Homography (mathematic) question

Hello everyone.

I have a small question that makes me crazy. I hope that someone, smartest than me, could help me to get out of it :).

I will try to explain my problem.

I compute the homography H between images A and B. I use findHomography for that.
No problem, it works well.
No I want to apply H. But in fact I want to apply H’ that corresponds to the Homography of the transformation between A and B’ without recomputing it. And B’ is just a flip of B. B’ is read bottom to up while B is read top down.

So I would like to know which transformation I need to apply to H.

At the beginning I was thinking that something like:

/* modify matrix for reverse Y axis */

Mat F = Mat::eye(3, 3, CV_64FC1);
F.at<double>(1,1) = -1.0;
F.at<double>(1,2) = height - 1.0;
H = F * H * F.inv();|

But it won’t work for image that have different size.
Well, I’m stuck with this problem. I know I just show you a small piece of code (code is too big to write a small standalone example) but as it is a mathematics issue, maybe someone could help.

So. Someone helped me … I think she found the solution. I need two flip matrix to handle the two different sizes:

    Mat F1 = Mat::eye(3, 3, CV_64FC1);
    F1.at<double>(1,1) = -1.0;
    F1.at<double>(1,2) = source_ry - 1.0;

    Mat F2 = Mat::eye(3, 3, CV_64FC1);
    F2.at<double>(1,1) = -1.0;
    F2.at<double>(1,2) = target_ry - 1.0;

    H = F2.inv() * H * F1;

Hi @Cyril_Richard ,

Warning, I’m not sure if I got you right!

F is the flip transformation, so

B' = B * F

H is the homography so

B = A * H

How to get B’ from A:

B' = B * F = A * H * F = A * H'


H' = H * F

Of course you will have to rewrite the formulas if you change the factors order, like

B' = F * B  instead of B' = B * F

Does it make sense?

Hello @Alejandro_Silvestri .

Thank you for your message.
I’m not sure to understand your message, however, my issue was due to the possible different size between the two images.
By setting two flip matrix if fixes the issue.

I’ve tested with a bunch of images and it looks very good so far.

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