6D pose estimation problem. How to estimate the rotation matrix?

asked 2014-01-13 21:36:26 -0600

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I have been trying to estimate the 6D pose of a moving camera (in 6D) using opencv library. My steps are:

  1. use goodFeaturesToTrack to get features.
  2. calculate optical flow
  3. track features
  4. get fundamental matrix
  5. get essential matrix
  6. check the combination of R or t to determine the true R & t using triangulation (could find a better way to do this.. had to combine R & t and then later extract R & t)
  7. Use the height information (from altitude sensor) to get scale factor and scale the translation
  8. Update the transformation
  9. Save the new feature set and image for next loop.

But the results are far from expectation. I verified that R is orthogonal. But cant figure out where I went wrong. My code is as follows:

void imageCb(const sensor_msgs::ImageConstPtr& image) 
        cv_ptr_prev_ = cv_bridge::toCvCopy(image, enc::BGR8);
        cv_ptr_new_ = cv_bridge::toCvCopy(image, enc::BGR8);
    catch (cv_bridge::Exception& e)
    ROS_ERROR("Conversion error: %s", e.what());
    image_prev_.create(cv_ptr_prev_->image.rows, cv_ptr_prev_->image.cols , CV_8UC1);
    image_new_.create(cv_ptr_new_->image.rows, cv_ptr_new_->image.cols , CV_8UC1);

    cvtColor( cv_ptr_new_->image, image_new_, CV_BGR2GRAY);
    cvtColor( cv_ptr_prev_->image, image_prev_, CV_BGR2GRAY);

    Mat F, E;                //Fundamental and Essential Matrices
    vector<Point2f> p1;      //common feature points in first image
    vector<Point2f> p2;      //common feature points in second image
    double f = 320;          //camera parameter focal length
    double Cx = 320.5;       //Resolution
    double Cy = 240.5;
    Mat K = (Mat_<double>(3,3) << f, 0 , -Cx, 0, f, -Cy, 0, 0, 1); //camera internal matrix

    Mat W = (Mat_<double>(3,3) << 0, -1, 0, 1, 0, 0, 0, 0 , 1);   //for SVD
    Mat Z = (Mat_<double>(3,3) << 0, 1, 0, -1, 0, 0, 0, 0 , 1);  //for SVD

    Mat w, u , vt; //SVD result of Essential Matrix
    Mat tx, Ra, Rb; //Temporary checking for R and t
    Mat t_unscaled, t_scaled, R;  //True R and t for the current step

    //Lucas Kanade Optical Flow

    vector<uchar> features_found;
    calcOpticalFlowPyrLK(image_prev_, image_new_, features_prev_, features_new_, features_found, noArray(),Size(win_size_*4+1, win_size_*4+1), 5,TermCriteria(CV_TERMCRIT_ITER|CV_TERMCRIT_EPS,20, 0.3));
        for( int i = 0; i < (int)features_new_.size(); i++ )
        circle(cv_ptr_prev_->image, features_new_[i],1,CV_RGB(255,255,255));
        circle(cv_ptr_prev_->image, features_prev_[i],1,CV_RGB(0,0,0));
        if( !features_found[i] )
        line(cv_ptr_prev_->image, features_prev_[i], features_new_[i], Scalar(255,0,0), 1, CV_AA);
        p1.push_back(Point2f(features_prev_[i].x-Cx, features_prev_[i].y-Cy));
        p2.push_back(Point2f(features_new_[i].x-Cx, features_new_[i].y-Cy));
    image_pub_.publish(cv_ptr_prev_->toImageMsg());          //publish image with features

    vector<uchar> inliers;

        F =  findFundamentalMat(p1, p2, CV_FM_RANSAC,1.0, 0.999, inliers);
        //      cout<<"Number of inliers is::::::"<<sum(inliers)<<endl;
        E = K.t()*F*K;
        SVD::compute(E, w, u, vt);

        tx = u * Z * u.t();     Ra = u * W * vt;        Rb = u * W.t() * vt;

        if(determinant(Ra)<=0)        Ra = -Ra;
        if(determinant(Rb)<=0)        Rb = -Rb;

        t_unscaled = (Mat_<double>(1,3) << tx.at<double>(2,1), tx.at<double>(0,2), tx.at<double>(1,0) );

        Mat tminus = -(t_unscaled.clone());

        //Check which Ra or Rb or t

        vector<Mat> R_vec ...
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Have you considered to use the function cv::findHomography()? Maybe I misunderstood your problem, but it might be a straight forward solution to use the with the keypoint matches your calculated. A good example for using it is the brief_match_test.cpp in the OpenCV samples....

Wolf gravatar image Wolf  ( 2014-01-14 02:06:10 -0600 )edit

I guess cvHomography is for feature points associated with a plane in the image. In my case the feature points are not necessarily associated with a single plane. In fact the image sequences are of rough terrain and camera has both translation and rotation.

espee gravatar image espee  ( 2014-01-14 17:33:23 -0600 )edit