Derivation of the Fixed Viewpoint Constraint Equation

Without loss of generality we can assume that the effective viewpoint $\mathbf{v}$ of the catadioptric system lies at the origin of a cartesian coordinate system. Suppose that the effective pinhole is located at the point $\mathbf{p}$. Then, again without loss of generality, we can assume that the z-axis $\hat\mathbf{z}$ lies in the direction $\vec{\mathbf{v}\mathbf{p}}$. Moreover, since perspective projection is rotationally symmetric about any line through $\mathbf{p}$, the mirror can be assumed to be a surface of revolution about the z-axis $\hat\mathbf{z}$. Therefore, we work in the 2-D cartesian frame $(\mathbf{v}, \hat\mathbf{r}, \hat\mathbf{z})$ where $\hat\mathbf{r}$is a unit vector orthogonal to $\hat\mathbf{z}$, and try to find the 2-dimensional profile of the mirror z(r) = z(x,y) where $r=\sqrt{x^{2}+y^{2}}$. Finally, if the distance from $\mathbf{v}$ to $\mathbf{p}$ is denoted by the parameter c, we have $\hat\mathbf{v}=(0,0)$ and $\hat\mathbf{p}=(0,c)$. See Figure 1 for an illustration of the coordinate frame.

 

Figure 1: The geometry used to derive the fixed viewpoint constraint equation. The viewpoint $\mathbf{v} = (0,0)$ is located at the origin of a 2-D coordinate frame $(\mathbf{v}, \hat\mathbf{r}, \hat\mathbf{z})$, and the pinhole of the camera $\mathbf{p} = (0,c)$is located at a distance c from $\mathbf{v}$ along the z-axis $\hat\mathbf{z}$.If a ray of light, which was about to pass through $\mathbf{v}$, is reflected at the mirror point (r,z), the angle between the ray of light and $\hat\mathbf{r}$ is $\theta=\tan^{-1}\frac{z}{r}$. If the ray is then reflected and passes through the pinhole $\mathbf{p}$, the angle it makes with $\hat\mathbf{r}$ is $\alpha = \tan^{-1} \frac{c-z}{r}$, and the angle it makes with $\hat\mathbf{z}$ is $\gamma = 90^{\circ} - \alpha$. Finally, if $\beta = \tan^{-1} \left(- \frac{\mathrm{d} z}{\mathrm{d} r}\right)$is the angle between the normal to the mirror at (r,z) and $\hat\mathbf{z}$, then by the fact that the angle of incidence equals the angle of reflection, we have the constraint that $\alpha + \theta + 2 \gamma + 2 \beta = 180^{\circ}.$ 

We begin the translation of the fixed viewpoint constraint into symbols by denoting the angle between an incoming ray from a world point and the r-axis by $\theta$. Suppose that this ray intersects the mirror at the point (z,r). Then, since we assume that it also passes through the origin $\mathbf{v} = (0,0)$ we have the relationship:  

$$\tan \theta \ = \ \frac{z}{r}.$$ (1)

If we denote the angle between the reflected ray and the (negative) r-axis by $\alpha$, we also have:  

$$\tan \alpha \ = \ \frac{c-z}{r}$$ (2)

since the reflected ray must pass through the pinhole $\mathbf{p} = (0,c)$.Next, if $\beta$ is the angle between the z-axis and the normal to the mirror at the point (r,z), we have:  

$$\frac{\mathrm{d} z}{\mathrm{d} r} \ = \ - \tan \beta.$$ (3)

Our final geometric relationship is due to the fact that we can assume the mirror to be specular. This means that the angle of incidence must equal the angle of reflection. So, if $\gamma$ is the angle between the reflected ray and the z-axis, we have $\gamma = 90^{\mathrm{o}} - \alpha$ and $\theta + \alpha + 2 \beta + 2 \gamma = 180^{\mathrm{o}}$.(See Figure 1 for an illustration of this constraint.) Eliminating $\gamma$ from these two expressions and rearranging gives:

$$2 \beta \ = \ \alpha - \theta.$$ (4)

Then, taking the tangent of both sides and using the standard rules for expanding the tangent of a sum:

$$\tan (\mathrm{A} \pm \mathrm{B}) \ = \ \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$$ (5)

we have:

$$\frac{2 \tan \beta}{1 - \tan^{2} \beta} \ = \ \frac{\tan \alpha - \tan \theta}{1 + \tan \alpha \tan \theta}.$$ (6)

Substituting from Equations (1), (2), and (3) yields the fixed viewpoint constraint equation:

$$\frac{-2 \frac{\mathrm{d} z}{\mathrm{d} r}} {1- \left( \frac... ...hrm{d} r} \right)^{2}} \ = \ \frac{(c-2z)r}{r^{2} + cz - z^{2}}$$ (7)

which when rearranged is seen to be a quadratic first-order ordinary differential equation:

$$r (c-2z) \left( \frac{\mathrm{d} z}{\mathrm{d} r} \right)^{2... ... - z^{2}) \frac{\mathrm{d} z}{\mathrm{d} r} + r (2z-c) \ = \ 0.$$ (8)