Jump to content
merjet

Aristotle's wheel paradox

Recommended Posts

Darrell,

You are a benevolent force in the OL universe, superficially rendered. But the reality still crushess if you manage to get it in under the door.

J and J et al. are at the gates of Troy (Tony) and here you come with a horse.

Shame!

--Brant

  • Like 1

Share this post


Link to post
Share on other sites
On 12/21/2018 at 3:09 AM, Darrell Hougen said:

Hi Tony,

There are basically three scenarios being discussed:

1. An ordinary wheel or tire that runs on a road or track.

2. A pair of adjacent wheels or gears that run on adjacent rails or tracks at appropriate levels.

3. A bottle or Dixie cup whose ends run on widely spaced rails or tracks --- widely spaced relative to the sizes of the ends.

Your explanation works perfectly fine in the first case, but it doesn't begin to explain the other two scenarios. You agreed that a cone shaped object would veer off to one side, but you haven't explained why you think that to be the case.

Darrell

 

 

 

Hi Darrell, 

I thought it too obvious to reply to, a cone veers off because it's er, conical. The equivalent for a wheel is cutting an oblique angle, camber, into the wheel's tread. Btw, to pertain here, the cone, to represent wheel motion - with an inner wheel and track - should not roll flat, but be suspended at the smaller end as you would and have done with the bottle graphics. The base end then rolls on its rim.

2. and 3. are  interesting on their own as theoretical-mechanical exercises. Not because I think they have anything to do with resolving the paradox, but as analogous 'mock-ups' of an extruded wheel, plus inner wheel, plus track, are quite fun to play with. Of course, this approach brings in factors such as extension, extra friction, drag and slip, not present in and superfluous to the paradox.

Creating a simple type of machine or any apparatus cannot be the explanation called for.

Putting my case simply, by the inherent nature of internal speeds, a wheel has built-in slippage protection.

Share this post


Link to post
Share on other sites
On 12/15/2018 at 11:32 AM, Jonathan said:

That doesn't get anywhere toward addressing the ancient eddheads' concerns. It only adds more burdens to the "paradox." They couldn't grasp how two wheels of different circumferences could unroll without each slipping on its own path of the same length as the other's. Informing them of cycloids would only provide them with another set of facts that contradicted their false premise. "Now we not only have a circumference that unrolls, without slipping, the other wheel's length instead of its own, but we also now have two differing cycloids, which despite not being the same length or shape, nevertheless travel the same horizontal distance and have one-to-one corresponding points at any and all moments during travel. Therefore, are there voids and other eggheaded infinity stuff?"

Cycloids are pretty, and they're interesting, but they don't address the actual problem of the "paradox." The actual problem is that it contains the mistaken premise of both wheels rolling freely over their surfaces with neither slipping.

That premise is just as mistaken as the premise in the three gear challenge that I posted. One cannot solve a "paradox" which contains a false premise without addressing and correcting the false premise. Observing that there are different cycloids while maintaining the false belief that there is no slippage of either wheel on its own surface is failure. The only success involved with cycloids would be to take the position that they are one of many means of confirming slippage and of confirming which single circumference is rolling freely and not slipping.

Every word or phrase I put in bold highlight Jonathan’s mutilating MechanicaThe ancient Greeks in Mechanica said nothing about two wheels, surfaces, or slipping or not slipping.

Jonathan has no solution without his crutch – a second surface, track, or support. He needs a crutch so he can claim slippage.

When the problem conditions are met, the smaller circle horizontally moves the circumference of the larger circle (or vice-versa) with or without a second surface, track, or support. The paradox does not appear and disappear depending on whether or not there is a second surface, real or imagined. My solutions hold with or without a second surface, track, or support. Jonathan has refused to acknowledge my solutions as solutions. He has even denied they are, like he did this time. No good reason given; it’s mere ego and stubbornness.

