Reverse of OpenCV projectPoints

Question

I have a camera facing the equivalent of a chessboard. I know the world 3d location of the points as well as the 2d location of the corresponding projected points on the camera image. All the world points belong to the same plane. I use solvePnP:

Matx33d camMat;
Matx41d distCoeffs;
Matx31d rvec;
Matx31d tvec;
std::vector<Point3f> objPoints;
std::vector<Point2f> imgPoints;
solvePnP(objPoints, imgPoints, camMat, distCoeffs, rvec, tvec);

I can then go from the 3d world points to the 2d image points with projectPoints:

std::vector<Point2f> projPoints;
projectPoints(objPoints, rvec, tvec, camMat, distCoeffs, projPoints);

projPoints are very close to imgPoints.

How can I do the reverse with a screen point that corresponds to a 3d world point that belongs to the same plane. I know that from a single view, it's not possible to reconstruct the 3d location but here I'm in the same plane so it's really a 2d problem. I can calculate the reverse rotation matrix as well as the reverse translation vector but then how can I proceed?

Matx33d rot;
Rodrigues(rvec, rot);
Matx33d camera_rotation_vector;
Rodrigues(rot.t(), camera_rotation_vector);
Matx31d camera_translation_vector = -rot.t() * tvec;

parametrize a line from your camera center through the pixel. parametrize the plane. find the parameters of the line where it hits the plane. — Micka, Dec 07 '15 at 20:18

score 3 · Answer 1 · answered Jun 24 '19 at 15:19

Suppose you calibrate your camera by objpoints-imgpoints pair. Note first is real world 3-d coordinate of featured points on calibration board, the second one is 2-d pixel location of featured points in each image. So both of them should be the list where it has the number of calibration board images element. After following line of Python code, you will have calibration matrix mtx, each calibration board's rotations rvecs, and its translations tvecs.

ret, mtx, dist, rvecs, tvecs = cv2.calibrateCamera(objpoints, imgpoints, gray.shape[::-1], None, np.zeros(5,'float32'),flags=cv2.CALIB_USE_INTRINSIC_GUESS )

Now we can find any pixel's 3D coordinate under the assumption. That assumption is we need to define some reference point. Let's assume our reference is 0th (first) calibration board, where its pivot point is at 0,0 where the long axis of the calibration board is x, and the short one is y-axis, also the surface of calibration board shows Z=0 plane. Here is how we can create a projection matrix.

# projection matrix
Lcam=mtx.dot(np.hstack((cv2.Rodrigues(rvecs[0])[0],tvecs[0])))

Now we can define any pixel location and desired Z value. Note since I want to project (100,100) pixel location on the reference calibration board, I set Z=0.

px=100
py=100
Z=0
X=np.linalg.inv(np.hstack((Lcam[:,0:2],np.array([[-1*px],[-1*py],[-1]])))).dot((-Z*Lcam[:,2]-Lcam[:,3]))

Now we have X and Y coordinate of (px,py) pixel, it is X[0], X[1] . the last element of X is lambda factor. As a result we can say, pixe on (px,py) location drops on X[0],X[1] coordinate on the 0th calibration board's surface.

score 1 · Accepted Answer · edited May 23 '17 at 12:34

1

This question seems to be a duplicate of another Stackoverflow question in which the asker provides nicely the solution. Here is the link: Answer is here: Computing x,y coordinate (3D) from image point

edited May 23 '17 at 12:34

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answered Dec 09 '15 at 18:52

gregoiregentil

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Reverse of OpenCV projectPoints

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