I'm trying to find a general form for a spiral that fits the following criteria: the inner radius is $N$, and for any point $q$ on the spiral, the arc length from the start of the spiral to $q$ is twice the change in radius from the beginning to that point. For example, if the arc length from the start to point $h$ is $p$, the radius should have change $p/2$ units from the start to $h$, and the radius at $h$ should be $N + p/2$. These properties need to hold for at least the first complete turn of the spiral. I would like to represent the spiral in terms of polar coordinates. How would I go about doing this? An explanation of why the equation is as it is would be great too. Thanks.
[Math] Defining a spiral in polar coordinates
calculusgeometrygraphing-functionspolar coordinates
Related Solutions
(too long for a comment)
What you need here is what's termed as a clothoidal spline; in this paper, Meek and Walton devise a (rather elaborate) algorithm that determines a clothoidal spline passing through two given points and matches slopes and curvatures ($G^2$ continuity), under suitable conditions (since not all combinations of slopes and curvatures will yield a feasible clothoidal segment). See also this previous paper for sundry algorithms. All of this assumes that your computing environment is equipped for numerically evaluating Fresnel integrals; there are otherwise simple approximations for computing them if you don't need too much accuracy.
Using to describe the spiral
$$ \rho = \rho_0 + \frac{\lambda}{2\pi}\theta $$
with $\lambda = \frac{\Delta\rho}{\pi^2}$
we have that
$$ ds = \sqrt{\rho^2+\left(\frac{d\rho}{d\theta}\right)^2}d\theta\to L = \int_0^{\theta_f}\sqrt{\rho^2+\left(\frac{d\rho}{d\theta}\right)^2}d\theta $$
Putting numbers
$$ \rho_0 = 25\\ \Delta\rho = 6\\ L = 1000 $$
we have
$$ \theta_f\approx 8.45 \pi $$
See plot attached
NOTE
$$ L = \int_0^{\theta_f}\sqrt{\left(\frac{3 \theta }{\pi }+25\right)^2+\frac{9}{\pi ^2}}d\theta = \frac{\sqrt{(3 \theta_f +25 \pi )^2+9} (3 \theta_f +25 \pi )+9 \sinh ^{-1}\left(\theta_f +\frac{25 \pi }{3}\right)}{6 \pi }-\left(\frac{25 \pi \sqrt{9+625 \pi ^2}+9 \sinh ^{-1}\left(\frac{25 \pi }{3}\right)}{6 \pi }\right) $$
Best Answer
Denote the spiral in polar coordinates by $r=r(\theta)$. Your literal conditions translate into the following equations:
$$r(0)=N $$ $$\int_0^\theta \sqrt{r(\phi)^2+r'(\phi)^2} \mathrm{d}\phi=2(r(\theta)-r(0))=2(r(\theta)-N) $$
(Clearly, there is nothing special about the choice $\theta=0$ as the closest point).
Differentiating the second equation gives $$\sqrt{r(\theta)^2+r'(\theta)^2}=2r'(\theta). $$ If we square both sides we can solve for $r'$: $$r'(\theta)^2=\frac{1}{3} r(\theta)^2 $$ taking square roots we find $$\frac{\mathrm{d} r}{\mathrm{d} \theta}=\frac{1}{\sqrt{3}} r, $$ so that $$r(\theta)=C \mathrm{e}^{\frac{1}{\sqrt{3}} \theta }.$$ The initial condition shows that $C=N$, and the final solution is $$r(\theta)=N \mathrm{e}^\frac{\theta}{\sqrt{3}}. $$