How can I use matlab to do homographies knowing the rotation angles?

I want to use matlab to do project/warp images using homographies. I've computed the homography matrix using the equation H = K * R1 * R0^-1 * K^-1, where H is the homography matrix, K is the camera intrinsic matrix, R1 is the final orientation matrix and R0 is the initial orientation matrix. The rotation matrix is computed based on where I want the camera to look at and where it was looking at when the photo was taken; I then use tform = maketform ('projective', H) and I_reg=imtransform(I, tform), where I is the image and I_reg is the registered image, but the results are not correct. What can I be doing wrong?

[David Young]

In what way are the results not correct? I've always found it very easy to get this wrong myself, but the solution depends on what the problem seems to be. By the way, you probably should use projective2d instead of maketform, and imwarp instead of imtransform, if you have up-to-date MATLAB. One simple possibility - have you tried transposing the matrix before giving it to maketform?

[Guilherme]

The image that comes out of the transform is really really tiny and all black or all white, depending on how I try to fix the problem.

What I have is the extrinsics of the camera relative to a checkerboard pattern plane and I'm trying to fix the perspective of the image so that the camera appears to be perpendicular to the checkerboard. So I have a 3*3 rotation matrix of the camera relative to the checkboard (R0) I determine the orientation that I want (R1), compute H and pass it to maketform. I get a 3*3 matrix in the end. I tried to transpose it, but nothing improved. I've also tried projective2d and imwarp and it didn't change.

My matrices are

    R0 = [.9997, .0116, .0233; 
          .0101, -.998, .0632;
          .0240, -.0629, -.997]

    R1 = [1,0,0;
          0,-1,0;
          0,0,-1];
.
Rf=R1*R0^-1 = [0.9996    0.0101    0.0240
             -0.0116    0.9979    0.0629
             -0.0233   -0.0632    0.9978]

I insert the matrix Rf in both maketform('projective', Rf) and in projective2d(Rf) and then use the output on both imtransform and imwarp.

The original and final images are the ones attached

inital figure

processed figure

[Guilherme]

I've tried to apply very small rotations around the y axis (0.01 degrees) and find out that two things are happening:

- the rotated image seems to have rotated much more than the angle I inserted; - The image size is, somehow, not updated to accommodate the new image size, so only a portion of it is shown.

What may be going wrong? What am I missing;

The rotation matrix for a 0.01 degree rotation around the y axis is

[1.0000, 0, 0.0002;

0;1.0000,0;

-0.0002,0,1.0000]

[Matt J]

I've computed the homography matrix using the equation H = K * R1 * R0^-1 * K^1

Do you mean K^-1 or K^1?

Also, is K the total camera matrix (3x4) or just the calibration matrix (3x3)? If the latter, shouldn't the camera center of projection appear in the equations somewhere?

[Guilherme]

I meant K^-1.

I got this procedure from Szeliski's "Computer Vision-Algorithms and Applications", Eq.2.72, page 51. In it, K is the 3*3 calibration matrix where the camera centre is included.

I have the focal lengths and the camera centre in pixels and K is an upper triangular matrix.

I'm very new to this field, I must say, so I'm not at all comfortable with the theory behind this.

[Matt J]

I mean the 3D world coordinates of the camera center. The calibration matrix doesn't normally contain any information about the position of the camera in world coordinates and it seems to me that you would need that in order to derive the homography.

[Guilherme]

I found out what I was doing wrong. I didn't realize that matlab uses the notation x*H and was using H*x. I had to transpose the homography matrix. Now it is working!

[he]

I don't konow what do you mean.Do you mean that we must use x*H ?I just wonder am I right?x is the cordinate of a point,its size is 3X1,and the size of H is 3X3,so x*H is wrong.Another question,there are a lot of method to get H,can H = K * R1 * R0^-1 * K^-1 be used to project one image to another?Thank you for read!

[Guilherme]

You must transpose x and H before multiplying by H. If you have H (3*3) and x (3*1), you to transpose H and get H' (3*3) and transpose x and get x' (1*3) and multiply x'*H'

Yes, H can be used to get to project and image into another image. If R0 is the current orientation of the image, if must compute R1 (the desired orientation of the image) and use H = K* R1 * * R0^-1* K^1.