findHomography()/3x3矩阵-如何从中获得旋转部分? [英] findHomography() / 3x3 matrix - how to get rotation part out of it?

问题描述

作为对findHomography()的调用的结果，我得到了一个3x3矩阵mtx [3] [3].该矩阵在mtx [0] [2]和mtx [1] [2]中包含转换部分.但是如何从3x3矩阵中获得旋转部分呢?

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?

不幸的是，我的目标系统使用了完全不同的计算，所以我无法直接重用3x3矩阵，而不得不从中提取旋转值，这就是为什么我问这个问题.

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.

推荐答案

同时为我自己找到了它:有一个OpenCV函数RQDecomp3x3()可用于从矩阵中提取部分转换.

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.

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