Open Rosa712 opened 10 months ago
void triangulation( const vector<KeyPoint> &keypoint_1, const vector<KeyPoint> &keypoint_2, const std::vector<DMatch> &matches, const Mat &R, const Mat &t, vector<Point3d> &points) { Mat T1 = (Mat_<float>(3, 4) << 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0); Mat T2 = (Mat_<float>(3, 4) << R.at<double>(0, 0), R.at<double>(0, 1), R.at<double>(0, 2), t.at<double>(0, 0), R.at<double>(1, 0), R.at<double>(1, 1), R.at<double>(1, 2), t.at<double>(1, 0), R.at<double>(2, 0), R.at<double>(2, 1), R.at<double>(2, 2), t.at<double>(2, 0) ); 请问这里的T2是第二幅图片的相机位姿吗？如果是的话为什么用求解出来的R21，t21赋值，当1是世界坐标系时不应该是T2=T21.inverse()吗？
void triangulation( const vector<KeyPoint> &keypoint_1, const vector<KeyPoint> &keypoint_2, const std::vector<DMatch> &matches, const Mat &R, const Mat &t, vector<Point3d> &points) { Mat T1 = (Mat_<float>(3, 4) << 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0); Mat T2 = (Mat_<float>(3, 4) << R.at<double>(0, 0), R.at<double>(0, 1), R.at<double>(0, 2), t.at<double>(0, 0), R.at<double>(1, 0), R.at<double>(1, 1), R.at<double>(1, 2), t.at<double>(1, 0), R.at<double>(2, 0), R.at<double>(2, 1), R.at<double>(2, 2), t.at<double>(2, 0) );
void triangulation( const vector<KeyPoint> &keypoint_1, const vector<KeyPoint> &keypoint_2, const std::vector<DMatch> &matches, const Mat &R, const Mat &t, vector<Point3d> &points) { Mat T1 = (Mat_<float>(3, 4) << 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0); Mat T2 = (Mat_<float>(3, 4) << R.at<double>(0, 0), R.at<double>(0, 1), R.at<double>(0, 2), t.at<double>(0, 0), R.at<double>(1, 0), R.at<double>(1, 1), R.at<double>(1, 2), t.at<double>(1, 0), R.at<double>(2, 0), R.at<double>(2, 1), R.at<double>(2, 2), t.at<double>(2, 0) );请问这里的T2是第二幅图片的相机位姿吗？如果是的话为什么用求解出来的R21，t21赋值，当1是世界坐标系时不应该是T2=T21.inverse()吗？