On Tue, Aug 5, 2008 at 10:05 AM, James Propp <jpropp@cs.uml.edu> wrote:
Let me make sure I understand Mike's idea for the level-one approximation to the Cantor set: you roll the hoop so that the point of contact goes from (0,0) to (1/3,0), then swing the hoop down pivoting on the point (1/3,0) until the hoop touches the Cantor set at both (1/3,0) and (2/3,0) (so that the bottom point of the hoop is at (1/2,s) for some s < 0), then swing the hoop up pivoting on the point (2/3,0) until the hoop's lowest point is (2/3,0), and then roll the hoop so that the point of contact goes from (2/3,0) to (1,0).
Have I got that right?
Yes
It's a cute idea, but I don't see how it applies to the Koch curve problem.
If r is big enough then I claim that rolling the hoop over the Koch curve is identical to rolling it over the Cantor set (once if you go across the bottom, twice if you go across the top), since the hoop can't reach the concave parts of the fractal.
Can you elaborate?
The intersection of the complete Koch curve with the convex hull of the curve (a triangle with height sqrt(3)/6 and base 1) consists of three copies of the Cantor set, the one on the bottom and two scaled versions on top. I claim that if r is large enough, the rest of the fractal doesn't matter because the hoop can never touch it--the gaps in the Cantor set are too big to let it through. For smaller r, you chop it up into subproblems and rescale. But you only need to rescale a fininte number of times before you reach that base case I was describing. -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com