> Verify that if this sphere has a radius of > phi/sqrt(2), that this shape has a volume of 7.5, > relative to our basic 'water cup' (above). > This is not such an easy problem, answer tomorrow > (you don't have to look). > > Kirby Urner > 4dsolutions.net
OK, so I delayed a little, giving ya'll more time to scribble.
The approach I take is to start with some calculations already published, taking us to a rhombic triacontrahedron of tetravolume 5, and then apply a simple rule, covered by state standards (e.g. California's): expanding or contracting linear dimensions by factor x, e.g. elongating all edges of a shape by factor of 2, results in areal dimension expanding as x * x, and volume as x * x * x, i.e. area and volume go up by factors of 4 and 8 respectively, consequent to linear doubling.
Ergo, a volume expansion from 5.0 to 7.5 i.e. a 1.5 scale increase in volume (150%), will occur as a consequence of scaling linear dimensions by a third root of 1.5, ergo if the radius obtained from published calculations (late 1970s), multiplied by this factor, gives a new radius of phi/sqrt(2), then we've done the necessary verification.
Published calcs for volume 5 rhombic triacontahedron peg its radius at 3rdroot((sqrt(2) * (2 + sqrt(5))/6) -- see link below.
Multiplying that by 3rdroot(3/2) does in fact give the right answer of phi/sqrt(2), simplifications left as an exercise, or check this scanned image file of the calcs: