### Miss Register, I Presume?

In my AIM 2010 paper, I describe how to obtain a projective transformation $\mathbf{H}$ from the image plane to the laser plane in a line laser 3D range imaging system. With the laser oriented vertically (i.e. perpendicular to the transport direction of the object being scanned), this allows for mapping of image coordinates directly to 3D coordinates.

Specifically, the $(u, v)$ coordinates of a point in the image map through $\mathbf{H}$ to yield the $(x, z)$ coordinates in the laser plane. Then, if $y_\Delta$ is the transport direction offset between profiles, the 3D coordinates of such a point in profile $i$ are $(x, i y_\Delta, z)$.

There is an additional step when the angle between the laser and the vertical axis is nonzero, as in the common configuration shown below, where the camera (rather than the laser) is orthogonal to the transport surface. In this case, knowing the angle $\beta$ is the key. Assuming the transport direction is positive away from the laser side and the object runs toward the laser, a point $(u, v)$ in the image maps to 3D coordinates $(x, i y_\Delta - z\sin\beta, z\cos\beta)$, where $x$ and $z$ are found as before.

I have a feeling that it is possible to derive $\beta$ from $\mathbf{H}$. More formally, given the $3 \times 3$ matrix representation $\mathbf{H}$ of a projective transformation between two planes $P_1$ and $P_2$, and supposing $L$ is the line of intersection of $P_1$ and $P_2$, what is the angle $\alpha$ between vectors $\vec{n}_1$ and $\vec{n}_2$ lying, respectively, in $P_1$ and $P_2$ and perpendicular to $L$?

Jun 4th, 2011