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Let's say I have $n$ circles of radius $r$ that are spaced with a nearest-neighbor distance of $\delta$. (i.e. the shortest distance between any two particles is $\delta$.)

It is trivial to determine the coordinates ($x_n$,$y_n$) of the centre of these circles if they are placed along a circle. (You can simple use the rotation matrix.) Here is a schematic of these circles placed around a circle:

Circles placed around a circle

If I want to place these circles around an ellipse of axes $a$ and $b$, how do I determine the coordinates of the centre of each circle? Approximations are OK.

lemontwist
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    Your use of “edge-to-edge” confuses me, as you're talking about distances immediately afterwards. What edges? Given the fact that even computing arc lengths for the ellipse requires elliptic integrals which cannot be expressed using elementary functions, this sounds like a hard problem. Do you need exact formulas (very likely using unevaluated elliptic integrals), or are numeric approximations acceptable? Also where do you measure $\delta$? On the shortest route connecting circle centers, or along the arc of the ellipse? – MvG Mar 05 '13 at 21:29
  • @MvG, just updated my question for clarity. – lemontwist Mar 05 '13 at 23:03

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The points on the ellipse satisfy the equation $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. $$ If a small circle is centered at $(x_0,y_0)$, then any neighbor at a distance $\delta$ will have center coordinates satisfying $$ (x-x_0)^2+(y-y_0)^2=\delta^2; $$ if the center is, in addition, constrained to lie on the ellipse, then you have two quadratic equations in the two variables $(x,y)$, which can be solved exactly. Iterating this procedure (starting at $(a,0)$, for instance) will give the coordinates of successive circle centers on the ellipse.

mjqxxxx
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    Hmmm... I guess it is also important to determine what the lengths $a$ and $b$ have to be in order to have $n$ equally spaced particles along the ellipse. – lemontwist Mar 06 '13 at 14:36
  • This question can help: https://math.stackexchange.com/q/72555/57054 – Ivan Bunin Sep 24 '18 at 12:51