Transform point position in trapezoid to rectangle position - matrix

Question

I am trying to find out how I can transform a coordinate Pxy within the green trapezoid below into the equivalent coordinate on the real ground plane.
I have the exact measures of the room, meaning I can exactly say how long A,B,C and D are in that room shown below.
Also I know how long A,B,C and D are in that green triangle (coordinate wise).
I have already been reading about homography and matrix transformation, but can't really wrap my head around it. Any input steering me into the right direction would be appreciated.
Thanks!

Answer 1

There is the code computes the affine transformation matrix using the library Opencv (it shows how to trasform your trapezoid to rectangle and how to find transformation matrix for futher calculations):
//example from book
// Learning OpenCV: Computer Vision with the OpenCV Library
// by Gary Bradski and Adrian Kaehler
// Published by O'Reilly Media, October 3, 2008
#include <cv.h>
#include <highgui.h>
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char* argv[])
{
IplImage *src=0, *dst=0;
// absolute or relative path to image should be in argv[1]
char* filename = argc == 2 ? argv[1] : "Image0.jpg";
// get the picture
src = cvLoadImage(filename,1);
printf("[i] image: %s\n", filename);
assert( src != 0 );
// points (corners of )
CvPoint2D32f srcQuad[4], dstQuad[4];
// transformation matrix
CvMat* warp_matrix = cvCreateMat(3,3,CV_32FC1);
// clone image
dst = cvCloneImage(src);
// define all the points
//here the coordinates of corners of your trapezoid
srcQuad[0].x = ??; //src Top left
srcQuad[0].y = ??;
srcQuad[1].x = ??; //src Top right
srcQuad[1].y = ??;
srcQuad[2].x = ??; //src Bottom left
srcQuad[2].y = ??;
srcQuad[3].x = ??; //src Bot right
srcQuad[3].y = ??;
//- - - - - - - - - - - - - -//
//coordinates of rectangle in src image
dstQuad[0].x = 0; //dst Top left
dstQuad[0].y = 0;
dstQuad[1].x = src->width-1; //dst Top right
dstQuad[1].y = 0;
dstQuad[2].x = 0; //dst Bottom left
dstQuad[2].y = src->height-1;
dstQuad[3].x = src->width-1; //dst Bot right
dstQuad[3].y = src->height-1;
// get transformation matrix that you can use to calculate
//coordinates of point Pxy
cvGetPerspectiveTransform(srcQuad,dstQuad,warp_matrix);
// perspective transformation
cvWarpPerspective(src,dst,warp_matrix);
cvNamedWindow( "cvWarpPerspective", 1 );
cvShowImage( "cvWarpPerspective", dst );
cvWaitKey(0);
cvReleaseMat(&warp_matrix);
cvReleaseImage(&src);
cvReleaseImage(&dst);
cvDestroyAllWindows();
return 0;
Hope it will be helpfull!

Answer 2

If I understand your question correctly, you are looking for the transform matrix that expresses the position and orientation (aka the "pose") of your camera in relation to the world. If you have this matrix - lets call it M - you could map any point from your camera coordinate frame to the world coordinate frame and vice versa. In your case you'll want to transform a rectangle onto the plane (0, 1, 0)^T + 0 in world coordinates.
There are several ways to derive this pose Matrix. First of all you'll need to know another matrix - K - which describes the internal camera parameters to convert positions in the camera coordinate frame to actual pixel positions. This involves a standard pinhole projection as well as radial distortion and a few other things.
To determine both K and M you have to calibrate your camera. This is usually done by taking a calibration pattern (e.g. a chessboard-pattern) for which the positions of the chessboard-fields are known. Then you can establish so called Point-Correspondences between the known positions on the pattern and the observed pixel-positions. Once you have enough of these point-pairs you can solve a Matrix H = KM. This is your Homography matrix you've mentioned already. Once you have that, you can reconstruct K and M.
So much for the theory. For the practical part I would suggest to have a look at the OpenCV-Documentations (e.g. you could start here: OpenCV Camera calibration and here: OpenCV Pose estimation).
I hope this will point you in the right directions ;)

Answer 3

Just for the sake of completion. I ended up looking at the thread suggested by #mmgp and implemented a solution that is equivalent to the one presented by Christopher R. Wren:
Perspective Transform Estimation
This turned out to work really well for my case, although there was some distortion from the camera.