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# Barycentric coordinates of circumcenter of tetrahedron

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2 replies to this topic

### #1eppo  Members

Posted 21 January 2014 - 09:40 AM

I'm looking for a way to express the center of a tetrahedron's circumcircle as a set of barycentric coordinates. The way I do it now is to first calculate the center coordinates and then in a second step convert those to their bc-equivalent.

On this blog the author seems to use a more efficient method: http://realtimecollisiondetection.net/blog/?p=20

I can't quite figure out how to translate that to the tetrahedral case though.

Thanks.

### #2apatriarca  Members

Posted 22 January 2014 - 08:10 AM

The steps in the derivation are very similar to the steps in the triangle case and you should get:

Dot(B-A,B-A)*s + Dot(B-A,C-A)*t + Dot(B-A,D-A)*u = (1/2)*Dot(B-A,B-A)
Dot(C-A,B-A)*s + Dot(C-A,C-A)*t + Dot(C-A,D-A)*u = (1/2)*Dot(C-A,C-A)

Dot(D-A,B-A)*s + Dot(D-A,C-A)*t + Dot(D-A,D-A)*u = (1/2)*Dot(D-A,D-A)

which you can then solve using Cramer's rule (or something else).

### #3eppo  Members

Posted 22 January 2014 - 08:39 AM

The steps in the derivation are very similar to the steps in the triangle case and you should get:

Yeah, I worked it out. Thanks.

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