Aristotle's wheel paradox

[merjet]

On 10/27/2017 at 2:24 PM, Max said:

The crux of the paradox is the implied -and false- suggestion that both wheels can turn without slipping on their respective supports (rail or road etc.). Both (concentric) wheels are part of a rigid body, so they have the same rotational velocity and the same translational velocity (of their common center). When the larger wheel makes one rotation without slipping, it travels over a distance of 2 π R. So does the smaller wheel, but if this wheel wouldn't slip, it would only travel over a distance 2 π r (r < R). However, it has to travel over a distance of 2 π R, so apart from its rotation it must also slip with respect to its support, to keep up with the larger wheel. Mutatis mutandis if it is the smaller wheel that rotates without slipping. It's all so very simple and trivial, so why should we have a discussion that now covers already 25 pages? It isn’t that difficult!

Clear, simple, and wrong four ways. There aren't two wheels and two supports. There is one of each for an ordinary wheel. The crux of the paradox is that the inner "wheel" moves farther than its circumference with one full rotation. You mention translation, then abandon it in favor of "slipping." This video (and many others) explain rolling without slipping (or skidding) and translation. What part of “without slipping” do you not understand? Likewise, the smaller "wheel" does not slip nor skid. An inner “wheel” slipping on an imaginary road is as silly as a person slipping on imaginary ice. Translation fully accounts for its moving the horizontal distance 2πR, like it does for its center and the wheel with radius R and the same center. The video makes that clear.

This article does not clearly distinguish between slipping and skidding, but it can be done. In essence slipping is rotation without translation, such as a wheel of a car on ice or stuck in snow does and the driver pushes hard on the accelerator pedal. In essence skidding is translation without rotation, such as a wheel of a car does on an icy road and the driver pushes hard on the brake pedal. Both are due to a lack of traction and affect the translation movement of the entire wheel uniformly. An inner "wheel" slipping on an imaginary road is as foolish as a person slipping on imaginary ice. Such foolishness implies translation movement is not uniform – a smaller inner "wheel" "slips" more than a larger inner "wheel."

This thread 25 is pages for at least the following:

- Whether or not there is a paradox.

- Different and conflicting meanings of slip.

- The motion of a wheel can be analyzed in more than one way.

- Jonathan’s obsession for making personal attacks.

The obnoxious, self-deluded ignoranus Jonathan fails to understand a wheel’s motion – especially translation – as describe above. He abuses the concepts slip and skid. That serves his highest aim, which is to bray, sneer, and ridicule.

On 10/27/2017 at 9:59 AM, Jonathan said:

The actual proof is much simpler. The smaller circle has a circumference whose length is shorter than the line segment which represents the length of the larger circle's circumference. Therefore, by definition, the smaller slips as it rolls. QED.

A straight stick/pencil partly submerged in water appears to be bent (link), but the stick/pencil is not bent in reality. It is a classic optical illusion. An inner concentric circle of a wheel may appear to slip, but it does not slip in reality if the wheel doesn’t slip. Jonathan abandons reality in favor of appearance and his scam/ruse. If my not being duped by his scam/ruse counts as stubbornness, I’m fine with that. Suppose John Doe says: “Look at the pencil in the water. It is bent. It is as obvious as hell, but you can’t or refuse to see it. You are visually incompetent and retarded!” And he says this as if light refraction is irrelevant. Analogously, translation is irrelevant in Jonathan’s pretentious “proof.”

The following proof is simple and correct.

The distance a circle moves translation-wise is always the same distance as its center moves. Since a wheel and any inner circle concentric with it have the same center, the wheel and said circle always move the same distance translation-wise. QED. This is true independent of any rotation, slipping, or skidding.

[Jonathan]

The idiot is back:

2 hours ago, merjet said:

A straight stick/pencil partly submerged in water appears to be bent (link), but the stick/pencil is not bent in reality. It is a classic optical illusion.

And a cycloid and curtate cycloid could therefore also be optical illusions. Any measurement that you’re postulating could be an illusion. You’re just stupidly selectively applying your illusion theory to others’ solutions but not to your own. Idiot.

