The rectifying transformation

In order to rectify - let's say - the left image, we need to compute the transformation mapping the image plane of $\tilde{\bf P}_{o1} = [ {\bf Q}_{o1} \vert \tilde {\bf q}_{o1} ]$ onto the image plane of $\tilde{\bf P}_{n1} = [ {\bf Q}_{n1} \vert \tilde {\bf q}_{n1}]$. It is useful to think of an image as the intersection of the image plane with the cone of rays between points in 3D space and the optical centre. We are moving the image plane while leaving fixed the cone of rays. We will see that the sought transformation is the collinearity (linear transformatiopn of the projective plane) given by the $3\times3$ matrix ${\bf T}_1 = {\bf Q}_{n1} {\bf Q}_{o1}^{-1}$. The same result applies to the right image.

For any 3-D point w we can write

$$\left\{ \begin{array}{l} \tilde{\bf m}_{o1} \simeq\tilde{\bf... ...n1} \simeq\tilde{\bf P}_{n1} \tilde{\bf w}. \end{array}\right.$$ (11)

According to (8), the equations of the optical rays are the following (since rectification does not move the optical center):

$$\left\{ \begin{array}{l} {\bf w} = {\bf c}_{1} + \lambda_o {\... ...}_{n1} \;\;\;\; \lambda_n \in \mathbb{R} ; \end{array}\right.$$ (12)

hence

$$\tilde {\bf m}_{n1} = \lambda {\bf Q}_{n1} {\bf Q}_{o1}^{-1} \tilde{\bf m}_{o1}\;\;\;\; \lambda \in \mathbb{R}$$ (13)

(note that $\lambda$ absorbes the arbitrary scale factor, hence we use =).

The transformation ${\bf T}_1$ is then applied to the original left image to produce the rectified image, as in Figure 3. Note that the pixels (integer-coordinate positions) of the rectified image correspond, in general, to non-integer positions on the original image plane. Therefore, the gray levels of the rectified image are computed by bilinear interpolation.

Reconstruction of 3-D points by triangulation can be performed from the rectified images directly, using Pn1,Pn2.