CODEWITHSUNDEEP
image-processinghomographyprojection-matrix
Kozolovska
Kozolovska

Reputation: 1119

Homography and projective transformation

im trying to write a code that will do projective transformation, but with more than 4 key points. i found this helpful guide but it uses 4 points of reference https://math.stackexchange.com/questions/296794/finding-the-transform-matrix-from-4-projected-points-with-javascript

i know that matlab uses has a function tcp2form that handles that, but i haven't found a way so far.

anyone can give me some guidance, on how to do so? i can solve the equations using (least squares), but i'm stuck since i have a matrix that is larger than 3*3 and i can't multiple the homogeneous coordinates.

Thanks

Upvotes: 1

Views: 997

Answers (1)

MvG
MvG

Reputation: 60908

If you have more than four control points, you have an overdetermined system of equations. There are two possible scenarios. Either your points are all compatible with the same transformation. In that case, any four points can be used, and the rest will match the transformation exactly. At least in theory. For the sake of numeric stability you'd probably want to choose your points so that they are far from being collinear.

Or your points are not all compatible with a single projective transformation. In this case, all you can hope for is an approximation. If you want the best approximation, you'll have to be more specific about what “best” means, i.e. some kind of error measure. Measuring things in a projective setup is inherently tricky, since there are usually a lot of arbitrary decisions involved.

What you can try is fixing one matrix entry (e.g. the lower right one to 1), then writing the conditions for the remaining 8 coordinates as a system of linear equations, and performing a least squares approximation. But the choice of matrix representative (i.e. fixing one entry here) affects the least squares error measure while it has no effect on the geometric meaning, so this is a pretty arbitrary choice. If the lower right entry of the desired matrix should happen to be zero, you'd computation will run into numeric problems due to overflow.

Upvotes: 0

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