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It's kind of an infamous problem in differential equations to find the correct road surface so that a car with square wheels (and an axle located in the center) keeps its axle level as it drives along. I hope I won't offend anybody by saying that one smooth piece of the solution (for a wheel with sides of length 2) is $y = -\cosh(x)$

If you actually take this solution and describe the position of the axle at any given point, unless I have calculated incorrectly you find that the axle is always positioned directly over the point where the wheel makes contact with the road. I've been unable to come up with a physical justification of this phenomenon and it seems fairly non-obvious to me.

Is there a straightforward reason why this must be true? Is it specific to this wheel shape?

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Very related: mathoverflow.net/questions/29988 –  J. M. Oct 20 '10 at 4:36
    
...and a web page (in French). –  J. M. Oct 20 '10 at 4:38
    
...and two interesting articles. –  J. M. Oct 27 '10 at 2:39
    
@J.M. I repeat, you're the master of links. –  Pedro Tamaroff Feb 20 '12 at 16:10

3 Answers 3

The physical justification is quite simple. The point of contact between the wheel and the road is instantaneously stationary. Since the wheel moves rigidly, the velocity of any point on the wheel is as though the wheel were rotating about the point of contact. In particular, the velocity of the axle is perpendicular to the line joining it to the point of contact. Since we require that the axle stays level, it velocity must be purely horizontal, so it has to be directly above the contact point.

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I hope this is not identical to J.M.'s answer; I apologize if I misjudged it. –  Rahul Oct 21 '10 at 4:17
    
Yours is physical, mine is a bit more mathematical. Having both here should be fine. –  J. M. Oct 21 '10 at 4:28
    
@J.M.: I really hope this isn't coming off as rude, but I couldn't see how your answer addressed the case of non-circular wheels. That's why I posted this one, which I thought would explain the phenomenon directly for arbitrary wheel shapes. Did I misunderstand your answer? –  Rahul Oct 21 '10 at 5:04
    
Not at all rude, Rahul. :) But that's why I appended a comment in my answer. If you're going the vector route, you have to modify the translation terms in the expression I gave, and the rotation matrix should be using the difference of the tangential angles of the road and the wheel to solve for. That's where the DE arises. –  J. M. Oct 21 '10 at 5:08

Quite an exposition is in Stan Wagon's paper If you look at Fig 13, it appears that the axle of the ellipse is not over the contact point.

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I think appearances may be deceptive in this case; the paper explicitly defines a wheel to be such that "the axle of the wheel... stays on the x-axis directly above the wheel-road contact point as the wheel rolls along the road" (first page, last paragraph). –  Rahul Oct 20 '10 at 4:29
    
You are quite right. –  Ross Millikan Oct 20 '10 at 4:32
    
Note that for the square wheel cart to work, you do not want the entire catenary as a road; rather, you cut up a piece of the catenary at the leftmost and rightmost points where the axle remains vertically fixed, and string together copies of this piece. –  J. M. Oct 20 '10 at 4:42
    
The question still remains of why Hall and Wagon defined it that way. –  Rahul Oct 20 '10 at 5:06

This is supposed to be in reply to Rahul's second comment, but it got too long.

I would suppose that the requirement of Hall and Wagon for the axle to have the same horizontal displacement as the contact point of the road and wheel is a generalization of the concept of a circle rolling on a line. In that situation it is obvious that the axle is indeed directly above the point of contact.

Remember that rolling without slip can be decomposed as a translation, followed by a rotation, and then another translation.

For the case of the wheel parametrically defined as $(x\qquad y)^T=(\sin\;t\qquad 1-\cos\;t)^T$, the axle is initially at $(0\;1)^T$. If we look at the expression

$$\bigl(\begin{smallmatrix}t\\0\end{smallmatrix}\bigr)+\bigl(\begin{smallmatrix}\cos\;t&\sin\;t\\-\sin\;t&\cos\;t\end{smallmatrix}\bigr)\cdot\left(\bigl(\begin{smallmatrix}0\\1\end{smallmatrix}\bigr)-\bigl(\begin{smallmatrix}\sin\;t\\1-\cos\;t\end{smallmatrix}\bigr)\right)$$

(in English: translate axle to the point on the circle that would be the future point of contact of the circle with the road, rotate by some angle, and then translate by an amount equal to the length of the arc subtended by said angle horizontally)

we find that this whole mess simplifies to the vector $(t\qquad 1)^T$ which indeed has the same x-coordinate as the current point of contact $(t\qquad 0)$.

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Of course, in the more general case considered by Hall/Wagon, the rotation is retained, but the translation parts are a bit more complicated in that the wheel should satisfy the two requirements that the wheel be tangent to the road and that the axle be directly above this point of tangency. These requirements lead to the differential equation the wheel should satisfy. –  J. M. Oct 20 '10 at 10:18
    
Yet another bonus: replacing the point $(0\qquad 1)^T$ with $(0\qquad 0)^T$ nets you the parametric equations of the cycloid. –  J. M. Oct 21 '10 at 4:27

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