# Can a chain of solid round links be twisted without getting shorter?

If I have two linked tori of equal size with $$r_{\rm major} >> r_{\rm minor}$$ that are tangent at one point, can one of the toruses be slightly rotated around the line through both centers without colliding? (the x-axis in the image)

I'm trying to figure out if a fully stretched chain of round links can be twisted at all without getting shorter, and once it does start getting shorter, how to calculate length as a function of twist angle.

Let me set $$r=r_\text{major}$$ and $$\delta=r_\text{minor}$$ and let $$2a$$ (with $$a) be the distance between the centres of the tori. If the tori are "orthogonal" to each other, then they are internally tangent for $$a=r-\delta$$.

We can set up a convenient coordinate system, such that the central circle of the first torus lies on the $$x-y$$ plane and is centred at $$(0,0,0)$$, while the second torus has its centre at $$(2a,0,0)$$ and its central circle lies on a plane passing through the $$x$$-axis, forming with the $$x-y$$ plane a dihedral angle $$\pi/2+\alpha$$ (i.e. tori are orthogonal for $$\alpha=0$$). By symmetry, I will consider only the case $$\alpha\ge0$$. The central circles of the tori have then parametric equations given by: $$(r\cos t,\ r\sin t,\ 0);\quad (2a-r\cos s,\ r\sin\alpha\sin s,\ r\cos\alpha\sin s).$$ The square of the distance between a point on the first circle and a point on the second circle is then a function $$f(s,t)$$ given by: $$f(s,t)=4a^2-4ar(\cos s+\cos t)+2r^2(1+\cos s\cos t-\sin\alpha\sin s\sin t).$$ We can search for extrema of $$f$$ solving the equations $$\partial f/\partial s=\partial f/\partial t=0$$. It turns out that the minimum is at

$$\cases{ s=t=0 & if \displaystyle2{a\over r}\ge1+\sin\alpha;\\ \displaystyle s=t=\pm\arccos{2a/r\over1+\sin\alpha} & if \displaystyle2{a\over r}<1+\sin\alpha.\\ }$$

Minimum distance is $$2(r-a)$$ in the first case and $$\sqrt{2r^2(1-\sin\alpha)-4a^2{1-\sin\alpha\over1+\sin\alpha}}$$ in the second case.

Tori are tangent and don't overlap only if that minimum distance is equal to $$2\delta$$. Hence we find that half the distance between the centres of the tori is given by

$$\cases{ a=r-\delta & for 0\le\sin\alpha\le1-2\delta/r; \\ \\ \displaystyle a=\sqrt{r^2\cos^2\alpha-2\delta^2(1+\sin\alpha)\over2(1-\sin\alpha)} & for 1-2\delta/r<\sin\alpha\le \sqrt{1-4\delta^2/r^2}. \\ }$$ The bound $$\sin\alpha\le\sqrt{1-4\delta^2/r^2}$$ (i.e. $$\cos\alpha\ge2\delta/r$$) is necessary to ensure that $$a\ge\delta$$ (for $$a<\delta$$ the tori would overlap). The limiting value $$\alpha_\max=\arccos{2\delta\over r}$$ represents then the maximum possible twist between two consecutive rings of the chain (see also the EDIT below).

In conclusion, a twisted chain gets shorter by the amount $$2(r-\delta-a)(n-1)$$, where the value of $$a$$ must be computed as above and $$n$$ is the number of rings in the chain.

In the figure below you can see two linked tori, with $$\delta/r=0.2$$, twist angle $$\alpha=50°$$ and tangency points in yellow.

EDIT.

After I had completed this answer I found this paper which deals with the limiting case $$\alpha=\alpha_\max$$ (see figure below). The author shows that in such case the tori touch each other not only at two points, but all along a closed curve. In addition, the central circle of each torus is a Villarceau circle of a torus having the same central circle as the other torus but minor radius $$2\delta$$.

• Thank you!!!!!! – Ash Sep 7 '19 at 0:16