I need to find the length of a spiral. The spiral start at a certain radius R1 (25mm) and ends at a larger radius R2(unknown). As the spiral spins outwards, the distance between each arm of the spiral remains constant at 6mm.
I tried using the formula for the Arc Length of a Curve in Polar Coordinates to find the length but plugging in the length and trying to solve for the polar coordinates however it seems to give me the wrong answer. Is there another equation to use or a better way I can solve?
$\endgroup$ 31 Answer
$\begingroup$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
$$ L = \int_0^{\theta_f}\sqrt{\left(\frac{3 \theta }{\pi }+25\right)^2+\frac{9}{\pi ^2}}d\theta = \frac{\sqrt{(3 \theta_f +25 \pi )^2+9} (3 \theta_f +25 \pi )+9 \sinh ^{-1}\left(\theta_f +\frac{25 \pi }{3}\right)}{6 \pi }-\left(\frac{25 \pi \sqrt{9+625 \pi ^2}+9 \sinh ^{-1}\left(\frac{25 \pi }{3}\right)}{6 \pi }\right) $$
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