Kissing number for equal ellipses on the plane

It's quite easy to determine maximum kissing number for circles (disks) on a plane. Since we have circumference $C=2 \pi r$, we just take the integer part of $2\pi$, which is $6$.

However, for ellipses the problem seems to be much more complicated. To get maximum kissing number we need to arrange the ellipses on 'pointy ends' around the central ellipse. Obviously, for large $a/b$ ratio, the kissing number will be large as well, approaching infinity when the ellipse becomes a line segment.

This is how (awkwardly) the packing is supposed to look:

The only approximation I could think of is to us the length of an ellipse formula, imagining the centers of surrounding ellipses lying on another ellipse.

Let $a,b$ be the major and minor axes (radii) of the ellipses we are packing. Then the kissing number will approximately be:

$$N \left( \frac{a}{b} \right) \approx \left[\frac{4A}{2b} E \left(1-\frac{B^2}{A^2} \right) \right]=\left[\frac{4a}{b} E \left(1-\frac{(a+b)^2}{4a^2} \right) \right]$$

Here $A=2a$ and $B=a+b$ are the major and minor axes of the 'external' ellipse, on which the other ellipses are arranged. $E$ is complete elliptic integral of the second kind and Mathematica notation is used for the parameter, i.e.:

$$E(m)=\int_0^{\pi/2} \sqrt{1-m \sin^2 x}dx$$

The approximation is very close to $5 a/b$, as we can see from the plot:

But how good is this approximation? And how to obtain a better estimate (or even exact value) for the maximum kissing number for arbitrary $a/b$ ratio?

Solution 1:

Playing around with GeoGebra one finds that your approximation tends to overestimate the number of kissing ellipses, when $a/b$ grows. A possibly better formula can be found as follows.

Consider two consecutive tangent ellipses. Their major axes lie on two normal lines, meeting at some point $O$: the distance between $O$ and the tangency points is approximately equal to the radius of curvature $\rho$ of the ellipse at those points. The centers of those two ellipses are at an approximate distance of $2b$ between them and of $\rho+a$ from $O$. It follows that the distance between the tangency points is approximately $d=2b\rho/(a+\rho)$.

This distance of course varies along the central ellipse, but we can compute its average value $\overline{d}$: $$ \overline{d}={1\over2\pi}\int_0^{2\pi}{2b\rho(t)\over a+\rho(t)}\,dt $$ where $t$ is the usual parameter and $$ \rho(t)={(a^2\sin^2t+b^2\cos^2t)^{3/2}\over ab}. $$ By dividing the length of the ellipse by $\overline{d}$ we thus get an estimate for the number $N$ of kissing ellipses: $$ N\approx{4 b E(1 - a^2/b^2)\over \overline{d}}. $$ Unfortunately I didn't succeed in finding a simple expression for $\overline{d}$, which must then be evaluated numerically. In the graph below I plotted the value of $N$ given above as a function of $a/b$ (blue). For comparison I also plotted your result $N\approx 5a/b$ (red).

I think this is a better approximation than yours, but don't know how good it is.

EDIT.

I made some experimentation with GeoGebra: for $a/b=2.87$ my formula gives $N\approx13.3$, while I managed to put at most $12$ tangent ellipses tightly packed:

For $a/b=5.61$ my formula gives $N\approx21.8$, and $20$ tangent ellipses can be drawn (I'm showing two different arrangements):

It seems that the above formula for $N$, while still overestimating the actual number of ellipses, is quite good for large values of $a/b$.

EDIT 2.

Consider a point $C$ on the extension of the major axis of an ellipse, at a distance $R$ from its nearest vertex. A straightforward calculation shows that the angle $\varphi$ between the major axis and a tangent to the ellipse passing through $C$ is given by $$ \varphi =\arctan{b\over\sqrt{R(R+2a)}}. $$ Consider now two consecutive "kissing" ellipses (see diagram below), touching the inner ellipse at their vertices $P_1$ and $P_2$ and tangent between them at $T$. Let $C$ be the intersection of their (produced) major axes: line $CT$ is (to a very good approximation) a common tangent to both ellipses, forming angles $\varphi_1$ and $\varphi_2$ with their major axes, which can be computed with the above formula.

Distances $R_1$ and $R_2$ can be computed from the coordinates of $P_1$ and $P_2$, as well as the angle $$ \angle P_1CP_2=\arccos{R_1^2+R_2^2-\overline{P_1P_2}^2\over2R_1R_2}. $$ But this angle is the sum of $\varphi_1$ and $\varphi_2$, so we have the equality: $$ \arccos{R_1^2+R_2^2-\overline{P_1P_2}^2\over2R_1R_2}= \arctan{b\over\sqrt{R_1(R_1+2a)}}+\arctan{b\over\sqrt{R_2(R_2+2a)}}. $$ From this equation one can, in principle, find $P_2$ once $P_1$ is given. This can be done numerically and can be used to build a routine which actually finds all kissing ellipses: starting with $P_1$ (which I took at one vertex of the inner ellipse) one can find $P_2$, then $P_3$ and so on, stopping with the last ellipse not intersecting the first one.

I wrote some Mathematica code which performs this task, thus computing the number of kissing ellipses. As an example, you can see below the case $a/c=20$, with $55$ tangent ellipses. The detail in the square is a magnification of the left vertex zone.

The plot below shows the number of kissing ellipses (black dots) as a function of $a/c$. The blue line is the approximate value found above.

For large values of $a/b$ I expect $N\approx 2a/b$, because a very thin ellipse should behave approximately like a $2a\times2b$ rectangle as $a/b\to\infty$, plus some tiny "border effect". This is confirmed by the numbers I got from Mathematica, which are plotted below:

I added for comparison the plot of my estimate (blue) and of $2a/b$ (red). As you can see, for very large values of $a/b$ the very simple rule $N\approx 2a/b$ is a better approximation. This trend is confirmed for even larger values: $N(1000)=2064$, $N(2000)=4081$, $N(3000)=6094$, $N(4000)=8105$, $N(5000)=10128$.

As a last note, the equation for $\varphi$ above could be used to refine the rough estimate given at the beginning: a more precise approximation for the distance $d$ between two consecutive vertices is given by $$ d=2\rho\arctan{b\over\sqrt{\rho(\rho+2a)}}. $$ A quick test however shows that the estimate one gets from that is better than the older one for small values of $a/b$, but isn't a great improvement if $a/c$ is large.