do it with your own transforms, without JT's help. Geesh.

I didn't just assume anything. If you understood my post, and if you understood Gron's work, you would realize that. I can run more calculations for you, if I feel like it, but then you'll just say you already knew that once you learn it and change the argument to something else you are confused about. In fact, you'll likely say everyone already knows that, even if you still don't.

I could run a calculation for each rim atom of the wheel per ground in the instant t, determine their locations, velocities, and contracted lengths then determine their rest lengths and sum them all (if I were so willing). It'll approximate gamma*2piR. Then, you'll ask how I knew where the atoms were in the first place. I'd then tell you via the LTs. You'd then ask how I know the atomic velocities about the rim. I'd say via the composition of velocities formula. You would then argue that the LTs cannot work, because only the translation velocity is used in the transforms and that the atoms move in the axle system too. I'd tell you that you don't understand relativity, and we'd back here at ground zero all over again after my efforts. You may run the analysis if you like, and I'll tell you where you are wrong. How's that sound cinci? Good?

Thank You,

27. JT: I thought you would recognize that as a straight horizontal chord, with endpoints on the circle. I also thought you knew how to transform those three points to a v-shaped chord, with endpoints on the ellipse.

cinci: Why don’t you do it for 1 degree or .1 degrees where the v shape is virtually identical to the chord? That way you can clearly see that the ellipse doesn’t give you the correct roll out.
***************************

28. SYA: I didn't just assume anything. If you understood my post, and if you understood Gron's work, you would realize that. I can run more calculations for you, if I feel like it, but then you'll just say you already knew that once you learn it and change the argument to something else you are confused about. In fact, you'll likely say everyone already knows that, even if you still don't.

I could run a calculation for each rim atom of the wheel per ground in the instant t, determine their locations, velocities, and contracted lengths then determine their rest lengths and sum them all (if I were so willing). It'll approximate gamma*2piR. Then, you'll ask how I knew where the atoms were in the first place. I'd then tell you via the LTs. You'd then ask how I know the atomic velocities about the rim. I'd say via the composition of velocities formula. You would then argue that the LTs cannot work, because only the translation velocity is used in the transforms and that the atoms move in the axle system too.

cinci: It is a blatant lie to say I have said the LTs cannot work. They work every time. What I have said is that if you put the wrong velocity in them you get the wrong answer. Yes, I know your sentence has a “because…” in it, but you phrased the sentence to imply that I don’t understand relativity. If you need to employ these tactics, you know you are wrong.
******************************

I'd tell you that you don't understand relativity, and we'd back here at ground zero all over again after my efforts. You may run the analysis if you like, and I'll tell you where you are wrong. How's that sound cinci? Good?

cinci: JT and I are only discussing the contact area. I ran the number for you and put it in a table. You say it’s wrong. Instead of making your own calculation to prove I’m wrong you are trying to change the subject? You know the ellipse shape can’t give the same roll out you calculate in the wheel frame. If it did you could prove it with a 5 minute calculation. Why not just get honest?
***************************

29. Originally Posted by cincirob
JT: I thought you would recognize that as a straight horizontal chord, with endpoints on the circle. I also thought you knew how to transform those three points to a v-shaped chord, with endpoints on the ellipse.

cinci: Why don’t you do it for 1 degree or .1 degrees where the v shape is virtually identical to the chord? That way you can clearly see that the ellipse doesn’t give you the correct roll out.
***************************
The v-shape does not approach the length of a straight horizontal line across the ellipse. As usual, you are forgetting about ROS. The end points of the curved chord are higher up on the ellipse, due to ROS. You can see that clearly in the picture I snagged from the Gron paper. Anyway, here are the calculations for a 2-degree chord near the bottom of the wheel, (1 degree on each side of the vertical axis). Be sure and let me know the lengths that you calculate, I'm interested to know how it works out:

THREE WHEEL-POINTS, AS MEASURED BY THE AXLE FRAME, ALL AT ONE INSTANT OF TIME :
LEFT__(x', y', t') = (-0.017452 -0.999848, 0.000000)
CENTER(x', y', t') = (0.000000, -0.999848, 0.000000)
RIGHT_(x', y', t') = (0.017452 -0.999848, 0.000000)

THE SAME THREE WHEEL-POINTS, AS MEASURED BY THE AXLE FRAME, AT THREE DIFFERENT TIMES:
LEFT__(x', y' ,t') = (-0.069582, -0.997576, 0.060258)
CENTER(x', y' ,t') = (0.000000, -0.999848, 0.000000)
RIGHT_(x', y' ,t') = (0.069582, -0.997576, -0.060258)

Transforms via the LT's to this:

THE SAME THREE WHEEL-POINTS, AS MEASURED BY THE ROAD FRAME, ALL AT ONE INSTANT OF TIME:
LEFT__(x, y, t) = (-0.034794, -0.997576, 0.000000)
CENTER(x, y, t) = (0.000000, -0.999848, 0.000000)
RIGHT_(x, y, t) = (0.034794, -0.997576, 0.000000)

30. cinci: Ok, let’s figure this out. Assume a 1 degree chord represents the contact length, Wc’, in the axle frame

Wc’ = 2*sin(.5/180*pi()) = 0.017453

This chord in the road frame if the wheel is a 2:1 ellipse is

Wc = Wc’/2 = 0.017453/2 = .008727

Roll out (calculated in road frame) = .008727*360 = 3.147 = ~piR

In the axle frame the roll out is 2piR on a road contracted by half so the roll out is

Roll out (calculated in axle frame) = 4piR.

Seems obvious, but I’ll say it anyway. Calculating roll out in the axle frame give a different answer than calculating it in the road frame using Gron’s elliptical wheel; that is, 4pi does not equal 1 pi.

Now tell me I didn’t include ROS. But if you think ROS is going to solve that problem, then have at it. I’m not wasting any time on it.
**********************************

31. Originally Posted by cincirob
It is a blatant lie to say I have said the LTs cannot work. They work every time. What I have said is that if you put the wrong velocity in them you get the wrong answer. Yes, I know your sentence has a “because…” in it, but you phrased the sentence to imply that I don’t understand relativity. If you need to employ these tactics, you know you are wrong.
If you understood relativity, you would already know that the translation velocity v is the only velocity one needs for determining where the wheel exists in the ground system at any ground time t, given the Gron scenario definition.

Originally Posted by cincirob
JT and I are only discussing the contact area. I ran the number for you and put it in a table. You say it’s wrong. Instead of making your own calculation to prove I’m wrong you are trying to change the subject? You know the ellipse shape can’t give the same roll out you calculate in the wheel frame. If it did you could prove it with a 5 minute calculation. Why not just get honest?
Very sorry, but the wheel rolls out a length of gamma*2piR upon the ground, and the wheel is elliptical per any ground POV. Relativity requires it. Gron's analysis supports that, and is the accepted analysis in mainstream relativity. Your assumption that a rolling wheel's shape is best determined by the motion of a pivoting wheel pinned to the ground, is, to put it nicely, ridiculous.

Thank You,

32. cinci: Ok, let’s figure this out. Assume a 1 degree chord represents the contact length, Wc’, in the axle frame

Wc’ = 2*sin(.5/180*pi()) = 0.017453

This chord in the road frame if the wheel is a 2:1 ellipse is

Wc = Wc’/2 = 0.017453/2 = .008727

Roll out (calculated in road frame) = .008727*360 = 3.147 = ~piR

In the axle frame the roll out is 2piR on a road contracted by half so the roll out is
Roll out (calculated in axle frame) = 4piR.

Seems obvious, but I’ll say it anyway. Calculating ROS in the axle frame give a different answer than calculating it in the road frame using Gron’s elliptical wheel; that is, 4pi does not equal 1 pi.

Now tell me I didn’t include ROS. But if you think ROS is going to solve that problem, the have at it. I’m not wasting any time on it.

OK, I wasted the time. The tips of the 1 degree chord accounting for ROS are 1.000742 degrees apart. Since this number of degrees shows up 360 times during a rotation, the roll out will be 360.267 degrees.

I’m a little rusty at these calculations so I’ll cop to an error if there is one. But until I see the error, this says the ellipse doesn’t come close to rolling out 4piR
********************************

33. SYA,

You are simply ignoring the facts.

cinci

34. Originally Posted by cincirob
cinci: Ok, let’s figure this out. Assume a 1 degree chord represents the contact length, Wc’, in the axle frame

Wc’ = 2*sin(.5/180*pi()) = 0.017453

This chord in the road frame if the wheel is a 2:1 ellipse is

Wc = Wc’/2 = 0.017453/2 = .008727

Roll out (calculated in road frame) = .008727*360 = 3.147 = ~piR

In the axle frame the roll out is 2piR on a road contracted by half so the roll out is
Roll out (calculated in axle frame) = 4piR.

Seems obvious, but I’ll say it anyway. Calculating ROS in the axle frame give a different answer than calculating it in the road frame using Gron’s elliptical wheel; that is, 4pi does not equal 1 pi.

Now tell me I didn’t include ROS. But if you think ROS is going to solve that problem, the have at it. I’m not wasting any time on it.

OK, I wasted the time. The tips of the 1 degree chord accounting for ROS are 1.000742 degrees apart. Since this number of degrees shows up 360 times during a rotation, the roll out will be 360.267 degrees.

