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Topic: Projects for Self Schoolers
Replies: 1   Last Post: Nov 16, 2008 4:46 PM

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Kirby Urner

Posts: 4,713
Registered: 12/6/04
Re: Projects for Self Schoolers
Posted: Nov 16, 2008 4:46 PM
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> 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

OK, so I delayed a little, giving ya'll more time to

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


Published calcs (recast using MathCad):

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