Hexagonal double gyroantiprismoid

The hexagonal double antiprismoid is a convex isogonal polychoron and the third member of the double antiprismoids that consists of 24 hexagonal antiprisms, 144 tetragonal disphenoids and 288 sphenoids obtained as the convex hull of two orthogonal 6-6 duoantiprisms. However, it cannot be made uniform.

Vertex coordinates
The vertices of a hexagonal double antiprismoid, assuming that the hexagonal antiprisms are regular of edge length 1, centered at the origin, are given by:
 * (0, ±1, 0, ±$\sqrt{1+√3}$)
 * (0, ±1, ±$\sqrt{3+3√3}$/2, ±$\sqrt{1+√3}$/2)
 * (±$\sqrt{3}$/2, ±1/2, 0, ±$\sqrt{1+√3}$)
 * (±$\sqrt{3}$/2, ±1/2, ±$\sqrt{3+3√3}$/2, ±$\sqrt{1+√3}$/2)
 * (±1, 0, ±$\sqrt{1+√3}$, 0)
 * (±1, 0, ±$\sqrt{1+√3}$/2, ±$\sqrt{3+3√3}$/2)
 * (±1/2, ±$\sqrt{3}$/2, ±$\sqrt{1+√3}$, 0)
 * (±1/2, ±$\sqrt{3}$/2, ±$\sqrt{1+√3}$/2, ±$\sqrt{3+3√3}$/2)
 * (0, ±$\sqrt{1+√3}$, 0, ±1)
 * (0, ±$\sqrt{1+√3}$, ±$\sqrt{3}$/2, ±1/2)
 * (±$\sqrt{3+3√3}$/2, ±$\sqrt{1+√3}$/2, 0, ±1)
 * (±$\sqrt{3+3√3}$/2, ±$\sqrt{1+√3}$/2, ±$\sqrt{3}$/2, ±1/2)
 * (±$\sqrt{1+√3}$, 0, ±1, 0)
 * (±$\sqrt{1+√3}$, 0, ±1/2, ±$\sqrt{3}$/2)
 * (±$\sqrt{1+√3}$/2, ±$\sqrt{3+3√3}$/2, ±1, 0)
 * (±$\sqrt{1+√3}$/2, ±$\sqrt{3+3√3}$/2, ±1/2, ±$\sqrt{3}$/2)