单目图像加单点测距，求解目标位
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单目相机通过对极约束来求解相机运动的位姿。参考了ORBSLAM中单目实现的代码，这里用opencv来实现最简单的位姿估计。
对极约束的概念可以参考我的这篇
Visual SLAM – 理解对极几何和约束19 赞同 · 0 评论文章
简单的说就是对于单目相机，在不同的位置下我们看到同一个物体P，在两个图像中分别是归一化像素P1（U1，V1）和P2（U2，V2），我们可以通过 P1TK(−1)TEK−1P2=0P_1TK{(-1)T}EK^{-1}P_2 = 0 来计算出E，E是本质矩阵。E=tXRE = t_XR , 这里的tx就是Transition的反对称矩阵。同时也就算出了T和R, 用的是8点法。
这里我通过opencv来计算E:
mLeftImg = cv::imread(lImg, cv::IMREAD_GRAYSCALE);
mRightImg = cv::imread(rImg, cv::IMREAD_GRAYSCALE);
cv::Ptr<ORB> OrbLeftExtractor = cv::ORB::create();
cv::Ptr<ORB> OrbRightExtractor = cv::ORB::create();OrbLeftExtractor->detectAndCompute(mLeftImg, noArray(), mLeftKps, mLeftDes);
OrbRightExtractor->detectAndCompute(mRightImg, noArray(), mRightKps, mRightDes);
Ptr<DescriptorMatcher> matcher = DescriptorMatcher::create(DescriptorMatcher::BRUTEFORCE_HAMMING);
matcher->match(mLeftDes, mRightDes, mMatches);
首先通过ORB特征提取，获取两幅图像的匹配度，我将其连线出来的效果:
可以看到，里面有很多的误匹配，先使用RANSAC算法，选出尽可能多的匹配准确的特征点，这样在之后的位姿计算时可以更精准。
RANSAC的算法原理可以google，很容易理解。先看看ORBSLAM中的实现：
bool Initializer::Initialize(const Frame &CurrentFrame, const vector<int> &vMatches12, cv::Mat &R21, cv::Mat &t21,vector<cv::Point3f> &vP3D, vector<bool> &vbTriangulated)
{// Fill structures with current keypoints and matches with reference frame// Reference Frame: 1, Current Frame: 2// Frame2 特征点mvKeys2 = CurrentFrame.mvKeysUn;// mvMatches12记录匹配上的特征点对mvMatches12.clear();mvMatches12.reserve(mvKeys2.size());// mvbMatched1记录每个特征点是否有匹配的特征点，// 这个变量后面没有用到，后面只关心匹配上的特征点mvbMatched1.resize(mvKeys1.size());// 步骤1：组织特征点对for(size_t i=0, iend=vMatches12.size();i<iend; i++){if(vMatches12[i]>=0){mvMatches12.push_back(make_pair(i,vMatches12[i]));mvbMatched1[i]=true;}elsemvbMatched1[i]=false;}// 匹配上的特征点的个数const int N = mvMatches12.size();// Indices for minimum set selection// 新建一个容器vAllIndices，生成0到N-1的数作为特征点的索引vector<size_t> vAllIndices;vAllIndices.reserve(N);vector<size_t> vAvailableIndices;for(int i=0; i<N; i++){vAllIndices.push_back(i);}// Generate sets of 8 points for each RANSAC iteration// **步骤2：在所有匹配特征点对中随机选择8对匹配特征点为一组，共选择mMaxIterations组 **// 用于FindHomography和FindFundamental求解// mMaxIterations:200mvSets = vector< vector<size_t> >(mMaxIterations,vector<size_t>(8,0));DUtils::Random::SeedRandOnce(0);for(int it=0; it<mMaxIterations; it++){vAvailableIndices = vAllIndices;// Select a minimum setfor(size_t j=0; j<8; j++){// 产生0到N-1的随机数int randi = DUtils::Random::RandomInt(0,vAvailableIndices.size()-1);// idx表示哪一个索引对应的特征点被选中int idx = vAvailableIndices[randi];mvSets[it][j] = idx;// randi对应的索引已经被选过了，从容器中删除// randi对应的索引用最后一个元素替换，并删掉最后一个元素vAvailableIndices[randi] = vAvailableIndices.back();vAvailableIndices.pop_back();}}// Launch threads to compute in parallel a fundamental matrix and a homography// 步骤3：调用多线程分别用于计算fundamental matrix和homographyvector<bool> vbMatchesInliersH, vbMatchesInliersF;float SH, SF; // score for H and Fcv::Mat H, F; // H and F// ref是引用的功能:http://en.cppreference.com/w/cpp/utility/functional/ref// 计算homograpy并打分thread threadH(&Initializer::FindHomography,this,ref(vbMatchesInliersH), ref(SH), ref(H));// 