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kyewong

Member Since 18 Oct 2011
Offline Last Active Dec 18 2013 08:56 PM

In Topic: modelview and projection matrices

20 November 2013 - 07:17 PM

In photogrammetry, this is a pretty standard problem, so we can consider it solved. ;-)

But as I already said: It is not simple.

You can probably get away with a reasonably good approximation by doing the following:

1) Solve the equation system for each pair of projected and unprojected points. This yields a transformation matrix for each pair.

2) Choose a good one from these by projecting all the points with each matrix and pick the one that produces the least error.

Do you mean I should compute such a transformation matrix which is one part of the modelview matrix? If yes, then the projection matrix and factors should be considered fixed for this problem, it is right?

In Topic: modelview and projection matrices

19 November 2013 - 06:14 PM

Yes, that's possible in principle, but it's not simple.

Basically you just solve the resulting system of linear equations and you are done.

What makes this hard is that  the system may be massively over-determined.This means, that some equations contradict each other, so you have to find a way of resolving those contradictions.

Google for suitable algorithms, if unsure.

Yes, it may probably be a over-determined problem, that's the reason why I said "optimal". My question now is, whether it is simple to solve for such a modelview matrix (or the translation/rotation/scale matrix), or such a projection matrix? don't know if there's any prior attempts on solving this problem...

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