I’m a little rusty at these calculations so I’ll cop to an error if there is one. But until I see the error, this says the ellipse doesn’t come close to rolling out 4piR
********************************
I don't know why you chose to use points which are 1 degree apart, considering I already calculated points which are 2 degrees apart. I don't feel like doing my analysis all over again (I had already done 10 degrees prior to this), so I'll just do your analysis for 2 degrees and see what I come up with:

First, for reference, here are the points I calculated for the 2 degree chord:

Originally Posted by JTyesthatJT
The v-shape does not approach the length of a straight horizontal line across the ellipse. As usual, you are forgetting about ROS. The end points of the curved chord are higher up on the ellipse, due to ROS. You can see that clearly in the picture I snagged from the Gron paper. Anyway, here are the calculations for a 2-degree chord near the bottom of the wheel, (1 degree on each side of the vertical axis). Be sure and let me know the lengths that you calculate, I'm interested to know how it works out:

THREE WHEEL-POINTS, AS MEASURED BY THE AXLE FRAME, ALL AT ONE INSTANT OF TIME :
LEFT__(x', y', t') = (-0.017452 -0.999848, 0.000000)
CENTER(x', y', t') = (0.000000, -0.999848, 0.000000)
RIGHT_(x', y', t') = (0.017452 -0.999848, 0.000000)

THE SAME THREE WHEEL-POINTS, AS MEASURED BY THE AXLE FRAME, AT THREE DIFFERENT TIMES:
LEFT__(x', y' ,t') = (-0.069582, -0.997576, 0.060258)
CENTER(x', y' ,t') = (0.000000, -0.999848, 0.000000)
RIGHT_(x', y' ,t') = (0.069582, -0.997576, -0.060258)

Transforms via the LT's to this:

THE SAME THREE WHEEL-POINTS, AS MEASURED BY THE ROAD FRAME, ALL AT ONE INSTANT OF TIME:
LEFT__(x, y, t) = (-0.034794, -0.997576, 0.000000)
CENTER(x, y, t) = (0.000000, -0.999848, 0.000000)
RIGHT_(x, y, t) = (0.034794, -0.997576, 0.000000)
In my first set of data (above), you can see that the 2 degree chord has a length in the axle frame of 2*0.017452=0.034904. I'm sure you would agree to this, since it is twice the size of your 1 degree chord. Now the next step you perform is to say that the ellipse is half as wide as the circle, so the 2 degree chord would only be 0.017452 long in the road frame. Your whole analysis is based on the idea that the horizontal chord in the axle frame transforms as an even shorter horizontal chord in the road frame. You say you did something with ROS, but you don't show the calcs, so I'm not sure what you did.

To see how ROS is done, please have a look at my second set of data (above). I chose axle-frame-times of t'=0.060258 for the left point, and t'=-0.060258 for the right point. The reason is because those times will transform to t=0 in the road frame. At those times, the left point is way over at x'=-0.069582 and the right point is way over at x'=0.069582. The y' coordinate is also higher up on the circle, not that that makes much difference. Anyway, the horizontal distance between those two points would be 2*0.069582=0.139164 and that is the distance that you should be using when you divide it by 2 to make it part of the ellipse. The end result is that that chord is about 0.069582 wide on the ellipse, even though it was only 0.034904 wide on the circle.

You're just doing Gron's analysis wrong, and then claiming the results are wrong. Try doing it the right way, and you will see the result is that the wheel rolls out gamma*2piR in the road frame. By the way, your pinned wheel model only rolls out 2piR in the road frame, and yet you thought that was correct for about 5 years. You probably still think it's correct. So please cut the crap about Gron's model not rolling out the correct distance in the road frame. You never cared about that before.

35. Originally Posted by cincirob
Ok, let’s figure this out. Assume a 1 degree chord represents the contact length, Wc’, in the axle frame

Wc’ = 2*sin(.5/180*pi()) = 0.017453

This chord in the road frame if the wheel is a 2:1 ellipse is

Wc = Wc’/2 = 0.017453/2 = .008727

Roll out (calculated in road frame) = .008727*360 = 3.147 = ~piR

In the axle frame the roll out is 2piR on a road contracted by half so the roll out is
Roll out (calculated in axle frame) = 4piR.

Seems obvious, but I’ll say it anyway. Calculating ROS in the axle frame give a different answer than calculating it in the road frame using Gron’s elliptical wheel; that is, 4pi does not equal 1 pi.

Now tell me I didn’t include ROS. But if you think ROS is going to solve that problem, the have at it. I’m not wasting any time on it.[/I]

OK, I wasted the time. The tips of the 1 degree chord accounting for ROS are 1.000742 degrees apart. Since this number of degrees shows up 360 times during a rotation, the roll out will be 360.267 degrees.

I’m a little rusty at these calculations so I’ll cop to an error if there is one. But until I see the error, this says the ellipse doesn’t come close to rolling out 4piR.
There are too many problems with your analysis here to bother noting. To make a long story short cincirob, consider the following ...

For a non-rotating purely translating wheel, the same ellipse shape exists, even though the location of the wheel atom's within that shape differ for rolling vs translating. That said, consider the following ...

1) non-rotating purely translating wheel ... Any horizontal wheel chord (rim to rim) in the axle system, is a horizontal chord in the ground system. And, the ground will hold that same string-of-wheel-atoms as length contracted by 1/γ. While the axle syste holds those atoms "in its instant", they are not synchronous in the ground system (each atom exists from a different time era of axle system time). There is no "rolling out" of any atoms of that chord, because that wheel does not roll.

2) rolling wheel ... Any horizontal wheel chord (rim to rim) in the axle system, is (virtually always) a curvilinear chord in the ground system ... assuming the chord in both systems is defined "by the very same wheel atoms". As such, the chord is virtually never horizontal in the ground system, if defined as horizontal in the axle system.

Make no mistake, this thread has never been about trying to help make cincirob's new relativistic wheel hypotheses workable from a relativistic standpoint. It's been only about showing cincirob that he does not fully understand relativity in very the first place, because if you did, you would then have no issues with the leading mainstream relativistic rolling wheel analysis, ie Gron's.

Thank You,

36. JT: In my first set of data (above), you can see that the 2 degree chord has a length in the axle frame of 2*0.017452=0.034904. I'm sure you would agree to this, since it is twice the size of your 1 degree chord. Now the next step you perform is to say that the ellipse is half as wide as the circle, so the chord would only be 0.034904 long in the road frame. Your whole analysis is based on the idea that the horizontal chord in the axle frame transforms as an even shorter horizontal chord in the road frame.

cinci: Why do you question this? Every x dimension in the road frame is half what it is in the wheel frame. I used 1 degree because I’m talking about the contact area which is very small. That’s why the 10 degree analysis was not a good idea. 2 degrees will give a similar result to 1 degree.
**************************

JT: You say you did something with ROS, but you don't show the calcs, so I'm not sure what you did.

cinci: I calculated the angle ROS gives you for that chord. You do it a different way than I so showing my analysis would just bring a barrage of questions. The important thing is that it doesn’t change the outcome: Doing this detail analysis of roll out using Gron’s model doesn’t give the same roll out that he gets in the wheel frame. That says to me his model is wrong.
**************************

JT: To see how ROS is done, please have a look at my second set of data (above).

cinci: I know how ROS is done. You and I settled this years ago.
******************************

JT: I chose axle-frame-times of t'=0.060258 for the left point, and t'=-0.060258 for the right point. The reason is because those times will transform to t=0 in the road frame. At those times, the left point is way over at x'=-0.069582 and the right point is way over at x'=0.069582. The y' coordinate is also higher up on the circle, not that that makes much difference. Anyway, the horizontal distance between those two points would be 2*0.069582=0.139164 and that is the distance that you should be using when you divide it by 2 to make it part of the ellipse. The end result is that that chord is about 0.069582 wide on the ellipse, even though it was only 0.034904 wide on the circle.

You're just doing Gron's analysis wrong, and then claiming the results are wrong. Try doing it the right way, and you will see the result is that the wheel rolls out gamma*2piR in the road frame.

cinci: I’ll recheck the numbers; I can’t tell what these are. It’s late and working around different messages on the site is still slow and frustrating for me.
*************************

JT: By the way, your pinned wheel model only rolls out 2piR in the road frame, and yet you thought that was correct for about 5 years. You probably still think its correct. So cut the crap about Gron's model not rolling out the correct distance in the road frame.

cinci: This has nothing to do with the pinned analysis. I still think it would and should roll out 2piR. But this discussion is about checking the consistency of Gron’s analysis. I think it gives different answers in the two frames. If I have made a mistake about Gron’s analysis in the road, I’ll admit it.

The thing that you need to think about is that you agreed that my way of doing the transformation gets the 4piR answer also.
**********************************

37. SYA: Make no mistake, this thread has never been about trying to help make cincirob's new relativistic wheel hypotheses workable from a relativistic standpoint. It's been only about showing cincirob that he does not fully understand relativity in very the first place, because if you did, you would then have no issues with the leading mainstream relativistic rolling wheel analysis, ie Gron's. Make no mistake, this thread has never been about trying to help make cincirob's new relativistic wheel hypotheses workable from a relativistic standpoint. It's been only about showing cincirob that he does not fully understand relativity in very the first place, because if you did, you would then have no issues with the leading mainstream relativistic rolling wheel analysis, ie Gron's.

cinci: Not true. I started this thread and I didn’t start it to create your personal arena to claim I don’t understand relativity. We already know you posted on this thread under the false pretense that you didn’t previously know me. Now we know you’re not really interested in the problem at hand but only here to level ad hominem attacks against me personally.

Why don’t you just start another thread where all you do is criticize me and stop cluttering up this one? Apparently you think there’s an audience for that.
****************************

38. Originally Posted by cincirob
JT: In my first set of data (above), you can see that the 2 degree chord has a length in the axle frame of 2*0.017452=0.034904. I'm sure you would agree to this, since it is twice the size of your 1 degree chord. Now the next step you perform is to say that the ellipse is half as wide as the circle, so the chord would only be 0.034904 long in the road frame. Your whole analysis is based on the idea that the horizontal chord in the axle frame transforms as an even shorter horizontal chord in the road frame.

cinci: Why do you question this? Every x dimension in the road frame is half what it is in the wheel frame.
**************************
I understand that every x dimension in the road frame is half of the corresponding x' dimension in the wheel frame. The thing I question is you applying that principle without also applying ROS. Applying ROS moves the x' dimensions farther apart for the chord near the contact point. You can see it in my numbers. Before ROS, the 2 degree chord has these endpoints LEFT__(x')=(-0.017452) and RIGHT_(x')=(0.017452). After ROS, the 2 degree chord has these endpoints LEFT__(x')=(-0.069582) and RIGHT_(x')=(0.069582). The "after ROS" numbers are the ones you are supposed to divide by 2 to get the x coordinates in the road frame.

Originally Posted by cincirob
cinci: I used 1 degree because I’m talking about the contact area which is very small. That’s why the 10 degree analysis was not a good idea. 2 degrees will give a similar result to 1 degree.
**************************
Understood. But I had already posted my results for the 2 degree chord, and then you went ahead and did your analysis with a 1 degree chord. For the sake of comparing your results to mine, it would have been nice if you had done your analysis with a 2 degree chord. No big deal...

