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***bo198214*** · (This post was last modified: 2007-02-22, 11:23 by bo198214.)

quickfur Wrote:But regarding polyhedral volumes... isn't it just a matter of dissecting the polyhedron into right pyramids and summing them up? This isn't too hard to do from the vertex coordinates.

But that dissection can not be really expressed in a formula (rather in an algorithm). To illustrate, consider the following.

Out of curiosity, I once computed the area of a triangle given by the length of its sides (you know it is determined by that data) and got the following (seemingly unknown) formula:

$\frac{1}{4} \sqrt{2(a^2b^2+b^2c^2+a^2c^2) - a^4 - b^4 - c^4}$

(Ok, I see thats not exactly what I asked before, but nevertheless the topic is general enough)

It seems for me that every simplex is exactly described by the set of its edge lengths. So how is the general formula for the n-simplex?

***bo198214*** · (This post was last modified: 2007-02-22, 11:31 by bo198214.)

Oh, I already found the answer by googeling, it seems quite a complicated formula involving tensors, described in this paper which was not published until 1996!

For a tetrahedron the direct formula is:
$144V^2 = \\(a^2+d^2)(-a^2d^2+b^2e^2 + c^2f^2)+\\ (b^2+e^2)(a^2d^2-b^2e^2+c^2f^2) +\\ (c^2+f^2)(a^2d^2+b^2e^2-c^2f^2)-\\a^2b^2c^2-a^2e^2f^2 - b^2d^2f^2 - c^2d^2e^2$

So what is the formula for the 4-simplex? If I counted right, the n-simlex should have $n+1$ vertices and so $\begin{pmatrix}{n+1}\\{2}\end{pmatrix}$ sides, so it should be a formula in $\begin{pmatrix}{5}\\{2}\end{pmatrix}=10$ variables.

quickfur · 2007-02-23, 05:39

Wow. It's amazing how the answers to the simplest questions are often extremely complicated.

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