 

2 hours ago, merjet said:

An inner concentric circle of a wheel may appear to slip, but it does not slip in reality if the wheel doesn’t slip. Jonathan abandons reality in favor of appearance and his scam/ruse.

No, the inner circle slips in reality despite your refusal to grasp its slippage. You’re visuospatially retarded, and you’re just petulantly refusing to believe that others do not share your retardation. Under the conditions specified in the initial setup, the inner circle slips in reality. As has been suggested several times, test it yourself in physical reality. It is not a scam or ruse. Everyone else here easily grasps what you’re failing to. Try it in reality! (I don't expect you to do so. I know that you're too stupid to even understand what you're being challenged to do.)

 

2 hours ago, merjet said:

If my not being duped by his scam/ruse counts as stubbornness, I’m fine with that. Suppose John Doe says: “Look at the pencil in the water. It is bent. It is as obvious as hell, but you can’t or refuse to see it. You are visually incompetent and retarded!” And he says this as if light refraction is irrelevant. Analogously, translation is irrelevant in Jonathan’s pretentious “proof.”

But, again, by the same standards, your translation/rotation/cycloid paths “proof” is also possibly a visual illusion. I mean, you're so retarded that you can't even draw the system at the level of a second grader, yet you expect us to just accept your say-so of what paths the points take in reality? So, since what you’re doing is eliminating vision as a reliable means of our interfacing with reality, therefore you would have to propose a method of proof which doesn’t rely in any way on vision, and your proposed solution doesn't meet that criterion. What’s left? Touch? If so, can’t we also be fooled via our tactile sense? In fact, aren’t tactile illusions even easier to pull off than visual ones?

By what means are you proposing that we know reality if you’re rejecting our senses as valid means of knowing it?

 

2 hours ago, merjet said:

The following proof is simple and correct.

The distance a circle moves translation-wise is always the same distance as its center moves. Since a wheel and any inner circle concentric with it have the same center, the wheel and said circle always move the same distance translation-wise. QED. This is true independent of any rotation, slipping, or skidding.

WTF??? No one has denied the above!!! Everyone understands that one of the conditions of the initial setup was that both circles share the same center point and therefore share a point which travels the same distance!!!! Do you not remember that there’s more to the setup than that?!!!

 

[Jonathan]

2 hours ago, merjet said:

Likewise, the smaller "wheel" does not slip nor skid.

Yes, it does slip/skid. It's a fact of reality that it slips/skids. If it did not do so, then the principle which drives bicycles would not work in reality. The principle does work in reality. What doesn't work in reality is your retarded belief that the inner wheel is not slipping.

2 hours ago, merjet said:

An inner “wheel” slipping on an imaginary road is as silly as a person slipping on imaginary ice.

Where did you get the wrongheaded idea that the road is "imaginary"? The scenario is that there are two circles or "wheels," and two lines or "roads" with which the "wheels" are in contact. That second line is not "imaginary" just because you don't like it and can't properly visualize its interaction with the inner wheel.

And after all this time, you're still insisting that the inner wheel doesn't slip in reality because you've interpreted the setup of the scenario as claiming that neither wheel slips when traveling over its line. You have decided to trust a mistaken assertion in the alleged "paradox's" setup.

The initial link that you posted to explains all of this:

"One way to understand the paradox of the wheel is to reject the assumption that the smaller wheel indeed traces out its circumference, without ensuring that it, too, rolls without slipping on a fixed surface. In fact, it is impossible for both wheels to perform such motion. Physically, if two joined concentric wheels with different radii were rolled along parallel lines then at least one would slip; if a system of cogs were used to prevent slippage then the wheels would jam. A modern approximation of such an experiment is often performed by car drivers who park too close to a curb. The car's outer tire rolls without slipping on the road surface while the inner hubcap both rolls and slips across the curb; the slipping is evidenced by a screeching noise."

Almost everyone other than you understands it. It is the simplest thing in the world, except to some people, people like you who are severely visuospatially retarded.

[Jonathan]

Here’s a another similar “paradox” for you, Merlin.

We have a small triangle whose sides are each 1 inch in length.

Inside of that triangle, we have a much larger triangle whose sides are 45 inches each, and which has four sides but is still a triangle.

