Bill Thurston writes:
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).
Thanks!
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.
As the slope of what line? I gather that this is a description of some rational map from R^2 to R^3, but I don't know which.
When you see the picture it becomes intuitively clear.
Is this written up somewhere (with the picture included)? Jim Propp