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# Re: [FT] [computer] 3D math

From: Nyrath the nearly wise <nyrath@c...>
Date: Thu, 07 Sep 2000 19:31:16 -0400
Subject: Re: [FT] [computer] 3D math
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"Barclay, Tom" wrote:
> - in 3 space
> - a sphere we shall call S1 located at coorinates (x1,y1,z1) of radius
r1.
> - a sphere we shall call S2 located at coorinates (x2,y2,z2) of radius
r2.
> - a point we shall call P3 located at coordinates (x3,y3,z3).
>
> B) What percentage of the volume of S2 is contained within S1? What
formula
> will determine this?

Got it.  The answer was at
http://mathworld.wolfram.com/Sphere-SphereIntersection.html

pi(R+r-d)^2 (d^2 + 2dr - 3r^2 + 2dR + 6rR - 3R^2)
V =  -------------------------------------------------
12d

where:
V = volume of the intersection of two spheres
pi = 3.14159265359......
R = radius of sphere A = r1
r = radius of sphere B = r2
d = distance between the centers of the two spheres
d = sqrt( (x1 - x2)^2 + (y1 - y2)^2 + (z1 - z2)^2 )

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