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Re: rational polygons
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Re: rational polygons - 01-31-2008, 02:26 PM
Quote:
Originally Posted by mkirkpatrick
snowflakes are all different in shape,although there does appear a certain symmetry with them.
You are correct. It is the hexagonal symmetry I'm interested in.

Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(t)·r(t)=c²
  
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