SRT from Matrix? - beginners question
Author
i know that _41,_42,_43 from a 16*16 matrix will give me an absolute position but where do i get the scale and rotate values? Unable to find this out?
Quote:Original post by jmgunn87The XYZ scale values are in _11, _22 and _33, but the rotation is a bit more awkward. Google has plenty of links on matrices, they're not DirectX specific. The direction (+Z), up (+Y) and right (+X) vectors are in (_11, _21, _31), (_12, _22, _32) and (_13, _23, _33) (Although I may have got the row and column the wrong way around there). Matrices don't store rotation directly.
i know that _41,_42,_43 from a 16*16 matrix will give me an absolute position but where do i get the scale and rotate values? Unable to find this out?
In a 4x4 matrix there are separate scales on each of the 3 axes that form the matrix. So, assuming row-major ordering, the length of the first row vector is the scale on the x axis, the length of the 2nd is the scale of y, and the 3rd is the scale of z. Evil S is right [smile]
Extracting rotation is more complicated and you shouldn't really need to do it. Assuming it's an orthogonal matrix (all 3 axes are perpendicular to each other) then the matrix will store the sum total of rotations by how much the coordinate system it represents is rotated from the default axes: (1,0,0) (0,1,0) (0,0,1). If you really need to extract a set of angles, google "Matrix to Euler". It's important to note that there are many possible ways you can rotate one coordinate system to become another, so there is not a single correct answer to this problem.
-me
Extracting rotation is more complicated and you shouldn't really need to do it. Assuming it's an orthogonal matrix (all 3 axes are perpendicular to each other) then the matrix will store the sum total of rotations by how much the coordinate system it represents is rotated from the default axes: (1,0,0) (0,1,0) (0,0,1). If you really need to extract a set of angles, google "Matrix to Euler". It's important to note that there are many possible ways you can rotate one coordinate system to become another, so there is not a single correct answer to this problem.
-me
You can use D3DXMatrixDecompose(), though the dx9.0 version has a bug.
I can't remember where it is, but there's substitute code for D3DXMatrixDecompose somewhere on the web. It's only about 40 lines of code and quite fast.
I can't remember where it is, but there's substitute code for D3DXMatrixDecompose somewhere on the web. It's only about 40 lines of code and quite fast.
Author
yea, i was using MatrixDecompose. Didn't realise there was an actual bug in it but stopped using it due to its unreliability.
I'll have a look at extracting rotation a bit further.
Cheers!
I'll have a look at extracting rotation a bit further.
Cheers!
Found a copy. I believe this is the corrected version.
HRESULT XAD3DXMatrixDecompose(D3DXVECTOR3 *poutscale, D3DXQUATERNION *poutrotation, D3DXVECTOR3 *pouttranslation, D3DXMATRIX *pm) { D3DXMATRIX normalized; D3DXVECTOR3 vec; /*Compute the scaling part.*/ //vec.x=(*pm)(0,0); //vec.y=(*pm)(0,1); //vec.z=(*pm)(0,2); vec.x=pm->_11; vec.y=pm->_12; vec.z=pm->_13; poutscale->x=D3DXVec3Length(&vec); vec.x=pm->_21; vec.y=pm->_22; vec.z=pm->_23; poutscale->y=D3DXVec3Length(&vec); vec.x=pm->_31; vec.y=pm->_32; vec.z=pm->_33; poutscale->z=D3DXVec3Length(&vec); /*Compute the translation part.*/ pouttranslation->x=pm->_41; pouttranslation->y=pm->_42; pouttranslation->z=pm->_43; /*Let's calculate the rotation now*/ if ( (poutscale->x == 0.0f) || (poutscale->y == 0.0f) || (poutscale->z == 0.0f) ) return D3DERR_INVALIDCALL; if( D3DXMatrixDeterminant(pm) < 0 ) poutscale->z *= -1; normalized._11=pm->_11/poutscale->x; normalized._12=pm->_12/poutscale->x; normalized._13=pm->_13/poutscale->x; normalized._21=pm->_21/poutscale->y; normalized._22=pm->_22/poutscale->y; normalized._23=pm->_23/poutscale->y; normalized._31=pm->_31/poutscale->z; normalized._32=pm->_32/poutscale->z; normalized._33=pm->_33/poutscale->z; D3DXQuaternionRotationMatrix(poutrotation,&normalized); return S_OK;}This topic is closed to new replies.
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