Gyroelongated pentagonal cupolarotunda

Gyroelongated pentagonal cupolarotunda
(3D model) (OFF file)
Rank	3
Type	CRF
Notation
Bowers style acronym	Gyepcuro
Elements
Faces	7×5 triangles 5 squares 1+1+5 pentagons
Edges	16×5
Vertices	7×5
Vertex figures	10 rectangles, edge lengths 1 and (1+√5)/2
	20 irregular pentagons, edge lengths 1, 1, 1, 1, (1+√5)/2
	5 isosceles trapezoids, edge lengths 1, √2, (1+√5)/2, √2
Measures (edge length 1)
Volume	$5{\frac {11+5{\sqrt {5}}+2{\sqrt {2-2{\sqrt {5}}+2{\sqrt {650+290{\sqrt {5}}}}}}}{12}}\approx 15.99110$
Dihedral angles	3–3 rotundaic join: $\arccos \left({\sqrt {\frac {5-2{\sqrt {5}}}{15}}}\right)+\arccos \left(-{\sqrt {\frac {11+4{\sqrt {5}}-2{\sqrt {50+22{\sqrt {5}}}}}{3}}}\right)\approx 174.43432^{\circ }$
	3–3 antiprismatic: $\arccos \left({\frac {1-{\sqrt {10+2{\sqrt {5}}}}}{3}}\right)\approx 159.18651^{\circ }$
	3–4 cupolaic: $\arccos \left(-{\frac {{\sqrt {3}}+{\sqrt {15}}}{6}}\right)\approx 159.09484^{\circ }$
	3–5 join: $\arccos \left({\frac {\sqrt {5}}{5}}\right)+\arccos \left(-{\sqrt {\frac {11+4{\sqrt {5}}-2{\sqrt {50+22{\sqrt {5}}}}}{3}}}\right)\approx 158.68159^{\circ }$
	4–5: $\arccos \left(-{\sqrt {\frac {5+{\sqrt {5}}}{10}}}\right)\approx 148.28253^{\circ }$
	3–5 rotundaic: $\arccos \left(-{\sqrt {\frac {5+2{\sqrt {5}}}{15}}}\right)\approx 142.62263^{\circ }$
	3–3 cupolaic join: $\arccos \left({\sqrt {\frac {5+2{\sqrt {5}}}{15}}}\right)+\arccos \left(-{\sqrt {\frac {11+4{\sqrt {5}}-2{\sqrt {50+22{\sqrt {5}}}}}{3}}}\right)\approx 132.62401^{\circ }$
	3–4 join: $\arccos \left(-{\sqrt {\frac {5+{\sqrt {5}}}{10}}}\right)+\arccos \left(-{\sqrt {\frac {11+4{\sqrt {5}}-2{\sqrt {50+22{\sqrt {5}}}}}{3}}}\right)\approx 126.96412^{\circ }$
Central density	1
Number of external pieces	47
Level of complexity	64
Related polytopes
Army	Gyepcuro
Regiment	Gyepcuro
Dual	Pentadeltodecapentapentagopentarhombirhombic triacontapentahedron
Conjugate	Gyroelongated retrograde pentagrammic cupolarotunda
Abstract & topological properties
Flag count	320
Euler characteristic	2
Surface	Sphere
Orientable	Yes
Genus	0
Properties
Symmetry	H2+×I, order 5
Flag orbits	64
Convex	Yes
Nature	Tame

The gyroelongated pentagonal cupolarotunda (OBSA: gyepcuro) is one of the 92 Johnson solids (J₄₇). It consists of 7×5 triangles, 5 squares, and 1+1+5 pentagons. It can be constructed by attaching a pentagonal cupola and a pentagonal rotunda to opposite bases of the decagonal antiprism.

It is one of five Johnson solids to be chiral.

Vertex coordinates

A gyroelongated pentagonal cupolarotunda of edge length 1 has the following vertices:

• $\left(\pm {\frac {1}{2}},\,-{\sqrt {\frac {5+2{\sqrt {5}}}{20}}},\,{\sqrt {\frac {5+2{\sqrt {5}}}{5}}}+H\right)$,
• $\left(\pm {\frac {1+{\sqrt {5}}}{4}},\,{\sqrt {\frac {5-{\sqrt {5}}}{40}}},\,{\sqrt {\frac {5+2{\sqrt {5}}}{5}}}+H\right)$,
• $\left(0,\,{\sqrt {\frac {5+{\sqrt {5}}}{10}}},\,{\sqrt {\frac {5+2{\sqrt {5}}}{5}}}+H\right)$,
• $\left(\pm {\frac {1+{\sqrt {5}}}{4}},\,{\sqrt {\frac {25+11{\sqrt {5}}}{40}}},\,{\sqrt {\frac {5+{\sqrt {5}}}{10}}}+H\right)$,
• $\left(\pm {\frac {3+{\sqrt {5}}}{4}},\,-{\sqrt {\frac {5+{\sqrt {5}}}{40}}},\,{\sqrt {\frac {5+{\sqrt {5}}}{10}}}+H\right)$,
• $\left(0,\,-{\sqrt {\frac {5+2{\sqrt {5}}}{5}}},\,{\sqrt {\frac {5+{\sqrt {5}}}{10}}}+H\right)$,
• $\left(\pm {\frac {1}{2}},\,\pm {\frac {\sqrt {5+2{\sqrt {5}}}}{2}},\,H\right)$,
• $\left(\pm {\frac {3+{\sqrt {5}}}{4}},\,\pm {\sqrt {\frac {5+{\sqrt {5}}}{8}}},\,H\right)$,
• $\left(\pm {\frac {1+{\sqrt {5}}}{2}},\,0,\,H\right)$,
• $\left(\pm {\frac {\sqrt {5+2{\sqrt {5}}}}{2}},\,\pm {\frac {1}{2}},\,-H\right)$,
• $\left(\pm {\sqrt {\frac {5+{\sqrt {5}}}{8}}},\,\pm {\frac {3+{\sqrt {5}}}{4}},\,-H\right)$,
• $\left(0,\,\pm {\frac {1+{\sqrt {5}}}{2}},\,-H\right)$,
• $\left(-{\sqrt {\frac {5+2{\sqrt {5}}}{20}}},\,\pm {\frac {1}{2}},\,-{\sqrt {\frac {5-{\sqrt {5}}}{10}}}+h\right)$,
• $\left({\sqrt {\frac {5-{\sqrt {5}}}{40}}},\,\pm {\frac {1+{\sqrt {5}}}{4}},\,{\sqrt {\frac {5-{\sqrt {5}}}{10}}}+H\right)$,
• $\left({\sqrt {\frac {5+{\sqrt {5}}}{10}}},\,0,\,{\sqrt {\frac {5-{\sqrt {5}}}{10}}}+H\right)$.

where $H={\sqrt {\frac {-4-2{\sqrt {5}}+{\sqrt {50+22{\sqrt {5}}}}}{8}}}$is the distance between the decagonal antiprism's center and the center of one of its bases.

External links

• Klitzing, Richard. "gyepcuro".
• Quickfur. "The Gyroelongated Pentagonal Cupolarotunda".

• Wikipedia contributors. "Gyroelongated pentagonal cupolarotunda".
• McCooey, David. "Gyroelongated Pentagonal Cupolarotunda"