Your estimates are not scale invariant, so I am trying to guess what you want from the picture.

A closed geodesic cuts your surface into two discs. Both have geodesic as a boundary, positive curvature and area $\le$ than area of your original surface. If geodesic is long, then (with the intrinsic metric) these discs look almost like segments. It has to have curvature near $\pi$ in concentrated form near the ends.

Thus if long geodesic exist then almost all curvature can be covered by 4 fingers on your surface...

For example,

• you can not have it if Gauss curvature $\ge 1$. (in In this case you can still have long shapes).shapes: say a doubling of a slice of unit shpere between meridians can be embedded into $\mathbb R^3$ as a convex surface, one can smooth singularities on the poles.)
• you can not have it on a polyhedral surface with more than 4 vertexes. If you have it for an arbitrary long simple geodesics on the surface of tetrahedral, the sum of angles around the vertexes each vertex has to be $=\pi$.
1

Your estimates are not scale invariant, so I am trying to guess what you want from the picture.

A closed geodesic cuts your surface into two discs. Both have geodesic as a boundary, positive curvature and area $\le$ than area of your original surface. If geodesic is long, then (with the intrinsic metric) these discs look almost like segments. It has to have curvature near $\pi$ in concentrated form near the ends.

Thus if long geodesic exist then almost all curvature can be covered by 4 fingers on your surface...

For example,

• you can not have it if Gauss curvature $\ge 1$ (in this case you can still have long shapes).
• you can not have it on a polyhedral surface with more than 4 vertexes. If you have it for the surface of tetrahedral, the sum of angles around the vertexes has to be $=\pi$.