plane and normal equations
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Ok can someone confirm the following to make sure I understand it.
I have a triangle, with three vectors V1, V2, V3 each with [x,y,z]
to get the normal, I first calculate
[V1-V2] * [V2-V3]
then get the length of the normal
= A * B
= (AyBz - ByAz)(BxAz - AxBz)(AxBy - BxAy)
= (x) (y) (z)
now we get the sqrt root of the length
= sqrt((x*x) + (y*y) + (z*z));
then we divide each coordinate of our length (x,y,z) by the sqr root of the length....
= (x/sqrt) (y/sqrt) (z/sqrt)
NOW we have the normal ......
thats part 1
PART 2 (oooo evil drumroll into a frenetic blast beat thanks to TRYM)
we have three points in space and the evil plane equation
The standard equation of a plane in 3 space is
Ax + By + Cz + D = 0
The normal to the plane is the vector (A,B,C).
using the three vertices -
A = y1 (z2 - z3) + y2 (z3 - z1) + y3 (z1 - z2)
B = z1 (x2 - x3) + z2 (x3 - x1) + z3 (x1 - x2)
C = x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)
- D = x1 (y2 z3 - y3 z2) + x2 (y3 z1 - y1 z3) + x3 (y1 z2 - y2 z1)
The sign of s = Ax + By + Cz + D determines which side of some random point (the x,y,z) lies with respect to the plane. If s > 0 then the point lies on the same side as the normal (A,B,C). If s < 0 then it lies on the opposite side, if s = 0 then the point (x,y,z) lies on the plane.
Is this correct ??
the plane equation is
A(random point.x)+B(random point.y)+C(random point.z)+D = 0
CHEERS
Im just a beginner!!!
http://www.actsofgord.com
Looks like you''ve nailed them all correctly. Here is a tip for simpler notation for part 2: Instead of storing A,B,C,D, store two vectors. A, a point on the plane, and n, the normal (of unit length).
Then, the signed distance (s in your post) of an arbitrary point B to this plane is (A-B) dot n. If you give the vectors appropriate coordinates and plug them in to this equation, you can verify it.
Then, the signed distance (s in your post) of an arbitrary point B to this plane is (A-B) dot n. If you give the vectors appropriate coordinates and plug them in to this equation, you can verify it.
quote:Original post by fatherjohn666
A = y1 (z2 - z3) + y2 (z3 - z1) + y3 (z1 - z2)
B = z1 (x2 - x3) + z2 (x3 - x1) + z3 (x1 - x2)
C = x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)
- D = x1 (y2 z3 - y3 z2) + x2 (y3 z1 - y1 z3) + x3 (y1 z2 - y2 z1)
this is the only part that i didn''t understood. Exept this, what i suppose to be correct (if you''ve calculated it right), all is correct.
But please, don''t nail down everything to the ground but stay with a little higher math level. Proceed like a waterfall, make functions for vector multiplication and don''t brake everything in pieces.
All your stuff can be resumed to:
normal = A x B
normal(a b c) => plane: ax + by + cz = -d
where -d can be found by replacing x,y,z by the coordinates of one of the triangle''s point.
Like this, there is no confusions, all stays simple & easy. ;-)
cheers
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