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I have a regular icosahedron. I'd like to evaluate a function of its vertices at all of their unique orientations. (Since there are infinitely many such orientations and this is a computational problem, I'll discretize in some way, probably by searching over rotational ranges with some step size.) This animation, for example, shows a partial set of orientations.

Let the icosahedron be orientated so that two of its vertices fall on the $z$-axis. The orientations of the icosahedron can then be described by angular displacements from this orientation, with respect to the vertex on the positive $z$-axis. Let $\phi$ be the polar angle, $\theta$ the azimuthal angle, and $\lambda$ the angle between the polar vertex and an arbitrarily chosen "equatorial vertex" (see spherical coordinates).

At the moment, I search over the following ranges:

$$ \begin{align*} \forall & \phi \in [0,63.4^\circ] \\ \forall & \theta \in [0,72^\circ] \\ \forall & \lambda \in [0,72^\circ] \\ &f(\phi,\theta,\lambda) \end{align*} $$

Per this article, $63.4^\circ$ degrees is the polar angle between a polar vertex and the nearest set of equatorial vertices, whereas $72^\circ$ is the angle between polar vertices. Given this, the above should search at least all orientations in which the polar vertex falls (after rotation) within one of the faces of the icosahedron.

But, upon inspection, some orientations are being left out while others seem to be heavily duplicated. Searching $\forall \theta \in [0,180^\circ]$ seems to capture all orientations, but seems to give greater duplication.

Is there a more effective method of enumerating the unique orientations of a regular icosahedron?

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  • $\begingroup$ I think that you should have a little idea about the icosahedron group (en.wikipedia.org/wiki/Icosahedral_symmetry) which should be easy as you say you have a background in physics, maybe in crystallography, $\endgroup$ – Jean Marie May 16 '17 at 18:58
  • $\begingroup$ I think that group is mappings of vertices to vertices, which doesn't say much about free rotation. $\endgroup$ – Richard May 16 '17 at 20:40

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