So we have a double "paradox" of a much larger triangle fitting into a smaller one, and a triangle that is a triangle despite not being one!

Solve the "paradox."

[Jonathan]

12 hours ago, merjet said:

In essence skidding is translation without rotation...

 

That would be complete skidding. There is also partial skidding as we've described on this thread, as well as presented in animations.

Translation plus the right amount of rotation in proportion to radius creates a proper cycloid, which equals rolling without any skidding. 

The same translation and rotation with a lesser radius creates a curtate cycloid, which equals rolling with skidding. The closer that the curve is to being a line, the more skidding there is in proportion to rotation.

The same translation and rotation with a greater radius creates a prolate cycloid, which equals spinning faster than rolling-contact, such as tires spinning and "burning out" at the beginning of a race when the car is moving forward but not yet getting full traction.

The specific form of the cycloid reveals whether the circle is rolling freely, or skidding or spinning, and, if it is skidding or spinning, how much.

Your own preferred method of "solving" the alleged "paradox"results in the inner circle creating a curtate cycloid, and therefore proves that the inner circle is skidding along the line that it contacts exactly to the degree that my animations have shown.

QED

If your brain power was a circle, and this thread was a line, it would create a curtate cycloid. Maybe even a line. Probably a line. Yeah, most definitely a line.

 

[BaalChatzaf]

2 hours ago, Jonathan said:

 

If your brain power was a circle, and this thread was a line, it would create a curtate cycloid. Maybe even a line. Probably a line. Yeah, most definitely a line.

 

nasty, nasty.   But a  very sophisticated insult.  Well done.

[Jonathan]

3 hours ago, BaalChatzaf said:

nasty, nasty.   But a  very sophisticated insult.  Well done.

Thank you.

[Max]

 

•  

  On 27-10-2017 at 8:24 PM, Max said:

The crux of the paradox is the implied -and false- suggestion that both wheels can turn without slipping on their respective supports (rail or road etc.). Both (concentric) wheels are part of a rigid body, so they have the same rotational velocity and the same translational velocity (of their common center). When the larger wheel makes one rotation without slipping, it travels over a distance of 2 π R. So does the smaller wheel, but if this wheel wouldn't slip, it would only travel over a distance 2 π r (r < R). However, it has to travel over a distance of 2 π R, so apart from its rotation it must also slip with respect to its support, to keep up with the larger wheel. Mutatis mutandis if it is the smaller wheel that rotates without slipping. It's all so very simple and trivial, so why should we have a discussion that now covers already 25 pages? It isn’t that difficult!

> Clear, simple, and wrong four ways. There aren't two wheels and two supports.

From the Wikipedia reference in your very first post in this thread: "There are two wheels, one within the other, whose rims take the shape of two circles with different diameters. The wheels roll without slipping for a full revolution." The accompanying figure clearly shows that those "two wheels, one within the other" move over their respective supports, so there are two wheels and two supports.

> There is one of each for an ordinary wheel. The crux of the paradox is that the inner "wheel" moves farther than its circumference with one full rotation. You mention translation, then abandon it in > favor of "slipping."

You should read better, I wrote " [both wheels...have] the same translational velocity (of their common center). I then show that the rim of at least one of the wheels must be slipping against its support.

>  This video (and many others) explain rolling without slipping (or skidding) andtranslation. What part of “without slipping” do you not understand?

This rather elementary video tells nothing that contradicts my text.

>  Likewise, the smaller "wheel" does not slip nor skid. An inner “wheel” slipping on an imaginary road is as silly as a person slipping on imaginary ice.

Aristoteles' example may be imaginary, but it can be easily and unambiguously realized in reality. Of course such realizations show the real slipping, predicted by the mathematical analysis.

 > Translation fully accounts for its moving the horizontal distance 2πR, like it does for its center and the wheel with radius R and the same center. The video makes that clear.

Yes, the movement of the common center can be described as a translation (as I did in my text), but the rim of at least one of the wheels must be slipping when that common center is translated.

