Elongated triangular orthobicupola

Infobox Polyhedron


Polyhedron_Type=Johnson
"J"₃₄ - J₃₅ - J₃₆
Face_List=2+6 triangles
2.3+6 squares
Edge_Count=36
Vertex_Count=18
Symmetry_Group="D"_3h|
Vertex_List=6(3.4.3.4)
12(3.4³)
Dual=-
Property_List=convex

In geometry, the elongated triangular orthobicupola is one of the Johnson solids ("J"₃₅). As the name suggests, it can be constructed by elongating a triangular orthobicupola ("J"₂₇) 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 "J"₃₅ can be calculated as follows:

"J"₃₅ 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 acube (of side $frac\{1\}\{sqrt\{2$, if tetrahedron has unit edges). The 4 triangularpyramids left if the tetrahedron is removed from the cube form half anoctahedron = 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:

$V\_\{J\_\{35\; =\; 4.9550988153084743549606507192748$

External links

* [http://mathworld.wolfram.com/JohnsonSolid.html Johnson Solid -- from MathWorld]