Toroidal blend of 20 triangular hebesphenorotundae

Toroidal blend of 20 triangular hebesphenorotundae
(3D model) (OFF file)
Rank	3
Type	Stewart toroid
Notation
Coxeter diagram	oxFx3xfox5fovo&#zxt
Elements
Faces	20+60+60+120 triangles, 60 pentagons, 20 hexagons
Edges	30+30+60+60+60+120+120+120
Vertices	60+60+60+60
Vertex figures	60 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	${\frac {150+70{\sqrt {5}}}{3}}\approx 102.17492$
Dihedral angles	3–3 at J92 join off of digonal-symmetry axis: $180^{\circ }+\arccos \left(-{\frac {\sqrt {5}}{3}}\right)\approx 318.18969^{\circ }$
	3–3 at J92 join on digonal-symmetry axis: $180^{\circ }+\arccos \left({\frac {\sqrt {5}}{3}}\right)\approx 221.81031^{\circ }$
	6–6: $180^{\circ }+\arccos \left({\frac {\sqrt {5}}{3}}\right)\approx 221.81031^{\circ }$
	3–5 rotundaic: $\arccos \left(-{\sqrt {\frac {5+2{\sqrt {5}}}{15}}}\right)\approx 142.62263^{\circ }$
	3–3 near J92 "tips": $\arccos \left(-{\frac {\sqrt {5}}{3}}\right)\approx 138.18969^{\circ }$
	3–6: $\arccos \left(-{\frac {\sqrt {5}}{3}}\right)\approx 138.18969^{\circ }$
	3–5 near J92 "tips": $\arccos \left(-{\sqrt {\frac {5-2{\sqrt {5}}}{15}}}\right)\approx 100.81237^{\circ }$
Central density	0
Related polytopes
Convex hull	Semi-uniform Tid, edge lengths 1 (triangles) and (1+√5)/2 (between dipentagons)
Abstract & topological properties
Flag count	2400
Euler characteristic	–20
Orientable	Yes
Genus	11
Properties
Symmetry	H3, order 120
Convex	No
Nature	Wild

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:

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

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.

• 

  The three "lunes" (arrangements of a square with triangles joining opposite edges of it) of a triangular hebesphenorotunda, and the congruent lunes in the small rhombicosidodecahedron.

• 

  Twenty small rhombicosidodecahedra can be outer-blended together in the same way as this toroid's parts.

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.

• 

  Plus 30 bilunabirotundae. This is related to a potential blend of the original non-enlarged toroid and 30 icosahedra. Also, note how the enlarged toroid can be blended with a great rhombicosidodecahedron placed in the center.

• 

  Plus 60 dodecahedra and 20 tridiminished icosahedra. Some coincident edges of the dodecahedra violate the diamond property here by being part of 4 faces.

• 

  Plus 60 more bilunabirotundae and 60 more tridiminished icosahedra. This takes care of the non-dyadicity. To fill in the gaps at the trigonal- and digonal-symmetry axes, one could use a polyhedron containing 36°-rhombus faces, and a polyhedron containing faces that are the inner blend of a pentagon and a golden rhombus, respectively.

• 

  The enlarged toroid will also admit 12 pentagonal rotundae. This specific blend puts 4 faces at some edges unless the aforementioned great rhombicosidodecahedron is also blended in.

External links

• Klitzing, Richard. "oxFx3xfox5fovo&#zxt".