# Bitetrahedral diacositetracontachoron

Bitetrahedral diacositetracontachoron
File:Bitetrahedral diacositetracontachoron.png
Rank4
TypeIsogonal
Notation
Bowers style acronymBittid
Elements
Cells1440 phyllic disphenoids, 720 rhombic disphenoids
Faces2880 scalene triangles, 1440 isosceles triangles
Edges480+480+1440
Vertices240
Vertex figureChiral ditriakis tetrahedron
Measures (based on 2 hexacosichora of edge length 1)
Edge lengths12-valence lacing (480: ${\displaystyle {\frac {\sqrt {12-7{\sqrt {2}}+4{\sqrt {5}}-3{\sqrt {10}}}}{2}}\approx 0.62408}$
Edges of hexacosichora (1440): 1
6-valence lacing (480): ${\displaystyle {\frac {\sqrt {12-5{\sqrt {2}}+4{\sqrt {5}}-3{\sqrt {10}}}}{2}}\approx 1.04718}$
Circumradius${\displaystyle {\frac {1+{\sqrt {5}}}{2}}\approx 1.61803}$
Central density1
Related polytopes
ArmyBittid
RegimentBittid
DualDitruncated-tetrahedral diacositetracontachoron
Abstract & topological properties
Euler characteristic0
OrientableYes
Properties
SymmetryB3●H3, order 2880
ConvexYes
NatureTame

The bitetrahedral diacositetracontachoron or bittid, also known as the octafold icosidodecaswirlchoron, is a convex isogonal polychoron that consists of 720 rhombic disphenoids and 1440 phyllic disphenoids. 12 rhombic disphenoids and 24 phyllic disphenoids join at each vertex. However, it cannot be made uniform. It is the second in a series of isogonal icosidodecahedral swirlchora.

It can be formed as the convex hull of two hexacosichora, oriented such that the hull of 2 icositetrachoric sets of 24 vertices from each is a bitetracontoctachoron.

The ratio between the longest and shortest edges is ${\displaystyle 1:{\frac {\sqrt {54+6{\sqrt {37+8{\sqrt {10}}}}}}{6}}\approx 1:1.67794}$.

## Vertex coordinates

Vertex coordinates for a bitetrahedral diacositetracontachoron, created from the vertices of a hexacosichoron of edge length 1, are given by all permutations of:

• ${\displaystyle \left(0,\,0,\,0,\,\pm {\frac {1+{\sqrt {5}}}{2}}\right),}$
• ${\displaystyle \left(\pm {\frac {1+{\sqrt {5}}}{4}},\,\pm {\frac {1+{\sqrt {5}}}{4}},\,\pm {\frac {1+{\sqrt {5}}}{4}},\,\pm {\frac {1+{\sqrt {5}}}{4}}\right),}$
• ${\displaystyle \left(0,\,0,\,\pm {\frac {{\sqrt {2}}+{\sqrt {10}}}{4}},\,\pm {\frac {{\sqrt {2}}+{\sqrt {10}}}{4}}\right),}$

as well as all even permutations of:

• ${\displaystyle \left(0,\,\pm {\frac {1}{2}},\,\pm {\frac {1+{\sqrt {5}}}{4}},\,\pm {\frac {3+{\sqrt {5}}}{4}}\right),}$

as well as all permutations and even sign changes of:

• ${\displaystyle \left({\frac {\sqrt {2}}{4}},\,{\frac {\sqrt {2}}{4}},\,{\frac {\sqrt {2}}{4}},\,{\frac {2{\sqrt {2}}+{\sqrt {10}}}{4}}\right),}$
• ${\displaystyle \left({\frac {{\sqrt {10}}-{\sqrt {2}}}{8}},\,{\frac {3{\sqrt {2}}+{\sqrt {10}}}{8}},\,{\frac {3{\sqrt {2}}+{\sqrt {10}}}{8}},\,{\frac {3{\sqrt {2}}+{\sqrt {10}}}{8}}\right),}$

as well as all permutations and odd sign changes of:

• ${\displaystyle \left({\frac {{\sqrt {2}}+{\sqrt {10}}}{8}},\,{\frac {{\sqrt {2}}+{\sqrt {10}}}{8}},\,{\frac {{\sqrt {2}}+{\sqrt {10}}}{8}},\,{\frac {5{\sqrt {2}}+{\sqrt {10}}}{8}}\right).}$

## Isogonal derivatives

Substitution by vertices of these following elements will produce these convex isogonal polychora: