User:Padex/hypercupola

Hypercupola

I've found that, in 3 dimensions, cupolas are formed by an "expansion" of pyramids.

So, in 4D, I've found 4 hypercupolas:

Hypercupolas
	Tetrahedral cupola		Cubic cupola		Octahedral cupola		Dodecahedral cupola
Type	Convex prismatoidal polychoron		Convex prismatoidal polychoron		Convex prismatoidal polychoron		Convex prismatoidal polychoron
Vertices	16		32		30		80
Edges	42		84		84		210
Faces	42	24 triangles 18 squares	80	32 triangles 48 squares	82	40 triangles 42 squares	194	80 triangles 90 squares 24 pentagons
Cells	16	1 tetrahedron 4 triangular prisms 6 triangular prisms 4 tetrahedra 1 cuboctahedron	28	1 cube 6 cubes 12 triangular prisms 8 tetrahedra 1 rhombicuboctahedron	28	1 octahedron 8 triangular prisms 12 triangular prisms 6 square pyramids 1 rhombicuboctahedron	64	1 dodecahedron 12 pentagonal prisms 30 triangular prisms 20 tetrahedra 1 rhombicosidodecahedron

They're composed of a {p,q} (all of the regular polyhedra, excepted the icosahedron) and a t0,2{p,q} (the cantellated polyhedron) linked by prisms and pyramids.

Cartesian coordinates

Tetrahedral cupola:

For the tetrahedral top:

(0, 0, √(6)/4, √(10)/4);

(±1/2, -1/(2√3), -√(2)/(4√3), √(5)/(2√2));

( 0, 1/√(3), -√2/(4√3), √(5)/(2√2));

For the cuboctahedral base:

the hexagon:

(±1, 0, 0, 0)

(±1/2, ±√(3)/2, 0, 0)

the triangles:

n°1

(±1/2, 1/(2√3), √(2/3), 0)

(0, -1/√3, √(2/3), 0)

n°2

(±1/2, -1/(2√3), -√(2/3), 0)

(0, 1/√3, -√(2/3), 0)

Cubic cupola:

(±1/2, ±1/2, ±1/2, τ);

(±1/2, ±1/2, ± (1/2 + τ), 0);

(±1/2, ± (1/2 + τ), ±1/2, 0);

(±(1/2 + τ), ±1/2, ±1/2, 0);

where τ = √2/2

Octahedral cupola:

( 0, 0 , ±τ, 1/2);

(0, ±τ, 0, 1/2);

(±τ, 0, 0, 1/2);

(±1/2, ±1/2, ± (1/2 + τ), 0);

(±1/2, ± (1/2 + τ), ±1/2, 0);

(± (1/2 + τ), ±1/2, ±1/2, 0);

where τ = √2/2