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Subject: Basic perspective/normal q



I'm still struggling with this basic 3d stuff!
I have a 3d rotation matrix, a set of planes, and a couple of normals
associated with these planes. To draw a plane I rotate the associated normal
and test if its z is positive (right-handed AFAIK). If so, the plane is
visible and should be drawn.
Drawing it requires a perspective transform, but that might render it
*invisible* when it should be visible -- and the other way around! Imagine
looking into an open box -- logically, both left and right inside walls
should be visible. But the normal transform tags one side visible and the
other one invisible. This calculation has its own logic (the opposite walls
have opposite normals), so I know must be the perspective transform that
messes up the (valid!) normals transformation.
Somehow the next step escapes me. Surely the perspective transform leaves
the origial z worthless, and the remaining x and y can't be used for
visibility tests?

[Jongware]



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