Is the recoverPose() function in OpenCV is left-handed?

Question

I run simple test for OpenCV camera pose estimation. Having a photo and the same photo scaled up (zoomed in) I use them to detect features, calculate essential matrix and recover camera poses.

Mat inliers;
Mat E = findEssentialMat(queryPoints, trainPoints, cameraMatrix1, cameraMatrix2,
                         FM_RANSAC, 0.9, MAX_PIXEL_OFFSET, inliers);

size_t inliersCount =
    recoverPose(E, queryGoodPoints, trainGoodPoints, cameraMatrix1, cameraMatrix2, R, T, inliers);

So when I specify the original image as the first one, and the zoomed image as the second one, I get translation T close to [0; 0; -1]. However the second camera (zoomed) is virtually closer to the object than the first one. So if Z-axis goes from image plane into the scene, the second camera should have positive offset along Z-axis. For the result I get, Z-axis goes from the image plane towards camera, which among with other axes (X goes right, Y goes down) forms left-handed coordinate system. Is that true? Why this result differs from the coordinate system illustrated here?

goodknight goodknight · Accepted Answer · 2017-08-16T21:01:22

According to the OpenCV document, the algorithm in the function recoverPose is based on the paper "Nistér, D. An efficient solution to the five-point relative pose problem, CVPR 2003." From equations in Section 2 in this paper, we know it uses the basic triangle relationship (see figure here):

x2 = R*x1 + t

Therefore, translation t is the vector from cam2 to cam1 in cam2 frame. This explains why you get the answer t close to [0; 0; -1].

Is the recoverPose() function in OpenCV is left-handed?

3 Answers