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We have a grid with red squares on it. Meaning we have an array of 3 squares (with angles == 90 deg) which as we know have same size, lying on the same plane and with same rotation relative to the plane they are lying on, and are not situated on same line on plane.

We have a projection of the space which contains the plane with squares.


We want to turn our plane projection with squares so that we would see it like it's facing us, in general we need a formula for turning each point of that original plane projection so that it would be facing us like on the image below.

What formulas can be used for solving such problem, how to solve it, has any one faced something like this before?


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closed as off-topic by Ricardo Andrade, Chris Godsil, abx, José Figueroa-O'Farrill, Yemon Choi Sep 28 '14 at 20:32

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If this question can be reworded to fit the rules in the help center, please edit the question.

Is English your native language? If it is, please fix your spelling and grammar. – Ricky Demer Sep 13 '10 at 23:58
@Ricky Demer: Sorry it is not. Could you please point me on to my errors - google docs does not see any.=( – Ole J Sep 14 '10 at 0:07
You fixed most of the errors. "relatively" should be "relative", "situated not" should be "not situated", "of space" should be "of the space", and "like its facing us" should be "like it's facing us". I think you left out the word "projection" in "We want to knowing all that turn our plane", if I am right it should be "We want to know all projections that turn out plane". "space in which is that" is technically correct, but something like "space which contains the" would sound better. – Ricky Demer Sep 14 '10 at 1:05
@Ricky Demer: thank you very much - I appreciate your help. – Ole J Sep 14 '10 at 22:34
You apparently realized this, but my using "turn out plane" was a misspelling, I should have typed "turn our plane". – Ricky Demer Sep 15 '10 at 4:32
up vote 2 down vote accepted

If $A,B,C,D$ are the four points determining your original square in space, and $a,b,c,d$ are the four points determining the projected quadrilateral, then there is a unique projective transformation mapping $(A,B,C,D)$ to $(a,b,c,d)$; I believe it is called a homography. Computing the projective transformation (sometimes called inversion) is a well-studied problem in computer vision. Some details may be found on Stack Overflow here. The computational issues might better be pursued on Stack Overflow.

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ok.. I see... thank you. – Ole J Sep 14 '10 at 0:26

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