计算fundamental matrix并打分thread threadF(&Initializer::FindFundamental,this,ref(vbMatchesInliersF), ref(SF), ref(F));// Wait until both threads have finishedthreadH.join();threadF.join();// Compute ratio of scores// 步骤4：计算得分比例，选取某个模型float RH = SH/(SH+SF);// Try to reconstruct from homography or fundamental depending on the ratio (0.40-0.45)// 步骤5：从H矩阵或F矩阵中恢复R,tif(RH>0.40)return ReconstructH(vbMatchesInliersH,H,mK,R21,t21,vP3D,vbTriangulated,1.0,50);else //if(pF_HF>0.6)return ReconstructF(vbMatchesInliersF,F,mK,R21,t21,vP3D,vbTriangulated,1.0,50);return false;
}
orbslam首先是从配对特征中随机迭代mMaxIterations次，每一次都从配对点中选出8个点用来计算homography和fundamental矩阵，都是用SVD来计算的，如下：
FindFundamental：
void Initializer::FindFundamental(vector<bool> &vbMatchesInliers, float &score, cv::Mat &F21)
{// Number of putative matchesconst int N = vbMatchesInliers.size();// 分别得到归一化的坐标P1和P2vector<cv::Point2f> vPn1, vPn2;cv::Mat T1, T2;Normalize(mvKeys1,vPn1, T1);Normalize(mvKeys2,vPn2, T2);cv::Mat T2t = T2.t();// Best Results variablesscore = 0.0;vbMatchesInliers = vector<bool>(N,false);// Iteration variablesvector<cv::Point2f> vPn1i(8);vector<cv::Point2f> vPn2i(8);cv::Mat F21i;vector<bool> vbCurrentInliers(N,false);float currentScore;// Perform all RANSAC iterations and save the solution with highest scorefor(int it=0; it<mMaxIterations; it++){// Select a minimum setfor(int j=0; j<8; j++){int idx = mvSets[it][j];vPn1i[j] = vPn1[mvMatches12[idx].first];vPn2i[j] = vPn2[mvMatches12[idx].second];}cv::Mat Fn = ComputeF21(vPn1i,vPn2i);F21i = T2t*Fn*T1; //解除归一化// 利用重投影误差为当次RANSAC的结果评分currentScore = CheckFundamental(F21i, vbCurrentInliers, mSigma);if(currentScore>score){F21 = F21i.clone();vbMatchesInliers = vbCurrentInliers;score = currentScore;}}
}
通过ComputeF21计算本质矩阵，
cv::Mat Initializer::ComputeF21(const vector<cv::Point2f> &vP1,const vector<cv::Point2f> &vP2)
{const int N = vP1.size();cv::Mat A(N,9,CV_32F); // N*9for(int i=0; i<N; i++){const float u1 = vP1[i].x;const float v1 = vP1[i].y;const float u2 = vP2[i].x;const float v2 = vP2[i].y;A.at<float>(i,0) = u2*u1;A.at<float>(i,1) = u2*v1;A.at<float>(i,2) = u2;A.at<float>(i,3) = v2*u1;A.at<float>(i,4) = v2*v1;A.at<float>(i,5) = v2;A.at<float>(i,6) = u1;A.at<float>(i,7) = v1;A.at<float>(i,8) = 1;}cv::Mat u,w,vt;cv::SVDecomp(A,w,u,vt,cv::SVD::MODIFY_A | cv::SVD::FULL_UV);cv::Mat Fpre = vt.row(8).reshape(0, 3); // v的最后一列cv::SVDecomp(Fpre,w,u,vt,cv::SVD::MODIFY_A | cv::SVD::FULL_UV);w.at<float>(2)=0; // 秩2约束，将第3个奇异值设为0 //强迫约束return u*cv::Mat::diag(w)*vt;
}
看到用的是线性SVD解。
通过重投影来评估本质矩阵的好坏。
float Initializer::CheckFundamental(const cv::Mat &F21, vector<bool> &vbMatchesInliers, float sigma)
