A rubber band around two cylinders

The two arcs on the cylinders are not parts of ellipses but of helices: If you unwrap the cylinders these arcs become straight line segments. In fact the completely unwrapped rubber band is a long straight line without corners at the points where the band takes off from the cylinders. Note that the band is mirror symmetric with respect to the line intersecting the axes of the cylinders orthogonally (red in your figure).

Consider a cylinder of radius $a$ with the $x$-axis as axis and a cylinder of radius $b$ with axis parallel to the $y$-axis at $z$-level $c>a+b$, see the following figure.

The helix on the first cylinder can be parametrized by $$s\mapsto{\bf p}(s):=(\lambda s, a\sin s,-a\cos s)\qquad(0\leq s\leq s_*)\ ,$$ and the helix on the second cylinder can be parametrized by $$t\mapsto{\bf q}(t):=(b\sin t,\mu t, c+b\cos t)\qquad(0\leq t\leq t_*)\ .$$ Here the parameters $\lambda$, $\mu>0$ and $s_*$, $t_*\in\bigl]{\pi\over2},\pi\bigr[\>$ have to be chosen such that the following main geometric condition is satisfied:

$\bullet \quad $ The vectors $\ {\bf p}'(s_*), \ {\bf q}'(t_*), \ {\bf q}(t_*)- {\bf p}(s_*)$ are parallel.

One obtains $${\bf p}'(s)=(\lambda, a\cos s, a\sin s),\quad{\bf q}'(t)=(b\cos t,\mu, -b\sin t)$$ and therefore $${\bf p}'(s)\times{\bf q}'(t)=\bigl(-a(\mu\sin s+b\cos s\sin t), b(a\cos t\sin s+\lambda\sin t),\ \ldots\bigr)\ .$$ At the solution point $S=(\lambda,\mu, s_*, t_*)$ we therefore have $$\lambda=-a\sin s\cot t,\qquad \mu=-b\sin t\cot s\ .\tag{1}$$ It follows that at $S$ both vectors ${\bf p}'(s_*)$, ${\bf q}'(t_*)$ are parallel to $${\bf z}=(-\sin s\cos t, \cos s\sin t,\sin s\sin t)\ ,$$ whereby I have omitted the $*$. Using $(1)$ we can say that at $S$ we also have $${\bf h}:={\bf q}(t)- {\bf p}(s)=(as\sin s\cot t+b\sin t, -a\sin s-bt\cot s\sin t,c+a\cos s+b\cos t)\ .$$ In order to obtain two more equations we now compute $${\bf h}\times{\bf z}=(-\sin t \> g_1, -\sin s\> g_2, \ \ldots)\ ,$$ where $$g_1:=a+\cos s(c+b\cos t+b t\sin t),\qquad g_2=b+\cos t(c+a\cos s+a s\sin s)\ .$$ Since ${\bf h}\times{\bf z}={\bf 0}$ at $S$ it follows that we obtain $s_*$, $t_*$ by solving $g_1=g_2=0$ for $s$ and $t$. Unfortunately these are transcendental equations. The following numerical example shows that we obtain indeed a unique solution in the required interval.

Interesting question describing a direct tangented rubber band between two (different diameter) parallel axis cylinders.. and subsequently twisted to make the axes perpendicular.

At the moment what appears to me as spatial determination/math geometric modeling of the connecting band is the following:

The bands of connecting cylinders run along helical geodesic paths. Else tension in the string veers off the band to slip away making it unstable. When rollers are in rotation and some solid lubricant eg French chalk is applied it becomes intuitively clear what the stable trajectories ought to be.

The taut connecting band in the air lies on a common tangent for edges of regression of tangential developable surfaces ( Gauss curvature $K=0$ ) out of /into either cylinder.

Towards solution of problem:

Parametric equation of Helix 1

$$ (x_1,y_1,z_1)= (a \cos u , a \sin u, p u) $$ where $u$ is reckoned from farthest point on Cylinder 1

Parametric equation of Helix 2

$$ (x_2,y_2,z_2)= (b \cos v +h, b \sin v , q v+k) $$ where $v$ is reckoned from farthest point on Cylinder 2

There are 4 unknowns $ (u,v, p,q)$ , known Cylinder radii $(a,b,h,k)$ including start points offset $(h,k)$.

To solve them the above are two equations.

Next two equations are determined from 3D alignment of direction cosines of common tangent

$$ \frac{x-x_1}{l}=\frac{y-y_1}{m}= \frac{z-z_1}{n} $$

$$\frac{x-x_2}{l} = \frac{y-y_2}{m}=\frac{z-z_2}{n}$$

An alternate modeling way (begin strategy at least) is to minimize total band length on the basis of same four independent variable parameters $(u,v,p,q)$ in

$$ u \sqrt{a^2+p^2} + v \sqrt{b^2+q^2} + \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+ (z_2^2-z_1)^2 }$$

by the usual methods of minimization setting derivatives to zero.

Between centers of helical arcs there is an ascending part and a descending part on either side:

As explained by Christian Blatter and Narasimham, the rubber band will wrap around the cylinders along a curve formed by two helices, joined by two tangent segments.

Let $PABQ$ be half this curve, with the other half being its reflection about line $PQ$, perpendicular to both axes (see diagram below), and with $A$, $B$ tangency points. Let then $\alpha$ be the central angle from $P$ to generatrix $AA'$ and $\beta$ be the central angle from $Q$ to generatrix $BB'$.

On the planes tangent to the cylinders at $A$ and $B$ we can then construct points $P'$ and $Q'$ such that $A'P'$ and $B'Q'$ are perpendicular to the tangency generatrices and have the same length as arcs $PA'$ and $QB'$. The curve is of minimum length if $\alpha$ and $\beta$ are such that points $P'ABQ'$ lie on a straight line.

To find $\alpha$ and $\beta$ we can consider the lateral view shown below, that is the projection of path $P'B$ on plane $PQB'$. We have: $$ \tag{1} -\cos\beta={r_B\over d+(\alpha\sin\alpha+\cos\alpha)r_A}, $$ where $r_A$, $r_B$ are the radii of the cylinders and $d$ is the distance between their axes. In the same way we can find the symmetric relation $$ \tag{2} -\cos\alpha={r_A\over d+(\beta\sin\beta+\cos\beta)r_B}. $$

These two equations can be solved for $\alpha$ and $\beta$, but we cannot express the solutions as simple expressions. Fortunately, the solutions can be obtained numerically with a simple iterative approach, starting with $\alpha=\beta=\pi/2$. Once $\alpha$ and $\beta$ are known, we can also compute $$ AA'=-\alpha \sin\alpha \cot\beta\, r_A, \quad BB'=-\beta \sin\beta \cot\alpha\, r_B. $$

EDIT.

An example. For $r_A=0.6$, $r_B=0.3$, $d=1$ (the same values chosen in Christian Blatter's answer) an iterative computation of $\alpha$ and $\beta$, using equations $(1)$, $(2)$ and starting with $\alpha=\beta=\pi/2$ quickly converges to the values shown below:

These results for $\alpha$ and $\beta$ seem to agree with the values found by Christian Blatter above (where they are named $s$ and $t$). In degrees, they amount to $\alpha=114.185°$ and $\beta=99.358°$. We then also get: $AA'=0.17976$ and $BB'=0.23053$.

EDIT 2.

I created a GeoGebra worksheet where the geometric parameters can be changed and the shape of the rubber band computed according to the construction explained above: https://www.geogebra.org/m/gngk7qzA