Gyroelongated square cupola

{{Infobox polytope The gyroelongated square cupola, or gyescu, is one of the 92 Johnson solids (J23). It consists of 4+4+4+8 triangles, 1+4 squares, and 1 octagon. It can be constructed by attaching an octagonal antiprism to the octagonal base of the square cupola.
 * type=CRF
 * img=Gyroelongated square cupola 2.png
 * off=Gyroelongated square cupola.off
 * 3d=J23 gyroelongated square cupola.stl
 * dim = 3
 * obsa = Gyescu
 * faces = 4+4+4+8 triangles, 1+4 squares, 1 octagon
 * edges = 4+4+4+8+8+8+8
 * vertices = 4+4+4+8
 * verf = 4 isosceles trapezoids, edge lengths 1, $\sqrt{2}$, $\sqrt{2}$, $\sqrt{2}$
 * verf2 = 8 irregular pentagons, edge lengths 1, 1, 1, 1, $\sqrt{2}$
 * verf3 = 8 isosceles trapezoids, edge lengths 1, 1, 1, $\sqrt{2+√2}$
 * coxeter = oxo8sox&#xt
 * army=Gyescu
 * reg=Gyescu
 * symmetry = BC2×I, order 8
 * volume = $$\frac{3+2\sqrt2+2\sqrt{4+2\sqrt2+2\sqrt{146+103\sqrt2}}}{3} ≈ 6.21077$$
 * dih= 3–3 antiprismatic: $$\arccos\left(\frac{1-2\sqrt{2+\sqrt2}}{3}\right) ≈ 153.96238°$$
 * dih2 = 3–3 join: $$\arccos\left(\frac{\sqrt3}{3}\right)+\arcos\left(-\sqrt{\frac{7+4\sqrt2-2\sqrt{20+14\sqrt2}}{3}}\right} ≈ 151.33013°$$
 * dih3 = 3–4 cupolaic: $$\arccos\left(-\frac{\sqrt6}{3}\right) ≈ 144.73561°$$
 * dih4 = 3–4 join: $$\arccos\left(\frac{\sqrt2}{2}\right)+\arccos\left(-\sqrt{\frac{7+4\sqrt2-2\sqrt{20+14\sqrt2}}}{3}}\right) ≈ 141.59451°$$
 * dih5 = 4–4: 135°
 * dih6 = 3–8: $$\arccos\left(-\sqrt{\frac{7+4\sqrt2-2\sqrt{20+14\sqrt2}}{3}}\right) ≈ 96.59451°$$
 * smm = Yes
 * dual = Tetradeltoctapentagonal hemitrapezohedron
 * conjugate = Gyroelongated retrograde square cupola
 * conv=Yes
 * orientable=Yes
 * nat=Tame}}

If a second cupola is attached to the other octagonal base of the antiprism, the result is the gyroelongated square bicupola.

Vertex coordinates
A gyroelongated square cupola of edge length 1 has the following vertices: where $$H=\sqrt{\frac{-2-2\sqrt2+\sqrt{20+14\sqrt2}}{8}}$$ is the distance between the octagonal antiprism's center and the center of one of its bases.
 * $$\left(±\frac12,\,±\frac12,\,\frac{\sqrt2+2H}{2}\right),$$
 * $$\left(±\frac12,\,±\frac{1+\sqrt2}2,\,H\right),$$
 * $$\left(±\frac{1+\sqrt2}2,\,±\frac12,\,H\right),$$
 * $$\left(0,\,±\sqrt{\frac{2+\sqrt2}2},\,-H\right),$$
 * $$\left(±\sqrt{\frac{2+\sqrt2}2},\,0,\,-H\right),$$
 * $$\left(±\frac{\sqrt{2+\sqrt2}}2,\,±\frac{\sqrt{2+\sqrt2}}2,\,-H\right),$$