Re: Finding edges of a given polyhedra in 3D

macio wrote:
I will be more specific (forgive me for that lack of some data).

Tha whole task is to compute the area of the planar polygon created by
the cross-sesction of the polyhedron.

The polyhedron is convex. I prepared a simple picture which depicts the
situation - it's here:
http://img131.imageshack.us/img131/5088/polyhedronao0.gif

I know coordinates of all the vertices of the convex polyhedron and I
can easily compute the coordinates of the points (blue and yellow)
where all the line-sections (created by the pairs of vertices of that
polyhedron - diagonals and edges) intercept that plane. Then I can
compute the cross-section area knowing the coords of the blue points.
The yellow points are redundant. So I have to know which line-section
is the diagonal and which is the edge of the polyhedra to choose right
points to compute the area of that polygon.


I disagree with your last sentence. It seems better to me to get down to
two dimensions as soon as possible. Doesn't the following do what you need?

1. For each pair A,B of vertices of the polyhedron, find the
intersection of the line connecting A and B with the plane. If the
intersection is A, B, or anywhere between them, add the point to a list L.

[List L contains the yellow and blue points].

2. Rotate and translate the points in list L so that they are in
coordinates relative to the plane, and need only two coordinates.

3. Apply any 2D convex hull algorithm to the transformed points. The
result is an ordered list of the coordinates of the vertices of the
planar polygon whose area you want.

[The convex hull vertex list contains only the blue points. The yellow
points are interior points of the convex hull.]

Patricia
.

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