> This article does not clearly distinguish between slipping and skidding, but it can be done. In essence slipping is rotation without translation, such as a wheel of a car on ice or stuck in > snowdoes and the driver pushes hard on the accelerator pedal. In essence skidding is translation without rotation, such as a wheel of a car does on an icy >road and the  driver pushes hard onthe brake pedal. Both are due to a lack of traction and affect the translation movement of the entire wheel uniformly. An inner "wheel" slipping on an >imaginary road is as foolish as a person slipping on imaginary ice. Such foolishness implies translation movement is not uniform – a smaller inner "wheel" "slips" more than a larger inner "wheel."

Translation of a rolling wheel (with or without slipping) is of course not uniform. The translation of a point on the rim is for example different from the translation of the center of the wheel. As you can see in the above-mentioned elementary video.

[..]

> The following proof is simple and correct.

> The distance a circle moves translation-wise is always the same distance as its center moves. Since a wheel and any inner circle concentric with it have the same center, the wheel andsaid circle > always move the same distance translation-wise. QED. This is true independent of any rotation, slipping, or skidding.

This formulation is rather vague: what is exactly the distance that a circle moves translation-wise? Different points on the circle/rim of the wheel move in an arbitrary time interval different distances translation-wise. But for two concentric rotating circles/wheels the distance the centers move is the same, as they have the same center.*) And that is exactly part of the argument I've used. It is only one half of the proof. In itself it doesn't tell us anything about the paradox. The second part of the proof is the realization that the paradox is created by the supposition that both wheels can rotate without slipping against their respective supports (see the Wikipedia article and the accompanying illustration). That is not possible, if one wheel rotates without slipping, the other one must be slipping (/skidding if you like, I make no distinction), not only in theory, but also in reality. If you realize such a system, the slipping is unavoidable and very real, it‘s not some kind of illusion, on the contrary, it is an essential part of solving the paradox.

*) In fact, this is part of description of the paradox (in the Wikipedia article:  "the two lines have the same length"). The paradox gives an argument that these lines must have the same length, and another argument that these lines have different lengths. That is a contradiction, and therefore we call this a paradox, but without inspecting both arguments, we don't know how this paradox can be solved. As I've said many times, the error is in the statement (Wikipedia:) "The wheels roll without slipping for a full revolution". From the other argument follows that this is impossible, both wheels cannot roll without slipping, at most one wheel can, and the other wheel must therefore roll with slipping. This non-slipping vs. slipping is not just some side-issue, it is the origin of the existence of this paradox. By identifying it, we solve the paradox.

 

Edited January 26, 2018 by Max
Footnote added

[Jonathan]

Here’s a challenge for Merlin:

What is the path made by a point on a wheel which is rolling but also skidding/slipping somewhat on a line which is longer than the wheel’s circumference?

Let’s make it precise. The wheel is 10 inches in diameter. The line/surface that it is traveling on is 60 inches long, and the wheel is having a specific amount of brake pressure applied to it so that it partially rotates while partially skidding/sliding at a steady, constant rate, resulting in the wheel making exactly one rotation over the translational distance of 60 inches.

Do the math, do the geometry, and identify and show the precise path that any one point on the wheel’s circumference will take under the conditions described above.

Which form does it take: a curtate cycloid, a common cycloid, or a prolate cycloid? Or none of the above? Prove your answer.

[Max]

Here is another graphic, illustrating the solution of Aristoteles' paradox.

The two concentric circles represent the corresponding wheels, with respective radius r and radius R, with r = 2/3 * R.

The outer wheel rolls without slipping over its support. Two instants are given: the start position and the position after a rotation over 3/4 pi radians. The blue spoke points north in the start position and southeast in the second position, the black spoke points northeast in the start position and south in the second position, etc.

If the small wheel would roll alone, without slipping over its support, its position would after a rotation over 3/4 pi radians be given by the grey circle, with its center at r*pi*3/4. But when it is fixed to the outer wheel and the outer wheel rolls without slipping, rotating over pi*3/4 radians, the smaller wheel is carried along by the large wheel and is translated over a distance R*pi*3/4. Its translation by rolling alone, rotating over the same angle, would result in a translation over only 2/3*R*pi*3/4. The difference (R-r)*pi*3/4= R*pi/4 must be made up by slipping of the smaller wheel over its support. The movement of the smaller wheel is thus a combination of rolling and slipping, in this case 2/3 rolling and 1/3 slipping. In the idealized case these proportions are fixed over the whole traject, valid for any time interval, no matter how small. In real life situations there would be tiny fluctuations, but over longer intervals the proportions would be the same.

