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To place objects equidistantly on an Archimedean (arithmetic) spiral, the arc length of the spiral has to increase linearly between the objects.

This is what I have so far: The length of a spiral is determined by $$ l = \frac{a}{2}\left[\varphi\cdot\sqrt{1+\varphi^2}+\ln \left(\varphi+\sqrt{1+\varphi^2} \right)\right] $$ I presume that solving this equation for $\varphi$ will give me what I need. But trying that with WolframAlpha leads to a timeout.

Is solving this equation for $\varphi$ really the right thing to do? If yes, how can I solve it?

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    $\begingroup$ Do you want the objects to have the same Euclidean distance from each other, or the same arc length along the spiral between them? $\endgroup$ – joriki Nov 13 '11 at 13:39
  • $\begingroup$ Yes, this one's quite related. $\endgroup$ – J. M. is a poor mathematician Nov 13 '11 at 15:45
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In order to place points equidistantly according to the arc-length, you should place points at $\phi_k$ determined by $$ x_k = k \frac{\Delta \ell}{2 a} = \phi_k \cdot \sqrt{1+\phi_k^2} + \operatorname{arcsinh}(\phi_k) $$ This equation admits no solution in simple functions, but can be easily solved numerically.

Also for large $x_k$, $\phi_k \sim \sqrt{x_k}$. More precisely: $$ \phi_k \sim \sqrt{ \frac{1}{2} W\left( \frac{1}{2} \mathrm{e}^{2x_k -1}\right)} $$ where $W(x)$ is Lambert W function. This gives rather good placement (red rod represent approximate locations, and centers of the blue circles represent exact solutions):

enter image description here

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