问题
As a result to a call to findHomography() I get back a 3x3 matrix mtx[3][3]. This matrix contains the translation part in mtx[0][2] and mtx[1][2]. But how can I get the rotation part out of this 3x3 matrix?
Unfortunaltely my target system uses completely different calculation so I can't reuse the 3x3 matrix directly and have to extract the rotation out of this, that's why I'm asking this question.
回答1:
Generally speaking, you can't decompose the final transformation matrix into its constituent parts. There are some certain cases where it is possible. For example if the only operation preceding the operation was a translation, then you can do arccos(m[0][0]) to get the theta value of the rotation.
回答2:
Found it for my own meanwhile: There is an OpenCV function RQDecomp3x3() that can be used to extract parts of the transformation out of a matrix.
回答3:
RQDecomp3x3 has a problem to return rotation in other axes except Z so in this way you just find spin in z axes correctly,if you find projection matrix and pass it to "decomposeProjectionMatrix" you will find better resaults,projection matrix is different to homography matrix you should attention to this point.
来源：https://stackoverflow.com/questions/13369069/findhomography-3x3-matrix-how-to-get-rotation-part-out-of-it