  • Haha 1

Share this post


Link to post
Share on other sites
On 12/15/2018 at 2:15 PM, Max said:

Because it doesn't address the problem of the paradox, namely how the points of the "forced" wheel are mapped onto the longer distance on its track. For the "forcing" wheel it's rather obvious: circumference and traveled distance are equal, so it's easy to construct an 1-1 map, while for the "forced" wheel the circumference and traveled distance are unequal. What the old guys didn't know, is that it is also possible to map a smaller segment 1-1 onto a larger segment, but if you do that, you'll have to take into account that the smaller wheel is slipping (supposing that the large wheel is the "forcing" wheel).

Well, that was obviously irony, the apparent result of Aristotle's paradox. Does that invalidate the article? And if Aristotle writes about two circles, the mechanical implementation would be two wheels, so nothing wrong with that either.

Except for "apparent",  I made the terms bold.

Max has no solution without his crutch – a second surface, track, or support. He needs a crutch so he can claim slippage.

When the problem conditions are met, the smaller circle horizontally moves the circumference of the larger circle (or vice-versa) with or without a second surface, track, or support. The paradox does not appear and disappear depending on whether or not there is a second surface, real or imagined. My solutions hold with or without a second surface, track, or support. Max has refused to acknowledge my solutions as solutions. He has even denied they are. No good reason given; it’s mere ego and stubbornness.

  • Haha 1

Share this post


Link to post
Share on other sites
2 hours ago, merjet said:

Max has no solution without his crutch – a second surface, track, or support. He needs a crutch so he can claim slippage.

When the problem conditions are met, the smaller circle horizontally moves the circumference of the larger circle (or vice-versa) with or without a second surface, track, or support. The paradox does not appear and disappear depending on whether or not there is a second surface, real or imagined. My solutions hold with or without a second surface, track, or support. Max has refused to acknowledge my solutions as solutions. He has even denied they are. No good reason given; it’s mere ego and stubbornness.

I'm stubborn in that I keep referring to the original formulation by Aristotle (or whoever it was). That seems to me to be appropriate if we're talking about Aristotle's wheel paradox. And in that original formulation the second track, ΗΘ, is an essential element. To fresh up your memory, from the original text (in translation): 

So that whenever the one shall have traversed a distance equal to ΗΘ, and the other to ΖΙ, and ΖΑ has again become perpendicular to ΖΛ, and ΑΗ has again to ΗΚ, the points Η and Ζ will again be in their original positions at Θ and Ι. As, then, nowhere does the greater stop and wait for the less in such a way as to remain stationary for a time at the same point (for in both cases both are moving continuously), and as the smaller does not skip any point, it is remarkable that in the one case the greater should travel over a path equal to the smaller, and in the other case the smaller equal to the larger.

It's obvious that Aristotle's essential problem was that he couldn't understand how the smaller circle could traverse the greater distance without skipping somewhere. The solution is that the smaller circle is not skipping but slipping, i.e. that the point of that circle in the 6 o'clock position has a translation speed > 0. We now know that a 1-1 mapping from a smaller circle or line segment onto a larger line segment is possible, so that is no longer an objection and with calculus we can give an exact description. The interaction between the circle and its track is the crux of the paradox, removing that track is not giving a solution, but explaining the paradox away. Your argument is like that of someone who "solves" Einstein's twin paradox by saying that there are no twins from whom one travels with extremely large speeds through space, they are just a crutch!

  • Like 1
  • Sad 1

Share this post


Link to post
Share on other sites
4 hours ago, merjet said:

Every word or phrase I put in bold highlight Jonathan’s mutilating MechanicaThe ancient Greeks in Mechanica said nothing about two wheels, surfaces, or slipping or not slipping.

Jonathan has no solution without his crutch – a second surface, track, or support. He needs a crutch so he can claim slippage.