Originally Posted by cincirob
JT: You say you did something with ROS, but you don't show the calcs, so I'm not sure what you did.

cinci: I calculated the angle ROS gives you for that chord. You do it a different way than I so showing my analysis would just bring a barrage of questions. The important thing is that it doesn’t change the outcome:
**************************
Are you saying that after you apply ROS, the chord near the contact point does not come out longer in the road frame than it was in the axle frame? If so, you are not doing ROS correctly. Can you please give me the chord length in x units, instead of degrees?

Originally Posted by cincirob
JT: To see how ROS is done, please have a look at my second set of data (above).

cinci: I know how ROS is done. You and I settled this years ago.
******************************
If you are doing ROS correctly, the chord near the contact point should come out longer in the road frame than it was in the axle frame.

Originally Posted by cincirob
JT: By the way, your pinned wheel model only rolls out 2piR in the road frame, and yet you thought that was correct for about 5 years. You probably still think its correct. So cut the crap about Gron's model not rolling out the correct distance in the road frame.

cinci: This has nothing to do with the pinned analysis. I still think it would and should roll out 2piR. But this discussion is about checking the consistency of Gron’s analysis. I think it gives different answers in the two frames. If I have made a mistake about Gron’s analysis in the road, I’ll admit it.
**********************************
I'm glad to hear you'll admit it if you find a mistake. I apologize for my above remark, I sound like a jerk, LOL.

Originally Posted by cincirob
cinci: The thing that you need to think about is that you agreed that my way of doing the transformation gets the 4piR answer also.
**********************************
I don't know what there is to think about. You finally started in the axle frame, built the rim on the fly, and came to the same conclusion that I did years ago. Namely, that the wheel should roll out gamma*2piR in the road frame.

39. Originally Posted by cincirob
Make no mistake, this thread has never been about trying to help make cincirob's new relativistic wheel hypotheses workable from a relativistic standpoint. It's been only about showing cincirob that he does not fully understand relativity in very the first place, because if you did, you would then have no issues with the leading mainstream relativistic rolling wheel analysis, ie Gron's. Make no mistake, this thread has never been about trying to help make cincirob's new relativistic wheel hypotheses workable from a relativistic standpoint. It's been only about showing cincirob that he does not fully understand relativity in very the first place, because if you did, you would then have no issues with the leading mainstream relativistic rolling wheel analysis, ie Gron's.

Originally Posted by cincirob
Not true. I started this thread and I didn’t start it to create your personal arena to claim I don’t understand relativity. We already know you posted on this thread under the false pretense that you didn’t previously know me. Now we know you’re not really interested in the problem at hand but only here to level ad hominem attacks against me personally.
Far as your misunderstandings of special relativity, that's your doing, by your own hand. If I argue that F=mv, I should expect that if everyone's correcting me saying it's F=ma, then maybe I have it wrong. Your problem, is that you don't do that. You (in analogy) just keep restating that F=mv again and again, then argue that no one else understands what mass is, or that everyone else is a liar liar pants on fire. Indeed you started this thread. You have stated that Gron's analysis is wrong, and you've tried your best to explain why. You were unsuccessful, and in your attempts, you have shown very clearly all your misunderstandings of the special theory. Just to name a few ...

Relativistic Rolling Wheel II

Now, you could take the stance that you're mistaken, given everyone agrees while only you do not. Take a happy pill, and spend a little time studying the many posts presented you for your behalf. It's been entertaining arguing your misunderstandings (for awhile), but no one here's going to make a career out of it. Folks will correct you when you misstate relativity though. You need more study regarding ROS, to learn the meaning of relativistic effects. You should cease on rolling wheels, and start with 2 (then 3) inertial bodies.

Markus already told you, that there is no law that one must use a user name from another forum site. I'm surprised you would never have known that actually, all the years you been foruming. I never pretended not to know anyone, and I didn't feel obligated to advertise if I did know anyone either. I let the posts stand for themselves. If it took you a few posts to figure out who I might be, that's your problem. You stuck with cincirob for your user name, so good for you then.

Thank You,

40. One casual glance at the 2 figures in this post...

link -> Relativistic Rolling Wheel II

explains why ...

Why a POV induces no stresses unto a rotating body.
Why a body can be transformed from an inertial frame in which it moves, to a differing inertial frame in which it also moves.
Why a rolling disk should attain the same perimeter shape as a purely translating disk (if same axle speed per ground).
Why a pivoting wheel pinned to the ground cannot produce the shape of a wheel rolling down the road.
Why an inertial rod and a rotating wheel-chord are NOT equivalent, even if considered in the instant per axle.

The following requires a little more than is presented in the figures, although this has already been completely explained in posts ...

Why Gron's analysis requires a gamma*2piR rollout for a single rotation, per ground. *

* If the axle holds the rotating round disk perimeter at 2piR, and the moving atoms of the perimeter length-contracted at 1/gamma wrt the circumferential axis of motion, then since the ground atoms at contact are also length-contracted at 1/gamma (per axle), then those wheel atoms must roll out the summation of their rest lengths per ground, which is gamma*2piR. Remember, the wheel is assumed not to slip as it rolls!

Supporting figures are found here ... link -> Relativistically Rolling Wheel

I recommend you study the many posts already presented you. One can do a single inertial body scenario, if you like. That would be the best idea, IMO. Compare the inertial POVs of 2 observers, one who moves relatively wrt the body, and one who comoves with the body. Then, we could add a 3rd observer whose inertial POV is midway between the other 2. This would present composition of velocities, and also explain how a body can be transformed from an inertial frame in which it moves, to a differing inertial frame in which it also moves. Then, the Gron analysis and solns will make sense.

Thank You,

41. I think there are a couple of typographical errors in the above post. I think it should say, "Why a pivoting wheel pinned to the ground can NOT better predict the shape of a wheel rolling down the road," and, "Why Gron's analysis DOES require a gamma*2piR rollout for a single rotation, per ground."

42. Yes, thanx JT. I made the corrections.

I originally wrote those bullets to point out cincirob's mistaken beliefs. It seems I did not reword them all, when I decided to instead state what relativity requires, that cincirob does not believe.

Thank You,

43. KJW,

Just 1 quick sidebar post ...

I contacted Markus, and he asked me to contact you, regarding some corrections to the locked thread ...

Minkowski Space and Proper Time (aka wristwatch time)

Only problem, it told me that you are not excepting PM msgs. Is there a way to get those corrections to you, and would you be able to edit that fine post accordingly, once you agree with the change requests? Being a good reference post, I figured the corrections would be desirable for future readers.

Thank You,

44. I saw an episode of Futurama recently that brought to mind a niggling inconsistency (re my graphics posted in the first thread).

Fry and Leela went into a 2D 'haunted house' or 'tunnel of love' type ride where they were translated into a 2D plane before disappearing into a slot like a slide going into a projector. Inside the 2D ride (from the perspective of someone viewing the tv) Fry says to Leela something like 'she is the most beautiful 2D person he has ever seen' and Leela replies that 'in a 2D plane he can only see a straight line' so Fry replies that 'she is the most beautiful straight line he has ever seen' as the scene changes to a side view so the tv viewer can see things from their 2D perspective.

The observer of the relativistic rolling wheel is in the same 2D plane as the wheel in both the wheel and road frames so eitherway the translation should only look like a line. If the observer is at a distance from the wheel and the road then the time taken for the light to travel to the observer should also be taken into consideration in the calculations.

So have I missed where the offset to the observer appears in the calculations or what?

45. Originally Posted by laurieag
The observer of the relativistic rolling wheel is in the same 2D plane as the wheel in both the wheel and road frames so eitherway the translation should only look like a line. If the observer is at a distance from the wheel and the road then the time taken for the light to travel to the observer should also be taken into consideration in the calculations.

So have I missed where the offset to the observer appears in the calculations or what?
The strange effects of special relativity, (e.g. length contraction, time dilation, etc.) are actually not related to how an object is "seen" by the eye. Some find this surprising because special relativity (SR) was built on the postulate that the speed of light is a constant in all inertial reference frames. To make matters worse, practically every text on the subject uses imprecise language such as, "An observer in the stationary frame sees the moving object as length contracted." I always try to avoid using words like "observer" and "sees" because that is actually not what SR is about at all.

To understand SR, one must first understand the way that reference frames are defined. A reference frame is just like a normal Cartesian coordinate system with x, y, and z axes, however there is also a time coordinate, thus (x,y,z,t). In order to understand the time coordinate, imagine that there are synchronized clocks at rest with every x,y,z point throughout the entire coordinate system. So, for example, if one says that an event occurred at coordinates (x,y,z,t)=(1,2,3,4) that means the spacial location of the event was (x,y,z)=(1,2,3) and the time on the clock at rest with that location displayed the time t=4 at the time of the event. If there is an observer located at a different coordinates such as (x,y,z,t)=(0,0,0,4) that person will not see the event yet, because it will take time for the light to reach their eyes. But once they do see the event, and they do the appropriate calculations to account for their distance from the event, they will conclude the event happened at (x,y,z,t)=(1,2,3,4) which was already given in the first place. So all of those visual-correction type calculations are usually left out.

46. Originally Posted by laurieag
I saw an episode of Futurama recently that brought to mind a niggling inconsistency (re my graphics posted in the first thread).

Fry and Leela went into a 2D 'haunted house' or 'tunnel of love' type ride where they were translated into a 2D plane before disappearing into a slot like a slide going into a projector. Inside the 2D ride (from the perspective of someone viewing the tv) Fry says to Leela something like 'she is the most beautiful 2D person he has ever seen' and Leela replies that 'in a 2D plane he can only see a straight line' so Fry replies that 'she is the most beautiful straight line he has ever seen' as the scene changes to a side view so the tv viewer can see things from their 2D perspective.

The observer of the relativistic rolling wheel is in the same 2D plane as the wheel in both the wheel and road frames so either way the translation should only look like a line. If the observer is at a distance from the wheel and the road then the time taken for the light to travel to the observer should also be taken into consideration in the calculations.

