Triangular double antiprismoid

The triangular double antiprismoid or tridiap is a convex isogonal polychoron and the second member of the double antiprismoid family. It consists of 12 triangular antiprisms, 36 tetragonal disphenoids, and 72 sphenoids. 2 triangular antiprisms, 4 tetragonal disphenoids, and 8 sphenoids join at each vertex. It can be obtained as the convex hull of two orthogonal triangular-triangular duoantiprisms or by alternating a hexagonal ditetragoltriate. However, it cannot be made uniform. It is the first in an infinite family of isogonal triangular antiprismatic swirlchora.

If the triangular antiprisms are regular octahedra, the triangular double antiprismoid can be vertex-inscribed into a bitetracontoctachoron. The edges of the base triangles of the 2 inscribed duoantiprisms have ratio $$1:\sqrt2 ≈ 1:1.41421$$.

Using the ratio method, the lowest possible ratio between the longest and shortest edges is 1:$$\sqrt{14+2\sqrt{17}}$$ ≈ 1:1.17915. For this variant, the ratio of the triangles of the duoantiprism is 1:$$\frac{3+\sqrt{17}}{4}$$ ≈ 1:1.78078.

Vertex coordinates
Coordinates for the vertices of a triangular double antiprismoid, assuming that the triangular antiprisms are regular octahedra of edge length 1, centered at the origin, are given by:
 * $$\left(0,\,\frac{\sqrt3}{3},\,0,\,\frac{\sqrt6}{3}\right),$$
 * $$\left(0,\,-\frac{\sqrt3}{3},\,0,\,-\frac{\sqrt6}{3}\right),$$
 * $$\left(0,\,\frac{\sqrt3}{3},\,±\frac{\sqrt2}{2},\,-\frac{\sqrt6}{6}\right),$$
 * $$\left(0,\,-\frac{\sqrt3}{3},\,±\frac{\sqrt2}{2},\,\frac{\sqrt6}{6}\right),$$
 * $$\left(±\frac12,\,-\frac{\sqrt3}{6},\,0,\,\frac{\sqrt6}{3}\right),$$
 * $$\left(±\frac12,\,\frac{\sqrt3}{6},\,0,\,-\frac{\sqrt6}{3}\right),$$
 * $$\left(±\frac12,\,\frac{\sqrt3}{6},\,±\frac{\sqrt2}{2},\,\frac{\sqrt6}{6}\right),$$
 * $$\left(±\frac12,\,-\frac{\sqrt3}{6},\,±\frac{\sqrt2}{2},\,-\frac{\sqrt6}{6}\right),$$
 * $$\left(0,\,\frac{\sqrt6}{3},\,0,\,-\frac{\sqrt3}{3}\right),$$
 * $$\left(0,\,-\frac{\sqrt6}{3},\,0,\,\frac{\sqrt3}{3}\right),$$
 * $$\left(0,\,\frac{\sqrt6}{3},\,±\frac12,\,\frac{\sqrt3}{6}\right),$$
 * $$\left(0,\,-\frac{\sqrt6}{3},\,±\frac12,\,-\frac{\sqrt3}{6}\right),$$
 * $$\left(±\frac{\sqrt2}{2},\,\frac{\sqrt6}{6},\,0,\,\frac{\sqrt3}{3}\right),$$
 * $$\left(±\frac{\sqrt2}{2},\,-\frac{\sqrt6}{6},\,0,\,-\frac{\sqrt3}{3}\right),$$
 * $$\left(±\frac{\sqrt2}{2},\,\frac{\sqrt6}{6},\,±\frac12,\,-\frac{\sqrt3}{6}\right),$$
 * $$\left(±\frac{\sqrt2}{2},\,-\frac{\sqrt6}{6},\,±\frac12,\,\frac{\sqrt3}{6}\right).$$

An alternate set of coordinates, assuming that the edge length differences are minimized, centered at the origin, are given by:
 * $$\left(0,\,\frac{\sqrt3}{3},\,0,\,\frac{3\sqrt3+\sqrt{51}}{12}\right),$$
 * $$\left(0,\,-\frac{\sqrt3}{3},\,0,\,-\frac{3\sqrt3+\sqrt{51}}{12}\right),$$
 * $$\left(0,\,\frac{\sqrt3}{3},\,±\frac{3+\sqrt{17}}{8},\,-\frac{3\sqrt3+\sqrt{51}}{24}\right),$$
 * $$\left(0,\,-\frac{\sqrt3}{3},\,±\frac{3+\sqrt{17}}{8},\,\frac{3\sqrt3+\sqrt{51}}{24}\right),$$
 * $$\left(±\frac12,\,-\frac{\sqrt3}{6},\,0,\,\frac{3\sqrt3+\sqrt{51}}{12}\right),$$
 * $$\left(±\frac12,\,\frac{\sqrt3}{6},\,0,\,-\frac{3\sqrt3+\sqrt{51}}{12}\right),$$
 * $$\left(±\frac12,\,\frac{\sqrt3}{6},\,±\frac{3+\sqrt{17}}{8},\,\frac{3\sqrt3+\sqrt{51}}{24}\right),$$
 * $$\left(±\frac12,\,-\frac{\sqrt3}{6},\,±\frac{3+\sqrt{17}}{8},\,-\frac{3\sqrt3+\sqrt{51}}{24}\right),$$
 * $$\left(0,\,\frac{3\sqrt3+\sqrt{51}}{12},\,0,\,-\frac{\sqrt3}{3}\right),$$
 * $$\left(0,\,-\frac{3\sqrt3+\sqrt{51}}{12},\,0,\,\frac{\sqrt3}{3}\right),$$
 * $$\left(0,\,\frac{3\sqrt3+\sqrt{51}}{12},\,±\frac12,\,\frac{\sqrt3}{6}\right),$$
 * $$\left(0,\,-\frac{3\sqrt3+\sqrt{51}}{12},\,±\frac12,\,-\frac{\sqrt3}{6}\right),$$
 * $$\left(±\frac{3+\sqrt{17}}{8},\,\frac{3\sqrt3+\sqrt{51}}{24},\,0,\,\frac{\sqrt3}{3}\right),$$
 * $$\left(±\frac{3+\sqrt{17}}{8},\,-\frac{3\sqrt3+\sqrt{51}}{24},\,0,\,-\frac{\sqrt3}{3}\right),$$
 * $$\left(±\frac{3+\sqrt{17}}{8},\,\frac{3\sqrt3+\sqrt{51}}{24},\,±\frac12,\,-\frac{\sqrt3}{6}\right),$$
 * $$\left(±\frac{3+\sqrt{17}}{8},\,-\frac{3\sqrt3+\sqrt{51}}{24},\,±\frac12,\,\frac{\sqrt3}{6}\right).$$