 Geodesics on a circular cylinder

In coordinates $(\theta ,z)$ on a circular cylinder of radius $a$, the induced metric is $a^{2}d{\theta }^2+d{z}^2$. Its coefficients are constant, so the Christoffel symbols vanish and the affinely parametrized geodesic equation is ${\theta }^{\prime \prime }=z^{\prime \prime }=0$. The nonconstant solutions are generators when $c=0$, circular parallels when $d=0$, and helices when both are nonzero. Unrolling the cylinder with coordinate $u=a\theta$ makes all these curves straight lines in the Euclidean metric. Unit-speed parametrization additionally requires $a^2{c}^2+d^2=1$.

For a worked helix on a circular cylinder of radius $1$, choose $\theta (t)=2\pi t,z(t)=2t$ for $0\le t\le 1$. Unrolling gives the straight segment $(u,z)=(2\pi t,2t)$, so its arc length is $\sqrt{4{\pi }^2+4}$. Its acceleration in space points normally to the surface, consistent with the geodesic equation.

This geodesic is not globally shortest between its endpoints: they share the same angular position, and the vertical generator has length $2$. The helix minimizes length among paths with the same endpoints and one full winding, because their unrolled lifts join $(0,0)$ to $(2\pi ,2)$. In general, geodesics minimize sufficiently short subarcs; changing winding can produce a shorter route. Keeping this local/global distinction makes unrolling useful without mistaking every straight lift for an unrestricted shortest path.