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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?

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I don't understand how you're using the "bending angle" to deform the icosahedron. Are you looking for the angle $\theta$ that each edge makes with the circumscribing sphere? – anon Jul 17 '11 at 23:49
@anon, sorry for the confusion! No I'm looking for the bending angle that places each edge on the surface of the sphere. – R.H. Jul 18 '11 at 3:04
Oh, you mean the angle formed by the arc which results from a radial projection of an edge onto the circumscribing sphere. – anon Jul 19 '11 at 12:15
up vote 2 down vote accepted

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 $11.6407^\circ$.
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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 $$\cos \theta=\frac{\frac{L}{2}}{\frac{L\sqrt{58+18\sqrt{5}}}{4}}=\frac{2}{\sqrt{58+18\sqrt{5}}}=\sqrt{\frac{29-9\sqrt{5}}{218}}$$ $$\implies \color{blue}{ \theta=\cos^{-1}\sqrt{\frac{29-9\sqrt{5}}{218}}\approx 78.35927686^{o}}$$

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