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Elongated triangular orthobicupola

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Edges
  
36

Vertex configuration
  
6(3.4.3.4) 12(3.4)

Vertices
  
18

Symmetry group
  
D3h

Elongated triangular orthobicupola

Type
  
Johnson J34 - J35 - J36

Faces
  
2+6 triangles 2.3+6 squares

In geometry, the elongated triangular orthobicupola is one of the Johnson solids (J35). As the name suggests, it can be constructed by elongating a triangular orthobicupola (J27) by inserting a hexagonal prism between its two halves. The resulting solid is superficially similar to the rhombicuboctahedron (one of the Archimedean solids), with the difference that it has threefold rotational symmetry about its axis instead of fourfold symmetry.

Contents

A Johnson solid is one of 92 strictly convex polyhedra that have regular faces but are not uniform (that is, they are not Platonic solids, Archimedean solids, prisms or antiprisms). They were named by Norman Johnson, who first listed these polyhedra in 1966.

Volume

The volume of J35 can be calculated as follows:

J35 consists of 2 cupolae and hexagonal prism.

The two cupolae makes 1 cuboctahedron = 8 tetrahedra + 6 half-octahedra. 1 octahedron = 4 tetrahedra, so total we have 20 tetrahedra.

What is the volume of a tetrahedron? Construct a tetrahedron having vertices in common with alternate vertices of a cube (of side 1 2 , if tetrahedron has unit edges). The 4 triangular pyramids left if the tetrahedron is removed from the cube form half an octahedron = 2 tetrahedra. So

V t e t r a h e d r o n = 1 3 V c u b e = 1 3 1 2 3 = 2 12

The hexagonal prism is more straightforward. The hexagon has area 6 3 4 , so

V p r i s m = 3 3 2

Finally

V J 35 = 20 V t e t r a h e d r o n + V p r i s m = 5 2 3 + 3 3 2

numerical value:

V J 35 = 4.9550988153084743549606507192748

The elongated triangular orthobicupola forms space-filling honeycombs with tetrahedra and square pyramids.

References

Elongated triangular orthobicupola Wikipedia