# Mean distance between N equidistributed points in a circle

I would like to calculate the mean distance depending on circle shape points,

This is a mean for calculating all possible distances between any two points

for $N=2$, line, there is only 1 distance.

for $N=3$, triangle, again, there is only 1 distance

for $N=4$, square, henceforth there are more than 1 distance.. in this case we would have four distances for sides and two distances in diagonal path, $\sqrt{2}$, then the mean distance would be

$D=(4+2\sqrt{2})/6=1.138..$

for $N=30$, it would be a "pixeled" circle .. and $N=\infty$ is a circle

How to calculate it for $N>4$ ? Is there any general formula or can it be derived?

Thanks!

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You could use the law of cosines to figure out the lengths of polygon chords... – J. M. Aug 14 '11 at 19:18
Actually, if the circle has unit radius, your example for N=4 has wrong distances: the long distances should be d=2. – leonbloy Aug 14 '11 at 20:03
I assume by "equidistributed" you mean the $n$ points are exactly the sides of a regular $n$-gon, as that's what your examples appear to intimate (although you seem to set the sides of the square of the $n=4$ case all equal to $1$, I will instead assume all points are located on the unit circle). In this case leonbloy's answer is the formula. – anon Aug 14 '11 at 20:16

The average distances among all points must be equal that the average distances from a given point. By geometry: we have that the distance is $d=2 \sin(\theta/2)$, so:

$$\bar d = \frac{2}{N-1} \sum_{k=1}^{N-1} \sin\left(\frac{\pi k}{N}\right)$$

On the limit, $N\to \infty$, you replace the sum by an integral and you get the limit by Christian Blatter: $\bar d \to 4/ \pi$

For example, for $N=30$:

 >>>  N=30;
>>>  r = 2* sin(pi*[1:N-1]/(N));
>>> sum(r)/(N-1)
ans =  1.3159
>>> 4/pi
ans =  1.2732


Update: If instead of having a unit circle (radius 1) we have that the distance among nearest neighbours is 1 (the question is not clear about this, and the example for N=4 only makes sense in this later case), we just divide the above result by $2\; \sin(\pi/N)$. In the limit, $\sin(\pi/N) \to \pi/N$ and so $\bar d \to N \; 2/\pi^2$

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Yes, the later I think Is the right answer, (because at least try to answer the right question!), I mean the real meaning of the question is for distance among nearest neighbours=1, so your answer would be $\bar d = \frac{1}{N-1} \frac{\sum_{k=1}^{N-1} \sin\left(\frac{\pi k}{N}\right)}{\sin(\pi/N)}$ Is it right? (I would like to have enough rep to upvote!) saludos desde Argentina! – Hernán Eche Aug 14 '11 at 23:44
@Hernan (tocayo y compatriota): Yes, that's right. – leonbloy Aug 15 '11 at 1:19

When two points $z_1$, $z_2$ are independently uniformly distributed on the unit circle $S^1$ then their mean distance $\bar d$ is ${4\over\pi}$. To prove this one may keep $z_1$ fixed and let $z_2$ have an angular distance $\phi\in[0,\pi]$ from $z_1$ which is uniformly distributed on $[0,\pi]$. The euclidean distance between these two points is given by $d(z_1,z_2)=2\sin{\phi\over2}$,whence $$\bar d={1\over\pi}\int_0^\pi 2\sin{\phi\over2} \ d\phi={4\over\pi}\ .$$ This can be interpreted as follows: If you have a polygon $P$ with $N\gg 1$ vertices independently and uniformly distributed on $S^1$ then the mean distance between these vertices will be approximately ${4\over\pi}$.

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It also not hard to derive the distribution for the distance: $f(d) = \frac{2}{\pi} \frac{1}{\sqrt{4-d^2}}$ where $0<d<2$. – Sasha Aug 14 '11 at 21:12