I haven't been following most of this discussion, but there's an easy picture for this: On Aug 26, 2008, at 9:22 PM, James Propp wrote:
Gene Salamin submitted a detailed algebraic solution, culminating in the conclusion that the rotation angle is 2 pi sqrt(1 + r^2). But isn't there a purely geometrical solution, where you somehow unroll the motion of the sphere to see a right triangle with legs of length 2 pi and 2 pi r, sort of like the way we measure the length of a geodesic on a cylinder by unrolling it, but different?
Here's a start towards such a picture: Stick a spit through the sphere that pierces it at the antipodal points (r+1,0,1) and (r-1,0,1) so that one end of the spit is at (0,0,1). Now spin the spit around (0,0,1), giving a little bit of a twirl as you do. (Think of the way you would have an ox walk in a circle so as to grind wheat at the center of the circle.) When the rolling sphere has completed one circuit, the spit is still going through it, so we can see that computing the axis of net rotation is easy. But how do we see what the rotation angle is?
Jim
For any circle on the sphere, just look at the cone tangent to the sphere along that circle. If you roll the sphere along the circle, it's the same as rolling the cone on this circle. The cone develops into the plane, so you see that the circle unrolls to a circle in the plane whose radius is its distance to the apex of the cone. This geometric picture is better than the algebraic formula, but of course the formula is immediate. The net angle is given by ratio of arc lengths, i.e. 2 pi * (radius of circle in space)/(distance to apex of cone). ---- Also, once I did (and recommend) the very easy computer experiment of drawing computer-generated Brownian curves equivalent to Brownian rolling of a sphere. One way to do it is take a Brownian path in the plane, then transform by using the same X-coordinates but use the Y- coordinate as a slope. When you see the picture it becomes intuitively clear. This is equivalent to Brownian rolling of a surface because all contact structures are locally isomorphic --- the only variable is the metric on the contact plane fields, which doesn't change the small- scale picture. Bill