Date: Wed, 4 Aug 1993 22:56:24 -1000
From: salanne@convex.csc.FI (Simo Salanne)
Message-Id: <1993Aug5.085624.24356@nic.funet.fi>
Organization: Finnish Academic and Research Network Project - FUNET
Subject: Soccer Ball Size - once more
Back to the basic geometry...
I already removed sqrt from my 1st solution, but it's
still wrong. I did my analysis without any math handbooks,
and calculated the hexagon/pentagon area wrong, as Ed Sarlls
in his private email to me pointed out.
For practical purposes Andrew's "5 times" is close enough.
Did we needed this proof?
Here's the whole derivation (by Ed):
>From CRC Standard Math Tables, 25th ed., Geometry Mensuration Formulas
>From p. 142: REGULAR POLYGONS
s = length of side, K = area
Polygon K
Pentagon 1.72048 (s^2)
Hexagon 2.59808 (s^2)
So area of 20 hexagons + 12 pentagons is:
A = 20 * 2.59808 (s^2) + 12 * 1.72048 (s^2)
A = 72.60736 (s^2)
>From p. 148: SPHERICAL FIGURES
D = diameter of sphere, S = surface area
S = pi * (D^2)
So to get the diameter of a soccer ball:
pi * (D^2) = 72.60736 (s^2)
(D^2) = (s^2) * 23.11164
D = s * 4.80746
Sanity check:
3" * 4.8 = 14.4", about 14" quoted by Marty above.
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Ed Sarlls, III (713) 964-6176 Western Geophysical Exploration Products
sarlls@wg2.waii.com WG2-1012 PO Box 2469 Houston, TX, 77252
Any opinions expressed above are mine and are not Western Geophysical's.
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