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?

  • $\begingroup$ There is a missing datum in the problem statement and I wonder how you could solve. Please explain. $\endgroup$ – Yves Daoust Jun 20 '18 at 11:58
  • $\begingroup$ I have this spiral going around a 25mm radius and an arm length between the spiral of 6mm and a full length of 250 mm. Is there anyway to calculate how many times the spiral goes around? $\endgroup$ – AHarrington Jun 20 '18 at 12:08
  • $\begingroup$ Welcome to MSE. Please read this text about how to ask a good question. $\endgroup$ – José Carlos Santos Jun 20 '18 at 12:22

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

enter image description here


$$ L = \int_0^{\theta_f}\sqrt{\left(\frac{3 \theta }{\pi }+25\right)^2+\frac{9}{\pi ^2}}d\theta = \frac{\sqrt{(3 \theta_f +25 \pi )^2+9} (3 \theta_f +25 \pi )+9 \sinh ^{-1}\left(\theta_f +\frac{25 \pi }{3}\right)}{6 \pi }-\left(\frac{25 \pi \sqrt{9+625 \pi ^2}+9 \sinh ^{-1}\left(\frac{25 \pi }{3}\right)}{6 \pi }\right) $$

  • $\begingroup$ Thank you! Can you show out how you solved the integral? The actual length I have is 250mm and I'm trying to write an equation in excel so I can input different values and it solves it automatically. $\endgroup$ – AHarrington Jun 20 '18 at 16:08
  • $\begingroup$ Also, in your equation, can you please explain what drho/dtheta is. Please and thanks! $\endgroup$ – AHarrington Jun 20 '18 at 16:16
  • $\begingroup$ See note attached. $\endgroup$ – Cesareo Jun 20 '18 at 16:19
  • $\begingroup$ Thank you! I'm just having issues of solving for theta f. When I write it all out i cant seem to get rid of the sinh^-1 because of the theta being in the parenthesis. Any chance you could either add another note or explain how you solved the rest of the process to me please. You're really doing me a big favor! $\endgroup$ – AHarrington Jun 20 '18 at 16:31

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