Archimedean spiral: arc length of coil

Question

Are the arc lengths of the coils - i.e. the parts $0-2\pi$, $2\pi-4\pi$, etc. - in arithmetic progression?

It would be nice if you could supply a formula for an Archimedean spiral. You might like to change the question tag: arc-length is very much differential geometry, it requires integrating certain derivatives.
$$L = \int_{t_1}^{t_2} \sqrt{\dot{x}+\dot{y}} , \operatorname{d}!t$$ — Fly by Night, Jun 19 '13 at 17:33
@Paul Stephenson For one rotation radius increases proportionally with $ \theta $ ...so naturally perimeter length also increases similarly. — Narasimham, Jul 25 '16 at 09:03

score 1 · Answer 1 · answered Jun 19 '13 at 18:13

Given the formula for an Archimedean spiral $r=a+b\theta$, the arc length of turn $n$ is $$\int_{(n-1)\pi}^{n\pi} \sqrt {\left(\frac {dr}{d\theta}\right)^2+ r^2 }d\theta \\=\int_{(n-1)\pi}^{n\pi}\sqrt {b^2+(a+b \theta)^2}d\theta \\=\int_{(n-1)\pi-\frac ab}^{n\pi-\frac ab}\sqrt{b^2+b^2\phi^2} d\phi \\=b\int_{(n-1)\pi-\frac ab}^{n\pi-\frac ab}\sqrt{1+\phi^2} d\phi \\=\left.\frac b2\left(\phi\sqrt{1+\phi^2}+\sinh^{-1}\phi\right)\right |_{(n-1)\pi-\frac ab}^{n\pi-\frac ab}$$ which Alpha shows is not linear.

Archimedean spiral: arc length of coil

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