 Shortest path on a circular cylinder

Unroll a circular cylinder of radius $R>0$ using the coordinate $u=R\varphi$. Its line element becomes $d{s}^2=d{u}^2+d{z}^2$, 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 $0\le \delta \le 2\pi$, put $m=\min (\delta ,2\pi -\delta )$. Then $d=\sqrt{R^2{m}^2+(z_1-z_0{)}^2}$. 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.