 

[Brant Gaede]

Merlin: what would it take to disprove your position?

--Brant

[Jon Letendre]

2 hours ago, Brant Gaede said:

Merlin: what would it take to disprove your position?

--Brant

A physical instantiation in his hands.

[Jonathan]

35 minutes ago, Jon Letendre said:

A physical instantiation in his hands.

No, he won't even accept that. He would claim that the physical slippage -- the grinding, screeching friction -- is a tactile and aural illusion. Nothing will change his mind. He is dedicated to refusing to accept reality. He's in the process of convincing himself that he's being brave and heroic in standing alone in his stupid stubbornness.

[merjet]

I don’t think of Aristotle’s wheel paradox as being about two wheels. One wheel or roll of tape is a better exemplar that satisfies cognitive economy. I believe the Wikipedia article is poorly written. I strongly agree with this comment about two wheels on the talk page. Note that the quotation from Mechanica describes the paradox as about two circles, not two wheels. (Even the topic is Aristotle’s wheel paradox, not “wheels paradox.“) Assuming it is about two wheels only muddles the problem. This thread attests to that. Max’s graphic as he described it depicts two wheels. Regardless, his graphic depicts one real wheel or roll of tape equally well.

On 1/25/2018 at 3:05 PM, Jonathan said:

WTF??? No one has denied the above!!! Everyone understands that one of the conditions of the initial setup was that both circles share the same center point and therefore share a point which travels the same distance!!!! Do you not remember that there’s more to the setup than that?!!!

No one else has constructed and presented my simple proof either, stupid. You merely saw it as a chance for ridicule. You failed to notice that (1) the proof is supported by a necessary property of translation, and (2) the proof elegantly solves the paradox.

Justifying them:

1. Every point of the two circles necessarily travels the same distance translation-wise with a full revolution.

2. Let R = the radius of the larger circle. If the center of the two conjoined circles is moved 2*pi*R, and hence both circles, too (like illustrated for one circle here), then the smaller circle is moved 2*pi*R. QED. It’s that simple. The proof shows implicitly that the circumference of the smaller circle is an irrelevant red herring.

Nor does Wikipedia include such a solution. The Wikipedia article says nothing about translation, translation necessitating that the smaller circle moves 2*pi*R, and the smaller circle’s circumference being irrelevant.

The proof occurred to me a few weeks ago. When I was first aware of this paradox -- about when I started this thread – I was distracted by the “trees” (the dynamics of rolling, cycloids) and missed seeing the “forest.” “The more original a discovery, the more obvious it seems afterwards.” - Arthur Koestler, The Act of Creation.

Suppose the radius of Max’s smaller circle is 2/3*R like he said, and the conjoined concentric circles are rolled one full rotation. The horizontal movement of the larger circle is 2*pi*R, whereas the circumference of the smaller circle is 2*pi*R*2/3. How can one explain the smaller circle moving 2*pi*R instead? One way is Max’s way of subtraction: the smaller circle slips – with scare quotes omitted – the distance 2*pi*R – 2*pi*R*2/3 = 2*pi*R*1/3.

The Wikipedia article alludes to another way using multiplication – the smaller circle’s “movement from any point to another can be calculated by using an inverse of their ratio.” Algebraically for a full rotation, (2*pi*R*2/3)*R/(R*2/3), which after canceling simplifies to 2*pi*R. So the circumference of the smaller circle is irrelevant, as in my simple proof. Unlike Max’s way, slipping – with or without scare quotes -- is also irrelevant.

On 1/25/2018 at 3:05 PM, Jonathan said:

So, since what you’re doing is eliminating vision as a reliable means of our interfacing with reality, therefore you would have to propose a method of proof which doesn’t rely in any way on vision .... By what means are you proposing that we know reality if you’re rejecting our senses as valid means of knowing it?

