From: israel@math.ubc.ca (Robert Israel)
Subject: Re: Tetrahedron-circumscribed sphere
Date: 3 Sep 1999 23:01:04 GMT
Newsgroups: sci.math
To: simha@udri.udayton.edu (Hari Simha)
Keywords: Cayley-Menger determinant
In article ,
simha@udri.udayton.edu (Hari Simha) writes:
> I need a simple formula to obtain the radius of a sphere
> circumscribed around an arbitrary tetrahedron. I need
> this for some 3D finite element work. Please also point
> to a referenc. Thanks.
Given n points P_1 to P_n in R^(n-1), let d(i,j) be the distance
from P_i to P_j. The volume of the convex hull of P_1 ... P_n is
then (-1)^n D/(2^(n-1) ((n-1)!)^2), where D is the Cayley-Menger
determinant
| 0 1 1 ... 1 |
| 1 0 d(1,2)^2 ... d(1,n)^2 |
D = det | 1 d(2,1)^2 0 ... d(2,n)^2 |
| .................................. |
| 1 d(n,1)^2 d(n,2)^2 ... 0 |
Consider the case n=5, where P_1,...,P_4 are the vertices of your tetrahedron
and P_5 is the centre of the circumscribed sphere (all in R^3 but considered as
embedded in R^4). Since they are in a three-dimensional hyperplane in R^4
the 4-dimensional volume is 0. So an equation for the radius R in
terms of the distances d(i,j) between the vertices is D = 0 where d(i,5)=R,
i.e.
| 0 1 1 1 1 1 |
| 1 0 d(1,2)^2 d(1,3)^2 d(1,4)^2 R^2 |
det | 1 d(2,1)^2 0 d(2,3)^2 d(2,4)^2 R^2 | = 0
| 1 d(3,1)^2 d(3,2)^2 0 d(3,4)^2 R^2 |
| 1 d(4,1)^2 d(4,2)^2 d(4,3)^2 0 R^2 |
| 1 R^2 R^2 R^2 R^2 0 |
What you will get is thus a quadratic equation in R^2. The
coefficients are rather complicated if written out in full
symbolically.
Reference: L.M. Blumenthal, "Distance Geometry", Chelsea 1970.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2