I have a square with the length of the sides being 1. This square is transformed by an unknown transformation matrix in the 3D space and then projected back to the plane (the projection is known). I know the coordinates of the 4 angles of the transformed, projected square.

How do I find the unknown transformation matrix?

UPDATE: sorry I forgot to add that the projection is perspective.

  • $\begingroup$ Just to clarify, you know the original coordinates of the square correct? $\endgroup$ – icurays1 Nov 22 '12 at 17:11
  • $\begingroup$ When you say "projected back to the plane" do you mean the original one in which the square started, or a different plane? I thought the first, but then wondered why you would add "the projection is known" unless it was projection to a different plane. $\endgroup$ – Tom Oldfield Nov 22 '12 at 17:14
  • $\begingroup$ @icurays1: yes, I can even suppose they are let's say (0,0) (0,1) (1,1), (1,0) $\endgroup$ – MrTJ Nov 23 '12 at 10:32
  • $\begingroup$ @Tom Oldfield yes, it's the original plane. And yes, you're right, it's a kind of redundant information. All in all the square is transformed first by the unknown matrix and then by the projection and I would be happy to know either the unknown matrix or the matrix of the concatenated transformations. $\endgroup$ – MrTJ Nov 23 '12 at 10:34
  • 1
    $\begingroup$ Do you have access to Hartley and Zisserman? This is explained there. $\endgroup$ – littleO Nov 26 '12 at 8:39

I would guess that you're supposed to use "homogeneous coordinates", and that the "3D" matrix you're supposed to compute is the matrix of a projective transformation. When we use homogeneous coordinates, a coordinate vector for a point in a plane has three components rather than two. This may seem unintuitive or unappealing, but there are advantages to homogeneous coordinates: 1) "points at infinity" can be assigned coordinates as easily as ordinary points ; 2) every projective transformation is given by multiplication by a $3 \times 3$ matrix. (There are other advantages too; mathematicians have discovered that homogeneous coordinates are a natural and elegant way to assign coordinates to points in a projective plane.)

Suppose we have a projective transformation that is represented by a matrix $H \in \mathbb R^{3 \times 3}$, so that point with coordinate vector $x$ maps to a point with coordinate vector $Hx$. If we have four points $x_1,\ldots,x_4$ in "general position" and we know these four points map to $y_1,\ldots, y_4$, then we can use this information to compute $H$.

We cannot simply say that $y_i = H x_i \, \,$, because one quirky thing about homogeneous coordinates is that if $y$ is a coordinate vector for a point in a projective plane, then any nonzero scalar multiple of $y$ is also a coordinate vector for the same point. This might seem weird but it's something we must get used to.

The most we can say is that $y_i$ is a (nonzero) scalar multiple of $H x_i$. An equivalent way to say the same thing is \begin{equation} y_i \times H x_i = 0 \end{equation} where $\times$ denotes the cross product.

Notice that this equation is linear in the components of $H$ ! For each $i$, we have three linear equations for the components of $H$. However, it turns out that one of those three equations is redundant, so each $i$ really only gives us two equations. All together, we have eight linear equations.

$H$ is found by finding a nonzero solution of this system of eight equations. Note that this only determines $H$ up to a scalar multiple. However, that's to be expected, because in homogeneous coordinates, any nonzero scalar multiple of $H$ represents the same projective transformation as $H$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.