When the problem conditions are met, the smaller circle horizontally moves the circumference of the larger circle (or vice-versa) with or without a second surface, track, or support. The paradox does not appear and disappear depending on whether or not there is a second surface, real or imagined. My solutions hold with or without a second surface, track, or support. Jonathan has refused to acknowledge my solutions as solutions. He has even denied they are, like he did this time. No good reason given; it’s mere ego and stubbornness.

So you agree that with his "crutch" Jonathan is correct?

--Brant

if this were a chess game your answer would be unnecessary

  • Like 1
  • Sad 1

Share this post


Link to post
Share on other sites
1 hour ago, Brant Gaede said:

So you agree that with his "crutch" Jonathan is correct?

Do you agree that all 3 solutions I put on Wikipedia are correct?

  • Haha 1

Share this post


Link to post
Share on other sites
2 hours ago, merjet said:

Do you agree that all 3 solutions I put on Wikipedia are correct?

These are not solutions. "Solution" 1 is nothing else than a short recapitulation of the paradox:

Aristotle: If I move the smaller circle I am moving the same centre, namely Α; now let the larger circle be attached to it [...] it will have invariably travelled the same distance [i.e. case 1: the smaller circle forces the larger circle to travel only the distance of the circumference of the small circle]

[...]Similarly, if I move the large circle and fit the small one to it [case 2: the large circle forces the smaller circle to travel the distance of the circumference of the large circle]

[...] nowhere does the greater stop and wait for the less in such a way as to remain stationary for a time at the same point [Aristotle doesn't understand how in case 1 the large circle is forced to travel the smaller distance]

the smaller does not skip any point [neither does he understand how in case 2 the smaller circle is forced to travel the larger distance]

[...]When, then, the large circle moves the small one attached to it, [in other words, when the large circle forces the small one]

the smaller one moves exactly as the larger one; when the small one is the mover, [that is, the small circle forces the large one]

the larger one moves according to the other's movement.


 

Compare that with your "solution" 1: If the smaller circle depends on the larger one (Case I), then the larger circle forces the smaller one to traverse the larger circle’s circumference. If the larger circle depends on the smaller one (Case II), then the smaller circle forces the larger one to traverse the smaller circle’s circumference. This is the simplest solution.

It is clear that this doesn't tell us anything new that Aristotle hadn't written already. 

That "solution" 2 and "solution" 3 are not solutions, I've already shown in earlier posts. In fact they are also just recapitulations of the paradox.

 

  • Like 2
  • Sad 1

Share this post


Link to post
Share on other sites
15 hours ago, Max said:

These are not solutions. "Solution" 1 is nothing else than a short recapitulation of the paradox:

[snip]

That "solution" 2 and "solution" 3 are not solutions, I've already shown in earlier posts. In fact they are also just recapitulations of the paradox.

 

LOL. Slippage is not a solution; it is a mere restatement of the paradox.

  • Haha 1

Share this post


Link to post
Share on other sites
55 minutes ago, merjet said:

LOL. Slippage is not a solution; it is a mere restatement of the paradox.

Now I'm confused. I didn't know slippage was in the paradox. I thought it was a consequence of real life use of the paradox.

--Brant

  • Sad 1

Share this post


Link to post
Share on other sites
18 hours ago, merjet said:

Do you agree that all 3 solutions I put on Wikipedia are correct?

Wikipedia is like a piece of soft clay.

--Brant

  • Sad 1

Share this post


Link to post
Share on other sites
29 minutes ago, Brant Gaede said:

Now I'm confused. I didn't know slippage was in the paradox. I thought it was a consequence of real life use of the paradox.

LOL. I didn't know cycloids were given by the paradox either. They are certainly not in Mechanica.  But somehow Max believes they are (by ESP, which fails to detect slippage?).

Share this post


Link to post
Share on other sites
1 hour ago, merjet said:

LOL. Slippage is not a solution; it is a mere restatement of the paradox.

