Specular reflection on the surface of a sphere: compass and ruler constructions

TL;DR

This paper presents a geometric ruler-and-compass algorithm to determine the specular reflection point on a sphere's surface given two external focal points, including numerical implementation and approximation analysis.

Contribution

It introduces an explicit geometric construction method for the reflection point on a sphere using only ruler and compass, with a first order approximation and error analysis.

Findings

Algorithm successfully computes reflection points for various cases.

First order approximation provides a closed-form expression with quantifiable error.

Numerical implementation validates the geometric construction.

Abstract

We provide an explicit geometric algorithm involving only ruler and compass constructions in order to specify the specular reflection point on the surface of a reflecting sphere of radius $r$ given two focal points $A$ and $B$ lying outside of it. By numerically implementing the algorithm we compute the point in question for a number of cases. We conclude by discussing how the first iteration of the algorithm constitutes a first order approximation to the real solution by providing a closed expression for it as well as the error involved in doing so, as a function of the distances of the two focal points from the origin and the angle formed between them.

Equations

$$\frac{r_A\sin (\theta -\varphi )}{r_A\cos (\theta -\varphi )-r}=\frac{r_B\sin (\theta +\varphi )}{r_B\cos (\theta +\varphi )-r}$$

$$(1+{\tan }^2\varphi ){\left[\left(\frac{r}{r_A}-\frac{r}{r_B}\right)\tan \theta +\left(\frac{r}{r_A}+\frac{r}{r_B}\right)\tan \varphi \right]}^2=4{\tan }^2\varphi (1+{\tan }^2\theta )$$

$$\frac{\theta -{\varphi }_{A^{\prime }}}{2}={\tan }^{-1}\frac{2E_B\sqrt{r_A^2{r}_{M_{0}B}^2-4E_B^2}-2E_A\sqrt{r_A^2{r}_{M_{0}B}^2-4E_A^2}}{r_A^2{r}_{M_{0}B}^2-4E_B^2-4E_A^2}$$

Full text

1 Introduction.

A point of specular reflection is defined as the unique point on a reflective surface such that the angles formed by the incident and reflecting rays from two focal points lying outside the surface with the perpedincular direction at that point become equal.

The problem of specifying the specular point in the case where the surface is that of a sphere appeared for the first time in [1]. This is of particular interest in the field of terrestrial and satellite communications, where knowledge of the exact point is necessary in order to avoid destructive interference between the direct signal sent over the air from transmitter to receiver and the one reflected from the surface of the earth. Although exact analytic solutions are known to exist [2], their complexity dramatically reduces our ability of obtaining a clear intuitive picture of the problem in contrast with a solution which would be based on geometric arguments.

In this letter we introduce such a solution by providing a geometric algorithm based on ruler and compass constructions. This approach constitutes a marked improvement to the analytic method especially when a simple answer is needed in order to make quick qualitative descriptions in applications, in this case the first iteration of the algorithm provides a satisfying approximation.

In Section 2 we briefly state the problem of specular reflection on the surface of a sphere and discuss how to construct the relevant equations. The main part consists of Section 3 where we explicitly describe the algorithm and provide a proof for it’s correctness as well as numerical results for a number of cases. Section 4 is concerned with the error made in approximating the exact solution with the first iteration of the algorithm for which a closed analytic expression is shown to exist while Section 5 includes closing remarks.

2 Analytic construction.

Let $A$ and $B$ be two focal points lying outside the surface of a sphere of radius $r$ centered at $O$ at distances $r_{A}$ and $r_{B}$ respectively from the origin forming an angle of $2\theta$ as in Figure 1 where without loss of generality we always assume that $r_{A}\leq r_{B}$. We are interested in specifying the point of specular reflection $M$ lying on the surface of the sphere such that the angles formed between $MA$, $MB$ and the ray connecting the origin with point $M$ become equal. Since this point lies on the reflection plane defined by the incident and reflected rays it suffices to restrict our attention to the circle of radius $r$ formed by the intersection of the sphere with this surface.