So have I missed where the offset to the observer appears in the calculations or what?
As JT put it, Einstein's model defines entities as they exist in spacetime systems. It was the 1959 work of physicists Terrell and Penrose that modeled how relativistically moving bodies are visually seen by the eye (or camera). They took Einstein's SR and considered light transit time to the eye, which produces an effect called Terrell rotation. It's not that Einstein or anyone else never thought about that, but rather that no one bothered to go through the trouble to publish a paper on it until 1959. I figure that Terrell and Penrose probably did it to lay the issue to rest, because so many folks trying the learn the theory are confused between "as it exists in spacetime systems" versus "as seen by the eye". As JT said, most folks tend to say "observer B sees the rocket at length L", when they really mean "the rocket exists at length L in observer B's spacetime system".

See ... http://en.wikipedia.org/wiki/Terrell_rotation

Thank You,

47. Accidental duplicate post, please disregard.

48. Just thinking aloud here ...

Regarding dimensions, here's my take on it, all things being equal. A 2D flatlander exists in 3 dimensional spacetime, ie 2-space plus time, but he visually sees a 2D body as a 1D line in the instant because he and the body exist within the same 2D plane of existence (curved or not). In analogy, we exist as 3D people in a 4 dimensional spacetime, ie 3-space plus time, but see a 3D body as a 2D surface in the instant. While bodies occupy a 3-space, the photons strike our retina (essentially a 2D surface), and so the 3D body (eg say Jennifer Lawrence) is seen as a 2D surface but with a 3D effect (in analogy to the image of a TV screen, presented on a 2D surface). We also have an improved sense of depth with binocular vision, and so 2D images are processed by the brain with an enhanced 3D effect. The 2D flatlander's retina is a 1D line, but with binocular vision should have an improved sense of depth, and so 1D images "could" be processed by the brain with an enhanced 2D effect. That is, while Fry sees Leela as a line, that line also has a depth. If Leela is facing Fry, Fry should see that the top of Leela's head is further away than her nose, even though he sees her as only a line.

Thank You,

49. Originally Posted by SinceYouAsked
It's not that Einstein or anyone else never thought about that, but rather that no one bothered to go through the trouble to publish a paper on it until 1959. I figure that Terrell and Penrose probably did it to lay the issue to rest, because so many folks trying the learn the theory are confused between "as it exists in spacetime systems" versus "as seen by the eye". As JT said, most folks tend to say "observer B sees the rocket at length L", when they really mean "the rocket exists at length L in observer B's spacetime system".

See ... Terrell rotation - Wikipedia, the free encyclopedia

Thank You,
Terrell's and Penrose's papers pointed out that although special relativity appeared to describe an "observed contraction" in moving objects, these interpreted "observations" were not to be confused with the theory's literal predictions for the visible appearance of a moving object. Thanks to the differential timelag effects in signals reaching the observer from the object's different parts, a receding object would appear contracted, an approaching object would appear elongated (even under special relativity) and the geometry of a passing object would appear skewed, as if rotated.[1][2]
For images of passing objects, the apparent contraction of distances between points on the object's transverse surface could then be interpreted as being due to an apparent change in viewing angle, and the image of the object could be interpreted as appearing instead to be rotated. A previously-popular description of special relativity's predictions, in which an observer sees a passing object to be contracted (for instance, from a sphere to a flattened ellipsoid), was wrong.[1][2]
Terrell's and Penrose's papers prompted a number of follow-up papers,[4][5][6][7][8] mostly in the American Journal of Physics, exploring the consequences of this correction. These papers pointed out that some existing discussions of special relativity were flawed and "explained" effects that the theory did not actually predict – while these papers did not change the actual mathematical structure of special relativity in any way, they did correct a misconception regarding the theory's predictions.
Thanks SYA, although it puts Gron's analysis under a bit of a cloud.

JT, it's interesting that you say that all of the visual correction type calculations are left out in SR. If you ignore the distance traveled in one complete rotation and instead make r unitary your actual visual calculations could be out by the 2 * Pi that was left out. As 2 * Pi is also the difference between the standard and reduced Compton wavelengths and the standard and reduced Planck lengths (basically the difference between rest mass and relative mass), not to mention the translation factor to spherical coordinant systems, leaving out the 2 * Pi for the analysis of astronomical data based on the apparent positions of rotating galactic sources could lead to a discrepancy in the calculated mass, to the magnitude of the current apparent discrepancy between visible matter and dark matter.

50. Originally Posted by laurieag
Thanks SYA, although it puts Gron's analysis under a bit of a cloud.
If you go to the OP's Gron reference here ...

http://areeweb.polito.it/ricerca/rel...los/gron_d.pdf

then see page 39, figure 9, part C.

Part C shows where each atom of the rolling disk was located in (say) the ground observer's system given light emanating from those disk-perimeter atoms all reach the ground observer's eyeball "at" the designated ground contact event. The disk appears rotated, and elongated, per said ground observer's eyeball. Those are optical effects (due to light transit time given the location of the observing POV) atop the relativistic effects. Keep in mind that this disk rolls, and is not purely translating. Also keep in mind, that the part C would appear somewhat different had the POV been that of a ground frame observer at the axle, versus at the ground contact point.

Thank You,

51. Originally Posted by laurieag
JT, it's interesting that you say that all of the visual correction type calculations are left out in SR.
Visual corrections are not "left out" in SR. There are virtual synchronized clocks at rest throughout the entire reference frame, so visual corrections are not necessary. As SinceYouAsked showed, the "visual result" can easily be deduced if someone happens to be curious about that.

52. Originally Posted by SinceYouAsked
then see page 39, figure 9, part C.
The legend for figure 9 clearly states "C: observed in 0 K at retarded points of time" so it can not be considered as what any natural observer would see.

53. Originally Posted by JTyesthatJT
Visual corrections are not "left out" in SR. There are virtual synchronized clocks at rest throughout the entire reference frame, so visual corrections are not necessary. As SinceYouAsked showed, the "visual result" can easily be deduced if someone happens to be curious about that.
An “optical appearance” does not equate to a "visual result" or point to easy solutions.

The positions of points on a rolling ring at retarded points of time were calculated with reference to 0 K by Ø. Grøn [111]. The result is shown in Fig. 9. Part C of the figure shows the “optical appearance” of a rolling ring, i.e. the positions of emission events where the emitted light from all the points arrives at a fixed point of time at the point of contact of the ring with the ground. In other words it is the position of the points when they emitted light that arrives at a camera on the ground just as the ring passes the camera.

54. Originally Posted by laurieag
The legend for figure 9 clearly states "C: observed in 0 K at retarded points of time" so it can not be considered as what any natural observer would see.
You have to look out for imprecise language in these texts. Note how the text also states that the light arrives at a camera on the ground just as the ring passes by the camera, which is one instant of time. What they are trying to say is that the light came from different times in the past, thus "retarded points in time". Everything we see is light from different times in the past.

I gave you an example where an event occurred at coordinates (x,y,z,t)=(1,2,3,4), and there was a visual observer located at (x,y,z)=(0,0,0). That person will not see the event until his own clock displays t=7.7416, even though the event actually occurred earlier, at t=4. The observer might say that t=4 is a "retarded point in time", because his own clocks displays t=7.7416 at the time when he sees the event.

Originally Posted by laurieag
An “optical appearance” does not equate to a "visual result". . .
I don't think there is any difference between "optical appearance" and "visual result". What do you think the difference is?

Originally Posted by laurieag
. . . or point to easy solutions.
The calculation I gave above was easy. It is just

55. Originally Posted by laurieag
The legend for figure 9 clearly states "C: observed in 0 K at retarded points of time" so it can not be considered as what any natural observer would see.
The visual image received by the eye via photons is a collection of photons from different events occurring at different times, that all reach the eye at the same instant. Gron has the model of a relativistically rolling wheel, and as such knows the location of any perimeter atom at any instant per ground. It's then easy to determine from where photons had to emanate (prior POE events) to reach the eye co-located at the specified ground-contact-event in the ground system, because light travels isotropically at c in any inertial system.

The elongated/rotated wheel (pg 39 Fig 9 Part C) shows what the ground observer at the ground contact event visually sees at that event (eye is at ground level, not above it). The length contracted ellipse (pg 39 Fig 9 Part B) shows how the wheel exists (not seen) in that ground observer's spacetime system upon the ground contact event. The round wheel (pg 39 Fig 9 Pact A) shows how the wheel exists in spacetime (not seen), per a non-rotating inertial observer co-moving with the axis-of-rotation, which Gron presents for reference. This is just to say that after Gron did his analysis of the relativistically rolling wheel (obtaining the relativistic effects), he also subsequently analyzed the wheel as per Terrell and Penrose (obtaining the visual effects).

See pg 39 Figure 9 here ...

Gron paper ... http://areeweb.polito.it/ricerca/rel...los/gron_d.pdf

Thank You,

56. Originally Posted by JTyesthatJT
What they are trying to say is that the light came from different times in the past, thus "retarded points in time". Everything we see is light from different times in the past.
...
I don't think there is any difference between "optical appearance" and "visual result". What do you think the difference is?
The only locations where the time will not be retarded for all points around the wheel is at the axle or any point perpendicular to the wheel plane, at the same height as the axle and comoving with the wheel.

57. Originally Posted by SinceYouAsked
The visual image received by the eye via photons is a collection of photons from different events occurring at different times, that all reach the eye at the same instant. Gron has the model of a relativistically rolling wheel, and as such knows the location of any perimeter atom at any instant per ground. It's then easy to determine from where photons had to emanate (prior POE events) to reach the eye co-located at the specified ground-contact-event in the ground system, because light travels isotropically at c in any inertial system.
You are only looking at small scales where the distances traveled is small when compared with c, like spinning a sparkler in a circle and taking a picture of it.

If I started photographing a sparkler around 6 and a bit feet away, the sparkler was being spun in a circle 2 feet in diameter and I captured the light from the spinning sparkler in one complete circle the ratio ( A ) of the distance between the rotating source and the observer over the diameter of rotation would be roughly equal to Pi.
In this case the ratio ( B ) of the distance between source and observer over the distance traveled by light in a year would be very small and the ratio ( C ) of the observation period over the time it takes for the sparkler to be rotated once will equal one. All observations should have a width of field that covers the complete diameter of rotation of the source being observed. D is the distance/time traveled from the rotating source to the observer.

