# Compound of two snub cubes

(Redirected from Disnub cuboctahedron)
Compound of two snub cubes Rank3
TypeUniform
SpaceSpherical
Notation
Bowers style acronymDisco
Coxeter diagramß4ß3ß (     )
Elements
Components2 snub cubes
Faces48 triangles, 16 triangles as 8 hexagrams, 12 squares as 6 stellated octagons
Edges24+48+48
Vertices48
Vertex figureFloret pentagon, edge lengths 1, 1, 1, 1, 2
Measures (edge length 1)
Volume≈ 15.77896
Dihedral angles3–3: ≈ 153.23459°
4–3: ≈ 142.98343°
Central density2
Number of external pieces160
Level of complexity26
Related polytopes
ArmySemi-uniform Girco
RegimentDisco
DualCompound of two pentagonal icositetrahedra
ConjugateCompound of two snub cubes
Convex coreCuboctatruncated disdyakis dodecahedron
Abstract & topological properties
Flag count480
OrientableYes
Properties
SymmetryB3, order 48
ConvexNo
NatureTame

The disnub cuboctahedron, disco, or compound of two snub cubes is a uniform polyhedron compound. It consists of 48 snub triangles, 16 further triangles, and 12 squares (the latter two can combine in pairs due to faces in the same plane). Four triangles and one square join at each vertex.

Its quotient prismatic equivalent is the snub cubic antiprism, which is four-dimensional.

## Measures

The circumradius R ≈ 1.34371 of the disnub cuboctahedron with unit edge length is the largest real root of

$32x^6-80x^4+44x^2-7.$ Its volume V ≈ 15.77896 is given by the largest real root of

$729x^6-182736x^4+310176x^2-798848$ .

Its dihedral angles can be given as acos(α) for the angle between two triangular faces, and acos(β) for the angle between a square face and a triangular face, where α ≈ –0.89286 equals the unique real root of

$27x^3-9x^2-15x+13,$ and β ≈ –0.79846 equals the unique negative real root of

$27x^6-99x^4+129x^2-49.$ ## Vertex coordinates

A disnub cuboctahedron of edge length 1 has coordinates given by all permutations of:

• c1, ±c2, ±c3),

where

• $c_1=\sqrt{\frac{1}{12}\left(4-\sqrt{17+3\sqrt{33}}-\sqrt{17-3\sqrt{33}}\right)},$ • $c_2=\sqrt{\frac{1}{12}\left(2+\sqrt{17+3\sqrt{33}}+\sqrt{17-3\sqrt{33}}\right)},$ • $c_3=\sqrt{\frac{1}{12}\left(4+\sqrt{199+3\sqrt{33}}+\sqrt{199-3\sqrt{33}}\right)}.$ 