{const int N = mvMatches12.size();const float f11 = F21.at<float>(0,0);const float f12 = F21.at<float>(0,1);const float f13 = F21.at<float>(0,2);const float f21 = F21.at<float>(1,0);const float f22 = F21.at<float>(1,1);const float f23 = F21.at<float>(1,2);const float f31 = F21.at<float>(2,0);const float f32 = F21.at<float>(2,1);const float f33 = F21.at<float>(2,2);vbMatchesInliers.resize(N);float score = 0;// 基于卡方检验计算出的阈值（假设测量有一个像素的偏差）const float th = 3.841; //置信度95%,自由度1const float thScore = 5.991;//置信度95%,自由度2const float invSigmaSquare = 1.0/(sigma*sigma);for(int i=0; i<N; i++){bool bIn = true;const cv::KeyPoint &kp1 = mvKeys1[mvMatches12[i].first];const cv::KeyPoint &kp2 = mvKeys2[mvMatches12[i].second];const float u1 = kp1.pt.x;const float v1 = kp1.pt.y;const float u2 = kp2.pt.x;const float v2 = kp2.pt.y;// Reprojection error in second image// l2=F21x1=(a2,b2,c2)// F21x1可以算出x1在图像中x2对应的线lconst float a2 = f11*u1+f12*v1+f13;const float b2 = f21*u1+f22*v1+f23;const float c2 = f31*u1+f32*v1+f33;// x2应该在l这条线上:x2点乘l = 0 const float num2 = a2*u2+b2*v2+c2;const float squareDist1 = num2*num2/(a2*a2+b2*b2); // 点到线的几何距离 的平方const float chiSquare1 = squareDist1*invSigmaSquare;if(chiSquare1>th)bIn = false;elsescore += thScore - chiSquare1;// Reprojection error in second image// l1 =x2tF21=(a1,b1,c1)const float a1 = f11*u2+f21*v2+f31;const float b1 = f12*u2+f22*v2+f32;const float c1 = f13*u2+f23*v2+f33;const float num1 = a1*u1+b1*v1+c1;const float squareDist2 = num1*num1/(a1*a1+b1*b1);const float chiSquare2 = squareDist2*invSigmaSquare;if(chiSquare2>th)bIn = false;elsescore += thScore - chiSquare2;if(bIn)vbMatchesInliers[i]=true;elsevbMatchesInliers[i]=false;}return score;
}
最后回到Initializer::Initialize，将单映矩阵和本质矩阵的得分进行比对，选出最合适的，就求出RT了。
ORBSLAM2同时考虑了单应和本质，SLAM14讲中也说到，工程实践中一般都讲两者都计算出来选择较好的，不过效率上会影响比较多感觉。
opencv实现就比较简单了，思路和上面的类似，只是现在只考虑本质矩阵。在之前获取到特征点之后，
/*add ransac method for accurate match*/30 vector<Point2f> vLeftP2f;31 vector<Point2f> vRightP2f;32 for(auto& each:mMatches)33 {34 vLeftP2f.push_back(mLeftKps[each.queryIdx].pt);35 vRightP2f.push_back(mRightKps[each.trainIdx].pt);36 }37 vector<unsigned char> vTemp(vLeftP2f.size());/*计算本质矩阵，用RANSAC*/38 Mat transM = findEssentialMat(vLeftP2f, vRightP2f, cameraMatrix,RANSAC, 0.999, 1.0, vTemp);39 vector<DMatch> optimizeM;40 for(int i = 0; i < vTemp.size(); i++)41 {42 if(vTemp[i])43 {44 optimizeM.push_back(mMatches[i]);45 }46 }47 mMatches.swap(optimizeM);48 cout << transM<<endl;49 Mat optimizeP;50 drawMatches(mLeftImg, mLeftKps, mRightImg, mRightKps, mMatches, optimizeP);51 imshow("output5", optimizeP);
看下结果：
确实效果好多了，匹配准确度比之前的要好，之后我们就可以通过这个本质矩阵来计算RT了。
Mat R, t, mask;
recoverPose(transM, vLeftP2f, vRightP2f, cameraMatrix, R, t, mask);
一个接口搞定。最后我们可以通过验证对极约束，来看看求出的位姿是否准确。
定义检查函数：
11 Mat cameraMatrix = (Mat_<double>(3,3) << CAM_FX, 0.0, CAM_CX, 0.0, CAM_FY, CAM_CY, 0.0, 0.0, 1.0);1213 bool epipolarConstraintCheck(Mat CameraK, vector<Point2f>& p1s, vector<Point2f>& p2s, Mat R, Mat t)14 {15 for(int i = 0; i < p1s.size(); i++)16 {17 Mat y1 = (Mat_<double>(3,1)<<p1s[i].x, p1s[i].y, 1);18 Mat y2 = (Mat_<double>(3,1)<<p2s[i].x, p2s[i].y, 1);19 //T 的凡对称矩阵20 Mat t_x = (Mat_<double>(3,3)<< 0, -t.at<double>(2,0), t.at<double>(1,0),21 t.at<double>(2,0), 0, -t.at<double>(0,0),22 -t.at<double>(1,0),t.at<double>(0,0),0);23 Mat d = y2.t() * cameraMatrix.inv().t() * t_x * R * cameraMatrix.inv()* y1;24 cout<<"epipolar constraint = "<<d<<endl;25 }26 }
最后可以看到结果都是趋近于0的，证明位姿还是比较准确的。