Any possible attempt of attacking the problem will inevitably require the solution of a quartic polynomial equation. For example, it should be immediately obvious from the above figure that the two triangles $(AMC)$ and $(BMD)$ formed by taking the distances of $A$ and $B$ to the ray connecting the origin to point $M$ are similar, if $\phi$ is it’s angular distance from the bisector of $\widehat{AOB}$ it follows that

[TABLE]

which after a little bit of algebra can be rewritten as

[TABLE]

Equation (1) is a quartic polynomial with respect to $\tan\phi$, although exact solutions can be found either with the use of a symbolic algebraic program, the general method for solving quartic equations developed by Cardano [3] or by direct analytic methods (see for instance [2] for the treatment of a similar quartic equation with complex coefficients), these tend to be extremely complicated (except for the case where $r_{A}=r_{B}$)222In this case the specular reflection point lies on the intersection of the circle of radius $r$ with the bisector of the angle between $A$ and $B$ $(\phi=0)$. and require the use of numerical methods. Moreover it can be shown that in general all four solutions are real and distinct (see Appendix A for a proof and a physical interpretation). Since it is virtually impossible to tell which one corresponds to point $M$ we are forced to calculate all four of them and then choose the one lying on the arc of the circle contained in $\widehat{AOB}$. As this seems highly impractical, an alternative solution to the problem is therefore needed.

3 Geometric construction.

We shall now explicitly describe a geometric algorithm involving only ruler and compass constructions for the specification of the specular reflection point. The algorithm consists of the following steps

1. Step 1:

Make an initial guess for the point of specular reflection on the sphere $M_{0}=(r,\phi_{0})$. 2. Step 2:

Connect this point to point $B$. Find the intersection $A^{\prime}$ of $M_{0}B$ with the circle of radius $r_{A}$ centered at $O$. 3. Step 3:

Replace $M_{0}$ with $M^{\prime}=(r,\phi^{\prime})$ where $\phi^{\prime}$ is such that the point lies on the bisector of the angle $\widehat{AOA^{\prime}}$. 4. Step 4:

Repeat steps 2 and 3 $n$ times.

Figure 2 presents the first iteration of the algorithm $(n=0)$.

We will now prove that in the limit of an infinite number of iterations $(n\to\infty)$, the sequence of points generated by the algorithm converges to $M$.

Proof.

Let $f(\phi)$ be the function that outputs the polar angle of the first iteration of the algorithm given an initial guess $\phi$. We note that the point of specular reflection is a fixed point of $f$, namely suppose $\phi_{M}$ is it’s polar angle then $f(\phi_{M})=\phi_{M}$. Furthermore Equation 1 suggests that it must lie on the arc of the circle bounded between the bisector of $\widehat{AOB}$ and the perpendicular distance of $AB$ from the origin. If we can thus establish that $f$ is a contraction mapping 333A contraction mapping is a function from a metric space $X$ to itself such that for any two elements $x,y\in X$ there exists a nonnegative real number $k<1$ such that $d(f(x),f(y))\leq kd(x,y)$ with $d$ the metric on $X$. on this interval then a simple application of the Banach fixed point theorem [4] suffices to prove that this is the unique fixed point.

To show that $f$ is a contraction suppose $d(x,y)=|x-y|$ is the metric on the unit circle with $x,y\in[-\pi,\pi]$ and let $\phi_{0}$ and $\psi_{0}$ be two initial guesses in it’s domain with $\phi_{0}>\psi_{0}$ as in Figure 3, with the bisector represented by the dashed and dotted line, from which it can also be seen that $f$ is necessarily a monotonically increasing function. It is now true that $f(\phi_{0})-f(\psi_{0})=\frac{\phi^{\prime}-\psi^{\prime}}{2}$, but because the distance between points $(r,\phi_{0})$, $(r_{A},\phi^{\prime})$ is greater than that of $(r,\psi_{0})$, $(r_{A},\psi^{\prime})$ it follows that $\psi_{0}-\psi^{\prime}<\phi_{0}-\phi^{\prime}$ so $f(\phi_{0})-f(\psi_{0})<\frac{\phi_{0}-\psi_{0}}{2}$ since the choice of $\phi_{0}$ and $\psi_{0}$ was arbitrary this must hold for any two choices in the interval. ∎

