extracting scale matrix from modelview matrix

Question

how do we extract scale matrix from model view matrix? Right now I am taking length of each coloumn, but it fails when the scale is negative. here is my code:

  float xs =
            matrix[0][0] * matrix[0][1] * matrix[0][2] * matrix[0][3] < 0 ?
                    -1 : 1;
    float ys =
            matrix[1][0] * matrix[1][1] * matrix[1][2] * matrix[1][3] < 0 ?
                    -1 : 1;
    float zs =
            matrix[2][0] * matrix[2][1] * matrix[2][2] * matrix[2][3] < 0 ?
                    -1 : 1;


    glm::vec3 new_scale;
    new_scale.x =  xs* glm::sqrt(
                    matrix[0][0] * matrix[0][0] + matrix[0][1] * matrix[0][1]
                            + matrix[0][2] * matrix[0][2]);
    new_scale.y =  ys* glm::sqrt(
                    matrix[1][0] * matrix[1][0] + matrix[1][1] * matrix[1][1]
                            + matrix[1][2] * matrix[1][2]);
    new_scale.z = zs* glm::sqrt(
                    matrix[2][0] * matrix[2][0] + matrix[2][1] * matrix[2][1]
                            + matrix[2][2] * matrix[2][2]);

For example:

float []mat={0.032254f, 0.000000f, 0.000000f, 0.000000f, 0.000000f, -0.0052254f, 0.000000f, 0.000000f, 0.000000f, 0.000000f, 0.4332254f, 0.000000f, 0.000000f, 0.000000f, 0.000000f, 1.000000f};

score 1 · Accepted Answer · edited May 23 '17 at 11:51

see homogenous transformation matrices. The way you are extracting scale is OK. So what about the sign? Your current approach will not work if your transformation contains rotations ...

Another problem is that you can not know which scale is negated and which not because if you negate single axis you can obtain the same result if you negate any other one and rotate to match the position. If you negate 2 axises you get the original matrix with different rotation.

The best you can do is detect if your matrix has inverted 1 or 3 axis:

Create sign table for original undistorted matrix

for example unit matrix but if you have different starting point use that
```
sz0=dot(cross(X0,Y0),Z0);
sy0=dot(cross(Z0,X0),Y0);
sx0=dot(cross(Y0,Z0),X0);
```
where X0,Y0,Z0 are extracted axises vectors from your start point matrix
compute the signs for your current matrix
```
sz1=dot(cross(X1,Y1),Z1);
sy1=dot(cross(Z1,X1),Y1);
sx1=dot(cross(Y1,Z1),X1);
```
where X1,Y1,Z1 are extracted axises vectors from your actual matrix
compare signs and deduce which axises scales are negative

if (sx0*sx1<0)||(sy0*sy1<0)||(sz0*sz1<0) then one or all 3 axises are negated but you can not know which ... Also all 3 sign comparisons should have the same result.

[edit1] clarifications

X=(matrix[0][0],matrix[0][1],matrix[0][2])
dot(a,b)=a.x*b.x+a.y*b.y+a.z*b.z is scalar multiplication of vectors (dot product)
c=cross(a,b) ... c.x=a.y*b.z+a.z*b.y c.y=a.z*b.x+a.x*b.z c.z=a.x*b.y+a.y*b.x is vector multiplication (cross product)

So when you compute cross of two vectors you get vector perpendicular to both operands. As matrix axis vectors should be perpendicular by multipliyng 2 axises you get the third. The dot product just compare if the original and computed third axis are in the same direction ... This way is invariant on rotations

can you please explain what is 1st point here? I just have modelview matrix, no other info — debonair, Nov 05 '15 at 08:31
It does not work if one of the axis is negative and other 2 are positive or vice-versa. — debonair, Nov 05 '15 at 18:22
@debonair In orthogonal coordinate systems you got only two option right or left handed axises ... you can not distinguish nothing more than that without additional info. as I wrote before does not matter if you negate 1 or 3 axises ... it is the same. And the second option is negated 0 or 2 axises ... There is no chance telling which or how many axises are negated. The best you can do is knowing if it is option 1 or 2 — Spektre, Nov 05 '15 at 20:18
@debonair if your matrix start point was unit matrix then the result for all `s?1` is negative which means 1 or 3 axises are negated. — Spektre, Nov 06 '15 at 08:29
Is it possible to detect if only one of them is negative or positive? — debonair, Nov 06 '15 at 16:21
@debonair not without some additional knowledge about the transforms inside matrix ... like rotations constraints etc ... for arbitrary matrix it is impossible — Spektre, Nov 06 '15 at 18:36

extracting scale matrix from modelview matrix

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