# 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!

• I thought you were looking for something like this. Commented Apr 4, 2012 at 21:42
• It's a slight variation of that link, which is actually quite useful and interesting. I apologize for not being clear. Commented Apr 5, 2012 at 6:52

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.

• +1 Thanks! How do I use this formula to find the (x, y) or (r, θ) coordinates of a points whose distance from the center is given? Commented Apr 1, 2012 at 10:33
• @AdamMatan In polar coordinates this is pretty easy: $(a\theta,\theta)$. Cartesian is left to you as an exercise... Commented Apr 1, 2012 at 17:33
• I can easily convert polar to Cartesian coordinate and vice versa. The part that I'm missing is the equation where s (arc length) is the input, and a coordinate is calculated from it. Doesn't your formula go the other way around? Commented Apr 1, 2012 at 17:48
• @AdamMatan hm, but then you would have to invert the formula for $s$...I'm not sure, if this will work. Commented Apr 1, 2012 at 17:58
• So how would you recommend drawing something like this: en.wikipedia.org/wiki/File:Sacks_spiral.svg ? traverse along the path in very small steps and accumulate the distance passed so far? Commented Apr 1, 2012 at 19:08

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?