Re: Water surface in hexahedron

I think hexahedron is a convex object. In that case the simplest
way is to tetrahedralize it and add the volumes of those
tetrahedrons.

Volume of a tetrahedron with 4 veritces (x_1, y_1, z_1), ..
(x_4, y_4, z_4) can be evaluated by computing the
determinant :


|1 x_1 y_1 z_1|
|1 x_2 y_2 z_2|
|1 x_3 y_3 z_3|
|1 x_4 y_4 z_4|

But if the body is not convex then you may have to add
sieiner points to tetrahedralize the object.

Thanks.

--- Pinaki
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macio wrote:
I think it is rather obvious situation but also very heavy task.

Given are the vertices of the irregular hexahedron, the volume of water
which is contained in it and the vector normal to the surface of the
water (flat surface).
How to compute the area of the water surface?

Maybe someone knows answer?

.