我想使用cv :: solvePnPRansac找到给定3D到2D对应关系的对象姿势。我的通信可能有异常值,所以我不想使用cv :: solvePnP。我使用了从opengl中的摄像头拍摄的快照的3D到2D对应关系。因此,在反投影后,我发现使用solvePnP后返回的旋转矩阵是正确的。但是,当我使用solvePnPRansac时,我得到一个单位矩阵和零转换向量。请帮忙!我使用了以下代码。提前谢谢!
// Parameters to solvePnP
Mat camera_mat(3, 3, CV_64FC1);
Mat distCoeffs(4, 1, CV_64FC1);
Mat rvec(3, 1, CV_64FC1);
Mat tvec(3, 1, CV_64FC1);
Mat d(3, 3, CV_64FC1);
vector<Point3f> list_points3d_model_match;
vector<Point2f> list_points2d_scene_match;
while(map >> u >> v >> x >> y >> z)
{
Point2f ip = Point2f(u, v);
Point3f sp = Point3f(x, y, z);
// cout << x << " " << y << " " << z << endl;
list_points3d_model_match.push_back(sp);
list_points2d_scene_match.push_back(ip);
}
camera_mat.at<double>(0, 0) = 600;
camera_mat.at<double>(1, 1) = 600;
camera_mat.at<double>(0, 2) = WIDTH / 2;
camera_mat.at<double>(1, 0) = HEIGHT / 2;
camera_mat.at<double>(3, 3) = 1;
// solvePnP( list_points3d_model_match, list_points2d_scene_match, camera_mat, distCoeffs, rvec, tvec,
// useExtrinsicGuess, CV_ITERATIVE);
tvec.at<float>(0) = 0;
tvec.at<float>(1) = 0;
tvec.at<float>(2) = 2;
d.at<float>(0, 0) = 1;
d.at<float>(1, 2) = -1;
d.at<float>(2, 1) = 1;
Rodrigues(d, rvec);
solvePnPRansac(list_points3d_model_match, list_points2d_scene_match, camera_mat, distCoeffs, rvec, tvec, false, CV_ITERATIVE);
Rodrigues(rvec, d);
double* _r = d.ptr<double>();
printf("rotation mat: \n %.3f %.3f %.3f\n%.3f %.3f %.3f\n%.3f %.3f %.3f\n",
_r[0],_r[1],_r[2],_r[3],_r[4],_r[5],_r[6],_r[7],_r[8]);
}
cout << "translation vec:\n" << tvec << endl;
运行代码后,我得到以下输出。
rotation mat:
1.000 0.000 0.000
0.000 1.000 0.000
0.000 0.000 1.000
translation vec:
[0; 0; 0]
答案 0 :(得分:3)
您的相机矩阵错误。 您的代码应为:
camera_mat.at<double>(0, 0) = 600;
camera_mat.at<double>(1, 1) = 600;
camera_mat.at<double>(0, 2) = WIDTH / 2;
camera_mat.at<double>(1, 2) = HEIGHT / 2;
camera_mat.at<double>(3, 3) = 1;
答案 1 :(得分:2)
之前我遇到过同样的问题,并不是因为相机矩阵输入错误 更改
的声明Mat camera_mat(3, 3, CV_64FC1);
Mat distCoeffs(4, 1, CV_64FC1);
到CV_32FC1
并且不要将初始尺寸设为Mat tvec
和Mat d
。