If I halve the exposure period I get half a circle and capture half as much light and when I double the exposure period I get 2 circles over each other and twice as much light in my photograph. If the sparkler is rotated twice as fast I would expect something that looked similar to when I doubled the exposure period but I would also expect to capture the same amount of light from only one rotation despite the doubling of the speed of rotation. If I taped two sparklers together I could halve the exposure time and double the speed of rotation to capture a similar amount of light from 1 sparkler doing 1 complete rotation. If the sparkler was moved at an angle to me I would observe an oval instead of a circle but the amount of light captured would remain the same as for a complete circle.

In this simplest base context D = Pi, B = tiny, C = 1 and the observer will capture one complete cycle. On any scale where C >= 1 the observer will capture at least one complete cycle despite the size of B.

On any scale where D = Pi * x, B >= 1 and C < 1 the observer will only capture the light from B * C = x of one rotation during any observation regardless of the speed of rotation of the source.

58. Originally Posted by SinceYouAsked
The visual image received by the eye via photons is a collection of photons from different events occurring at different times, that all reach the eye at the same instant. Gron has the model of a relativistically rolling wheel, and as such knows the location of any perimeter atom at any instant per ground. It's then easy to determine from where photons had to emanate (prior POE events) to reach the eye co-located at the specified ground-contact-event in the ground system, because light travels isotropically at c in any inertial system.

Originally Posted by Laurieg
You are only looking at small scales where the distances traveled is small when compared with c,
Why would you compare a distance to a rate (eg. c)?

In Gron's figures, the scale is determined by the distance light must travel from any and all points of the rolling disk's perimeter relative to the point of image reception at the designated ground contact point. There is no other scale that matters, as the question being addressed here is specific.

Originally Posted by Laurieg
You are only looking at small scales where the distances traveled is small when compared with c, like spinning a sparkler in a circle and taking a picture of it.

If I started photographing a sparkler around 6 and a bit feet away, the sparkler was being spun in a circle 2 feet in diameter and I captured the light from the spinning sparkler in one complete circle, the ratio ( A ) of the distance between the rotating source and the observer over the diameter of rotation would be roughly equal to Pi.
Assuming "a bit of feet" means ~2 feet, I'd agree.

Originally Posted by Laurieg
In this case the ratio ( B ) of the distance between source and observer over the distance traveled by light in a year would be very small and the ratio ( C ) of the observation period over the time it takes for the sparkler to be rotated once will equal one. All observations should have a width of field that covers the complete diameter of rotation of the source being observed. D is the distance/time traveled from the rotating source to the observer.

If I halve the exposure period I get half a circle and capture half as much light and when I double the exposure period I get 2 circles over each other and twice as much light in my photograph. If the sparkler is rotated twice as fast I would expect something that looked similar to when I doubled the exposure period but I would also expect to capture the same amount of light from only one rotation despite the doubling of the speed of rotation. If I taped two sparklers together I could halve the exposure time and double the speed of rotation to capture a similar amount of light from 1 sparkler doing 1 complete rotation. If the sparkler was moved at an angle to me I would observe an oval instead of a circle but the amount of light captured would remain the same as for a complete circle.

In this simplest base context D = Pi, B = tiny, C = 1 and the observer will capture one complete cycle. On any scale where C >= 1 the observer will capture at least one complete cycle despite the size of B.

On any scale where D = Pi * x, B >= 1 and C < 1 the observer will only capture the light from B * C = x of one rotation during any observation regardless of the speed of rotation of the source.
OK, so you are considering what a non-rotating camera co-moving with the sparkler's axis-of-rotation would capture, from differing ranges from the sparkler (orthogonal to its plane of sweep), for different shutter speeds. IOW, you are no longer talking about Gron's work which considered everything in only the instant per ground. So instead of considering the cycloid produced by a ground POV, you are considering the circular arc produced by he who co-moves with (essentially) the axle.

What is the purpose of your scenario specifications and summary here, in so far as how it relates to relativity theory or Gron? Can you summarize in a brief statement, the point you are trying to convey, and as to why you think it is important?

Thank You,

59. Originally Posted by laurieag
The only locations where the time will not be retarded for all points around the wheel is at the axle or any point perpendicular to the wheel plane, at the same height as the axle and comoving with the wheel.
I don't think you understand. The camera is located on the road. The wheel rolls up to the camera as if it is getting ready to roll over it. The camera takes a picture when the bottom of the wheel (the point labeled 1) is in virtually the same place as the camera. Therefore the light from the bottom point of the wheel travels the shortest distance, essentially zero. Thus, the bottom point (labeled 1) is the only photographed point on the wheel which did not come from a retarded point in time.

60. SYA and JT, have you tried to locate the position of the road at each emission event from Gron's Figure 9 part C?

The positions of points on a rolling ring at retarded points of time were calculated with reference to 0 K by Ø. Grøn [111]. The result is shown in Fig. 9. Part C of the figure shows the “optical appearance” of a rolling ring, i.e. the positions of emission events where the emitted light from all the points arrives at a fixed point of time at the point of contact of the ring with the ground. In other words it is the position of the points when they emitted light that arrives at a camera on the ground just as the ring passes the camera.

61. Originally Posted by laurieag
SYA and JT, have you tried to locate the position of the road at each emission event from Gron's Figure 9 part C?
Personally, I have not. I don't see it as a problem of any sort. Should be straight forward once one has the transformations solns from axle frame to ground frame. I presume you have made the attempt, and obtained solns that make no sense to you, for otherwise I would venture you would not ask. If you did run the calculations, what solns did you attain, and what do you think is wrong with those solns?

If you can show your method of calculation, that would be more beneficial, because if that process were insufficient, then the solns would be improper.

Thank You,

62. Originally Posted by laurieag
SYA and JT, have you tried to locate the position of the road at each emission event from Gron's Figure 9 part C?
I haven't either, but I think I know of a good way to check:

Beginning with figure B we see that the 9:00 o'clock point location on the ellipse is between the points labeled 14 and 15, so I'd say it's about 14.5. In figure C, the point labeled 10 happens to be located at the same height as the axle, so an ellipse can be drawn there, and the 9:00 o'clock point would be the point labeled 10. Thus the ellipse has rotated (14.5-10)/16 of one rotation, or 4.5/16 of one rotation. Since the wheel rolls out gamma2piR per rotation, we can label (in blue) the distance the wheel rolled as 4.5gamma2piR/16. We know the light must travel a longer path than the wheel travels by a factor of 1/v, so we can label (in red) the distance the light traveled as 4.5gamma2piR/16v. The rest of the labels (in green and orange) should be self explanatory. Now all we have to do is solve for the hypotenuse of the right triangle using the blue, green, and orange labels, and compare the result to the red label. If all is well, they should be the same.

EDIT: I've checked it now, and it seems to work out pretty nicely for v=0.8c gamma=1.666. Gron must have changed his value of v from figure 8 which says v=0.7c in the opposite direction. Also, 4.511 works better than 4.5.

63. Considering that both A and B have points 9 and 1 at common heights i.e. wheel, road and camera are in the same plane so 9 to 1 = diameter of the wheel from top to bottom, I just projected downwards the distance from each emission point on A and B to the 'road' and joined all of the lines together.

Do you get different variations for the different points or is the variation consistent across all points in your analysis?

While the geometry is obviously skewed (with respect to the road) and the wheel is obviously elongated, according to Terrell rotation should it be both approaching and passing at the same time?

Terrell rotation - Wikipedia, the free encyclopedia

Thanks to the differential timelag effects in signals reaching the observer from the object's different parts, a receding object would appear contracted, an approaching object would appear elongated (even under special relativity) and the geometry of a passing object would appear skewed, as if rotated.[1][2]

64. JT, could you plot Gron's lengths from point 1 to each emission point from part C onto the respective horizontal y position for the point given in part B (I don't have a compass)?

The resulting plot should just be skewed (using part A y positions doesn't look right).

65. Originally Posted by laurieag
...using part A y positions doesn't look right.
For the visual/optical analysis, there is no reason to use part A at all, (that is just shown for reference). The points to use are the points labeled 1-16 in part C which represent the same points as those labeled in part B, but with the wheel rolled back to earlier times. You have to imagine rolling the wheel backwards (to the left) to find out where it was at some time in the past. When you roll it backwards, it still remains an upright ellipse, but the labeled points change position around the perimeter.

In the diagram I posted above, the point labeled 10 in part C is located on the same y position as the point labeled 14.5 (if it were to exist) in part B. That is because the wheel has rolled 4.5/16 of a rotation between the two ellipses shown. Does that make sense?

66. Originally Posted by laurieag
JT, could you plot Gron's lengths from point 1 to each emission point from part C onto the respective horizontal y position for the point given in part B (I don't have a compass)?
I think this is half of what you asked for -- these are the light rays that all enter the camera at the same instant of time:

I don't know what else you are asking me to do. As I said, the y positions of the labeled points in part B are not applicable to any of these earlier times. The wheel had to be virtually rolled backward to find these earlier positions.

67. JTyesthatJT,

I cannot put the time into redrafting Gron's Part C from my own figures at present, but just a couple notes about the Gron figures ...

I am supposing the figures were published many years ago, and so the accuracy of his circles and ellipse are not perfect. I'm sure they are very close though. Based on his ellipse and circle, graphically, I estimate his velocity in Figure 9 should be about v=0.8216c, and so gamma ~ 1.754 for an approximate 57% length-contraction. Yet, it makes more sense he'd have used rounded velocities, eg 0.8c, I agree. I think he just posted ball park figures for sake of point, and hard to know how editors might have altered his own manuscript's geometric figures in that respect. His part C should reach all the way up to the height of 2R, so the height of point 9 of parts A & B. So I'm not sure if his rotated part C figure is (slightly) compressed vertically, or if the whole Part C is (slightly) "uniformly contracted" more than it should be? Were those figures done back in 1975, and copied into this more recent paper by chance? All in all, his figures are accurate enough, in so far as getting the key points across nicely.

Indeed, Gron changed his velocity from what was used i Figure 8 (v=0.7), when making his considerations for what he presented in Figures 9A, 9B, & 9C.

You're light rays and general response for laurieag look in order, IMO.

************************************************** ********

Laurieag,

Never can any part of the disk be below the height of the bottom of the wheel, for the same reason no atom of a rolling wheel can ever be below the height of the road. Your figure shows EM emission (or reflection) events below the bottom of the wheel, in only your Part C redraft. Why do you think the perimeter atoms of Gron's disk could exist below the bottom of the wheel wrt that depicted by Gron's parts A & B?

