# Compound of four triangular prisms

(Redirected from Rhomboctahedron)
Compound of four triangular prisms
Rank3
TypeUniform
Notation
Bowers style acronymRo
Elements
Components4 triangular prisms
Faces8 triangles, 12 squares
Edges12+24
Vertices24
Vertex figureIsosceles triangle, edge length 1, 2, 2
Measures (edge length 1)
Circumradius${\displaystyle {\frac {\sqrt {21}}{6}}\approx 0.76376}$
Volume${\displaystyle {\sqrt {3}}\approx 1.73205}$
Dihedral angles4–4: 60°
4–3: 90°
Central density4
Number of external pieces56
Level of complexity26
Related polytopes
ArmyNon-uniform Snic, edge lengths ${\displaystyle {\sqrt {\frac {2-{\sqrt {2}}}{3}}}}$ (squares), ${\displaystyle {\sqrt {\frac {3-{\sqrt {2}}}{3}}}}$ (equilateral triangles), ${\displaystyle {\sqrt {\frac {4-2{\sqrt {2}}}{3}}}}$ (between scalene triangles)
RegimentRo
DualCompound of four triangular tegums
ConjugateCompound of four triangular prisms
Convex coreRhombic dodecahedron
Abstract & topological properties
Flag count144
OrientableYes
Properties
SymmetryB3+, order 24
ConvexNo
NatureTame

The rhomboctahedron, ro, or compound of four triangular prisms is a uniform polyhedron compound. It consists of 12 squares and 8 triangles, with one triangle and two squares joining at a vertex.

Its quotient prismatic equivalent is the triangular prismatic tetrahedroorthowedge, which is six-dimensional.

## Vertex coordinates

The vertices of a rhomboctahedron of edge length 1 are given by all even sign changes and even permutations, plus all odd sign changes and odd permutatoins, of:

• ${\displaystyle \left({\frac {{\sqrt {3}}+{\sqrt {6}}}{6}},\,{\frac {{\sqrt {6}}-{\sqrt {3}}}{6}},\,{\frac {\sqrt {3}}{6}}\right)}$.

## Related polyhedra

This compound is chiral. The compound of the two enantiomorphs is the disrhomboctahedron.