I realized last night that what I called a "no-slip" condition should properly be called a "no-spin" condition. Think of a billiard ball that's been hit off-center: it travels in a straight line with a spin. This point of contact with the table does not trace out a great circle, but some other sort of curve (can anyone tell me what family of curves on a sphere arises in this situation?). This is *not* the sort of thing we want when we're rolling a ball over a fractal curve. I now believe that if a nowhere-differentiable curve is sufficiently close to a straight line, there's a well-defined way to roll a ball along it. Specifically, use a variant of the Koch construction with free parameters epsilon_1, epsilon_2, epsilon_3, ..., where at the nth stage of the construction, all existing segments get little equilateral bumps on them, where the ratio of the size of the bump to the size of the segment is epsilon_n. I think one can show that if the epsilons go to 0 sufficiently fast, there's a well-defined limiting behavior if one rolls a ball along a polygonal approximation to the curve of mesh delta and sends delta to 0. And I'm fairly sure that this modified Koch construction gives a nowhere-differentiable curve, though not 100% positive. However, I'm fairly sure such a curve would have fractal dimension 1, and in fact have finite length. Indeed, I find Josh's argument

This seems easy - since the limit of the length is infinite, you could only get a fixed point as you go from level n to n+1 if there's some very nice (rational) relationship between the size of the ball and the length of the level n snowflake.

In other words, I think the answer to all your questions are "no, it doesn't make sense to talk about rolling a ball on a fractal".

quite compelling. At first I found Rich's argument

The difference between rolling the ball along one edge of the polygonal path, and the 4 smaller edges of the next iteration, should be O(edge^2). Even after multiplying by the number of edges, this is O(edge). This sum converges. So we'd expect the sphere-rolling-along-snowflake-path to converge.

quite compelling too, which put me in the position of holding two contradictory beliefs! But I think what Rich's argument misses is the non-commutativity of the situation. Imagine one ball whose point of contact with the plane travels from (0,0) to (epsilon,epsilon) to (2*epsilon,0) and on to (N,0), and a second ball that rolls directly from (0,0) to (N,0). Even if epsilon is minuscule, the diference in orientation of the two balls can be sizeable when they arrive at (N,0), provided N is big enough, since the ball that goes from (0,0) to (2*epsilon,0) directly is already oriented differently from the ball that goes via (epsilon,epsilon), and this tiny difference gets magnified as the balls continue to roll. My current guess (I won't call it a belief yet) is that the class of rectifiable curves coincides with the class of curves that one can roll a ball along in a well-defined way. What do you all think? Jim Propp