[vtkusers] Re: Volume of a tetrahedron
anast.jm at pg.com
anast.jm at pg.com
Wed Jun 6 13:59:12 EDT 2001
The definition I have (CRC Standard Math Tables and Formulae) states volume of a
tetrahedron
with one vertex at the origin and the other vertices at x1y1z1 x2y2z2
x3y3z3
| x1 y1 z1 |
v= 1/6 | x2 y2 z2 |
| x3 y3 z3 |
so it seems to me the W matrix in Peter's simplified equation below should be
W=[p2-p1, p3-p1, p4-p1] so as to translate all points relative to p1 as the
origin.
I'm a little rusty so perhaps I missed something???
...john
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From: Peter.Vanroose at esat.kuleuven.ac.be@public.kitware.com on 06/06/2001 06:25
PM ZE2
Peter.Vanroose at esat.kuleuven.ac.be@public.kitware.com To: vtkusers at public.kitware.com
Cc: ferrant at tele.ucl.ac.be (bcc: John
Anast-JM/PGI)
Sent by: vtkusers-admin at public.kitware.com Subject: [vtkusers] Re: Volume of a
06/06/2001 12:25 PM tetrahedron
Pearu Peterson wrote:
> If a tetrahedron is defined by its vertices p1,p2,p3,p4, then its volume is
>
> V = sqrt(abs(determinant(W * W^T)))/2
>
> where W is matrix defined as
>
> W = [p2-p1,p3-p2,p4-p3]
>
> and p1,p2,p3,p4 are column vectors.
If I'm not mistaken, the denominator should be 6 instead of 2.
(Correct me if not; I just checked this for a simple case.)
Also, I think you can avoid taking sqrt, since determinant(W * W^T)
is just the square of determinant(W), which simplifies the expression to:
V = abs(determinant(W)) / 6.
and you may of course also take p1 etc as row vectors instead of as column
vectors.
-- Peter Vanroose
Electrotechnical Department (ESAT/PSI)
K.U.Leuven, Belgium.
Peter.Vanroose at esat.kuleuven.ac.be
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