Moving Line Segment Problem

Question

So I was contemplating the following problem: Given two circles of radius $r$, which will position vertically so that their centers are separated by a distance of $2r+d$ (and thus, at the nearest points, the circles are $d$ units apart). Suppose we have a line segment which is also positioned vertically so that its endpoints each lie a different circle to the right of those circle's centers. The question is the following: For what lengths $l$ can this line segment exist (easy question), but more importantly can also be moved to the analogous position on to the left of the centers by only moving its endpoints along the circles that they each began on and without changing the line segment's length?

I have reasoned informally that the answer is that any length $l$ that can exist can also be moved to the equivalent position in the required manner, which means that $d<l<4r+d$, but my question is how to actually, formally, prove it? And what of the more general question being alluded to, when can we move a line segment with endpoints on distinct curves to another position with endpoints on the curves by moving said endpoints along the respective curves? (I realize this latter question is really broad, I'm most interested in what, if anything, is known about it.)

Edit: Here is a photo of a free-hand sketch of the situation (apologies for my crappy drawing and handwriting) https://goo.gl/photos/cZnb6znt5Ju8mdfh8

Answer 1

Assume $0<r<1$, and let $$s\mapsto z(s):=(r\sin s,1- r\cos s),\qquad t\mapsto w(t):=(r\sin t,-1+r\cos t)$$ be the two circles. The upper circle is then drawn counterclockwise, the lower clockwise; and $s=0$, resp. $t=0$, produce the two points where these two circles come nearest to each other.

Any choice of a point $(s,t)$ on the Torus ${\mathbb T}=({\mathbb R}/2\pi)^2$ determines a pair of points on the two circles, and the (euclidean) distance between these two points is $\ell(s,t):=\sqrt{f(s,t)}$, whereby $f(s,t)$ computes to $$f(s,t)=|z(s)-w(t)|^2=2\bigl(2+r^2-2r\cos s-2r\cos t+r^2\cos(s+t)\bigr)\ .$$ The following images show contour plots of the function $\ell:{\mathbb T}\to{\mathbb R}$, the first of them having $(0,0)$ as center, and the second of them having $(\pi,\pi)$ as center. We can see that all contour lines with one exception are loops, and these loops are symmetric with respect to $(0,0)$, resp. $(\pi,\pi)$. This implies that one may walk from any point $(s_0,t_0)$ on such a contour line to the opposite point $(-s_0,-t_0)$. But this means that the linkage configuration corresponding to the point $(s_0,t_0)$ can be continuously transformed into the configuration corresponding to $(-s_0,-t_0)$.

The contour line corresponding to the level $\ell=2$ consists of two loops intersecting at the points corresponding to $(0,\pi)$ and $(\pi,0)$. One of these loops appears as straight line $\phi+\psi=\pi\ {\rm mod}\ 2\pi$ and parametrizes the rotary motion of a vertical bar of length $2$.

It will be computationally difficult to convert these pictures into formulas which prove that the level lines for $\ell<2$ and for $\ell>2$ are indeed loops. But it is easily checked that all four critical points $(0,0)$, $(0,\pi)$, $(\pi,0)$, and $(\pi,\pi)$ of $f$ are nondegenerate. Morse theory then tells us that all level lines for $\ell_{\min}<\ell<2$ as well as all level lines for $2<\ell<\ell_{\max}$ have to be loops.

Answer 2

Imagine the point $(0,0)$ is in the center of the line between the two circles. Hence, the center of the top circle is $(0,r+\frac{d}{2})$ and the center of the bottom circle is $(0,-r-\frac{d}{2})$. The line on the right with the length $l$ connect two points, say $A$ from the top and $B$ from the bottom, with coordinates $(x,y_A)$ and $(x,y_B)$. Note that the line on the right is vertical, hence $A$ and $B$ share the same $x$. In addition $y_A=-y_B$.

In this case, the distance between $A$ and $B$ is simply $l = 2\times y_A$. Since the point $A$ lies on the top circle, hence satisfying its equation, we get $$ y_A = \pm \sqrt{x^2 - r^2} +r+\frac{d}{2} $$ Hence, $$ l = \pm 2\sqrt{x^2 - r^2} +2r+d $$

To maximize, looking at the right side where $x=0 \to r$, we take the derivative of $l$ and we will get $x=0$. Hence, $y_A = \frac{d}{2}$ or $y_A = 2r+\frac{d}{2}$.

Clearly, $y_A=2r+\frac{d}{2}$ is for the maximum and $y_A=\frac{d}{2}$ is for the minimum. Hence, $d\leq l \leq 4r+d$.