Compound of eight triangular prisms

(Redirected from Disrhomboctahedron)
Compound of eight triangular prisms
Rank3
TypeUniform
Notation
Bowers style acronymDro
Elements
Components8 triangular prisms
Faces16 triangles as 8 hexagrams, 24 squares as 12 stellated octagons
Edges24+48
Vertices48
Vertex figureIsosceles triangle, edge length 1, 2, 2
Measures (edge length 1)
Circumradius${\displaystyle {\frac {\sqrt {21}}{6}}\approx 0.76376}$
Volume${\displaystyle 2{\sqrt {3}}\approx 3.46140}$
Dihedral angles4–3: 90°
4–4: 60°
Central density8
Number of external pieces152
Level of complexity26
Related polytopes
ArmySemi-uniform Girco, edge lengths ${\displaystyle {\frac {{\sqrt {6}}-{\sqrt {3}}}{3}}}$ (octagons), ${\displaystyle {\frac {\sqrt {3}}{3}}}$ (ditrigon-rectangle)
RegimentDro
DualCompound of eight triangular tegums
ConjugateCompound of eight triangular prisms
Convex coreRhombic dodecahedron
Abstract & topological properties
Flag count288
OrientableYes
Properties
SymmetryB3, order 48
Flag orbits6
ConvexNo
NatureTame

The disrhomboctahedron, dro, or compound of eight triangular prisms is a uniform polyhedron compound. It consists of 24 squares and 16 triangles (all of which pair up due to lying in the same plane), with one triangle and two squares joining at a vertex.

It is the result of combining the two chiral forms of the rhomboctahedron.

Its quotient prismatic equivalent is the triangular prismatic octaexoorthowedge, which is ten-dimensional.

Vertex coordinates

The vertices of a disrhomboctahedron of edge length 1 are given by all permutations of:

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