Re: Water surface in hexahedron

Paul E. Black wrote:
On Sunday 26 November 2006 18:46, macio wrote:

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?

I don't think the inputs match up.

The irregular hexahedron is a planar figure, right?

How does the hexahedron relate to the volume of the water and the
normal? Do you mean a hexagonal prism outlined by the hexahedron? If
so, the volume is irrelevant.

-paul-

A hexahedron is a solid with six faces. A cube is a regular hexahedron.
I think you are confusing it with a hexagon.

Patricia
.

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