decomposing a 4x4 transformation matrix

If you are using DirectX, you can use D3DXMatrixDecompose or check the math here:decomposing-matrices

Thnx il look into it

I remember seeing this topic in one of the 5 graphics gems books.

Quote:

Original post by EternityZA how would i decompose this matrix back into the origanel 3 matrixes?

You can't get the original matrices back, but you can get equivalents. Given a matrix M:

a b c de f g hi j k lm n o pvec3 translate(m,n,o);float scaleX = length( vec3(a,b,c) )float scaleY = length( vec3(e,f,g) )float scaleZ = length( vec3(i,j,k) )vec3 tempZ = vec3(a,b,c).cross(vec3(e,f,g))if( tempZ.dot( vec3(i,j,k) ) < 0 ){  scaleX = -scaleX;  a = -a;   b = -b;   c = -c;}normalise(a,b,c);normalise(e,f,g);normalise(i,j,k);quat rotation;rotation.fromMatrix( M );

\edit fixed obvious flaw in the code....

[Edited by - RobTheBloke on March 27, 2009 5:45:35 AM]

Thnx!

EDIT: what does your length() function do?

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Original post by EternityZA EDIT: what does your length() function do?

calculates the length of a vector. The X,Y and Z axes of the matrix (i.e. [a,b,c], [e,f,g] and [i,j,k]) store the rotation and scale components of the matrix. The length of those axes give the scale, but you need to know if it's got a negative scale in an axis or not. That's what the tempZ and dot/cross product are for.

Great! i got it working.