No, it isn't. Artistotle wrote: "nowhere does the greater stop and wait for the less in such a way as to remain stationary for a time at the same point" and "the smaller does not skip any point", so he considers only stopping of the large circle and skipping of the small circle as possible explanations for the problem. As he rejects these possibilities, he cannot solve the paradox, because he is not aware of the possibility of slipping (forwards for the small circle, forced by the large one, and backwards for the large circle, forced by the small one), which enables a continuous movement that explains the problem. 

  • Like 2
  • Sad 1

Share this post


Link to post
Share on other sites
32 minutes ago, merjet said:

LOL. I didn't know cycloids were given by the paradox either. They are certainly not in Mechanica.  But somehow Max believes they are (by ESP?).

As I've shown before, cycloids are a completely unnecessary element added by you, allegedly "proving" that both wheels travel the same distance. Well, that they do, Aristotle already knew, you can read that in his text. So in that regard you don't prove anything that isn't already in Aristotle's text. The cycloids are just an irrelevant extra.

  • Like 2
  • Sad 1

Share this post


Link to post
Share on other sites
11 minutes ago, Max said:

No, it isn't. 

Yes, it is. Saying "it slips" is merely another way of saying the smaller circle rolls further than its own circumference.

  • Haha 1

Share this post


Link to post
Share on other sites
9 minutes ago, merjet said:

Yes, it is. Saying "it slips" is merely another way of saying the smaller circle rolls further than its own circumference.

We now know that this is the mechanism behind the further rolling, but Aristotle didn't understand it, as I've shown in one of my previous posts. Therefore it is no longer a real puzzle for us, while it was an enigma for those guys in the past. But I'm glad to know that you now have also been converted to the Slipping School.

  • Like 2
  • Sad 1

Share this post


Link to post
Share on other sites
5 hours ago, merjet said:

Yes, it is. Saying "it slips" is merely another way of saying the smaller circle rolls further than its own circumference.

Semantics.

--Brant

  • Sad 1

Share this post


Link to post
Share on other sites

There's that elephant in the room, again, which everyone is slipping past, going unrecognized.

You cannot even begin to consider the option, sliding/slip, until you first factor in the (proportionately) reduced speeds of inner radii (e.g. an inner wheel). That is a given. (An inner track is not).

 

Ask yourself, could it be the case that its ~lesser~ rotating velocity (Vtangential) *cancels out* any potential slippage, for the inner wheel? Shouldn't this inherent property of the wheel be considered way above all else?

If so, a track and slippage will become a redundant exercise. (And that was premised on an equal rotating velocity, at every radius and concentric circle, throughout the main wheel).

-----

Following, one erroneous explanation that I found - "Satisfying Explanation..." - by one fellow who comprehends the apparent 'problem', but tied himself in complicated knots and got his conclusion wrong, because he took into account only angular velocity.

E.g. "...both rotate with equal velocity..." [!]

"Therefore, this 'paradox' is actually a proof by contradiction that the two wheels cannot simultaneously stay concentric and roll without sliding". [!]

 "...the small circle will slide to catch up with the pace of the big circle".

"In conclusion, the smaller the smaller circle is, the greater it slides."

(And his statements would be logically true -- if his premises were valid).

Share this post


Link to post
Share on other sites
3

The paradox:

enter image description here

We have a circle and there is another circle with smaller radius. They are co-centeric.

If circle make full turn without sliding, both smaller and bigger circle make full turn too. If we assume that the passed road is equal to the circumference of circles. We have got smaller circle's radius is equal to bigger one's.

Unsatisfying Solutions I found:

  1. "Do not assume that smaller circle's circumference is equal to passed road since the surface that contacts to the ground is bigger one. " // Okey but it does not explain the paradox, it explains just what is the wrong assumption (even does not explain why it is a wrong assumption.)
  2. It is undeniable that every point on both smaller and bigger circle will contact exactly one and only one point on their path. Therefore we can think that this is a bijective maps and smaller circle is isomorphic to bigger one. (Okey but ....)

