# Toroidal blend of 20 triangular hebesphenorotundae

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Toroidal blend of 20 triangular hebesphenorotundae
Rank3
TypeStewart toroid
Notation
Coxeter diagramoxFx3xfox5fovo&#zxt
Elements
Faces20+60+60+120 triangles, 60 pentagons, 20 hexagons
Edges30+30+60+60+60+120+120+120
Vertices60+60+60+60
Vertex figures60 rectangles, edge lengths 1 and (1+5)/2
60 isosceles trapezoids, edge lengths 1, 1, 1, (1+5)/2
60 (3.3.5)2
60 (34.62)
Measures (edge length 1)
Volume${\displaystyle {\frac {150+70{\sqrt {5}}}{3}}\approx 102.17492}$
Dihedral angles3–3 at J92 join off of digonal-symmetry axis: ${\displaystyle 180^{\circ }+\arccos \left(-{\frac {\sqrt {5}}{3}}\right)\approx 318.18969^{\circ }}$
3–3 at J92 join on digonal-symmetry axis: ${\displaystyle 180^{\circ }+\arccos \left({\frac {\sqrt {5}}{3}}\right)\approx 221.81031^{\circ }}$
6–6: ${\displaystyle 180^{\circ }+\arccos \left({\frac {\sqrt {5}}{3}}\right)\approx 221.81031^{\circ }}$
3–5 rotundaic: ${\displaystyle \arccos \left(-{\sqrt {\frac {5+2{\sqrt {5}}}{15}}}\right)\approx 142.62263^{\circ }}$
3–3 near J92 "tips": ${\displaystyle \arccos \left(-{\frac {\sqrt {5}}{3}}\right)\approx 138.18969^{\circ }}$
3–6: ${\displaystyle \arccos \left(-{\frac {\sqrt {5}}{3}}\right)\approx 138.18969^{\circ }}$
3–5 near J92 "tips": ${\displaystyle \arccos \left(-{\sqrt {\frac {5-2{\sqrt {5}}}{15}}}\right)\approx 100.81237^{\circ }}$
Central density0
Related polytopes
Convex hullSemi-uniform Tid, edge lengths 1 (triangles) and (1+5)/2 (between dipentagons)
Abstract & topological properties
Flag count2400
Euler characteristic–20
OrientableYes
Genus11
Properties
SymmetryH3, order 120
Flag orbits20
ConvexNo
NatureWild

The toroidal blend of 20 triangular hebesphenorotundae is a Stewart toroid. It can be obtained by outer-blending twenty triangular hebesphenorotundae together at their square faces, leaving no squares behind. It has these squares, as well as pentagons and star pentambi, as pseudo-faces.

## Vertex coordinates

The vertices of a toroidal blend of 20 triangular hebesphenorotundae, centered at the origin and with unit edge length, are given by all even permutations of:

• ${\displaystyle \left(0,\,\pm {\frac {1}{2}},\,\pm 3{\frac {1+{\sqrt {5}}}{4}}\right)}$,
• ${\displaystyle \left(\pm {\frac {1}{2}},\,\pm {\frac {5+{\sqrt {5}}}{4}},\,\pm {\frac {1+{\sqrt {5}}}{2}}\right)}$,
• ${\displaystyle \left(\pm {\frac {1+{\sqrt {5}}}{4}},\,\pm 1,\,\pm {\frac {2+{\sqrt {5}}}{2}}\right)}$,
• ${\displaystyle \left(\pm {\frac {{\sqrt {5}}-1}{4}},\,\pm {\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {5+3{\sqrt {5}}}{4}}\right)}$,
• ${\displaystyle \left(0,\,\pm {\frac {2+{\sqrt {5}}}{2}},\,\pm 3{\frac {1+{\sqrt {5}}}{4}}\right)}$,
• ${\displaystyle \left(\pm {\frac {1}{2}},\,\pm {\frac {5+{\sqrt {5}}}{4}},\,\pm {\frac {3+{\sqrt {5}}}{2}}\right)}$,
• ${\displaystyle \left(0,\,\pm {\frac {1}{2}},\,\pm {\frac {7+3{\sqrt {5}}}{4}}\right),}$
• ${\displaystyle \left(\pm {\frac {3+{\sqrt {5}}}{2}},\,\pm {\frac {2+{\sqrt {5}}}{2}},\,\pm {\frac {1+{\sqrt {5}}}{4}}\right)}$,
• ${\displaystyle \left(\pm {\frac {5+3{\sqrt {5}}}{4}},\,\pm {\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {3+{\sqrt {5}}}{4}}\right)}$,
• ${\displaystyle \left(\pm {\frac {1+{\sqrt {5}}}{4}},\,0,\,\pm {\frac {3+2{\sqrt {5}}}{2}}\right)}$,
• ${\displaystyle \left(\pm {\frac {1+{\sqrt {5}}}{2}},\,\pm {\frac {1}{2}},\,\pm {\frac {7+3{\sqrt {5}}}{4}}\right)}$,
• ${\displaystyle \left(\pm {\frac {2+{\sqrt {5}}}{2}},\,\pm {\frac {5+{\sqrt {5}}}{4}},\,\pm {\frac {3+{\sqrt {5}}}{2}}\right)}$.

## Relations

Twice the 4–6 dihedral angle of a triangular hebesphenorotunda, plus the 6–6 dihedral angle of a truncated icosahedron, is 360°. This can be taken as an explanation of why the toroid forms: it is possible to blend the toroid with a truncated icosahedron placed in the center. Doing this will remove all tunnels (making the genus 0) and hexagonal faces.

The triangular hebesphenorotunda's relation to the small rhombicosidodecahedron can also provide some insight on the formation of the toroid. The two polyhedra have certain arrangements of faces in common.

If one inserts 30 cubes in between the triangular hebesphenorotundae, the resulting toroid will admit many outer-blends in outward directions that contribute to almost-complete space-fillings, like a less-strict version of an aperiodic tiling. The gaps in these partial space-fillings can take the form of polyhedra with irregular but equilateral faces such as rhombi.