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iLikeCarrots

Invert Transform Matrix [Solved]

8 posts in this topic

I'm having trouble inverting my object matrix with non-uniform scaling. I have scanned the web and written the multiplication out by hand, but I haven't been able to find my error. I am creating a matrix for OpenGL, and I combine my 4x4 matrix in the order Translation, Rotation, Scale. Where could I be going wrong?

I find the position X,Y,Z in the last column
I find the scaling X,Y,Z as the lengths the first three columns
I use the upper left 3x3 matrix as the rotation
[CODE]
~ Inv Scale ~ ~ Inv Rot ~ ~ Inv Pos ~
[1/X 0 0 0] [\.......] [1 0 0 -X]
[ 0 1/Y 0 0] [..flip..] [0 1 0 -Y]
[ 0 0 1/Z 0] [diagonal] [0 0 1 -Z]
[ 0 0 0 1] [.......\] [0 0 0 1]
[/CODE]

After combining the above matrices in that order, I come up with the following code with matrix A as the result:
[CODE]
(GL Matrix Indices)
[0 4 8 12]
[1 5 9 13]
[2 6 10 14]
[3 7 11 15]

void invertMatrix (float A[], float B[]) {
float sclX = lengthVector(&B[0]);
float sclY = lengthVector(&B[4]);
float sclZ = lengthVector(&B[8]);

A[0] = B[0]/sclX; A[4] = B[1]/sclX; A[8] = B[2] /sclX;
A[1] = B[4]/sclY; A[5] = B[5]/sclY; A[9] = B[6] /sclY;
A[2] = B[8]/sclZ; A[6] = B[9]/sclZ; A[10] = B[10]/sclZ;
A[3] = 0.0; A[7] = 0.0; A[11] = 0.0;

float posX = -B[12],
posY = -B[13],
posZ = -B[14];

A[12] = A[0]*posX + A[4]*posY + A[8] *posZ;
A[13] = A[1]*posX + A[5]*posY + A[9] *posZ;
A[14] = A[2]*posX + A[6]*posY + A[10]*posZ;
A[15] = 1.0;
}
[/CODE]

To test my inverse, I created a matrix with the following transform:
[i]Position[/i] x=0, y=0, z=0
[i]Rotation[/i] x=1, y=0, z=0
[i]Scaling[/i] x=1, y=1, z=3
I multiply the inverse times the matrix, which always returns the scaling on the diagonal.
[CODE]
-- Object Matrix -- -- Inverse Matrix -- -- Inverse * Object --
[1.0 0.0 0.0 0.0] [1.0 0.0 0.0 -0.0] [1.0 0.0 0.0 0.0]
[0.0 0.5 -2.5 0.0] [0.0 0.5 0.8 -0.0] [0.0 1.0 -0.0 0.0]
[0.0 0.8 1.6 0.0] [0.0 -0.8 0.5 0.0] [0.0 0.0 3.0 0.0]
[0.0 0.0 0.0 1.0] [0.0 0.0 0.0 1.0] [0.0 0.0 0.0 1.0]
[/CODE]
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When you take the inverse of a product of matrices, you need to compute the product of the inverses [b]in reverse order [/b]. That might explain some of what you are seeing.

However, even if you don't get back to the identity matrix, you should get a matrix with determinant 1, and yours has determinant 3, so there's probably some other mistake.

Can you be more explicit about the matrices you are multiplying? In particular, I don't know what "rot(1,0,0)" means.
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[quote]you need to compute the product of the inverses [b]in reverse order[/b][/quote]
I'm assuming that I did this part correctly - I performed Scaling * Rotation * Translation. I flipped the 3x3 rotation section along the diagonal, so the transformed indices (OpenGL) would be..
[CODE]

[0 1 2]
[4 5 6]
[8 9 10]
[/CODE]

[quote]I don't know what "rot(1,0,0)" means.[/quote]
I modified the description of my test. They are the values that I use in my object matrix, which is combined in the order Translation, Rotation, Scale.

Also for completeness, this is how I create the matrix. Position, rotation, and scale are arrays with three floats.
[CODE]
mat[12] = node->position[0];
mat[13] = node->position[1];
mat[14] = node->position[2];

float A = cos(node->rotation[1]), B = sin(node->rotation[1]);
float C = cos(node->rotation[0]), D = sin(node->rotation[0]);
float E = cos(node->rotation[2]), F = sin(node->rotation[2]);

mat[0] = ( A*E + B*D*F) * node->scale[0];
mat[1] = ( C*F ) * node->scale[0];
mat[2] = (-B*E + A*D*F) * node->scale[0];

mat[4] = ( B*D*E - A*F) * node->scale[1];
mat[5] = ( C*E ) * node->scale[1];
mat[6] = ( B*F + A*D*E) * node->scale[1];

mat[8] = ( B*C ) * node->scale[2];
mat[9] = (-D ) * node->scale[2];
mat[10]= ( A*C ) * node->scale[2];
[/CODE]
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I use the invertMatrix function from the first post. Argument A is the resulting inverted matrix, and B would be the object's matrix.
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In your original example, how did you end up with a 4.0 at index 14? If your position vector is (0,0,0) that shouldn't happen... Also, can you print out those matrices at the bottom of the first post with more precision?
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[quote]how did you end up with a 4 in position 14?[/quote]
Sorry - those are remains from an earlier test that I didn't change correctly.

[quote]Also, can you print out those matrices at the bottom of the first post with more precision?[/quote]
I'll post directly from the console to prevent typos..
[CODE]
Object
1.000000 0.000000 0.000000 0.000000
0.000000 0.540302 -2.524413 0.000000
0.000000 0.841471 1.620907 0.000000
0.000000 0.000000 0.000000 1.000000
Inverse
1.000000 0.000000 0.000000 -0.000000
0.000000 0.540302 0.841471 -0.000000
0.000000 -0.841471 0.540302 0.000000
0.000000 0.000000 0.000000 1.000000
Inverse * Object
1.000000 0.000000 0.000000 0.000000
0.000000 1.000000 -0.000000 0.000000
0.000000 0.000000 3.000000 0.000000
0.000000 0.000000 0.000000 1.000000
[/CODE]

Note: I wonder if the inverse of scale (1/scale) should be squared? Perhaps I should get another piece of paper and work through the math.

Also, here is code that compiles to demonstrate the problem. [url="http://pastebin.com/ks07zNm2"]http://pastebin.com/ks07zNm2[/url]
Name the file as Matrix.c then compile with the command 'gcc -o Matrix Matrix.c -lm'
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Since the problem seems to be with the non-uniform scaling, I would write a test with no rotation first.

I think you are correct, and you need to divide by the square of the lengths (which will save you computing the square roots).
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Thank you, it appears to be working. I'm assuming that 1/scale will remove scaling on the inverse, but that alone won't remove the scaling multiplied with the original matrix, which is why I have to square it. I just hope that this is correct and won't end up with bugs later! [img]http://public.gamedev.net//public/style_emoticons/default/rolleyes.gif[/img]

Here is the fixed code: [url="http://pastebin.com/EMu7zjcF"]http://pastebin.com/EMu7zjcF[/url]
And the compile command: [b]gcc -o Matrix Matrix.c -lm[/b]
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