At what angles of the solution are the geodesics on the cone self-intersecting I have a circular cone. I'm trying to understand how self-intersecting of geodesics depends on the angle at the vertex of the cone. I tried to create the unfolded cone, but it didn't help me to solve the task. I counted the geodesics in an explicit form, but it did not help me at all
 A: Let the unfolding $C'$ of the cone $C$ take place in the $(x,y)$-plane with the vertex at the origin. All geodesics are straight lines in the unfolding. A geodesic $\gamma$ that is not a generatrix of $C$ has a culmination point $P$. We may choose the $x$-axis of the unfolding through $P'$. The unfolded geodesic $g$ will then be parallel to the $y$-axis at distance $a>0$ from the origin. Draw rays from the origin corresponding to generatrices going through $P$ or its  mirror image with respect to the axis of the cone. If the generatrices of $C$ make an angle $\alpha$ with the axis of $C$ then the angle $\beta$ between two successive such rays is given by $\beta=\pi\sin\alpha$. Whenever $g$ intersects two opposite (with respect to the $x$-axis) such rays we have a selfintersection of $\gamma$. The required angles can then be computed using elementary trigonometry. 
A: When developing the cone you see how the geodesics bend  and how the intersection points map in both the views. $P,Q$ map to the same point in bending but they are at opposite ends of unbent chord.
Important it is to understand that the semi sector angle $ PON $ reduces to cone semi-vertical angle $MON$ in projection, a result obtained by comparing the 3D situation of cone bending deformation.
$$ PON = 180^{0}\, \sin MON $$
The angle $\psi= OPG $  made by geodesic line (red) to slant cone generator $OP,OQ$ at geodesic intersection point is given by geodesic law of Clairaut, remains unchanged in bending due to isometric dependendency of this angle on First Fundamental Form coefficients.
$$ QM \sin \psi = AG $$

