opencv triangulatePoints，奇怪的結果

Question

我在opencv中使用triagulatePoints函數。 最后，我在絕望之后讓它工作，但結果看起來不對。 我已經閱讀了其他一些問題，但我仍然不明白！

我在跑步：

cv::Mat Q,P1,P2,LP,RP,D1,D2,M1,M2;
    char *CALIB_FILE = (char *)"extrinsics.xml";
    FileStorage fs(CALIB_FILE, FileStorage::READ);
    fs["Q"] >> Q;
    fs["P1"] >> P1;
    fs["P2"] >> P2;


    cv::Mat cam0pnts(1, 5, CV_64FC2);  //681 432 479    419 550 320 682 274 495 254
    cv::Mat cam1pnts(1, 5, CV_64FC2); //800 466 587 451 657 352 791 311 592 283

    cv::Mat points_3D(1, 5, CV_64FC4);   


    cv::triangulatePoints(P1, P2, cam0pnts, cam1pnts, points_3D);

P1和P2是來自stereo_calib函數的計算外部函數。

有5個點，一個粗糙的正方形，每個角落有一個點，中間有一個點。

結果矩陣是：

 [-0.6620691274599629, 0.6497615623177577, -0.6585234150236594, 0.6529909432980171, -0.6604373884239706;
  -0.7091492226203088, 0.7208075295879011, -0.7119285643550911, 0.7174438199266364, -0.710244308941275;
  0.242429054072024, -0.2413429417514131, 0.2439357048056051, -0.2426462227979475, 0.2436708320163396;
  -6.52928664505207e-005, -4.348043360405063e-005, -5.515313727475824e-005, -6.149577656504346e-005, -5.668087253108842e-005]

當在3d中繪制時，如果完全縮放錯誤，則給出兩個看起來幾乎正確的位置，然后是這兩個中的三個重復。

我在哪里錯了？ 我是否需要對生成的矩陣執行某些操作才能獲得xyz坐標？ 或者我是否錯誤地實現了該功能？

Answer 1

取消，我設法通過忽略cv :: triangulate函數並在此處使用此方法來實現：

http://www.morethantechnical.com/2012/01/04/simple-triangulation-with-opencv-from-harley-zisserman-w-code/

通過一些小改動修復一些錯誤的代碼......

Mat_<double> IterativeLinearLSTriangulation(Point3d u,    //homogenous image point (u,v,1)
    Matx34d P,          //camera 1 matrix
    Point3d u1,         //homogenous image point in 2nd camera
    Matx34d P1          //camera 2 matrix
    ) {

    double wi = 1, wi1 = 1;
    Mat_<double> X(4, 1);



    for (int i = 0; i < 10; i++) { //Hartley suggests 10 iterations at most
        Mat_<double> X_ = LinearLSTriangulation(u, P, u1, P1);
    X(0) = X_(0); X(1) = X_(1); X(2) = X_(2); X(3) = 1.0;
        //recalculate weights
        double p2x = Mat_<double>(Mat_<double>(P).row(2)*X)(0);
        double p2x1 = Mat_<double>(Mat_<double>(P1).row(2)*X)(0);

        //breaking point
        if (fabsf(wi - p2x) <= EPSILON && fabsf(wi1 - p2x1) <= EPSILON) break;

        wi = p2x;
        wi1 = p2x1;

        //reweight equations and solve
        Matx43d A((u.x*P(2, 0) - P(0, 0)) / wi, (u.x*P(2, 1) - P(0, 1)) / wi, (u.x*P(2, 2) - P(0, 2)) / wi,
            (u.y*P(2, 0) - P(1, 0)) / wi, (u.y*P(2, 1) - P(1, 1)) / wi, (u.y*P(2, 2) - P(1, 2)) / wi,
            (u1.x*P1(2, 0) - P1(0, 0)) / wi1, (u1.x*P1(2, 1) - P1(0, 1)) / wi1, (u1.x*P1(2, 2) - P1(0, 2)) / wi1,
            (u1.y*P1(2, 0) - P1(1, 0)) / wi1, (u1.y*P1(2, 1) - P1(1, 1)) / wi1, (u1.y*P1(2, 2) - P1(1, 2)) / wi1
            );
        Mat_<double> B = (Mat_<double>(4, 1) << -(u.x*P(2, 3) - P(0, 3)) / wi,
            -(u.y*P(2, 3) - P(1, 3)) / wi,
            -(u1.x*P1(2, 3) - P1(0, 3)) / wi1,
            -(u1.y*P1(2, 3) - P1(1, 3)) / wi1
            );

        solve(A, B, X_, DECOMP_SVD);
        X(0) = X_(0); X(1) = X_(1); X(2) = X_(2); X(3) = 1.0;
    }

    return X;
}

還有這個：

Mat_<double> LinearLSTriangulation(Point3d u,       //homogenous image point (u,v,1)
    Matx34d P,       //camera 1 matrix
    Point3d u1,      //homogenous image point in 2nd camera
    Matx34d P1       //camera 2 matrix
    )
{
    //build matrix A for homogenous equation system Ax = 0
    //assume X = (x,y,z,1), for Linear-LS method
    //which turns it into a AX = B system, where A is 4x3, X is 3x1 and B is 4x1
    Matx43d A(u.x*P(2, 0) - P(0, 0), u.x*P(2, 1) - P(0, 1), u.x*P(2, 2) - P(0, 2),
        u.y*P(2, 0) - P(1, 0), u.y*P(2, 1) - P(1, 1), u.y*P(2, 2) - P(1, 2),
        u1.x*P1(2, 0) - P1(0, 0), u1.x*P1(2, 1) - P1(0, 1), u1.x*P1(2, 2) - P1(0, 2),
        u1.y*P1(2, 0) - P1(1, 0), u1.y*P1(2, 1) - P1(1, 1), u1.y*P1(2, 2) - P1(1, 2)
        );
    Mat_<double> B = (Mat_<double>(4, 1) << -(u.x*P(2, 3) - P(0, 3)),
        -(u.y*P(2, 3) - P(1, 3)),
        -(u1.x*P1(2, 3) - P1(0, 3)),
        -(u1.y*P1(2, 3) - P1(1, 3)));

    Mat_<double> X;
    solve(A, B, X, DECOMP_SVD);

    return X;
}