接上一篇。
看函数
第一个参数和第二个参数与上一篇类似,不过上一篇的是多幅图的多组点,类型为双重vector,这里是一副图中的一组点,类型为单重vector。
vector<Point3f> calcBoardCornerPositions(int gridW, int gridH, float squareSize)
{vector<Point3f> objectPoints;for (int i = 0; i < gridH; i++)for (int j = 0; j < gridW; j++)objectPoints.push_back(Point3f(float(j*squareSize), float(i*squareSize), 0));return objectPoints;
}vector<Point3f> objectPoints = calcBoardCornerPositions(grids.width, grids.height);
第二个参数
vector<Point2f> corners;bool found = findChessboardCorners(images, grids, corners, CALIB_CB_ADAPTIVE_THRESH + CALIB_CB_NORMALIZE_IMAGE + CALIB_CB_FAST_CHECK);
第三四个参数既是上篇的结果,读入之。
再后两个参数是我们要得到的结果,声明后带入即可。
再其他参数默认就好。
那么得到rvecs, tvecs之后怎么得到棋盘的空间姿态呢,姿态坐标系又是什么呢?
rvecs是旋转向量,使用Rodrigues转成旋转矩阵。原理百度可查。
现在得到了旋转矩阵和平移矩阵了,那么根据标定公式就可以得到世界坐标了。
接下来求解三轴旋转角。原理如下:
这个矩阵对应与旋转矩阵,由旋转矩阵即可反解欧拉角。
最终效果:
X,Y,Z是以镜头为原点的世界坐标单位0.1mm,ang_X,Y,Z是棋盘沿三轴的旋转角度,单位度。
代码实时运行,速度超过20FPS,精度1cm以内。
完整代码:
#include <cv.h>
#include <highgui.h>
#include <iostream>
#include <math.h>
#define pi 3.14159265358979323846
using namespace std;
using namespace cv;
vector<Point3f> calcBoardCornerPositions(int gridW, int gridH, float squareSize)
{vector<Point3f> objectPoints;for (int i = 0; i < gridH; i++)for (int j = 0; j < gridW; j++)objectPoints.push_back(Point3f(float(j*squareSize), float(i*squareSize), 0));return objectPoints;
}
vector<Point3f> calcBoardCornerPositions(int gridW, int gridH)
{vector<Point3f> objectPoints;for (int i = 0; i < gridH; i++)for (int j = 0; j < gridW; j++)objectPoints.push_back(Point3f(float(j * 89), float(i * 83), 0));return objectPoints;
}
int main()
{VideoCapture cap(1);if (!cap.isOpened()) return -1;//cap.set(CV_CAP_PROP_FRAME_WIDTH, 1920);//cap.set(CV_CAP_PROP_FRAME_HEIGHT, 1080);Mat images, gray;Size grids(9, 6);int Grids_Size = 260;float Ang_X, Ang_Y, Ang_Z;float X, Y, Z;char key; int i = 0;//float A[][3] = { { 644.8137843176841, 0, 302.6526363184274 },{0, 649.3562275395091, 286.5283609342574 },{0, 0, 1 }};//float B[] = { 0.01655500980525433, 0.1901812353222618, 0.003461616464410258, 0.002455084197033077, -1.444734184159016 };float A[][3] = { { 988.74755, 0, 309.709197 }, { 0, 988.2410178, 239.85705 }, { 0, 0, 1 } };float B[] = { -0.41287433, 1.80600373, 0.00250586, 0.0013610796, -7.6232044988 };Mat rvecs(3, 1, CV_32F), tvecs(3, 1, CV_32F), cameraMatrix(3, 3, CV_32F), distCoeffs(1, 5, CV_32F), R(3, 3, CV_32FC1);for (int i = 0; i < 3; i++)for (int j = 0; j < 3; j++){cameraMatrix.at<float>(i, j) = A[i][j];R.at<float>(i, j) = 0;}for (int i = 0; i < 5;i++)distCoeffs.at<float>(0, i) = B[i];namedWindow("chessboard", 0);namedWindow("qqq", 0);cap >> images;while (1){cap >> images;imshow("qqq", images);vector<Point2f> corners;bool found = findChessboardCorners(images, grids, corners, CALIB_CB_ADAPTIVE_THRESH + CALIB_CB_NORMALIZE_IMAGE + CALIB_CB_FAST_CHECK);if (found) {cvtColor(images, gray, CV_BGR2GRAY);cornerSubPix(gray, corners, Size(11, 11), Size(-1, -1),TermCriteria(CV_TERMCRIT_EPS + CV_TERMCRIT_ITER, 30, 0.1));//0.1为精度drawChessboardCorners(images, grids, corners, found);vector<Point3f> objectPoints = calcBoardCornerPositions(grids.width, grids.height, Grids_Size);//vector<Point3f> objectPoints = calcBoardCornerPositions(grids.width, grids.height);solvePnP(objectPoints, corners, cameraMatrix, distCoeffs, rvecs, tvecs);Rodrigues(rvecs, R);Ang_X = asin(R.at<double>(1, 0) / cos(asin(-R.at<double>(2, 0)))) / pi * 180;Ang_Y = asin(-R.at<double>(2, 0)) / pi * 180;Ang_Z = asin(R.at<double>(2, 1) / cos(asin(-R.at<double>(2, 0)))) / pi * 180;X = R.at<double>(0, 0) *objectPoints[22].x + R.at<double>(0, 1) * objectPoints[22].y + R.at<double>(0,2) * objectPoints[22].z + tvecs.at<double>(0,0);Y = R.at<double>(1, 0) *objectPoints[22].x + R.at<double>(1, 1) * objectPoints[22].y + R.at<double>(1, 2) * objectPoints[22].z + tvecs.at<double>(1, 0);Z = R.at<double>(2, 0) *objectPoints[22].x + R.at<double>(2, 1) * objectPoints[22].y + R.at<double>(2, 2) * objectPoints[22].z + tvecs.at<double>(2, 0);putText(images, "X:"+to_string(X), { 1, 50 }, 0, 1.0f, CV_RGB(255, 0, 0), 2);putText(images, "Y:"+to_string(Y), { 1, 150 }, 0, 1.0f, CV_RGB(0, 255, 0), 2);putText(images, "Z:"+to_string(Z), { 1, 250 }, 0, 1.0f, CV_RGB(0, 0, 255), 2);putText(images, "Ang_X:" + to_string(Ang_X), { 300, 50 }, 0, 1.0f, CV_RGB(255, 0, 0), 2);putText(images, "Ang_Y:" + to_string(Ang_Y), { 300, 150 }, 0, 1.0f, CV_RGB(0, 255, 0), 2);putText(images, "Ang_Z:" + to_string(Ang_Z), { 300, 250 }, 0, 1.0f, CV_RGB(0, 0, 255), 2);imshow("chessboard", images);}key=waitKey(20);if (key == ' ')break;}return 0;
}
注意,代码中取objectPoints[22]计算是求解棋盘较为中间的点的坐标,若取0则是计算的棋盘第一个点坐标,这个不重要。
完了。
接下来如果有空的话把双目的也写上来。还有单目标定也可以用MATLAB工具箱,结果差不多。
注意:本博文采用的是HPR转角系统;同时此处计算的X,Y,Z并不是摄像机在世界坐标系下的位置(即像片外方位线元素),而是世界坐标系下的某点的坐标在摄像机坐标系下的坐标
转载自:https://blog.csdn.net/calch/article/details/54985056
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