finding the real world coordinates of an image point

Question

I am searching lots of resources on internet for many days but i couldnt solve the problem.

I have a project in which i am supposed to detect the position of a circular object on a plane. Since on a plane, all i need is x and y position (not z) For this purpose i have chosen to go with image processing. The camera(single view, not stereo) position and orientation is fixed with respect to a reference coordinate system on the plane and are known

I have detected the image pixel coordinates of the centers of circles by using opencv. All i need is now to convert the coord. to real world.

http://www.packtpub.com/article/opencv-estimating-projective-relations-images in this site and other sites as well, an homographic transformation is named as:

p = C[R|T]P; where P is real world coordinates and p is the pixel coord(in homographic coord). C is the camera matrix representing the intrinsic parameters, R is rotation matrix and T is the translational matrix. I have followed a tutorial on calibrating the camera on opencv(applied the cameraCalibration source file), i have 9 fine chessbordimages, and as an output i have the intrinsic camera matrix, and translational and rotational params of each of the image.

I have the 3x3 intrinsic camera matrix(focal lengths , and center pixels), and an 3x4 extrinsic matrix [R|T], in which R is the left 3x3 and T is the rigth 3x1. According to p = C[R|T]P formula, i assume that by multiplying these parameter matrices to the P(world) we get p(pixel). But what i need is to project the p(pixel) coord to P(world coordinates) on the ground plane.

I am studying electrical and electronics engineering. I did not take image processing or advanced linear algebra classes. As I remember from linear algebra course we can manipulate a transformation as P=[R|T]-1*C-1*p. However this is in euclidian coord system. I dont know such a thing is possible in hompographic. moreover 3x4 [R|T] Vector is not invertible. Moreover i dont know it is the correct way to go.

Intrinsic and extrinsic parameters are know, All i need is the real world project coordinate on the ground plane. Since point is on a plane, coordinates will be 2 dimensions(depth is not important, as an argument opposed single view geometry).Camera is fixed(position,orientation).How should i find real world coordinate of the point on an image captured by a camera(single view)?

EDIT I have been reading "learning opencv" from Gary Bradski & Adrian Kaehler. On page 386 under Calibration->Homography section it is written: q = sMWQ where M is camera intrinsic matrix, W is 3x4 [R|T], S is an "up to" scale factor i assume related with homography concept, i dont know clearly.q is pixel cooord and Q is real coord. It is said in order to get real world coordinate(on the chessboard plane) of the coord of an object detected on image plane; Z=0 then also third column in W=0(axis rotation i assume), trimming these unnecessary parts; W is an 3x3 matrix. H=MW is an 3x3 homography matrix.Now we can invert homography matrix and left multiply with q to get Q=[X Y 1], where Z coord was trimmed.

I applied the mentioned algorithm. and I got some results that can not be in between the image corners(the image plane was parallel to the camera plane just in front of ~30 cm the camera, and i got results like 3000)(chessboard square sizes were entered in milimeters, so i assume outputted real world coordinates are again in milimeters). Anyway i am still trying stuff. By the way the results are previosuly very very large, but i divide all values in Q by third component of the Q to get (X,Y,1)

FINAL EDIT

I could not accomplish camera calibration methods. Anyway, I should have started with perspective projection and transform. This way i made very well estimations with a perspective transform between image plane and physical plane(having generated the transform by 4 pairs of corresponding coplanar points on the both planes). Then simply applied the transform on the image pixel points.

Answer 1

You said "i have the intrinsic camera matrix, and translational and rotational params of each of the image.” but these are translation and rotation from your camera to your chessboard. These have nothing to do with your circle. However if you really have translation and rotation matrices then getting 3D point is really easy.

Apply the inverse intrinsic matrix to your screen points in homogeneous notation: C^-1*[u, v, 1], where u=col-w/2 and v=h/2-row, where col, row are image column and row and w, h are image width and height. As a result you will obtain 3d point with so-called camera normalized coordinates p = [x, y, z]^T. All you need to do now is to subtract the translation and apply a transposed rotation: P=R^T(p-T). The order of operations is inverse to the original that was rotate and then translate; note that transposed rotation does the inverse operation to original rotation but is much faster to calculate than R^-1.