单应矩阵分解为旋转矩阵和平移向量

Question

我正在使用opencv 2.4.4为android开发一个增强现实应用程序，在单应性分解方面遇到了一些问题。我们知道，单应矩阵定义为H=A.R t，其中A是本征相机矩阵，R是旋转矩阵，t是平移向量。我想用图片来估计相机的视角，以及3d房间中相机的方向。

单应矩阵我可以估计与opencv函数: findHomography，我认为它的工作！下面是我是如何做到的：

static Mat mFindHomography(MatOfKeyPoint keypoints1, MatOfKeyPoint keypoints2, MatOfDMatch matches){
    List<Point> lp1 = new ArrayList<Point>(500);
    List<Point> lp2 = new ArrayList<Point>(500);

    KeyPoint[] k1 = keypoints1.toArray();
    KeyPoint[] k2 = keypoints2.toArray();

    List<DMatch> matchesList = matches.toList();

    if (matchesList.size() < 4){
        MatOfDMatch mat = new MatOfDMatch();
        return mat;
    }

    // Add matches keypoints to new list to apply homography
    for(DMatch match : matchesList){
        Point kk1 = k1[match.queryIdx].pt;
        Point kk2 = k2[match.trainIdx].pt;
        lp1.add(kk1);
        lp2.add(kk2);
    }

    MatOfPoint2f srcPoints = new MatOfPoint2f(lp1.toArray(new Point[0]));
    MatOfPoint2f dstPoints  = new MatOfPoint2f(lp2.toArray(new Point[0]));

    Mat mask = new Mat();

    Mat homography = Calib3d.findHomography(srcPoints, dstPoints, Calib3d.RANSAC, 10, mask); // Finds a perspective transformation between two planes. ---Calib3d.LMEDS Least-Median robust method 

    List<DMatch> matches_homo = new ArrayList<DMatch>();
    int size = (int) mask.size().height;
    for(int i = 0; i < size; i++){          
        if ( mask.get(i, 0)[0] == 1){
            DMatch d = matchesList.get(i);
            matches_homo.add(d);
        }
    }

    MatOfDMatch mat = new MatOfDMatch();
    mat.fromList(matches_homo);

    matchesFilterdByRansac = (int) mat.size().height;
    return homography;
}

然后，我想分解这个单应矩阵并计算euler角。正如我们知道的那样，我将单应矩阵乘以相机内部矩阵的逆矩阵: H.A^{-1} =R。所以，我想要在旋转和平移中分解R，并从旋转矩阵计算欧拉角。但它并没有起作用。这是怎么了？！！

if(!homography.empty()){ // esstimate pose frome homography 
Mat intrinsics = Mat.zeros(3, 3, CvType.CV_32FC1);  // camera intrinsic matrix 
intrinsics.put(0, 0, 890);
intrinsics.put(0, 2, 400);
intrinsics.put(1, 1, 890);
intrinsics.put(1, 2, 240);
intrinsics.put(2, 2, 1);

// Inverse Matrix from Wolframalpha
double[] inverseIntrinsics = { 0.001020408, 0 , -0.408163265,
        0, 0.0011235955, -0.26966292,
        0, 0 , 1 }; 

// cross multiplication 
double[] rotationTranslation = matrixMultiply3X3(homography, inverseIntrinsics);

Mat pose = Mat.eye(3, 4, CvType.CV_32FC1);  // 3x4 matrix, the camera pose
float norm1 = (float) Core.norm(rotationTranslation.col(0));
float norm2 = (float) Core.norm(rotationTranslation.col(1));
float tnorm = (norm1 + norm2) / 2.0f;       // Normalization value  ---test: float tnorm = (float) h.get(2, 2)[0];// not worked

Mat normalizedTemp = new Mat();
Core.normalize(rotationTranslation.col(0), normalizedTemp);
normalizedTemp.convertTo(normalizedTemp, CvType.CV_32FC1);
normalizedTemp.copyTo(pose.col(0)); // Normalize the rotation, and copies the column to pose

Core.normalize(rotationTranslation.col(1), normalizedTemp);
normalizedTemp.convertTo(normalizedTemp, CvType.CV_32FC1);    
normalizedTemp.copyTo(pose.col(1));// Normalize the rotation and copies the column to pose

Mat p3 = pose.col(0).cross(pose.col(1)); // Computes the cross-product of p1 and p2
p3.copyTo(pose.col(2));// Third column is the crossproduct of columns one and two

double[] buffer = new double[3];
rotationTranslation.col(2).get(0, 0, buffer);
pose.put(0, 3, buffer[0] / tnorm);  //vector t [R|t] is the last column of pose
pose.put(1, 3, buffer[1] / tnorm);
pose.put(2, 3, buffer[2] / tnorm);

float[] rotationMatrix = new float[9];
rotationMatrix = getArrayFromMat(pose);

float[] eulerOrientation = new float[3];
SensorManager.getOrientation(rotationMatrix, eulerOrientation); 

// Convert radian to degree
double yaw = (double) (eulerOrientation[0]) * (180 / Math.PI));// * -57;
double pitch = (double) (eulerOrientation[1]) * (180 / Math.PI));
double roll = (double) (eulerOrientation[2]) * (180 / Math.PI));}

请注意，OpenCV3.0有一个同源分解函数(here)，但我使用的是opencv 2.4.4 for android！在java中有包装器吗？

第二个问题是欧拉角中旋转矩阵的分解。是否存在以下问题：

    float[] eulerOrientation = new float[3];
SensorManager.getOrientation(rotationMatrix, eulerOrientation); 

我也用了这个link，但效果并不好！

double pitch = Math.atan2(pose.get(2, 1)[0], pose.get(2, 2)[0]);
double roll = Math.atan2(-1*pose.get(2, 0)[0], Math.sqrt( Math.pow(pose.get(2, 1)[0], 2) + Math.pow(pose.get(2, 2)[0], 2)) );
double yaw = Math.atan2(pose.get(1, 0)[0], pose.get(0, 0)[0]);

非常感谢您的回复

Answer 1

我希望这个答案能帮助那些今天正在寻找解决方案的人。

我的答案是使用c++和opencv 2.4.9。我复制了opencv 3.0中的分解同形映射函数。在计算单应性之后，我使用复制的函数来分解单应性。要过滤单应矩阵并从4个分解中选择正确答案，请查看my answer here。

要从旋转矩阵中获得欧拉角，可以参考this。用这种方法我能得到很好的结果。