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User:Padex/hypercupola

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Hypercupola

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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

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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