Finding a point on Archimedean Spiral by its path length I've been curious about Archimedean Spirals and their relations to Sacks Spirals and prime numbers. 

I would like to draw some visualizations of the points with a given distance from the center, across the spiral path. Is there a formula for that?
Clarification:

I'm looking for a function that, given a distance along the path, fives the coordinates of a point along that path.
Consider the example image above. $f(0)=(0,0)$, because a distance of 0 along the path is the origin of axis. In this (rough) sketch, $f(1)≃(1, -3)$, $f(2)≃(0.8, 1.1)$, $f(3)≃(-1.3, 1.1)$ and so on.
Thanks a lot for your help!
 A: The arc length for $r=a\theta^{1/n}$ is given by
$$
s(\theta)=a\theta^{1/n} {_2F_1}\left(-\frac{1}{2},\frac{1}{2n}; 1+\frac{1}{2n}; -n^2\theta^2\right),
$$ 
where $_2F_1(a,b;c;x)$ is a hypergeometric function. 
See here.
For $n=1$ (your case) this boils down to 
$$
    \frac{a}{2}\left[\theta\cdot\sqrt{1+\theta^2}+\ln \left(\theta+\sqrt{1+\theta^2} \right)\right]. 
$$
You get there by using $\vec r= (a\cdot \theta\cdot\cos\,\theta,a\cdot \theta\cdot\sin\,\theta)$ and the formula for the arc length
$$
\begin{eqnarray}
s(\theta)&=&\int_{0}^{\theta} \sqrt { [r_x'(\varphi)]^2 + [r_y'(\varphi)]^2 }\, d\varphi\\
&=&\int_{0}^{\theta} \sqrt {1+x^2} \, dx .\\
\end{eqnarray}
$$
You'll find some ways to solve similar things like the last integral here.
A: The answer should be quite clear given the post above. All you need to do is to find the inverse function of that
$$
s(\theta)=\frac{P}{2\pi}[\frac{\theta}{2}\sqrt{1+\theta^{2}}+\frac{1}{2}\ln(\theta+\sqrt{1+\theta^{2}})]
$$
where P is pitch length. Once you get $\theta(s)$, you can find $x(\theta)$ and $y(\theta)$ fairly easily.
If you can't solve $\theta(s)$ analytically, you solve it numerically. As given in the comment of the previous post, you may have a look at this one:
How to place objects equidistantly on an Archimedean spiral?
