Question about square-wheeled cars 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?
 A: 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.
A: 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.
A: 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 $(t\qquad 0)$.
