Rich's comment

... The sum of the edge lengths goes to infinity (a not-absolutely-convergent series), while the sum of the edge vectors is finite (in effect, a convergent series). ...

is quite right. Moreover, note that if you've got a fractal curve on a sphere of radius 1, no matter how wild it is, there's a unique way to "contact-transfer" the fractal to another sphere of radius 1 by rolling the second sphere along the first sphere so that the point of contact follows the fractal. The way to see this is to change to a coordinate system in which the point of contact and the plane of tangency between the two spheres stays fixed, and both spheres just rotate in place, with each doing the opposite of what the other one does. Both spheres will execute fractal jiggling, but when the jiggling ends, the second sphere has a mirror image of the fractal on the first sphere. So I believe that hoops are a red herring, and my remark

"How can you roll a ball along a Koch curve when you can't even roll a *hoop* along a Koch curve?" was a non sequitur (or rather, the implicit argument --- "A hoop is a subset of a ball, so if you can't roll the hoop you can't roll the ball" --- is a non sequitur), sort of on a par with "Since the volume of revolution obtained from revolving y = 1/x (with x greater than or equal to 1) about the x-axis has finite volume, you can fill it with paint, so you can paint it, so it has finite surface area".

The reason it's a non sequitur is that the way you roll a hoop is different from the way you roll a ball. This is easiest to see if you roll a hoop and a ball along a polygonal curve. When you come to a corner, you rotate the hoop by some positive angle, but you don't rotate the ball at all --- you just switch to rolling it along a different great circle (that is, along a different hoop).

Rolling a short distance multiplies this by another matrix I+A, where (the entries in) A are bounded by D, the distance rolled. The difference between (I+A)(I+B) and (I+B)(I+A) is AB-BA, which is bounded by twice the product of the two distances. Any change in contact point, or orientation, is also bounded by the product. For the snowflake, going down a level produces 4x as many edges of 1/3 the length. The difference between rolling along an edge, versus rolling along 4 smaller edges, is proportional to the square of the edge length. The effect of successive levels is multiplied by roughly 4/9 for each level, giving a convergent series.

This now looks right to me. Is everyone else as convinced as I am that you can contact-transfer the Koch curve to a sphere of any radius? Jim