从基本矩阵中找到同形异义矩阵 [英] Find Homography matrix from Fundamental matrix

问题描述

在给定一组对应关系和基本矩阵F的情况下，我试图计算同构矩阵H.

根据对极几何学原理，我知道这可以通过解决方案

首先，请注意，您在链接中提到的方程式(C29)使用的线l与e _ij 具有相同的坐标，是图像j中的子极.因此，l = e _ij 不是一条上界，因为dot(e _ij ，e _ij )== norm(e _ij )²== 1！= 0.

如果我坚持使用您链接中给出的符号，则可以将对数极子e _ij 计算为F _ij 的左空矢量.您可以通过在F_ij的转置上调用cv::SVD::solveZ来获取它.然后，如您的链接中所述，可以将单应性H _ij (将点从图像i映射到图像j)计算为H _ij = [e _ij ] _x F _ij ，其中[e _ij ] _x 表示法是3x3斜对称算子.可以在有关跨产品的Wikipedia文章中找到该符号的定义.. >

但是，请注意，这样的单应性H _ij 使用与e _{ij相同的坐标线，通过从图像j反投影的平面定义了从图像i到图像j的映射.子>.通常，这将提供与cv::findHomography返回的结果非常不同的结果，在这种情况下，所得的单应性是通过观察到的场景中的主平面从图像i到图像j的映射.因此，您将能够使用cv::findHomography返回的单应性近似注册两个图像，但是通常对于使用上述方法获得的单应性来说并非如此.}

I'm trying to compute Homography matrix H given a set of correspondences and the Fundamental matrix F.

From the principle of Epipolar geometry I know this can be done by cross product of epiline and F from Epipole Geometry

[e_ij] x F_ij = H_ij

I'm using OpenCV for finding Fundamental matrix F from set of matches between two views using cv::findFundamentalMat().

My question is that how can I find e_ij and how to use it in order to compute H. In OpenCV there is a function cv::computeCorrespondEpilines() that finds epilines corespond to each given point.

It's worth mentioning that I'm not interested in computing H directly from set of matches but only from computed Fundamental matrix.

Thanks

解决方案

First, notice that the equation (C29) you mentioned from your link uses a line l with same coordinates as e_ij, which is the epipole in image j. Therefore, l=e_ij is not an epiline, since dot( e_ij , e_ij ) == norm(e_ij)² == 1 != 0.

If I stick to the notations given in your link, you can compute the epipole e_ij as the left null-vector of F_ij. You can obtain it by calling cv::SVD::solveZ on the transpose of F_ij. Then, as is mentioned in your link, the homography H_ij (which maps points from image i to image j) can be computed as H_ij = [e_ij]_x F_ij, where the notation [e_ij]_x refers to the 3x3 skew-symmetric operator. A definition of this notation can be found in the Wikipedia article on cross-product.

However, be aware that such an homography H_ij defines a mapping from image i to image j via the plane backprojected from image j using the line with same coordinates as e_ij. In general, this will give a result which is very different from the result returned by cv::findHomography, where the resulting homography is a mapping from image i to image j via the dominant plane in the observed scene. Hence, you will be able to approximately register the two images using the homography returned by cv::findHomography, but in general this will not be the case for the homography obtained using the method above.

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