Tuesday, February 16, 2010
Kirby Urner: in the calcs below, the assumption is always tetravolumes, meaning we've calibrated to the concentric hierarchy with the volume 20 cuboctahedron.
The icosahedron with that same edge, of volume ~18.51 is the icosahedron of volume 5 Φ2 √2 mentioned below.
The pentagonal dodecahedron is its partner in the rhombic triacontahedron, inside of which one finds these unit-edge cubes (but not of unit volume in this schema).
Icosahedron edge = 1 in tetra volumes (blue)
= 5 Φ2 √2
Pentagonal dodecahedron edge Φ-1 or .618034 (orange)
= (Φ2 + 1) 3√2
Rhombic triacontahedron diagonals 1 and Φ-1 (red)
Cube edge = 1 (purple)
Posted by Kirby Urner at 12:14 AM