Given a Torus $T$ with major and minor radius $R$ and $r$, respectively, I can obtain a circle lying in $T$ by cutting $T$ with a bi-tangential plane.

Now I don't want circles, but Tori with major radius $R$ and minor radius $r' \ll r$.

Given $R,r,r'$ and assuming that $T$ lies in the $X-Y$ plane:

What is the minimum angle I have to rotate the cutting plane around the $Z$-axis, such that I can place two $(R,r')$ tori as close together as possible?


  • 1
    $\begingroup$ I think you might need to explain this more clearly. (At least I can't understand what this question is about at all.) $\endgroup$ – Hans Lundmark May 18 '11 at 20:10
  • $\begingroup$ Cutting a torus with a bi-tangential planes gives so called villarceau circle. I can now rotate the cutting planes by an infinitesimal angle and thus associating each point of the torus with exactly one such villarceau circle. Now I want to replace the circles with thin tori and I'm interested in how much i have to rotate the cutting plane at least s.t. no 2 consecutive thin tori intersect $\endgroup$ – stefan May 18 '11 at 20:16
  • $\begingroup$ OK, now I see (I think). When you say "replace", you mean you're "adding flesh" to the infinitesimally thin Villarceau circle so that it thickens into a torus with (small but finite) radius $r'$, right? $\endgroup$ – Hans Lundmark May 18 '11 at 22:13

The Villarceau circles can be parametrized by

$$\vec{x}=\left(\begin{array}{ccc} \cos\phi & \sin\phi & \\ -\sin\phi & \cos\phi & \\ &&1 \end{array} \right) \left( \begin{array}{c} \rho\sin\theta\\ r+R\cos\theta\\ r\sin\theta \end{array} \right)\;, $$

where $\theta$ is the angle along the circle, $\phi$ is the orientation of the circle around the $z$-axis and $\rho=\sqrt{R^2-r^2}$. At $\phi=0$, the derivatives with respect to $\theta$ and $\phi$ are

$$\frac{\partial\vec{x}}{\partial\theta}= \left( \begin{array}{c} \rho\cos\theta\\ -R\sin\theta\\ r\cos\theta \end{array} \right)\;, \;\;\; \frac{\partial\vec{x}}{\partial\phi}= \left( \begin{array}{c} r+R\cos\theta\\ -\rho\sin\theta\\ 0 \end{array} \right)\;. $$

We want to know where changing $\phi$ will be least effective in getting us away from the circle. This is determined on the one hand by the magnitude of the derivative with respect to $\phi$ and on the other hand by the angle between the two derivatives; the rate at which we move away from the circle as we change $\phi$ is the product of that magnitude and the sine of that angle:

$$ \begin{eqnarray} \frac{\mathrm ds}{\mathrm d\phi} &=& \left|\frac{\partial\vec{x}}{\partial\phi}\right| \sqrt{1-\left(\frac{\frac{\partial\vec{x}}{\partial\phi}\cdot\frac{\partial\vec{x}}{\partial\theta}}{\left|\frac{\partial\vec{x}}{\partial\phi}\right|\left|\frac{\partial\vec{x}}{\partial\theta}\right|}\right)^2} \\ &=& \sqrt{\left|\frac{\partial\vec{x}}{\partial\phi}\right|^2-\left(\frac{\frac{\partial\vec{x}}{\partial\phi}\cdot\frac{\partial\vec{x}}{\partial\theta}}{\left|\frac{\partial\vec{x}}{\partial\theta}\right|}\right)^2} \\ &=& \sqrt{(r+R\cos\theta)^2+\rho^2\sin^2\theta-\left(\rho(r\cos\theta+R)/R\right)^2}\;. \end{eqnarray} $$

We need to find the minimum of this expression with respect to $\theta$. Setting the derivative of the radicand with respect to $\cos\theta$ to zero yields


which simplifies to


Since $R> r$, this has no solution, so the extrema occur on the boundary at $\cos\theta=\pm1$. At these points, $|\partial\vec{x}/\partial\theta|$ is parallel to the projection of $|\partial\vec{x}/\partial\phi|$ into the cutting plane, so the result is the same as would have been obtained by using the angle between $|\partial\vec{x}/\partial\phi|$ and the plane instead:

$$ \frac{\mathrm ds}{\mathrm d\phi}=(R\pm r)\frac{r}{R}\;\;\text{for}\;\;\cos\theta=\pm1\;. $$

The factor $R\pm r$ is $|\partial\vec{x}/\partial\phi|$, and the factor $r/R$ is the sine of the angle between $\partial\vec{x}/\partial\phi$ and the plane. The lesser of these is the inner one at $\cos\theta=-1$, and the angle by which you need to rotate to make two tori with minor radii $r'$ fit is thus (to first order for $r'\ll r$)

$$\Delta\phi=\frac{2r'}{\mathrm ds/\mathrm d\phi}=\frac{2r'R}{(R-r)r}\;.$$

  • $\begingroup$ @stefan: If I understand correctly what you're doing, the volume should indeed go to zero, since you're only filling an outer shell of the torus of thickness of order $r'$ with the little tori? $\endgroup$ – joriki May 29 '11 at 6:25

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.