In Figure 4 we plot the exact point of specular reflection as a function of $\theta$ for a number of distances of the two focal points, obtained from a numerical implementation of the geometric algorithm as a simple loop in a Python script. As was already mentioned in Section 2, angular distances are measured with respect to the bisector of $\widehat{AOB}$, this means that $0\leq\phi\leq\theta+\cos^{-1}\frac{r^{2}_{AB}+r^{2}_{A}-r^{2}_{B}}{2r_{A}r_{AB}}-\frac{\pi}{2}$ where $r_{AB}$ is the distance between the two focal points.

Figures 5 (a)-(d) display the output of the first iteration $f(\phi)$ for fixed values of $r_{A}$, $r_{B}$ and $\theta$ from which the specular reflection point is easily determined by simply demanding that $f(\phi)=\phi$. Note that since for a fixed value of the distances, the angle formed between them cannot increase more than the value it will attain when the distance of point $A$ from $B$ ($r_{AB}$) is tangent to the sphere, $\theta$ will range between $0\leq 2\theta\leq\cos^{-1}\frac{r^{2}-\sqrt{(r_{A}^{2}-r^{2})(r_{B}^{2}-r^{2})}}{r_{A}r_{B}}$. Numerical results suggest that the sequence of approximations to the specular point converge linearly to the real point.

4 First order approximation.

Let $M_{0}=(r,\phi_{0})$ be our initial guess and $A^{\prime}=(r_{A},-\phi_{A^{\prime}})$, $M^{\prime}=(r,\frac{\theta-\phi_{A^{\prime}}}{2})$ the points generated after the first iteration of the algorithm $(n=0)$ as in Figure 6 where as in Figure 3 the bisector is represented by the dashed and dotted line. Since the area $E_{B}$ of $(OM_{0}B)$ is equal to that of $(OM_{0}A^{\prime})$ and $(OA^{\prime}B)$ it follows that $rr_{B}\sin(\theta+\phi_{0})=r_{A}r_{B}\sin(\theta-\phi_{A^{\prime}})+rr_{A}sin(\phi_{0}+\phi_{A^{\prime}})$. Solving for $\theta-\phi_{A^{\prime}}$ we find that

[TABLE]

where $E_{A}$ is the area of $(OM_{0}F)$, $F$ being the intersection of $OB$ with the circle of radius $r$ and $r_{M_{0}B}$ the distance of the initial guess from point $B$. Point $M^{\prime}$ can be considered as a first order approximation to the specular reflection point.

In Figure 7 we plot the relative error to the first order approximation for the polar angle of the specular reflection point for an initial guess of $\phi_{0}=0$ as a function of the angle $\widehat{AOB}$ for varying distances of the two focal points. We note that in these cases the approximation is accurate within three orders of magnitude. Equation 2 constitutes therefore a satisfying analytic estimation in cases where a quick and qualitative solution is needed.

In Figures 8 (a)-(d) we present the relative error as a function of the initial guess. As is expected this decreases the closer our initial guess is to the actual point of specular reflection.

5 Conclusions.

It was shown that it is possible in principle to determine the point of spherical specular reflection by implementing a geometric algorithm involving only ruler and compass constructions. Numerical results for the error involved in approximating the actual solution with the first iteration of the algorithm provided convincing evidence for it’s use in applications.

Acknowledgments.

The author wishes to thank K. Blekos for helpful discussions and remarks as well as G. Dasios for pointing out the problem of the spherical specular reflection point.

Appendix A Solutions to the quartic equation.

In Figure 9 we present the points on the surface of the sphere corresponding to the solutions of Equation 1 where we have also included the original solution for completeness. Considerations similar to those in Section 2 will convince that in each case $(AMC)$ and $(BMD)$ are similar triangles and lead to the same equation. The designations internal-external refer to the cases where the reflection angles with respect to the ray connecting the origin to $M$ lie within and outside the circle of radius $r$ respectively. Note how in cases (a) and (c) the angles at $O$ sum to $2\theta$ while those in cases (b) and (d) to $\pi-2\theta$.