I have the length of the spiral while the spiral starts at a certain radius R1 and ends at a larger radius R2. As the spiral spins outwards, the distance between each arm of the spiral remains constant.
I tried using the archemedian spiral equation (r=a+bθ) and making b and a function of r (where a is start radius and 2πb is the increase in radius per revolution),and then using a standard spiral length integral formula but this seems to give the wrong answer. I think that the formula above doesn't apply to my situation, would anyone know a better way to model it?
Thanks!
Wolfram gives the formula $$s(\theta) = \frac b2\left[\theta\sqrt{1+ \theta^2} + \ln\left(\theta + \sqrt{1+\theta^2}\right)\right]$$
for the arclength along the spiral $r = b\theta$. If you let $a = R_1 - b\theta_1$ in your formula, we get the same spiral, but crossing the $R_1$ circle at $\theta_1$. So effectively, we have changed the unknown value $a$ in your version for the unknown value $\theta_1$. Similarly, let $\theta_2$ be the angle where it crosses $R_2$, and let $L$ be the total length.
Then you have three unknowns: $b, \theta_1, \theta_2$, and three equations:
$$b\theta_1 = R_1\\b\theta_2 = R_2\\L = \frac b2\left[\theta_2\sqrt{1+ \theta_2^2} - \theta_1\sqrt{1+ \theta_1^2} + \ln\left(\theta_2 + \sqrt{1+\theta_2^2}\right) - \ln\left(\theta_1 + \sqrt{1+\theta_1^2}\right)\right]$$
Substituting the first two into the third gives $$L = \frac b2\left[\frac{R_2}b\sqrt{1+ \left(\frac {R_2}b\right)^2} - \frac{R_1}b\sqrt{1+ \left(\frac {R_1}b\right)^2} + \ln\left(\frac{R_2}b + \sqrt{1+ \left(\frac {R_2}b\right)^2}\right) - \ln\left(\frac{R_1}b + \sqrt{1+ \left(\frac {R_1}b\right)^2}\right)\right]\\L=\frac 1{2b}\left[R_2\sqrt{b^2 + R_2^2} - R_1\sqrt{b^2 + R_1^2} + b^2\ln\left(\frac{R_2 + \sqrt{b^2 + R_2^2}}{R_1 + \sqrt{b^2 + R_1^2}}\right)\right]$$
Which gives you an equation with $b$ as the only unknown, but not a very pretty one. Still, once solved for $b$, you can easily obtain $\theta_1, \theta_2$.
The archimedean spiral is not the only possible spiral that could be used. You could choose any spiral sufficiently slow growing for your initial circling after leaving $R_1$. Then when you've gone completely around, add some fixed constant $q$ for every additional round. So your conditions do not uniquely specify a solution.