TITLE: How to place objects equidistantly on an Archimedean spiral?
QUESTION [12 upvotes]: 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?

REPLY [14 votes]: 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):