What is the equation for the radius of a polar coordinate for an Archimedean spiral with the arc length known relative to theta?
arc length: L = $\int_{\theta_0}^{\theta_1}\sqrt{(r)^2+(dr/d\theta)^2}d\theta$ (mathworld.wolfram)
Archimedean Spiral: $r=a+b\theta^{1/c}$
Some known information for one example:
- $\theta$={$0$ to $\pi$}, L=8
- $\theta$={$0$ to 2$\pi$}, L=32
- $\theta$={$0$ to 3$\pi$}, L=72
- $\theta$={$0$ to 4$\pi$}, L=128
- $\theta$={$\pi$ to 2$\pi$}, L=24
- $\theta$={2$\pi$ to 3$\pi$}, L=40
- $\theta$={3$\pi$ to 4$\pi$}, L=56
so: L{$\theta_0$ to $\theta_1$} $=8*(\theta_1/\pi)^2 - 8*(\theta_0/\pi)^2$
and: L{$0$ to $\theta$} $=8*(\theta/\pi)^2$
but how do I get back to $r$ ? What are $a$, $b$, and $c$ ?
I was trying to use Wolfram-Alpha, but could not figure out the antiderivative to continue semi-manually, or the solver to get to a formula for $r$ directly.
The problem does not look feasible (at least to me) if $c\neq 1$ (the generalized Archimedean spiral) .
If $c=1$ and then $r=a+b\,\theta$, $$L=\int\sqrt{r(\theta )^2+r'(\theta )^2}\,d\theta=\frac{\sqrt{(a+b \theta )^2+b^2} (a+b \theta )}{2 b}+\frac{1}{2} b \log \left(\sqrt{(a+b \theta )^2+b^2}+a+b \theta \right)$$ So, if you know two arc lengths, you need to solve two equations for the two unknowns $a$ and $b$; I do not think that the problem could be solve without numerical methods.
Using the first two values you gave, I did find solutions and I had to turn the problem into an optimization problem. The norm was minimized for $a=-1.70012$, $b=1.88222$. Recomputing the arc lengths gives $L_1=8.53$ and $L_2=31.86$.