Thank You,

68. Originally Posted by SinceYouAsked
I am supposing the figures were published many years ago, and so the accuracy of his circles and ellipse are not perfect. I'm sure they are very close though. Based on his ellipse and circle, graphically, I estimate his velocity in Figure 9 should be about v=0.8216c, and so gamma ~ 1.754 for an approximate 57% length-contraction. Yet, it makes more sense he'd have used rounded velocities, eg 0.8c, I agree. I think he just posted ball park figures for sake of point, and hard to know how editors might have altered his own manuscript's geometric figures in that respect. His part C should reach all the way up to the height of 2R, so the height of point 9 of parts A & B. So I'm not sure if his rotated part C figure is (slightly) compressed vertically, or if the whole Part C is (slightly) "uniformly contracted" more than it should be? Were those figures done back in 1975, and copied into this more recent paper by chance? All in all, his figures are accurate enough, in so far as getting the key points across nicely.

Indeed, Gron changed his velocity from what was used i Figure 8 (v=0.7), when making his considerations for what he presented in Figures 9A, 9B, & 9C.
When I first tried to solve for the velocity he might have used in Figure 9, I tried using the length-contracted shape of the wheel. That gave me results which did not match the geometry of part C, so I decided to use the "right-triangle-method" instead. When I saw that gave a nice value of v=0.8c, I felt satisfied. However, I have now discovered something in Figure 9 which cannot be explained by any choice of velocity:

In blue, notice how the 9:00 o'clock position rolls from a point labeled 10 to a point that is approximately half-way between the points labeled 14 and 15, hence 14.5. Since both of these are measured at the same location, (the 9:00 o'clock position), there can be no dispute that the wheel would have had to have rolled approximately (14.5-10)/16 or 4.5/16 of a rotation over that duration.

However, in red, notice how the approximately 10:00 o'clock position rolls from a point labeled 7 to a point labeled 13. Since both of these are measured at the same location, (approximately the 10:00 o'clock position), there can be no dispute that the wheel would have had to have rolled (13-7)/16 or 6/16 of a rotation over that duration.

But there is no way the wheel can have rolled (6-4.5)/16 or 1.5/16 of a rotation between the two ellipses which are so close together on the left. That is way out of scale to the distance between those ellipses the the ellipse in part B So, I conclude something is wrong with the optical part of this analysis. It may be that the graph would have had to have extended far off the page in order to show the correct results, or there just might be an error in the calculations. However, I do agree that he made his point about the optical appearance in general.

69. Since we all seem to be in agreement that Gron's optical/visual analysis is perhaps less than ideal, I have decided to do one of my own:

This is based on v=0.866025c, gamma=2.000, and 12 points on the rim, (equally spaced in the axle frame).

The wheel rolls 1/12 of one rotation between each of the adjacent ellipses, thus a distance of 4piR/12 along the road. The green lines extending from point 1 to the closest adjacent ellipse have a length of 4piR/12v to match the distance that light would travel. Likewise, the green lines extending from point 1 to the second closest ellipse have a length of 2(4piR/12v), the green lines extending from point 1 to the third closest ellipse have a length of 3(4piR/12v), etc.

The smooth curve was created by a "spline" function, and is thus only approximate.

70. JTyesthatJT,

My spreadsheet analysis considered the wheel over full rotations, starting with the 0 deg spoke tip at contact at the 6:00 position. As such, I think my spreadsheet is already designed for easy analysis of Gron's Part C figure. I'll try and take a look at it, see how easy (and thus quick) it might be to verify Gron's figure, which it should.

EDIT: Not so easy as I hoped. As such, I may not bother with this for awhile being time restricted at present. I'll put it on the sideline for future consideration, if it turns out I don't wish to attack it at present. Your figure has the same shape as Gron's, so it looks consistent.

Thank You,

71. Originally Posted by SinceYouAsked
JTyesthatJT,

My spreadsheet analysis considered the wheel over full rotations, starting with the 0 deg spoke tip at contact at the 6:00 position. As such, I think my spreadsheet is already designed for easy analysis of Gron's Part C figure. I'll try and take a look at it, see how easy (and thus quick) it might be to verify Gron's figure, which it should.

EDIT: Not so easy as I hoped. As such, I may not bother with this for awhile being time restricted at present. I'll put it on the sideline for future consideration, if it turns out I don't wish to attack it at present. Your figure has the same shape as Gron's, so it looks consistent.

Thank You,
Aw come on, you can do it.

72. JTyesthatJT,

Looks like your light path "departing spoke tip 5" has a counter part (of same length) that departs at 6.9 (or thereabouts) on that same contracted ellipse, which would be the most-leftest-point of the Part C figure. Yes?

You spaced your contracted ellipses by the amount of time it takes a spoke tip to rotate 1 hr axle-time t' (like 11:00 to 12:00), that duration as transformed into ground system as time t=γt', given a translation velocity of v=0.866c? So you determined that delta time t, then separate the contracted ellipses (axle to axle) by that spatial amount of x = vt = γvt' ?

Thank You,

73. I can see now how Gron got his curve.

The camera would require a very fast shutter speed to just capture the part C curve.

The curve consists of 16 different wheel locations, one wheel location for each emission point and the spacings between each wheel location are not the same.

74. Originally Posted by laurieag
I can see now how Gron got his curve.

The camera would require a very fast shutter speed to just capture the part C curve.

The curve consists of 16 different wheel locations, one wheel location for each emission point and the spacings between each wheel location are not the same.
Sounds about right laurieag, far as Gron's Figure 9 Part C figure goes. Indeed, the shutter speed would have to be near light speed itself to catch that photo. But the point is, in theory, the photons "in the instant" would be presented as such. That would have to be a mighty fast instant.

In reality, bodies moving at relativistic speed would be streaks passing you by, assuming the light are not redshifted out of detection, and you would not even make out the body without a wonder shutter speed. I think engineers would have to develop "relativistic speed lanes" with inertial buoys stationed along side for referencing, with lanes free of space debris, if one wanted to safely travel from star system to star system. They would also have to speed limits slow enough such that no one ages too much wrt inertial others as you go.

I might add, it's a good feeling when someone takes their time and learns what a model (or figure) published by a great mind means, instead of arguing about it for years and years and years in circles. Thank You.

Thank You,

75. Originally Posted by JTyesthatJT
However, I have now discovered something in Figure 9 which cannot be explained by any choice of velocity:

In blue, notice how the 9:00 o'clock position rolls from a point labeled 10 to a point that is approximately half-way between the points labeled 14 and 15, hence 14.5. Since both of these are measured at the same location, (the 9:00 o'clock position), there can be no dispute that the wheel would have had to have rolled approximately (14.5-10)/16 or 4.5/16 of a rotation over that duration.

However, in red, notice how the approximately 10:00 o'clock position rolls from a point labeled 7 to a point labeled 13. Since both of these are measured at the same location, (approximately the 10:00 o'clock position), there can be no dispute that the wheel would have had to have rolled (13-7)/16 or 6/16 of a rotation over that duration.

But there is no way the wheel can have rolled (6-4.5)/16 or 1.5/16 of a rotation between the two ellipses which are so close together on the left. That is way out of scale to the distance between those ellipses the the ellipse in part B So, I conclude something is wrong with the optical part of this analysis. It may be that the graph would have had to have extended far off the page in order to show the correct results, or there just might be an error in the calculations. However, I do agree that he made his point about the optical appearance in general.
In the ground system, the spoke-tips rotate much faster (angularly) in lower parts of the wheel than upper parts of the wheel, while they move (v_trans) much faster on upper parts of the wheel than lower parts. That 1.5 degrees is the first portion of the 6 degrees of rotation, where the spoke tip resides between the 9:00 and 10:00 O'clock positions. Spoke tips in that region move much faster angularly than when they later (from 1.5 hr thru 6 hr later) reside nearer the top of the wheel, where they rotate very slowly with passing ground time. The closer to the 12:00 position, the slower the angular rotation of a spoke tip, per ground. Yes?

Thank You,

76. Originally Posted by SinceYouAsked
Looks like your light path "departing spoke tip 5" has a counter part (of same length) that departs at 6.9 (or thereabouts) on that same contracted ellipse, which would be the most-leftest-point of the Part C figure. Yes?
Maybe, but I seem to recall there was only one point on the left-most wheel that "fit" the light line. I'm away for the weekend, but I can check that on Monday. If it's not there, it just means it would be on a different ellipse which would be located very close to the left-most ellipse.

Originally Posted by SinceYouAsked
You spaced your contracted ellipses by the amount of time it takes a spoke tip to rotate 1 hr axle-time t' (like 11:00 to 12:00), that duration as transformed into ground system as time t=γt', given a translation velocity of v=0.866c? So you determined that delta time t, then separate the contracted ellipses (axle to axle) by that spatial amount of x = vt = γvt' ?
I just spaced the ellipses 4piR/12 center-to-center. That distance represents 1/12 of one rotation, because one rotation displaces 4piR of road for gamma=2.

Originally Posted by SinceYouAsked
In the ground system, the spoke-tips rotate much faster (angularly) in lower parts of the wheel than upper parts of the wheel, while they move (v_trans) much faster on upper parts of the wheel than lower parts. That 1.5 degrees is the first portion of the 6 degrees of rotation, where the spoke tip resides between the 9:00 and 10:00 O'clock positions. Spoke tips in that region move much faster angularly than when they later (from 1.5 hr thru 6 hr later) reside nearer the top of the wheel, where they rotate very slowly with passing ground time. The closer to the 12:00 position, the slower the angular rotation of a spoke tip, per ground. Yes?
Look at my diagram and you will see that when the bottom point changes from (say) 10 to 11, the top point changes from 4 to 5, and every other point also advances by one. You are correct that the angular velocity is different for different parts of the wheel, but there should be a linear relationship between the wheel rollout distance and the fraction of rotation. Gron has the wheel rolling out almost the same distance for 4.5/16 of a rotation as it does for 6/16 of a rotation. The 1.5/16 difference should be one third the length of the 4.5/16 rollout shown in blue, but it is much shorter than that.