Question: What is the true answer? What is wrong with the definition of circumference of a circle and relationship with its taken path.

asked Jun 10 at 13:42
7839aef0ec5974772303251269a54a56?s=32&d=
user2312512851
1,183521
  • 1
    It is impossible for both wheels to roll without slipping. You can prove that using vectors, for instance. – Aretino Jun 10 at 13:55

3 Answers

2
 

The velocity of any point PP on a wheel can be written as the sum of two velocities: the velocity VV→ of the center OO and the velocity ω×OPω→×OP→ of rotation about the center, where ωω→ is angular velocity (perpendicular to the plane of the wheel).

A wheel turns without sliding with respect to a given path if the velocity of the contact point between wheel and path vanishes. Let then CC and 0043.png?V=2.7.42032.png?V=2.7.4C′ be the contact points of the two wheels. We have

0076.png?V=2.7.42192.png?V=2.7.40043.png?V=2.7.4003D.png?V=2.7.40056.png?V=2.7.42192.png?V=2.7.4002B.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42192.png?V=2.7.40061.png?V=2.7.4006E.png?V=2.7.40064.png?V=2.7.40076.png?V=2.7.42192.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.4003D.png?V=2.7.40056.png?V=2.7.42192.png?V=2.7.4002B.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.42192.png?V=2.7.4v→C=V→+ω→×OC→andv→C′=V→+ω→×OC′→
If 0076.png?V=2.7.42192.png?V=2.7.40043.png?V=2.7.4003D.png?V=2.7.40030.png?V=2.7.4v→C=0 then 0056.png?V=2.7.42192.png?V=2.7.4003D.png?V=2.7.42212.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42192.png?V=2.7.4V→=−ω→×OC→ and
0076.png?V=2.7.42192.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.4003D.png?V=2.7.42212.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42192.png?V=2.7.4002B.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.42192.png?V=2.7.4003D.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.40028.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.42192.png?V=2.7.42212.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42192.png?V=2.7.40029.png?V=2.7.4003D.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.40028.png?V=2.7.40043.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.42192.png?V=2.7.40029.png?V=2.7.4002E.png?V=2.7.4v→C′=−ω→×OC→+ω→×OC′→=ω→×(OC′→−OC→)=ω→×(CC′→).
This cannot vanish, unless 0043.png?V=2.7.4003D.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.4C=C′. So the assumption that both circles turn without sliding is false.
answered Jun 10 at 14:12
svg+xml,%3Csvg%20enable-background%3D%22
OI0Gw.jpg?s=32&g=1
Aretino
22.6k21442
3

Starting with three assumptions

  1. Both wheels stay concentric all the way,

  2. Both make full circle without sliding,

  3. At the end the distance is equal to the perimeter

you get a contradictory result (both wheels have same perimeter). The principle of proof by contradictions tells us that the conjunction of your assumptions is erroneous. Of course, as you pointed out, it makes no sense to reject assumption 3 because one can always go on rolling the wheel until it crosses the right distance.

Therefore, this "paradox" is actually a proof by contradiction that the two wheels cannot simultaneously stay concentric and roll without sliding. If they are to stay concentric all the way, at least one has to slide.

answered Jun 10 at 13:47
d3f15e0b666613021fc08b1a6035ee2f?s=32&d=
Régis
658310
  • 1
    This restates the first explanation given by OP, which was rejected. – Ross Millikan Jun 10 at 13:50
  •  
    No, it rejects another assumption. If you refuse to reject erroneous assumptions, there is no way not to deduce erroneous conclusion. Actually, this "paradox" is a proof by contradiction that both circle cannot make full circle without sliding - otherwise they would have same perimeter, which is a contradiction. – Régis Jun 10 at 14:06
1

Fix two aligned points (green and red) on the circles. And consider another circle identical to the smaller circle that is attached on top of the bigger circle. Refer to the figure below:

enter image description here

Case 1: If we consider the big circle and top small circle, both rotate with equal velocity and the top small circle will make full round at the red point (at this point the big circle has not made full round yet). Note that for convenience we may consider that the small circle rotates opposite to the big circle.