More lies and hogwash from the obnoxious, dishonest, inept, self-deluded buffoon and con artist. Go to your barn and eat some hay, jackass. My analogy with a “bent” pencil spotlighted his con game. His reply was wholly futile and incompetent.   “Few things are harder to put up with than the annoyance of a good example.” – Mark Twain.

[merjet]

On 1/27/2018 at 10:57 AM, Brant Gaede said:

Merlin: what would it take to disprove your position?

Prove that when my car’s 4 tires roll on a real road, there exists another parallel, invisible surface tangent to the bottom of the metal rim of each wheel. Also, the rim really rolls on, slips on, and is really supported by said invisible surface.

Prove that when I roll a roll of tape on a table top, there is another parallel, invisible surface tangent to the bottom of the hole. Also, the hole really rolls on, slips on, and is really supported by said invisible surface.

Perhaps something else, but not con art.

[merjet]

The Millstone

Jonathan and Bubba went for a stroll. They discover an old millstone shaped like this one. They measured the diameter and hole. The diameter was 50 inches and the square hole was 10 inches on each side. Despite its great weight they somehow managed to roll it one rotation, a distance of approximately157 inches.

Bubba: Isn’t that a paradox? We moved the millstone 157 inches. However, the hole rotated only once and its perimeter is only 40 inches. How did that happen?

Jonathan: It slipped.

Bubba: What are you talking about?

Jonathan: The supporting rail a fraction of an inch below the hole. Didn’t you see the hole slip on it, idiot? Didn’t you hear the screeching and grinding of friction from the hole on the rail, nitwit? Do I need to make one of my illustrious animated videos with sound effects to prove it to you, stupid?

Bubba: I didn’t see or hear any such thing.

Jonathan: Then you are visually and aurally retarded!! Hahahaha.

 

[merjet]

On 1/26/2018 at 1:15 PM, Max said:

This formulation is rather vague: what is exactly the distance that a circle moves translation-wise? Different points on the circle/rim of the wheel move in an arbitrary time interval different distances translation-wise.

My proposition was about a circle as a whole, not each individual point on it. For example, let P1 denote the point at 11:00 o’clock on the circle before movement. There is a point P2 at 11:00 o’clock on the circle after movement. A vector drawn from P1 to P2 is parallel to, and the same length as, the center's translation vector. Like I said, this is true independent of any rotation, slipping, or skidding. If the circle is rotated exactly 0, 1, 2, 3, …. times – which is true for Aristotle’s wheel paradox -- then the vector from P1 to P2 is P1's translation vector, and every point on the circle moves the same distance translation-wise.

Since I will have so little free time the next several weeks, it’s time for another long hiatus.

 

[Jonathan]

8 minutes ago, merjet said:

I don’t think of Aristotle’s wheel paradox as being about two wheels. One wheel or roll of tape is a better exemplar that satisfies cognitive economy.

It doesn't matter if it's two wheels glued together, or one wheel with a hole (like a roll of tape), or one wheel with a protrusion (as in my animation), it's all the same. They are all valid options on showing the same thing. And you don't understand it. The inner circle slips on its line as it rolls.

14 minutes ago, merjet said:

Prove that when my car’s 4 tires roll on a real road, there exists another parallel, invisible surface tangent to the bottom of the metal rim of each wheel. Also, the rim really rolls on, slips on, and is really supported by said invisible surface.

Prove that when I roll a roll of tape on a table top, there is another parallel, invisible surface tangent to the bottom of the hole. Also, the hole really rolls on, slips on, and is really supported by said invisible surface.

Perhaps something else, but not con art.

Dipshit, the original "paradox" includes the line/surface that you're angry about and trying to ridicule us for acknowledging and addressing in our solutions. It's not "invisible" just because you've decided to ignore it and wish it out of existence because you've arbitrarily chosen to visualize a roll of tape which doesn't contact a second line. Apparently you, in your state of visuospatial retardation, can't grasp anything more than the roll of tape image, so you're stuck on it, and therefore you need to eliminate the vital part of the "paradox" setup -- the inner circle's contact with the line -- that your brain can't grasp.

