Resolution of a Catadioptric Sensor

  In this section, we assume that the conventional camera used in the catadioptric sensor has a frontal image plane located at a distance u from the pinhole, and that the optical axis of the camera is aligned with the axis of symmetry of the mirror. See Figure 8 for an illustration of this scenario. Then, the definition of resolution which we will use is the following. Consider an infinitesimal area $\mathrm{d} A$ on the image plane. If this infinitesimal pixel images an infinitesimal solid angle $\mathrm{d} \nu$ of the world, the resolution of the sensor as a function of the point on the image plane at the center of the infinitesimal area $\mathrm{d} A$ is:

$$\frac{\mathrm{d} A}{\mathrm{d} \nu}.$$ (29)

  

Figure 8: The geometry used to derive the spatial resolution of a catadioptric sensor. Assuming the conventional sensor has a frontal image plane which is located at a distance u from the pinhole and the optical axis is aligned with the z-axis $\hat\mathbf{z}$, the spatial resolution of the conventional sensor is $\frac{\mathrm{d} A}{\mathrm{d} \omega} = \frac{u^{2}}{\cos^{3} \psi}$. Therefore the area of the mirror imaged by the infinitesimal image plane area $\mathrm{d} A$ is $\mathrm{d} S = \frac{(c-z)^{2} \cdot \cos \psi}{u^{2} \cos \phi} \cdot \mathrm{d} A$. So, the solid angle of the world imaged by the infinitesimal area $\mathrm{d} A$ on the image plane is $\mathrm{d} \nu = \frac{(c-z)^{2} \cdot \cos \psi} {u^{2} (r^{2}+z^{2})} \cdot \mathrm{d} A$.Hence, the spatial resolution of the catadioptric sensor is $\frac{\mathrm{d} A}{\mathrm{d} \nu} = \frac{u^{2} (r^{2}+z^{2})}{(c-z)^{2} \cdo... ...frac{r^{2}+z^{2}}{r^{2}+(c-z)^{2}} \cdot \frac{\mathrm{d} A}{\mathrm{d} \omega}$since $\cos^{2} \psi = \frac{(c-z)^{2}}{(c-z)^{2} + r^{2}}$.

If $\psi$ is the angle made between the optical axis and the line joining the pinhole to the center of the infinitesimal area $\mathrm{d} A$ (see Figure 8), the solid angle subtended by the infinitesimal area $\mathrm{d} A$ at the pinhole is:

$$\mathrm{d} \omega \ = \ \frac{\mathrm{d} A \cdot \cos \psi }... ...2} \psi} \ = \ \frac{\mathrm{d} A \cdot \cos^{3} \psi }{u^{2}}.$$ (30)

Therefore, the resolution of the conventional camera is:  

$$\frac{\mathrm{d} A}{\mathrm{d} \omega} \ = \ \frac{u^{2}}{\cos^{3} \psi}.$$ (31)

Then, the area of the mirror imaged by the infinitesimal area $\mathrm{d} A$ is:

$$\mathrm{d} S \ = \ \frac{\mathrm{d} \omega \cdot (c-z)^{2}}{... ...{\mathrm{d} A \cdot (c-z)^{2} \cdot \cos \psi}{u^{2} \cos \phi}$$ (32)

where $\phi$ is the angle between the normal to the mirror at (r,z) and the line joining the pinhole to the mirror point (r,z). Since reflection at the mirror is specular, the solid angle of the world imaged by the catadioptric camera is:

$$\mathrm{d} \nu \ = \ \frac{\mathrm{d} S \cdot \cos \phi}{r^{... ...rm{d} A \cdot (c-z)^{2} \cdot \cos \psi} {u^{2} (r^{2}+z^{2})}.$$ (33)

Therefore, the resolution of the catadioptric camera is:

$$\frac{\mathrm{d} A}{\mathrm{d} \nu} \ = \ \frac{u^{2} (r^{2}... ...psi }{(c-z)^{2}}\right] \frac{\mathrm{d} A}{\mathrm{d} \omega}.$$ (34)

But, since:

$$\cos^{2} \psi \ = \ \frac{(c-z)^{2}}{(c-z)^{2} + r^{2}}$$ (35)

we have:  

$$\frac{\mathrm{d} A}{\mathrm{d} \nu} \ = \ \left[\frac{r^{2}+... ...-z)^{2} + r^{2}}\right] \frac{\mathrm{d} A}{\mathrm{d} \omega}.$$ (36)

Hence, the resolution of the catadioptric camera is the resolution of the conventional camera used to construct it multiplied by a factor of:  

$$\frac{r^{2}+z^{2}}{(c-z)^{2} + r^{2}}$$ (37)

where (r,z) is the point on the mirror being imaged.

The first thing to note from Equation (38) is that for the planar mirror $z=\frac{c}{2}$, the resolution of the catadioptric sensor is the same as that of the conventional sensor used to construct it. This is as expected by symmetry. Secondly, note that the factor in Equation (39) is the square of the distance from the point (r,z) to the effective viewpoint $\mathbf{v} = (0,0)$, divided by the square of the distance to the pinhole $\mathbf{p} = (0,c)$. Let $d_{\mathbf{v}}$ denote the distance from the viewpoint to (r,z) and $d_{\mathbf{p}}$ the distance of (r,z) from the pinhole. Then, the factor in Equation (39) is $\frac{d_{\mathbf{v}}^{2}}{d_{\mathbf{p}}^{2}}.$For the ellipsoid, $d_{\mathbf{p}} + d_{\mathbf{v}} = K_{e}$ for some constant $K_{e} \gt d_{\mathbf{p}}$. Therefore, for the ellispoid the factor is:

$$\left(\frac{K_{e}}{d_{\mathbf{p}}} - 1\right)^{2}$$ (38)

which increases as $d_{\mathbf{p}}$ decreases and $d_{\mathbf{v}}$increases. For the hyperboloid, $d_{\mathbf{p}} - d_{\mathbf{v}} = K_{h}$ for some constant $0 < K_{h} < d_{\mathbf{p}}$. Therefore, for the hyperboloid the factor is:

$$\left(1 - \frac{K_{h}}{d_{\mathbf{p}}}\right)^{2}$$ (39)

which increases as $d_{\mathbf{p}}$ increases and $d_{\mathbf{v}}$increases. So, for both ellipsoids and hyperboloids, the factor in Equation (39) increases with r. Hence both hyperboloidal and ellipsoidal catadioptric sensors constructed with a uniform resolution conventional camera will have their highest resolution around the periphery, a useful property for certain applications such as teleconferencing.