Does anyone know of a fast way to tell if a matrix has non-uniform scaling?

**1**

# Detect non-uniform scaling in matrix

###
#2
Members - Reputation: **656**

Posted 12 January 2013 - 11:46 AM

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.

**Edited by Buckshag, 12 January 2013 - 11:48 AM.**

###
#4
Members - Reputation: **549**

Posted 13 January 2013 - 04:30 AM

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 ;)

###
#5
Crossbones+ - Reputation: **15264**

Posted 13 January 2013 - 01:01 PM

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;

**Edited by Álvaro, 13 January 2013 - 01:07 PM.**

###
#6
Members - Reputation: **549**

Posted 13 January 2013 - 02:24 PM

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?

###
#7
Crossbones+ - Reputation: **15264**

Posted 13 January 2013 - 07:57 PM

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).

**Edited by Álvaro, 13 January 2013 - 07:58 PM.**