Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The idea comes from a recreational math problem-- Place two identical coins side by side and roll one along the circumference of another without slipping, how many revolutions will the rolling coin undergo before it returns to its original position? Counterintuitively the answer is 2.

Then I had this hunch that given two points $\mathbf{A}$ and $\mathbf{B}$ and a circle of radius 1, there might be a curve that connects the two points such that it would take the least rotations for the circle to roll from $\mathbf{A}$ to $\mathbf{B}$.

It wouldn't be a straight line as far as curvature is concerned.

As the first step that I think might be in the right direction, we have the curve

$\gamma =(x(t),y(t))$

Curvature of the curve

$\kappa=\frac{x'(t)y''(t)-y'(t)x''(t)}{(x'(t)^2+y'(t)^2)^{3/2}} $

The number of rotations

$\int \left( 1-\kappa \right) \sqrt{y'(t)^2+x'(t)^2}dx$=$\int \left( 1-\frac{x'(t)y''(t)-y'(t)x''(t)}{(x'(t)^2+y'(t)^2)^{3/2}} \right) \sqrt{y'(t)^2+x'(t)^2}dx$

Divided by $2\pi$

However, the Euler-Lagrange equation appears to give only one stationary point that is of curvature 0. I know there are other conditions that should be considered, e.g. The curvature at any point shouldn't be larger than 1, there shouldn't be intersection, etc.

And the problem starts to seem beyond reach, but I do have a hunch that there would be some other stationary functions, Any idea?

share|cite|improve this question
But isn't the total rotated angle the same as the length of the curve it rolls on? Why should the optimal curve not be a straight line segment? – Thomas Feb 7 '13 at 9:27
No, that's the tricky part, the rotated angle is the same as the length$*(1-r*\kappa)$, where r is the radius of the circle and $\kappa$ is the curvature of the curve it rolls on, so it rotates less when $\kappa$ is larger, yet not so large that $1-r*\kappa $ is negative, and it rotates more if it rolls on negative curvature, as shown in the rolling-coin brainteaser above. – Lafix Feb 7 '13 at 10:52
Oh, right, I see, sorry. Interesting question! – Thomas Feb 7 '13 at 11:08

This is not a complete answer, but maybe you can use optimal control theory and its minimum principle on that problem. I would suggest taking position and unit tangent as states and curvature $u$ as control, ranging in $]-\infty, 1]$. Your Lagrangian simplifies significantly to the integral of $1-u$. For just minimizing length, the bounded curvature case in three dimensions was treated by Héctor Sussmann in SHORTEST 3-DIMENSIONAL PATHS WITH A PRESCRIBED CURVATURE BOUND, maybe you can get some inspiration from that paper.

share|cite|improve this answer
It helps, I haven't figured out a solution yet though, thank you! – Lafix Feb 11 '13 at 15:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.