Platonic solids' dimensionless pattern

Guatama

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For a tetrahedron, volume=(2^.5/12)a^3, surface area=(3^.5)a^2, edge=a, corners=4

For a cube, volume =a^3, surface area=6a^2, edge=a, corners=8

For an octahedron, volume=(2^.5/3)a^3, surface area=2(3^.5)a^2, edge=a, corners=6

For a dodecahedron, volume=(7.663...)a^3, surface area=(20.646...)a^2, edge=a, corners=20

For an icosahedron, volume=(2.182...)a^3, surface area=(8.660...)a^2, edge=a, corners=12

volume/(surface area * edge * corners)=P

For a tetrahedron_____.01701...=P

For a cube_____.02083...=P

For an octahedron_____.02268...=P

For a dodecahedron_____.01856...=P

For an icosahedron_____.02100...=P

Can you explain this cluster of values given P for Platonic solid geometries?
 
For a tetrahedron, volume=(2^.5/12)a^3, surface area=(3^.5)a^2, edge=a, corners=4

For a cube, volume =a^3, surface area=6a^2, edge=a, corners=8

For an octahedron, volume=(2^.5/3)a^3, surface area=2(3^.5)a^2, edge=a, corners=6

For a dodecahedron, volume=(7.663...)a^3, surface area=(20.646...)a^2, edge=a, corners=20

For an icosahedron, volume=(2.182...)a^3, surface area=(8.660...)a^2, edge=a, corners=12

volume/(surface area * edge * corners)=P

For a tetrahedron_____.01701...=P

For a cube_____.02083...=P

For an octahedron_____.02268...=P

For a dodecahedron_____.01856...=P

For an icosahedron_____.02100...=P

Can you explain this cluster of values given P for Platonic solid geometries?
May be because their geometric properties are very similar i.e.

1. The vertices of Solid all lie on a sphere.


2. All the dihedral angles are equal.


3. All the vertex figures are regular polygons.


4. All the solid angles are equivalent.


5. All the vertices are surrounded by the same number of faces.
 
Fine qualitative argument regarding Platonic solids' definition in general. However, despite their characteristic volumes varying up to 65-fold, their surface areas 12-fold, and vertices 5-fold (given equal edges) the measure P remains relatively constant.
 
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