# Torus: Circle cut

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?

Thanks.

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I think you might need to explain this more clearly. (At least I can't understand what this question is about at all.) –  Hans Lundmark May 18 '11 at 20:10
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 –  stefan May 18 '11 at 20:16
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? –  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

$$R(r+R\cos\theta)-\rho^2\cos\theta-r\rho^2(r\cos\theta+R)/R^2=0\;,$$

which simplifies to

$$\cos\theta=-\frac{R}{r}\;.$$

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 $r'\ll r$)

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

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@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? –  joriki May 29 '11 at 6:25