Elongated triangular orthobicupola

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Elongated triangular orthobicupola
Elongated triangular orthobicupola
Type Johnson
J34 - J35 - J36
Faces 2+6 triangles
2.3+6 squares
Edges 36
Vertices 18
Vertex configuration 6(3.4.3.4)
12(3.43)
Symmetry group D3h
Dual -
Properties convex

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.

The 92 Johnson solids were named and described by Norman Johnson in 1966.

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 \frac{1}{\sqrt{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_{tetrahedron} = \frac{1}{3} V_{cube} = \frac{1}{3} \frac{1}{{\sqrt{2}}^3} = \frac{\sqrt{2}}{12}


The hexagonal prism is more straightforward. The hexagon has area 6 \frac{\sqrt{3}}{4}, so

V_{prism} = \frac{3 \sqrt{3}}{2}

Finally

V_{J_{35}} = 20 V_{tetrahedron} + V_{prism} = \frac{5 \sqrt{2}}{3} + \frac{3 \sqrt{3}}{2}

numerical value:

V_{J_{35}} = 4.9550988153084743549606507192748

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