I can derive the Pythagoreom Theorem for ya... If you do derive those great functions, let me know what those labels really mean. That''ll mean an interesting post!

-Lewis [m80]
Play QUADz MX @
www.m80produxions.com

quote:

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U and N are normalized.

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My bad, you don''t want to normalize U. I was misreading some sphere mapping documentation.

Actually, please disregard everything I said about U This is what happens when I rely on documentation in a rush... take a look at this:


U is defined as the Displacement vector, or (Intersection - Point) in the drawing. This means that you have to know where the ray and the mirror will intersect.

Thanx man... cuz to be honest with you, I couldn''t figure out what you gave me. One thing with me is that I have to know how things work before I use them. I actually am doing something very similar to that figure you posted. I''m using the paramatic form of the line equation to figure out the point of contact. I do have a question for you though, since I am new to this forum. How in the world do you upload those pix and HMTL forming withing the replies? Is this BOLD?

-Lewis [m80]
Play QUADz MX @
www.m80produxions.com

Ignore my last HTML question... sleep can me a person dumb.

-Lewis [m80]
Play QUADz MX @
www.m80produxions.com

And to get the matrix representing a reflection in an arbitar plane through orgio..

M = I - 2* |n><n|

I = identifymatrix
n = plane normal (normalized)

<n| = rowvector
|n> = colum vector

|n><n| = transpose(n1 n2 n3) * (n1 n2 n3)


so

1 0 0 n1*n1 n1*n2 n1*n3
M = 0 1 0 - 2* n2*n1 n2*n2 n2*n3
0 0 1 n3*n1 n3*n2 n3*n3


If you want to reflect through a plane not through origo, you need to bring it in to origo first..

That is performed by translating an arbitary point on the plane to origo, then perform the reflection, and translate back...

One more try... The blanks in the matrix formula didn''t appear..


| 1 0 0 n1*n1 n1*n2 n1*n3
|M = 0 1 0 - 2* n2*n1 n2*n2 n2*n3
| 0 0 1 n3*n1 n3*n2 n3*n3