I'm trying to compute the turning number of a closed plane curve, which should be a really straight forward computation. But my result is not a whole number, i.e. the total curvature is not a multiple of $2\pi$, which is impossible so I'm making a mistake somewhere. I know from plotting the curve that the turning number should be $-2$. I've been looking for hours but cannot seem to find it. Maybe someone can spot the mistake... Here is my work:

Let $\beta(t) = ((1+2\cos(t))\sin(t), (1+2\cos(t))\cos(t))$ be a regular closed plane curve. To compute the turning number $n_{\beta}$ I want to use the following:

$$n_{\beta} = \frac{1}{2\pi}\int_{0}^{L}K_{\beta}(t)dt\, ,$$

where $L$ is the period of $\beta$ and $K_{\beta}(t)$ is the curvature. I can compute the curvature like this:

$$K_{\beta} = \frac{1}{||\dot{\beta}(t)||^3}\cdot\det({\dot{\beta}(t), \ddot{\beta}(t)})$$


$$\dot{\beta}(t) = (\cos t + 2\cos 2t, -\sin t -2\sin 2t)$$ $$\ddot{\beta}(t) = (-\sin t - 4 \sin 2t, -\cos t - 4 \cos 2t)$$ $$||\dot{\beta}(t))|| = (4\cos t + 5)^{1/2}$$ $$\det({\dot{\beta}(t),\ddot{\beta}(t)}) = -6 \cos t - 9$$

This is giving me the curvature

$$K_{\beta}(t) = \frac{-6 \cos t -9}{(4 \cos t + 5)^{3/2}}$$

At this point I integrated the curvature with the help of Wolframalpha, which did it numerically but I'm pretty sure it's $-2\pi-1$, when from the picture it should be $-4\pi$. Oddly enough, if I integrate the curvature without the power of $\frac{3}{2}$ in the denominator i actually get the correct result. Can anyone help me spot my mistake? Thanks a lot!

Edit: Here's a link to see the graph: https://www.desmos.com/calculator/vcxbpj2ios


I found my mistake! Posting it as an answer, so future viewers can see what is happening. The issue was not the computation, but misuse of a formula. The thing is that the formula I wanted to use for the turning number, i.e.

$$n_{\beta} = \frac{1}{2\pi} \int_{0}^{L} K_{\beta}(t)dt$$

may only be used for curves that are parametrized by arclength. Since curvature is invariant under reparametrization I just assumed that it would work the same way. But the reparametrization-invariance actually gives us

$$K_{\beta \circ \varphi}(t) = K_{\beta}(\varphi(t)),$$

where $\beta \circ \varphi$ is a reparametrization of $\beta$ by arclength. So what I ignored was the $\varphi(t)$ as the argument of the curvature. If we account for this, by a change of variable we get

$$n_{\beta} = \frac{1}{2\pi} \int_{0}^{L} K_{\beta}(\varphi(t))\, dt = \frac{1}{2\pi} \int_{0}^{L} K_{\beta}(t)\cdot|\varphi'(t)|\,dt = \frac{1}{2\pi} \int_{0}^{L} K_{\beta}(t)\cdot||\dot{\beta}(t)||\,dt,$$

where for the last equality we used the fact that the reparametrization is given by $\varphi(s) = \int_{s_0}^{s} ||\dot{\beta}(\tau)||\,d\tau$, so $\varphi'(s) = ||\dot{\beta}(s)||$ by the fundamental theorem of calculus. Therefore, in my exercise one must integrate the function

$$K_{\beta}(t)\cdot||\dot{\beta}(t)|| = \frac{-6\cos t-9}{4\cos t + 5}$$

which evaluates to $-4\pi$, giving the turning number of -2, which is exactly what it should be.

  • $\begingroup$ Nicely done (+1). $\endgroup$ – Severin Schraven Jan 29 at 9:59

Is this your curve, given by your equations?

enter image description here

  • $\begingroup$ Yes, this picture looks right! $\endgroup$ – noam.szyfer Jan 28 at 20:07

The signed curvature being $$k=-\frac{6 \cos (t)+9}{(4 \cos (t) + 5)^{\frac 32}}$$ the curvature is $$\kappa=\frac{|6 \cos (t)+9|}{(4 \cos (t) + 5)^{\frac 32}}$$ $$\int \kappa \,dt=F\left(\frac{t}{2}|\frac{8}{9}\right)+E\left(\frac{t}{2}|\frac{8}{9}\right)-\frac{4 \sin (t)}{3 \sqrt{4 \cos (t)+5}}$$ where appear elliptic integrals. $$n = \frac{1}{2\pi}\int_{0}^{2\pi}\kappa \,dt=\frac{K\left(\frac{8}{9}\right)+E\left(\frac{8}{9}\right)}{\pi }\sim 1.15940$$

  • $\begingroup$ Yes, but my problem was that it must be an integer. I found the mistake (see my own answer), but thanks a lot for taking the time to write this answer! $\endgroup$ – noam.szyfer Jan 29 at 7:51
  • $\begingroup$ @noam.szyfer. Good to see that you found the solution ! $\endgroup$ – Claude Leibovici Jan 29 at 8:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.