It is still very entertaining watching you make a fool of yourself. You have no doubt at all, despite everything that you've been presented with on this thread! It's absolutely fascinating.

[Jonathan]

34 minutes ago, merjet said:

I strongly agree with this comment about two wheels on the talk page. Note that the quotation from Mechanica describes the paradox as about two circles, not two wheels. (Even the topic is Aristotle’s wheel paradox, not “wheels paradox.“) Assuming it is about two wheels only muddles the problem. This thread attests to that. Max’s graphic as he described it depicts two wheels.

The "paradox" is actually nothing more than a perfect accidental litmus test for visuospatial retardation. It's a very efficient test for mechanical reasoning incompetence. And it's a nice little peek into the limitations of past thinkers. There were some otherwise bright men of the past who were clearly dunces when it came to this very simple visuospatial mechanical reasoning scenario. Lacking in some basic cognitive abilities, they tried to egghead it with their cognitive strengths, which happened to not be suited to allowing them to grasp it. But fragile egos couldn't accept that reality, so instead of identifying the reality that they were visuospatially retarded, they misidentified their internal limitations as an external "paradox."

[Jonathan]

Repeating the challenge that Merlin didn't address:

What is the path made by a point on a wheel which is rolling but also skidding/slipping somewhat on a line which is longer than the wheel’s circumference?

Let’s make it precise. The wheel is 10 inches in diameter. The line/surface that it is traveling on is 60 inches long, and the wheel is having a specific amount of brake pressure applied to it so that it partially rotates while partially skidding/sliding at a steady, constant rate, resulting in the wheel making exactly one rotation over the translational distance of 60 inches.

Do the math, do the geometry, and identify and show the precise path that any one point on the wheel’s circumference will take under the conditions described above.

Which form does it take: a curtate cycloid, a common cycloid, or a prolate cycloid? Or none of the above? Prove your answer.

-----

Come on, genius. Solve it. Do you understand the relevance of this challenge? Can you get your brain around it? I doubt it.

 

[Jon Letendre]

1 hour ago, merjet said:

I don’t think of Aristotle’s wheel paradox as being about two wheels.

Yet the paradox asks us to do exactly that.

Get back with us when your obstinacy retreats and you feel ready to discuss the paradox.

[Max]

It doesn't matter whether you talk about wheels or about circles. One is a physical description and the other one the mathematical equivalent. Translation from one description to the other one is no problem: the rims of the wheels are the circles and the supports (road/rail/etc.) are the lines. The important condition is what in the original article is called "unrolling the line", "tracing out the circumference", and in mechanical terms "rolling, i.e. rotating without slipping". The origin of the paradox is the supposition that both wheels (that form one rigid body with a common center) can rotate without slipping/can trace out their circumference. 

Suppose the large wheel/circle rolls without slipping. After 1 period in time T the center of the circle is translated over a distance 2*pi*R, with a uniform translation speed of its center v= 2*pi*R/T. The point at the top of the circle is translated with speed 2*v and the bottom (that touches the line (=support) has translation speed zero. The translation speed of the point at the top of the smaller circle ≡ v2 = 2*pi*(R+r)/T. This can be checked by substituting r=R and r=0. Similarly, the translation speed of the point at the bottom of the smaller circle ≡v3= 2*pi*(R-r)/T > 0 for r < R. So we see that for the smaller circle and its tangent (support) the condition for tracing out the circumference is not met. That the bottom point of the smaller circle has a translation speed > 0 is the mathematical equivalent of saying that the smaller wheel is rotating and slipping.

 

So the notion of slipping is essential to the solution of the paradox. If you don’t like the word, you can say it in mathematical terms: it is not possible that the bottom points of both circles during rotation have zero translation speed. But it is just the same as saying that it is not possible that both wheels rotate without slipping.

 

[Jon Letendre]

All confusion ends, even the most stubborn cases, as soon as a physical example is in the hands.

[Jon Letendre]

I like your ad absurdum route, Max.

[jts]

What does anyone hope to accomplish by continuing this "wheel paradox" discussion so long? Do you all have nothing better to do or discuss?