= Shortest path on a circular cylinder
{title2=$d=\min_{k\in\mathbb Z}\sqrt{R^2(\phi_1-\phi_0+2\pi k)^2+(z_1-z_0)^2}$}
Unroll a <circular cylinder> of radius $R>0$ using the coordinate $u=R\phi$. Its <line element> becomes $ds^2=du^2+dz^2$, so a <geodesic> lifts to a <straight line> in the <plane>. An endpoint has infinitely many lifts differing by $2\pi Rk$. Minimize the ordinary <Euclidean distance> over those lifts to obtain the unrestricted shortest path. A fixed lift instead specifies a winding class and need not give the global shortest path.
For angular difference $0\leq\delta\leq2\pi$, put $m=\min(\delta,2\pi-\delta)$. Then $d=\sqrt{R^2m^2+(z_1-z_0)^2}$. At $\delta=\pi$, two different <geodesics> have the same shortest <length>. This is the source of the corner in the distance-versus-angle graph at the opposite generator of the <circular cylinder>.
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