= Riemannian distance on a circular cylinder
{c}
{title2=$d=\min_{k\in\mathbb Z}\sqrt{R^2(\Delta\theta+2\pi k)^2+(\Delta z)^2}$}
On the lateral <surface> of an infinite <circular cylinder> of radius $R$, the <induced metric> is $R^2d\theta^2+dz^2$. Lifting a <curve> to the unrolled plane $(x,z)=(R\theta,z)$ preserves its <arc length>. If the endpoints differ by angular coordinate $\Delta\theta$ and height $\Delta z$, the possible lifted endpoint displacements are $(R(\Delta\theta+2\pi k),\Delta z)$ for <integers> $k$. Each straight segment gives a <geodesic>, and the shortest of their <lengths> is
$$
d_g(p,q)=\min_{k\in\mathbb Z}
\sqrt{R^2(\Delta\theta+2\pi k)^2+(\Delta z)^2}.
$$
Choose the representative of $\Delta\theta$ with smallest absolute value modulo $2\pi$; if its absolute value is $\pi$, the two angular directions tie. This minimum is attained because the candidate <lengths> tend to infinity as $|k|\to\infty$. The formula concerns travel on the lateral surface; it does not allow a shortcut through the cylinder's interior or across end caps.
For $R=1$, take $p=(1,0,0)$ and $q=(0,1,3)$ in surrounding <Euclidean space>. The angular difference is $\pi/2$ and height difference is $3$. The $k=0$ route has <length> $\sqrt{9+\pi^2/4}\approx3.386$. The $k=-1$ route goes the other way around and has <length> $\sqrt{9+9\pi^2/4}\approx5.586$. Both are among the <geodesics on a circular cylinder>, but only the first is a <minimizing geodesic> between these endpoints.
\Image[/cylinder-geodesics-unrolled.png]
{title=Cylinder distance compares straight segments to different lifts of the same endpoint}
This is a global consequence of <intrinsic flatness of a circular cylinder>: locally everything is Euclidean, yet the periodic angular coordinate leaves several competing routes. A shortest-route problem on the curved surface becomes a family of ordinary planar distance calculations.
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