Altered disnub icosahedron

The altered disnub icosahedron, addasi, or compound of twenty octahedra with rotational freedom is a uniform polyhedron compound. It consists of 40+120 triangles, with 4 triangles joining at each vertex.

This compound has rotational freedom, represented by an angle θ. The 20 octahedra can each be thought of as triangular antiprisms, and we rotate them in pairs in opposite directions. This compound goes through the following phases as θ increases:


 * θ = 0°: Double-cover of the snub icosahedron
 * 0° < θ < 14.33033°: General phase sometimes known as the outer disnub icosahedron or oddasi
 * θ = $$\arccos\left(\sqrt{\frac{-1+3\sqrt5+3\sqrt{10\sqrt5-22}}{8}}\right) ≈ 14.33033°$$: Vertices coincide by pairs to form the disnub icosahedron
 * 14.33033° < θ < 37.76124°: General phase sometimes known as the inner disnub icosahedron or iddasi
 * θ = $$\arccos\left(\frac{\sqrt{10}}{4}\right) ≈ 37.76124°$$: Octahedra coincide by four, forming a quadruple-cover of the small icosicosahedron
 * 37.76124° < θ < 60°: General phase known as the great disnub icosahedron or giddasi
 * θ = 60°: Double-cover of the great snub icosahedron

Vertex coordinates
The vertices of an altered disnub icosahedron of edge length 1 and rotation angle θ are given by all even permutations of:
 * $$\left(±\frac{\sqrt3\sin(\theta)}{3},\,±\frac{\sqrt{10}-\sqrt2+2(1+\sqrt5)\cos(\theta)}{12},\,±\frac{\sqrt2+\sqrt{10}-2(\sqrt5-1)\cos(\theta)}{12}\right),$$
 * $$\left(±\frac{2\sqrt2-(3-\sqrt5)\cos(\theta)+(\sqrt{15}-\sqrt3)\sin(\theta)}{12},\,±\frac{\sqrt2+\sqrt5\cos(\theta)+\sqrt3\sin(\theta)}{6},\,±\frac{2\sqrt2+(3-\sqrt5)\cos(\theta)-(\sqrt3+\sqrt{15})\sin(\theta)}{12}\right),$$
 * $$\left(±\frac{\sqrt{10}-\sqrt2-(1+\sqrt5)\cos(\theta)-(\sqrt3+\sqrt{15})\sin(\theta)}{12},\,±\frac{\sqrt2+\sqrt{10}+(\sqrt5-1)\cos(\theta)+(\sqrt{15}-\sqrt3)\sin(\theta)}{12},\,±\frac{3\cos(\theta)-\sqrt3\sin(\theta)}{6}\right),$$
 * $$\left(±\frac{\sqrt2-\sqrt{10}+(1+\sqrt5)\cos(\theta)-(\sqrt3+\sqrt{15})\sin(\theta)}{12},\,±\frac{\sqrt2+\sqrt{10}+(\sqrt5-1)\cos(\theta)-(\sqrt{15}-\sqrt3)\sin(\theta)}{12},\,±\frac{3\cos(\theta)+\sqrt3\sin(\theta)}{6}\right),$$
 * $$\left(±\frac{-2\sqrt2+(3+\sqrt5)\cos(\theta)+(\sqrt{15}-\sqrt3)\sin(\theta)}{12},\,±\frac{\sqrt2+\sqrt5\cos(\theta)-\sqrt3\sin(\theta)}{6},\,±\frac{2\sqrt2+(3-\sqrt5)\cos(\theta)+(\sqrt3+\sqrt{15})\sin(\theta)}{12}\right).$$