Imagine that I have a truncated icosahedron consisting of 60 identical vertices, each of degree $deg(v) = 3$, and fixed edge length $L$. I'd like to assign some constant curvature or bending angle $\theta$ to each edge s.t. I can deform the truncated icosahedron into its circumscribing sphere.
As a function of the edge length $L$, what value of $\theta$ allows me to properly perform this deformation?
Fix $L=1$. Then the radius of the circumscribed sphere is $r=\frac{1}{4} \sqrt{58+18 \sqrt{5}} \approx 2.478$. Now look at the isosceles triangle formed by the center of the sphere and one edge. It has sides of length $r$ and base length 1. So the angles at either end of the base are $\cos^{-1} (1/(2r)) \approx 78.3593^\circ$. The angle between the tangent to the sphere at one endpoint and the edge is then about $11.6407^\circ$.