Case 2: Now if we consider the big circle and the cocentric small circle, the small circle will slide to catch up with the pace of the big circle.

In conclusion, the smaller the smaller circle is, the greater it slides.

Share this post


Link to post
Share on other sites

As for the track, in practice. Consequent to the above "...the small circle will slide to catch up with the pace of the big circle" - is that 'sliding' in any part of the wheel(s) disrupts the function of the whole wheel - the large wheel will either slide, too, or simply stop. Comparable to your bike's wheel hitting an oily patch - or someone shoving a stick into your bicycle spokes. 

This was best seen with the different sized cogs. Attempting to place two concentric cog wheels onto two flat gears (at adjusted planes) simultaneously.

Neither cog can 'slip', but turning at disparate velocities, both will jam solid.

Share this post


Link to post
Share on other sites
5 hours ago, anthony said:

There's that elephant in the room, again, which everyone is slipping past, going unrecognized.

You cannot even begin to consider the option, sliding/slip, until you first factor in the (proportionately) reduced speeds of inner radii (e.g. an inner wheel). That is a given. (An inner track is not).

Nobody ignores the fact that a point on the smaller wheel has a lower tangential speed than a point on the larger wheel. And an inner track is given, read the original text. 

Quote

Ask yourself, could it be the case that its ~lesser~ rotating velocity (Vtangential) *cancels out* any potential slippage, for the inner wheel? Shouldn't this inherent property of the wheel be considered way above all else?

No, it does not cancel out slippage, as I've shown in my proof (you should really read that some time). I've given the exact amounts of the various speeds, instead of your vague conjecture.

Quote

Following, one erroneous explanation that I found - "Satisfying Explanation..." - by one fellow who comprehends the apparent 'problem', but tied himself in complicated knots and got his conclusion wrong, because he took into account only angular velocity.

E.g. "...both rotate with equal velocity..." [!]

He is correct, they do rotate with equal velocity, rotation velocity or angular velocity is the vector ω, with magnitude = number of revolutions/ time unit. The tangential velocity = ω  and the tangential speed is ω*r.

With boldface I indicate vectors, instead of trying to put an arrow over the variable (see also below). "x" indicates a vector product.

 

5 hours ago, anthony said:

The velocity of any point PP on a wheel can be written as the sum of two velocities: the velocity VV→ of the center OO and the velocity ω×OPω→×OP→ of rotation about the center, where ωω→ is angular velocity (perpendicular to the plane of the wheel).

A wheel turns without sliding with respect to a given path if the velocity of the contact point between wheel and path vanishes. Let then CC and 0043.png?V=2.7.42032.png?V=2.7.4C′ be the contact points of the two wheels. We have

0076.png?V=2.7.42192.png?V=2.7.40043.png?V=2.7.4003D.png?V=2.7.40056.png?V=2.7.42192.png?V=2.7.4002B.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42192.png?V=2.7.40061.png?V=2.7.4006E.png?V=2.7.40064.png?V=2.7.40076.png?V=2.7.42192.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.4003D.png?V=2.7.40056.png?V=2.7.42192.png?V=2.7.4002B.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.42192.png?V=2.7.4v→C=V→+ω→×OC→andv→C′=V→+ω→×OC′→
If 0076.png?V=2.7.42192.png?V=2.7.40043.png?V=2.7.4003D.png?V=2.7.40030.png?V=2.7.4v→C=0 then 0056.png?V=2.7.42192.png?V=2.7.4003D.png?V=2.7.42212.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42192.png?V=2.7.4V→=−ω→×OC→ and
0076.png?V=2.7.42192.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.4003D.png?V=2.7.42212.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42192.png?V=2.7.4002B.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.42192.png?V=2.7.4003D.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.40028.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.42192.png?V=2.7.42212.png?V=2.7.4004F.png?V=2.7.40043.png?V=2.7.42192.png?V=2.7.40029.png?V=2.7.4003D.png?V=2.7.403C9.png?V=2.7.42192.png?V=2.7.400D7.png?V=2.7.40028.png?V=2.7.40043.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.42192.png?V=2.7.40029.png?V=2.7.4002E.png?V=2.7.4v→C′=−ω→×OC→+ω→×OC′→=ω→×(OC′→−OC→)=ω→×(CC′→).
This cannot vanish, unless 0043.png?V=2.7.4003D.png?V=2.7.40043.png?V=2.7.42032.png?V=2.7.4C=C′. So the assumption that both circles turn without sliding is false.
 
wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==

The formatting has gone awry, but the explanation is correct, it is in fact the same as my proof, only using velocity vectors instead of speed scalars+direction. I've reformatted it, avoiding those pesky arrows by indicating vectors with boldface (and "x" for the vector product).

vC = V + ω x OC and vC’= V + ω x OC’  [I see that I now somehow have used different fonts, ignore that; quirks of this editor.]

if vC = 0 then

V = - ω x OC [the translation velocity and the tangential velocity at the contact point cancel when there is no slipping] and

vC’ = - ω x OC + ω x OC’ = ω x (OC’ - OC) = ω x CC’

This cannot vanish, unless C = C’. So the assumption that both circles turn without sliding is false.

[ vC and vC’ are the velocity vectors of the contact points (6 o’clock positions) of respectively the larger and the smaller circle. vC = 0 is the condition that the large circle rotates without slipping, vC’ = 0 is the condition that the small circle rotates without slipping.]

 

Edited by Max
reformatted formulas added
  • Like 1

Share this post


Link to post
Share on other sites
3 hours ago, Max said:

Nobody ignores the fact that a point on the smaller wheel has a lower tangential speed than a point on the larger wheel. And an inner track is given, read the original text. 

No, it does not cancel out slippage, as I've shown in my proof (you should really read that some time). I've given the exact amounts of the various speeds, instead of your vague conjecture.

 

 

 

 

 

11

Max: "...an inner track *is* given".

But I'm quite happy to accept an inner track, and such.  

The problems it brings in become greater than the original 'problem'. As my last post, this wheel isn't going anywhere, let alone solve anything.

For there to be 'slippage' on a track, one must infer a degree of drag on the track by the inner wheel (it has to be in contact). Never mind that it is ~meant~ to go faster "to catch up with the pace of the big wheel"...

So we'll get rolling by the large wheel - accompanied by frictional resistance by the small one. These opposing forces mean that 'something has to give'. The wheel stops (like putting your foot on the gas pedal and the brake, together), or everything skids forwards, nothing rolls. Or somewhere in between.

I don't mind your track and slippage, it is just a dead-end route to try resolving the paradox. The theory doesn't meld with the practice.

Happy New Year, all!

Share this post


Link to post
Share on other sites
1 hour ago, Brant Gaede said:

Have we entered Crazy Land?

--Brant

Yes. Merlin started the thread from Crazy Land. He really likes cycloids. He's cuckoo for them. When he heard of the Aristotle's Wheel Paradox, he wanted cycloids to be the answer. He came here to be revered as a great teacher and bringer of cycloids to the unwashed. He wanted it so badly that he altered the formulation of the problem, first here, and then at Wikipedia. He eliminated some things, and imposed some arbitrary new rules, but still didn't end up with anything that is solved with cycloids. But he forced his cycloids onto the problem anyway. His one success is that he was able to fool/confuse Tony, a fellow resident of Crazy Land. Neither of them can keep the entire scenario straight. Both need to drop certain conditions in order to try to believe that they've gotten their minds around it. Even then, they fail. They don't have the visuospatial/mechanical reasoning abilities to grasp it.

J

  • Like 1
  • Sad 1

Share this post


Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...

×
×
  • Create New...