77. Originally Posted by SinceYouAsked
Originally Posted by laurieag
I can see now how Gron got his curve.
I might add, it's a good feeling when someone takes their time and learns what a model (or figure) published by a great mind means, instead of arguing about it for years and years and years in circles. Thank You.
I agree. Thank you, laurieag. Maybe you can teach cincirob how to think things through. Just kidding, I don't think anybody can do that.

78. Originally Posted by SYA
In the ground system, the spoke-tips rotate much faster (angularly) in lower parts of the wheel than upper parts of the wheel, while they move (v_trans) much faster on upper parts of the wheel than lower parts. That 1.5 degrees is the first portion of the 6 degrees of rotation, where the spoke tip resides between the 9:00 and 10:00 O'clock positions. Spoke tips in that region move much faster angularly than when they later (from 1.5 hr thru 6 hr later) reside nearer the top of the wheel, where they rotate very slowly with passing ground time. The closer to the 12:00 position, the slower the angular rotation of a spoke tip, per ground. Yes?

Originally Posted by JTyesthatJT
Look at my diagram and you will see that when the bottom point changes from (say) 10 to 11, the top point changes from 4 to 5, and every other point also advances by one. You are correct that the angular velocity is different for different parts of the wheel, but there should be a linear relationship between the wheel rollout distance and the fraction of rotation. Gron has the wheel rolling out almost the same distance for 4.5/16 of a rotation as it does for 6/16 of a rotation. The 1.5/16 difference should be one third the length of the 4.5/16 rollout shown in blue, but it is much shorter than that.
Yes, the linear relation is between the angular rotation rate of the center of the axis-of-rotation (ie axle) and the translation velocity (vx) of the axle wrt ground. The ground just records the axle's rotation rate decreased by the factor of gamma. The axle and ground systems always agree on the angular orientation of the axle, although they disagree on its rate of rotation since they disagree on the rate of time. The other atoms of the spoke (away from axle including the spoke tip) cannot rotate linearly per ground, because the steady angular rotation of (say) the spoke tip in advancing horizontal planes-of-simultaneity per axle, are angularly sliced by a single simultaneity-plane of the ground system (eg at t = 0). See ...

whereby one can see that the steady rotation rate per axle, must become unsteady per the ground, because of the spatial offset wrt the axle per the time LT ... t = γ(t'+vx'/c²). So the rotation rate per ground must vary (since t varies) with a change in horizontal distance (x') of the spoke atom wrt the axle-center. Make sense?

Gron's figure 9 part C looks in order IMO, although it was slightly contracted in height for some strange reason, as it should at one point (at its peak) attain the height of a 12:00 spoke tip.

Thank You,

79. Originally Posted by SinceYouAsked
Gron's figure 9 part C looks in order IMO, although it was slightly contracted in height for some strange reason, as it should at one point (at its peak) attain the height of a 12:00 spoke tip.
Well then, let me ask you this:

On part B, the contact point is labeled 1, and the point located around the 9:00 o'clock point is approximately labeled 14.5 (please refer to the right endpoint of the BLUE line in my diagram).

At the left end of that blue line, on part C, there is a point labeled 10. Unfortunately the 1 is obscured so it looks like 0, but I assure you it is a 10. I have drawn an ellipse in that location, with the point labeled 10 at the very same 9:00 o'clock point. Now, can you tell me what labeled point would you suppose is in contact with the ground for that ellipse?

..........

Next question, on part B, the contact point is still labeled 1, and the point located around the 10:00 o'clock point is labeled 13 (please refer to the right endpoint of the RED line in my diagram).

At the left end of that red line, on part C, there is a point labeled 7. I have drawn an ellipse in that location, with the point labeled 7 at the very same 10:00 o'clock point. Now, can you tell me what labeled point would you suppose is in contact with the ground for that ellipse?

80. Originally Posted by JTyesthatJT
Well then, let me ask you this:

On part B, the contact point is labeled 1, and the point located around the 9:00 o'clock point is approximately labeled 14.5 (please refer to the right endpoint of the BLUE line in my diagram).

At the left end of that blue line, on part C, there is a point labeled 10. Unfortunately the 1 is obscured so it looks like 0, but I assure you it is a 10. I have drawn an ellipse in that location, with the point labeled 10 at the very same 9:00 o'clock point. Now, can you tell me what labeled point would you suppose is in contact with the ground for that ellipse?
about ... spoke-tip 12.5 @ ground contact point, leftmost contracted ellipse.

Originally Posted by JTyesthatJT
Next question, on part B, the contact point is still labeled 1, and the point located around the 10:00 o'clock point is labeled 13 (please refer to the right endpoint of the RED line in my diagram).

At the left end of that red line, on part C, there is a point labeled 7. I have drawn an ellipse in that location, with the point labeled 7 at the very same 10:00 o'clock point. Now, can you tell me what labeled point would you suppose is in contact with the ground for that ellipse?
spoke tip 11 @ ground contact point, second contracted ellipse from the left.

Then 12.5 - 11 = 1.5 (which of course is in agreement with your prior 6-4.5 = 1.5) so 1.5/16 of a full rotation of the axle, for the ground time interval defined by leftmost contracted ellipse to the contracted ellipse immediately to the right of it.

yes?

Thank You,

81. Originally Posted by SinceYouAsked
about ... spoke-tip 12.5 @ ground contact point, leftmost contracted ellipse.

spoke tip 11 @ ground contact point, second contracted ellipse from the left.

Then 12.5 - 11 = 1.5, so 1.5/16 of a full rotation of the axle
Correct. So you agree with me that, for the two ellipses I drew which are very close together, Gron has the wheel rotating a huge 1.5/16 of a rotation. Meanwhile the very long blue line represents 4.5/16 of a rotation. Note that this is based on the contact point alone now. Do you see the problem?

82. Originally Posted by JTyesthatJT
Correct. So you agree with me that, for the two ellipses I drew which are very close together, Gron has the wheel rotating a huge 1.5/16 of a rotation. Meanwhile the very long blue line represents 4.5/16 of a rotation. Note that this is based on the contact point alone now. Do you see the problem?
Ahh Huh. I'll need to look at it a wee bit, see if your "16" denominator is accurate.

Thank You,

83. Originally Posted by SinceYouAsked
Ahh Huh. I'll need to look at it a wee bit, see if your "16" denominator is accurate.
The 1.5/16 length is too short to match the 4.5/16 length. It's out of scale. What does it matter if you change the denominator to (say) 1.5/100 and 4.5/100. It's still out of scale.

84. Agreed, no difference. I was reconsidering your approach overall, but looking at it again, it looks in order to me. So the rotated part C figure appears in need of contraction wrt x, and very slightly stretched vertically. It would seem the part C figure may have been stretched to declutter and separate the numbers wrt each figure maybe? I wonder if it were contracted to conform, if all the other spoke-tips would work out if they were considered instead?

I'll look at it some more, just to be certain we're not missing something.

Thank You,

85. Originally Posted by SinceYouAsked
Agreed, no difference. I was reconsidering your approach overall, but looking at it again, it looks in order to me. So the rotated part C figure appears in need of contraction wrt x, and very slightly stretched vertically. It would seem the part C figure may have been stretched to declutter and separate the numbers wrt each figure maybe? I wonder if it were contracted to conform, if the spoke-tips would work out if they were considered instead?
I think some numbers might have been moved to declutter, or just to "fudge" the data so people could see every labeled point. Notice, in my diagram, I did not even try to label every point, (only those which were easily determined).

Originally Posted by SinceYouAsked
I'll look at it some more, just to be certain we're not missing something.
Nothing wrong with that, and I will continue to do the same.

I think it's more important that Gron and I agree on the optical shape. That's the main point, after all. Cheers!

86. Originally Posted by JTyesthatJT
Look at my diagram and you will see that when the bottom point changes from (say) 10 to 11, the top point changes from 4 to 5, and every other point also advances by one. You are correct that the angular velocity is different for different parts of the wheel, but there should be a linear relationship between the wheel rollout distance and the fraction of rotation. Gron has the wheel rolling out almost the same distance for 4.5/16 of a rotation as it does for 6/16 of a rotation. The 1.5/16 difference should be one third the length of the 4.5/16 rollout shown in blue, but it is much shorter than that.
I tried copying part B so that it went through each emission point in part C and came to similar conclusions so I did the following breakdowns to try to get the correct firing sequence. The wheel is rolling from the left to the right so the longest ray distance traveled is the start position of the wheel and the emission point is that shown in part C.

The first direction describes the direction a ray from the current emission point travels with respect to the previous point in normal order and the second is the direction of the ray from the current emission point with respect to the previous emission point in Time Firing Order. Note that points 6 and 8 travel the same distance to the camera at point 1.

