单应性opencv的旋转和平移矩阵 [英] rotation and translation matrix from homography opencv

问题描述

我正在用Java开发一个android应用程序，我使用findHomography()用opencv计算了单应性，如何找到包含通过单应性旋转和平移的相机姿势矩阵

I'm working on an android application in java,I calculated homography with opencv using findHomography(), how I can find the matrix of the camera pose containing the rotation and translation through homography

 Mat homography = Calib3d.findHomography(ReferencePoints2, ReferencePoints1,0,Calib3d.RANSAC);

推荐答案

您需要知道相机的固有参数.

You need to know the intrinsic parameters of the camera to do that.

考虑z = 0平面.重点

Considers the z=0 plane. The point

 X=(x,y,0,1)'

投影到图像上

p=P*X.

现在使用分解

P=K[R t],

其中，K是校准矩阵，[R t]是外部参数.由于z = 0，R的第三列向量乘以零.现在，我们可以删除第三列以获取

where K is the calibration matrix and [R t] are extrinsic parameters. Since z=0, the third column vector of R is multiplied by zero. We can now drop the 3rd column to get

p=K*[r1 r2 t]*(x,y,1)=H*(x,y,1),

其中H是平面单应性.

您已经根据例如已知点.现在可以恢复R的第一和第二列以及向量t

You have already computed H from e.g. known points. The first and second column of R and the vector t can now be recovered

[r1 r2 t]=inv(K)*H.

确保r1和r2是单位长度，然后t是正确的转换向量. R的第三列向量可以恢复，因为R是正交的，例如使用叉积.

Make sure that r1 and r2 are unit length, then t is the correct translation vector. The third column vector of R can be recovered because R is orthogonal, for example using the cross product.

r3=cross(r1,r2).

由于H是测量值，因此您计算出的r1和r2不精确.您可以使用SVD获得最接近测量值的旋转矩阵.然后，您可以组成一个投影矩阵

Since H is a measurement, the r1 and r2 you computed are not exact. You can use the SVD for obtaining the closest rotation matrix to a measurement. You can then compose a projection matrix

P=K[r1 r2 r3 t]

根据您的同构2D坐标系在坐标系中投影任何3D点.

which projects any 3D point in the coordinate frame based on your 2D coordinate system of the homograohy.

这里有一些课程材料，描述了这种情况.

Here is some course material, which describes this situation.

https://www.dropbox.com/s/qkulg4j64lyn0qa /2018_proj_geo_for_cv_projcv_assignment.pdf?dl=0

这是一个相关的问题. 使用基于4个共面的单应矩阵计算相机姿态点

Here is a related question. Computing camera pose with homography matrix based on 4 coplanar points

正如@nbsrujan(感谢)指出的那样，对于使用OpenCV的用户，有一个

As @nbsrujan (thanks) pointed out, for those using OpenCV, there is a function which can decompose a homography into translation and rotation matrices given the intrinsics.

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