Does anyone know of a fast way to tell if a matrix has non-uniform scaling?
Detect non-uniform scaling in matrix
You can probably get the length of the basis vectors of the matrix (for example the length of the first, second and third row). This gives you the x, y and z scale. If they are not equal to each other there is non-uniform scaling.
Not sure if this is the best/safest or fastest method though, but it is definitely faster than doing full polar/spectral decomposition.
If a matrix is the composition of a rotation and a uniform scaling, its three columns should be orthogonal to each other and have the same length.
Author
Thanks for the answer. This is the formula I came up with finally:
// note: this is opengl type matrix
bool Matrix::IsUnformScaling() const
{
const float THRESHOLD = 0.01f;
float xLen = m[0]*m[0] + m[1]*m[1] + m[2]*m[2];
float yLen = m[4]*m[4] + m[5]*m[5] + m[6]*m[6];
float zLen = m[8]*m[8] + m[9]*m[9] + m[10]*m[10];
return fabsf(xLen - yLen) < xLen*THRESHOLD && fabsf(xLen - zLen) < xLen*THRESHOLD;
}
I tested it and it seems to work ;)
I can construct a matrix that has non-uniform scaling in it and passes your test, though...
EDIT: More explicitly,
EDIT: More explicitly,
m[0] = 0.8; m[1] = 0.6; m[2] = 0.0;
m[4] = 0.6; m[5] = 0.8; m[6] = 0.0;
m[8] = 0.0; m[9] = 0.0; m[10] = 1.0;
Author
I can construct a matrix that has non-uniform scaling in it and passes your test, though...
EDIT: More explicitly,m[0] = 0.8; m[1] = 0.6; m[2] = 0.0; m[4] = 0.6; m[5] = 0.8; m[6] = 0.0; m[8] = 0.0; m[9] = 0.0; m[10] = 1.0;
Ok, my bad.
do you have a solution perchance?
I posted that already, although perhaps my description was too succinct. You also need to verify that those three vectors are orthogonal to each other (check all three pairs).
EDIT: To be complete, you should also verify that the transformation preserves orientation (its determinant is positive).
bool are_orthogonal(Vector3 v, Vector3 w) {
static float const tolerance = 1e-6;
float dp = dot_product(v,w);
return dp * dp < dot_product(v,v) * dot_product(w,w) * tolerance;
}
EDIT: To be complete, you should also verify that the transformation preserves orientation (its determinant is positive).
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