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I have an intuition that given a rectangle of arbitrary width and height rotated arbitrarily in 3D space and perspective-projected onto the 2D plane, that not all arbitrary sets of resulting 2D points are possible.

That is to say that I intuit that there is a constrained relationship between the points due to the original rectangle having the constraints of right angles and parallel sides.

Is my intuition right or wrong?

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    $\begingroup$ What sorts of projections did you have in mind? $\endgroup$
    – amd
    Aug 9, 2017 at 1:15
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    $\begingroup$ Do you mean parallel projections or perspective projections? If parallel, then only parallelograms are possible. If perspective, then I'm not sure but I'd guess all convex quadrilaterals are possible. $\endgroup$
    – user856
    Aug 9, 2017 at 1:50
  • $\begingroup$ Indeed: see math.stackexchange.com/a/13409/856 and then geometrictools.com/Documentation/PerspectiveMapping.pdf $\endgroup$
    – user856
    Aug 9, 2017 at 1:54
  • $\begingroup$ I'm only interested in perspective projections. As in analysing or interacting with images from cameras. Let me clarify the question text. $\endgroup$ Aug 9, 2017 at 2:02
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    $\begingroup$ If you just look at planar perspective projections, the only restriction is that colinearity is preserved—a rectangle can be mapped to any nondegenerate quadrilateral. As for the 3D projections that you’re interested in, they can collapse a rectangle into a line segment, so even the nondegeneracy restriction is removed. $\endgroup$
    – amd
    Aug 9, 2017 at 7:57

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I've done a lot of searching and reading on the web since posting the question but I don't understand all the concepts in all the articles so I may have overlooked something...

It turns out my intuition was only "slightly right", but mostly wrong.

There are only two constraints if the 4 points are not considered to have an intrinsic order:

  1. The four points must form a convex quadrilateral.
  2. No three points can lie on the same line. (The degenerate case where the points actually form a triangle. Not illustrated.)

types of quadrilaterals

I didn't think about this when posting the question, but if the points are considered to have an intrinsic order then there is a constraint that no two edges may cross one another. But this is just one kind of non-convex quadrilateral anyway, so is already covered.

My intuition was that there would be a lot more to constrain acceptable sets of four points than this, so in that I was wrong.

I now believe that for any given set of four points forming a convex quadrilateral there are exactly two squares in 3D space than can map to them via some perspective projection onto the 2D plane, and an infinite number of rectangles. (I did not fully understand every concept in the articles I read to reach this conclusion though, so I may still be in error.)

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    $\begingroup$ I think this would have been more useful as a clarification (edit) to the Question. Given that a 3D rectangle is a convex quadrilateral, its point-perspective projections onto a plane will also be convex (even in the degenerate cases where the projection is a line segment). Posting an Answer that ends, without mathematical arguments, in "I may still be in error" is not the sort of content we are trying to collect and curate. $\endgroup$
    – hardmath
    Aug 19, 2017 at 16:30
  • $\begingroup$ Sometimes people that don't know much about mathematics have questions and answers that involve mathematics and we just do the best we can in the absence of answers from experts that we can understand. It does seem obvious when looked at this way but when a computer programmer faces such a mathematical problem he has to tell the computer to check for such things since the computer has no common sense. In that realm we need to check for sane inputs which, while not what I thought I was asking about, hadn't occurred to me when I asked either. I very much welcome better answers! $\endgroup$ Aug 19, 2017 at 22:55

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