在Opencv中仅使用平移，旋转和缩放计算单应性 [英] Calculate a Homography with only Translation, Rotation and Scale in Opencv

问题描述

我确实有两组点，我想找到它们之间的最佳变换. 在OpenCV中，您具有以下功能:

I do have two sets of points and I want to find the best transformation between them. In OpenCV, you have the following function:

Mat H = Calib3d.findHomography(src_points, dest_points);

，使用RANSAC返回3x3同形矩阵.我现在的问题是，我只需要平移和旋转(可能还有缩放)，不需要仿射和透视.

that returns you a 3x3 Homography matrix, using RANSAC. My problem is now, that I only need translation and rotation (& maybe scale), I don't need affine and perspective.

问题是，我的点仅以2D表示.

The thing is, my points are only in 2D.

(1)是否有计算单应性但自由度较小的函数?

(1) Is there a function to compute something like a homography but with less degrees of freedom?

(2)如果不存在，是否可以从3x3单应矩阵提取仅进行平移和旋转的3x3矩阵?

(2) If there is none, is it possible to extract a 3x3 matrix that does only translation and rotation from the 3x3 homography matrix?

在此先感谢您的帮助！

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推荐答案

OpenCV

OpenCV estimateRigidTransform function is exactly what you need: it returns Translation, Rotation and Scale (use false value for fullAffine flag). And it DOES use RANSAC (see source code to be sure of it).

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