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I have a closed curve $g$ as a map from $\mathbb{R}\rightarrow\mathbb{R}^3$ defined as: $$\vartheta \mapsto \left( \begin{array}{c} x\\ y\\ z\\ \end{array} \right) = \left( \begin{array}{c} (r_1+r_2\cos(n \vartheta))\sin(\vartheta)\\ (r_1+r_2\cos(n \vartheta))\cos(\vartheta)\\ r_2\sin(n \vartheta)\\ \end{array} \right),$$ for $0 \le \vartheta < 2\pi,$ with $r_1,r_2 \in \mathbb{R^+}$ and $n \in \mathbb{N}^+$.

The curve thus looks like a coil wound $n$ times around a torus with larger and smaller radius of $r_1$ and $r_2$.

Question 1: Does this curve have a special specific name apart from "coil"?

Question 2: How long is it?

To answer 2 as far as I understand you have to calculate $$ l = \int_{\vartheta = 0}^{2 \pi} \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}\; d\vartheta.$$ I have tried three times and get different results (for the term in the integral) each time (no CAS available :-(). But I have thought about an geometric approach to the solution. If one thinks like in topology about a torus as being represented by a rectangle of side lengths $2r_1\pi$ and $2r_2\pi$. When you draw the coil into the rectangle you get stripes of total lengths $$l = n\sqrt{(2 \pi \frac{r_1}{n})^2 + (2\pi r_2)^2}.$$ Might that be correct?

Raphael J.F. Berger
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    Maple says the integral simplifies to $$l = \int_{0}^{2\pi} \sqrt{n^2 r_2^2 + r_1^2 + 2 r_1 r_2 \cos(n\varphi) + r_2^2 \cos(n\varphi)^2} d\varphi$$without direct solutions; specific solutions for integer $n$ like $n=2$, $n=3$, etc. do yield closed expressions, but with lots of terms including various elliptic integrals. The problem with your geometric approach is that the metric is not the same on the sheet and the torus; you cannot wrap a pliable but un-stretchy paper onto a torus. Points drawn on the paper will have different distances on the torus, in other words. – Nominal Animal Jun 04 '16 at 14:41
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    In general, torus knots/toroidal spirals such as yours will have arclength functions that are (hyper)elliptic integrals; in this case, speaking as somebody who deals with elliptic integrals, numerical integration is far more practical for evaluating your arclength than forcing through the closed-form expression. – J. M. ain't a mathematician Jun 04 '16 at 15:04
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    Related: Tight approximation of a Torus Knot length. – Andrew D. Hwang Jun 04 '16 at 15:24
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    @AndrewD.Hwang's answer he linked to is exactly the integral here, but with $q=1$, $p=n$, $R = r_1$, and $r = r_2$. So, those limits say that $$2\pi\sqrt{n^2 r_2^2 + (r_1-r_2)^2} \le l \le 2\pi\sqrt{n^2 r_2^2 + (r_1+r_2)^2}$$Funnily enough, OP's geometric estimate,$$l \approx 2\pi\sqrt{n^2 r_2^2 + r_1^2}$$is pretty well in the middle there. (I must admit it surprised the heck out of me to realize how well OP's geometric estimate fits into the limits!) – Nominal Animal Jun 04 '16 at 15:38
  • So it is a torus knot? Is there maybe even another term? (knots on 2D manifolds are at times a bit odd). – Raphael J.F. Berger Jun 04 '16 at 15:46
  • I wonder if $2\pi\sqrt{(n^2 + \frac{1}{4})r_2^2 + r_1^2}$ would be a slightly closer approximation to $l$ (using the "average" of $\cos$ and $\cos^2$)? – Raphael J.F. Berger Jun 05 '16 at 14:08

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