Let $\Lambda$ be a cone with apex $a$ and apex angle $\alpha$. Draw a simple (non-self-intersecting) curve $C=(x,y)$ on $\Lambda$, and then develop it to a curve $\overline{C}$ on a plane by rolling $\Lambda$ on that plane, starting at $x$ and ending at $y$. I am interested in the conditions under which $C$ develops to a simple curve $\overline{C}$ in the plane. If $C$ is composed of geodesic segments, then $\overline{C}$ is a drawing in the plane of segments of the same lengths, and meeting at the same angles, as they do on $\Lambda$. Intuitively, if $C$ crosses each ray from $a$ on $\Lambda$ at most once, and if $C$ is drawn with wet ink, $\overline{C}$ is the drawing that results by rolling $\Lambda$ on the plane, transferring the ink from $x$ to $y$ along $C$.

I know various conditions on $C$ that ensure that $\overline{C}$ is simple, e.g., if $C$ is a convex curve on $\Lambda$, having $\le \pi$ to one side at every point. One can impose various conditions on $C$, e.g., that $C$ be closed and encircle the apex $a$. But perhaps the simplest version is when $C$ is a simple, open curve, in which case the development from $x$ to $y$ is determined up to rigid motions.

Here is an example that shows that not all curves $C$ on a cone develop without intersection.
Here $\alpha=\frac{3}{4}\pi$, and the cone is shown cut opened by two different generators
in (a) and (b) below. The development $\overline{C}$ is shown in (c).

_{(This figure is adapted from one in the paper,
"Conical Existence of Closed Curves on Convex Polyhedra," arXiv:1102.0823.)}

I suspect the analogous problem on a sphere is difficult, but I am hoping that on a cone it is much simpler. I would appreciate any even tangential references or ideas. Thanks!

thinkthis is the same as your definition... – Joseph O'Rourke Jun 24 '12 at 12:50developmentof $C$. – Misha Jun 24 '12 at 13:04