Deforming a truncated icosahedron into its circumscribing sphere 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?
 A: 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 $11.6407^\circ$.

          

A: In general the circumscribed radius ($R$) of a truncated icosahedron, having 12 congruent regular pentagonal faces & 20 congruent regular hexagonal faces each with edge length $L$, is given by the generalized expression (derived in Mathematical analysis of truncated icosahedron by HCR) $$\bbox[4pt, border: 1px solid blue;]{\color{blue}{R}=\color{red}{\frac{L\sqrt{58+18\sqrt{5}}}{4}}\color{purple}{\approx 2.478018659\space L}}$$ 
Now, join one of end-points & the mid-point of the edge to the center of truncated icosahedron, we get a right triangle with base $\color{blue}{\frac{L}{2}}$ & hypotenuse $\color{blue}{\frac{L\sqrt{58+18\sqrt{5}}}{4}}$ 
Hence, the angle $\theta$ between the edge & the radius is given as $$\cos \theta=\frac{\frac{L}{2}}{\frac{L\sqrt{58+18\sqrt{5}}}{4}}=\frac{2}{\sqrt{58+18\sqrt{5}}}=\sqrt{\frac{29-9\sqrt{5}}{218}}$$ $$\implies \color{blue}{ \theta=\cos^{-1}\sqrt{\frac{29-9\sqrt{5}}{218}}\approx 78.35927686^{o}}$$
