Hexagonal double antiprismoid

The hexagonal double antiprismoid or hidiap is a convex isogonal polychoron and the fifth member of the double antiprismoid family. It consists of 24 hexagonal antiprisms, 144 tetragonal disphenoids, and 288 sphenoids. 2 hexagonal antiprisms, 4 tetragonal disphenoids, and 8 sphenoids join at each vertex. It can be obtained as the convex hull of two orthogonal hexagonal-hexagonal duoantiprisms or by alternating the dodecagonal ditetragoltriate. However, it cannot be made uniform. It is the first in an infinite family of isogonal hexagonal antiprismatic swirlchora.

Using the ratio method, the lowest possible ratio between the longest and shortest edges is 1:$$\sqrt{\frac{11+4\sqrt3-\sqrt{41+24\sqrt3}}{8}}$$ ≈ 1:1.05128. For this variant the edges of the hexagons of the inscribed duoantiprisms have ratio 1:$$\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{4}$$ ≈ 1:1.54171. A variant with uniform hexagonal antiprisms also exists; this variant is based on a duoantiprism with hexagons with edge length ratio 1:$$\sqrt{1+\sqrt3}$$ ≈ 1:1.65289.

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:
 * $$\left(0,\,±1,\,0,\,±\sqrt{1+\sqrt3}\right),$$
 * $$\left(0,\,±1,\,±\frac{\sqrt{3+3\sqrt3}}{2},\,±\frac{\sqrt{1+\sqrt3}}{2}\right),$$
 * $$\left(±\frac{\sqrt3}{2},\,±\frac12,\,0,\,±\sqrt{1+\sqrt3}\right),$$
 * $$\left(±\frac{\sqrt3}{2},\,±\frac12,\,±\frac{\sqrt{3+3\sqrt3}}{2},\,±\frac{\sqrt{1+\sqrt3}}{2}\right),$$
 * $$\left(±1,\,0,\,±\sqrt{1+\sqrt3},\,0\right),$$
 * $$\left(±1,\,0,\,±\frac{\sqrt{1+\sqrt3}}{2},\,±\frac{\sqrt{3+3\sqrt3}}{2}\right),$$
 * $$\left(±\frac12,\,±\frac{\sqrt3}{2},\,±\sqrt{1+\sqrt3},\,0\right),$$
 * $$\left(±\frac12,\,±\frac{\sqrt3}{2},\,±\frac{\sqrt{1+\sqrt3}}{2},\,±\frac{\sqrt{3+3\sqrt3}}{2}\right),$$
 * $$\left(±\sqrt{1+\sqrt3},\,0,\,0,\,±1\right),$$
 * $$\left(±\sqrt{1+\sqrt3},\,0,\,±\frac{\sqrt3}{2},\,±\frac12\right),$$
 * $$\left(±\frac{\sqrt{1+\sqrt3}}{2},\,±\frac{\sqrt{3+3\sqrt3}}{2},\,0,\,±1\right),$$
 * $$\left(±\frac{\sqrt{1+\sqrt3}}{2},\,±\frac{\sqrt{3+3\sqrt3}}{2},\,±\frac{\sqrt3}{2},\,±\frac12\right),$$
 * $$\left(0,\,±\sqrt{1+\sqrt3},\,±1,\,0\right),$$
 * $$\left(0,\,±\sqrt{1+\sqrt3},\,±\frac12,\,±\frac{\sqrt3}{2}\right),$$
 * $$\left(±\frac{\sqrt{3+3\sqrt3}}{2},\,±\frac{\sqrt{1+\sqrt3}}{2},\,±1,\,0\right),$$
 * $$\left(±\frac{\sqrt{3+3\sqrt3}}{2},\,±\frac{\sqrt{1+\sqrt3}}{2},\,±\frac12,\,±\frac{\sqrt3}{2}\right).$$

An alternate set of coordinates, assuming that the edge length differences are minimized, centered at the origin, are given by:
 * $$\left(0,\,±1,\,0,\,±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{4}\right),$$
 * $$\left(0,\,±1,\,±\frac{3+2\sqrt3+\sqrt{12\sqrt3-3}}{8},\,±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{8}\right),$$
 * $$\left(±\frac{\sqrt3}{2},\,±\frac12,\,0,\,±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{4}\right),$$
 * $$\left(±\frac{\sqrt3}{2},\,±\frac12,\,±\frac{3+2\sqrt3+\sqrt{12\sqrt3-3}}{8},\,±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{8}\right),$$
 * $$\left(±1,\,0,\,±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{4},\,0\right),$$
 * $$\left(±1,\,0,\,±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{8},\,±\frac{3+2\sqrt3+\sqrt{12\sqrt3-3}}{8}\right),$$
 * $$\left(±\frac12,\,±\frac{\sqrt3}{2},\,±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{4},\,0\right),$$
 * $$\left(±\frac12,\,±\frac{\sqrt3}{2},\,±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{8},\,±\frac{3+2\sqrt3+\sqrt{12\sqrt3-3}}{8}\right),$$
 * $$\left(±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{4},\,0,\,0,\,±1\right),$$
 * $$\left(±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{4},\,0,\,±\frac{\sqrt3}{2},\,±\frac12\right),$$
 * $$\left(±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{8},\,±\frac{3+2\sqrt3+\sqrt{12\sqrt3-3}}{8},\,0,\,±1\right),$$
 * $$\left(±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{8},\,±\frac{3+2\sqrt3+\sqrt{12\sqrt3-3}}{8},\,±\frac{\sqrt3}{2},\,±\frac12\right),$$
 * $$\left(0,\,±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{4},\,±1,\,0\right),$$
 * $$\left(0,\,±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{4},\,±\frac12,\,±\frac{\sqrt3}{2}\right),$$
 * $$\left(±\frac{3+2\sqrt3+\sqrt{12\sqrt3-3}}{8},\,±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{8},\,±1,\,0\right),$$
 * $$\left(±\frac{3+2\sqrt3+\sqrt{12\sqrt3-3}}{8},\,±\frac{2+\sqrt3+\sqrt{4\sqrt3-1}}{8},\,±\frac12,\,±\frac{\sqrt3}{2}\right).$$