8
$\begingroup$

I've been curious about Archimedean Spirals and their relations to Sacks Spirals and prime numbers.

enter image description here

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:

enter image description here

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!

$\endgroup$
2
  • $\begingroup$ I thought you were looking for something like this. $\endgroup$
    – draks ...
    Commented Apr 4, 2012 at 21:42
  • $\begingroup$ It's a slight variation of that link, which is actually quite useful and interesting. I apologize for not being clear. $\endgroup$
    – Adam Matan
    Commented Apr 5, 2012 at 6:52

2 Answers 2

Reset to default
5
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ +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? $\endgroup$
    – Adam Matan
    Commented Apr 1, 2012 at 10:33
  • $\begingroup$ @AdamMatan In polar coordinates this is pretty easy: $(a\theta,\theta)$. Cartesian is left to you as an exercise... $\endgroup$
    – draks ...
    Commented Apr 1, 2012 at 17:33
  • 1
    $\begingroup$ 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? $\endgroup$
    – Adam Matan
    Commented Apr 1, 2012 at 17:48
  • $\begingroup$ @AdamMatan hm, but then you would have to invert the formula for $s$...I'm not sure, if this will work. $\endgroup$
    – draks ...
    Commented Apr 1, 2012 at 17:58
  • $\begingroup$ 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? $\endgroup$
    – Adam Matan
    Commented Apr 1, 2012 at 19:08
0
$\begingroup$

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?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .