next up previous contents
Next: The image of the Up: Self-calibration Previous: A counting argument   Contents

Geometric interpretation of constraints

In this section a geometric interpretation of a camera projection matrix is given. It is seen that constraints on the internal camera parameters can easily be given a geometric interpretation in space.

A camera projection plane defines a set of three planes. The first one is parallel to the image and goes through the center of projection. This plane can be obtained by back-projecting the line at infinity of the image (i.e. $[0 0 1]^\top$). The two others respectively correspond to the back-projection of the image $x$- and $y$-axis (i.e. $[0 1 0]^\top$ and $[1 0 0]^\top$ resp.). A line can be back-projected through equation (3.9):

\begin{displaymath}
{\tt\Pi} \sim {\bf P}^\top{\tt l} \sim \left[\begin{array}{c...
...x{-}{\tt t}^\top{\bf R}\end{array}\right] {\bf K}^\top {\tt l}
\end{displaymath} (F3)

Let us look at the relative orientation of these planes. Therefore the rotation and translation can be left out without loss of generality (i.e. a camera centered representation is used). Let us then define the vectors ${\tt v}_x$, ${\tt v}_y$ and ${\tt v}_i$ as the first three coefficients of these planes. This then yields the following three vectors:
\begin{displaymath}
{\tt v}_x = \left[ \begin{array}{c} 0 \\ f_y \\ c_y \end{arr...
...v}_i = \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right]
\end{displaymath} (F4)

The vectors coinciding with the direction of the $x$ and the $y$ axis can be obtained by intersections of these planes:
\begin{displaymath}
{\tt l}_x = {\tt v}_x \times {\tt v}_i = \left[\begin{array}...
...[\begin{array}{c} s \\ -f_x \\ 0 \end{array}\right] \enspace .
\end{displaymath} (F5)

The following dot products can now be taken:
\begin{displaymath}
{\tt l}_x.{\tt l}_y= sf_y \, , \, {\tt v}_x.{\tt v}_i=c_y \mbox{ and } {\tt v}_y.{\tt v}_i=c_x
\end{displaymath} (F6)

Equation (6.6) proves that the constraint for rectangular pixels (i.e. $s=0$), and zero coordinates for the principal point (i.e. $c_x=0$ and $c_y=0$) can all be expressed in terms of orthogonality between vectors in space. Note further that it is possible to pre-warp the image so that a known skewF1 or known principal point parameters coincide with zero. Similarly a known focal length or aspect ratio can be scaled to one.

The AC is also possible to give a geometric interpretation to a known focal length or aspect ratio. Put a plane parallel with the image at distance $d$ from the center of projection (i.e. $Z=d$ in camera centered coordinates). In this case a horizontal motion in the image of $f_x$ pixels will move the intersection point of the line of sight over a distance $d$. In other words a known focal length is equivalent to knowing that the length of two (typically orthogonal) vectors are equal. If the aspect ratio is defined as the ratio between the horizontal and vertical sides of a pixel (which makes it independent of $s$), a similar interpretation is possible.


next up previous contents
Next: The image of the Up: Self-calibration Previous: A counting argument   Contents
Marc Pollefeys 2002-11-22