Code:
```Point	Direction	Reverse Firing Order	x Firing Order	Time firing Order Direction	Actual Emission Point Firing Order

1	start		1			1		dn,right			15
2	up, left	3			2		dn,right			13
3	up, left	6			5		up,right			10
4	dn, left	8			7		up,right			8
5	dn, left	13			10		up,right			3
6	dn, left	14/15			12		up,right			2
7	dn, left	16			14/15		start				1
8	dn, left	14/15			16		dn,left				2
9	dn, right	12			14/15		dn,right			4
10	dn, right	11			13		dn,right			5
11	dn, right	10			11		dn,right			6
12	dn, right	9			9		dn,right			7
13	dn, right	7			8		dn,right			9
14	dn, right	5			6		dn,right			11
15	dn, right	4			4		dn,right			12
16	dn, right	2			3		dn,right			14```
This is the only Firing Order that will allow the light from all emission points to reach the camera at the same instant in time. Can this order work with a consistent angular velocity?

87. I tested points 2, 3, 4, and 5:

The contact points do not even maintain a consistent order! They are supposed to be ascending, (at least until they reach point 1), but they go 15.9, 14, 12.7, 13.1. And look how big the gap is between 12.7 and 13.1 -- that looks a lot bigger than 0.4/16 of a rotation!

And how about the enormous gap between contact point 15.9 and contact point 1, which represents only 1.1/16 of a rotation!

88. Originally Posted by JTyesthatJT
The contact points do not even maintain a consistent order! They are supposed to be ascending, (at least until they reach point 1), but they go 15.9, 14, 12.7, 13.1. And look how big the gap is between 12.7 and 13.1 -- that looks a lot bigger than 0.4/16 of a rotation!

And how about the enormous gap between contact point 15.9 and contact point 1, which represents only 1.1/16 of a rotation!
JT, I tried to overlay part B over the emission points but did not get very far as points 7 and 8 lie on the same part B wheel and points 9 and 6 also lie on the same part B wheel. This makes things very difficult as points 6 and 8 should be the only points on the same part B wheel but that's geometrically impossible.

I wonder if another plane is involved, a plane skewed between x and y, either that or part B is somehow skewed for each different emission point.

89. Ya, that's what I was concerned about as well JT, as I looked at that last night too. Looked fishy even at a glance. Unless we are missing something, it does look amiss. It would seem that we should model the snapshot ourselves, and hopefully we can attain the same shape doing it independently. Then, maybe we can better determine the likeliest reason the part C figure is amiss in the way it seems to be.

I'm going to rethink it first, with a clean slate, and if no bananas then I'll model it in spreadsheet ... soon as I can put the required time in (of course, which is difficult at the moment). I would normally prefer to do v=0.8660125, however Gron's contracted ellipse (although it and the round circle are distorted slightly) best fit a velocity of v=0.82164 (1/gamma = ~57%), so it may be better to run that velocity for direct comparison to Gron's actual part A & B figures.

Thank You,

90. Originally Posted by laurieag
I wonder if another plane is involved, a plane skewed between x and y, either that or part B is somehow skewed for each different emission point.
Also, maybe Gron assumed the camera lens was far from the ground contact point wrt y, in his analysis? That would change things a bit as well, if far enough way wrt y=y'=0 upon ground contact. At first thought, I don't suspect that's the issue here though. The paper does say the camera lens is AT the ground contact point though, so I would figure the analysis was done for "virtually at".

Thank You,

91. Originally Posted by JTyesthatJT
I tested points 2, 3, 4, and 5:

The contact points do not even maintain a consistent order! They are supposed to be ascending, (at least until they reach point 1), but they go 15.9, 14, 12.7, 13.1. And look how big the gap is between 12.7 and 13.1 -- that looks a lot bigger than 0.4/16 of a rotation!

And how about the enormous gap between contact point 15.9 and contact point 1, which represents only 1.1/16 of a rotation!
First problem, is that the figure must be stretched a tad bit wrt the vertical, such that spoke-tip 3's emission event is (virtually) at the 12:00 location at y=2R. That would mean the blue line extends from spoke-tip 3's emission event (at about 12:03 O-clock) to just right of spoke-tip 9 (so at about 12:03 O'clock) upon the specified ground contact event. That's roughly speaking. Your spoke-tip 4 and 5 emission events, would exist on the left side of the contracted wheel at each their respective events. The spoke-tip 2 emission event would be on the right side of the contracted wheel, at a subsequent event. It's hard to do your estimations with the figure being off-scale, as the events near the highest part of the part C figure would have larger errors due to the "vertical off-scaling". I think redoing the analysis might be quicker than trying to figure out his with the presumed off-scaling.

Thank You,

92. Originally Posted by SinceYouAsked
First problem, is that the figure must be stretched a tad bit wrt the vertical...
I'll work on stretching part C vertically, just to see if the numbers work out any better. I don't really expect them to, though.

Originally Posted by SinceYouAsked
I think redoing the analysis might be quicker than trying to figure out his with the presumed off-scaling.
I agree, which is why I already re-did the analysis in post #169.

Originally Posted by JTyesthatJT
Since we all seem to be in agreement that Gron's optical/visual analysis is perhaps less than ideal, I have decided to do one of my own:

This is based on v=0.866025c, gamma=2.000, and 12 points on the rim, (equally spaced in the axle frame).

The wheel rolls 1/12 of one rotation between each of the adjacent ellipses, thus a distance of 4piR/12 along the road. The green lines extending from point 1 to the closest adjacent ellipse have a length of 4piR/12v to match the distance that light would travel. Likewise, the green lines extending from point 1 to the second closest ellipse have a length of 2(4piR/12v), the green lines extending from point 1 to the third closest ellipse have a length of 3(4piR/12v), etc.

The smooth curve was created by a "spline" function, and is thus only approximate.
Can you replicate any of the Gron problems with my graphic? I tend to think not, but I would be very impressed if you found any errors in mine.

93. Originally Posted by JTyesthatJT
I'll work on stretching part C vertically, just to see if the numbers work out any better. I don't really expect them to, though.
Here's the one I did, maybe it'll help ...

Originally Posted by JTyesthatJT
I agree, which is why I already re-did the analysis in post #169...

Can you replicate any of the Gron problems with my graphic? I tend to think not, but I would be very impressed if you found any errors in mine.
No, your's is good. The only point I would make (a minor point), is that when you estimate the spoke-tip point, it's not linear between the displayed numbers. For example, where you show 10.6 it's more like 10.4. Where you show 11.8 it's more like 11.6.

I would be very impressed if you could derive a single formula that defines the location of the contracted ellipse that possesses the spoke-tip from where the longest lightpath emits from. That would be nice to have. It would also be nice to have the right most emission event calculated. To automate that.

Thank You,

94. Originally Posted by SinceYouAsked
The only point I would make (a minor point), is that when you estimate the spoke-tip point, it's not linear between the displayed numbers. For example, where you show 10.6 it's more like 10.4. Where you show 11.8 it's more like 11.6.
I understand that it's not linear, but I'm just making my best approximation. I don't see any value in trying to apply a bias in either direction.

Originally Posted by SinceYouAsked
I would be very impressed if you could derive a single formula that defines the location of the contracted ellipse that possesses the spoke-tip from where the longest lightpath emits from. That would be nice to have. It would also be nice to have the right most emission event calculated. To automate that.
Those things can be calculated for a non-rotating disk, and they would still be the same for a rotating one.

95. lol, for obvious reasons...

lol, for obvious reasons...

96. Here's your figure again JT (for future readers), with the same projections by the same method on your part C that you have been doing on Gron's part C. Yours is good ...

With the rightmost contracted ellipse representing the disk at the specified ground contact event.

leftmost contracted ellipse ... horiz line projects from spoketip 5 to spoketip 9, so the wheel rolls 4 degrees.
next contracted ellipse rightward ... horiz line projects from spoketip 3.4 to spoketip 6.4, so the wheel rolls 3 degrees.
next contracted ellipse rightward ... horiz line projects from spoketip 2.5 to spoketip 4.5, so the wheel rolls 2 degrees.
next contracted ellipse rightward ... horiz line projects from spoketip 1.8 to spoketip 2.8, so the wheel rolls 1 degree.

Note: Minor point, but I used your fractional spoketip numbers, which assumed a linearity of tip separation between "round-numbered spoketips", although the separation of round-numbered spoketips are not linear.

As such, your figure does not suffer from that which Gron's part C figure does.

Now, it would be nice to have a simplified way in which to determine each of 16 equally spaced spoketips per axle in the snapshot per ground (as Gron did). Alas, I'm not gonna have the time to do it at the moment here.

Thank You,

97. Originally Posted by JTyesthatJT
lol, for obvious reasons...

lol, for obvious reasons...
I mentioned this before, but your leftmost contracted ellipse should intersect the green line on the upper left of the ellipse, not the upper right.

Thank You,

98. Originally Posted by SinceYouAsked
Here's your figure again JT (for future readers), with the same projections by the same method on your part C that you have been doing on Gron's part C. Yours is good ...

With the rightmost contracted ellipse representing the disk at the specified ground contact event.

leftmost contracted ellipse ... horiz line projects from spoketip 5 to spoketip 9, so the wheel rolls 4 degrees.
next contracted ellipse rightward ... horiz line projects from spoketip 3.4 to spoketip 6.4, so the wheel rolls 3 degrees.
next contracted ellipse rightward ... horiz line projects from spoketip 2.5 to spoketip 4.5, so the wheel rolls 2 degrees.
next contracted ellipse rightward ... horiz line projects from spoketip 1.8 to spoketip 2.8, so the wheel rolls 1 degree.

Note: Minor point, but I used your fractional spoketip numbers, which assumed a linearity of tip separation between "round-numbered spoketips", although the separation of round-numbered spoketips are not linear.

As such, your figure does not suffer from that which Gron's part C figure does.
Thank you, SYA. That was very helpful.

By the way, you were correct on this:

Originally Posted by SinceYouAsked
Looks like your light path "departing spoke tip 5" has a counter part (of same length) that departs at 6.9 (or thereabouts) on that same contracted ellipse, which would be the most-leftest-point of the Part C figure. Yes?
Yes, good catch!

You must have good eyes.

EDIT: I'll post a new image after I run the spline function again. Even though its only one additional point, the spline needs to be re-calculated. Cheers!

99. Originally Posted by SinceYouAsked
I mentioned this before, but your leftmost contracted ellipse should intersect the green line on the upper left of the ellipse, not the upper right.
Ah, you're right. Because point 4 is to the left of the highest point, it must come from the left side of the wheel.

Okay I fixed it. That helps somewhat, but the lengths are still out of scale. The distance between the blue 10.8 and the red 12.5 is not much different than the distance between the red 12.5 and the black 1. The former represents 1.7/16 of a rotation, and the latter represents 4.5/16 of a rotation. The latter should be over twice the length of the former.

100. Yes, Grons part C figure looks inconsistent with his parts A & B.

It seems the Gron fig 9 part C is inconsistent with the contracted ellipse and its designated 16 spoketip placements. I get the same numbers you do. In fact, this was my crack at it ...

1) Again, his contracted ellipse is geometrically valid for ... v=0.82164c ... 57% contraction, and so ... gamma = 1.754.
2) I'm not sure how accurate his 16 spoke tips are on the contracted ellipse?
3) After vertical stretching, Gron's part C figure seems the correct height, but the width I am not sure yet. I left that unaltered.
4) Gron's part C looks to be the same general shape as (so consistent with) JT's part C.

Seems that either the 16 spoke-tips are improperly located on part C, or the 16 spoke-tips are not properly placed on the contracted ellipse. Maybe it was an editor problem, versus Gron. Maybe part C was drafted for a different velocity that parts A & B, but they were superposed anyway to convey a relation in general.

We're getting there. Anyway, still looking at it.

Thank You,