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Barycentric coordinates of circumcenter of tetrahedron

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#1 eppo   Members   

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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.



#2 apatriarca   Members   

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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). 



#3 eppo   Members   

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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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