Background: A polygonal billiards table $P$ with rational angles gives rise to a flat structure $S(P)$ in a standard way, described here. Curves of constant argument on $S(P)$ which start and end at a zero of the $1$-form are called saddle connnections, and are important because (among other things) for any fixed direction $\theta$, periodic trajectories on $S(P)$ with argument $\theta$ occur in cylinders bounded by saddle connections. Unfortunately, I do not know any way to compute saddle connections, or even their lenghts, other than high school trigonometry, which is all but useless except in the most trivial cases.
Question: Is there a standard technique which, given a flat structure $S$, angle $\theta$, and zeros $p,q\in S$ of the $1$-form on $S$ allows one to determine the saddle connections from $p$ to $q$ with argument $\theta$, and estimate their lengths? Even better, is there a way to compute the decomposition of $S$ into cylinders in the direction $\theta$?
Context: Consider the isosceles triangles $T_{nk}$ with base angles $\frac{2^n}{2^{n+k}-1}\pi$, where $k\ge 4$. These triangles arise naturally in the study of periodic trajectories on nearly isosceles triangles. I would like to show that as $n\to \infty$, the minimal length of saddle connections (excluding certain relatively easy-to-describe exceptions) between the vertex point and itself in $S(T_{nk})$ with argument $\pi/2$ or $\pi/2-\frac{2^n}{2^{n+k}-1}\pi$ diverges, or alternatively that the lengths of the cylinders do. I have (indirect) theoretical and computational reasons to believe this is true. This should imply a conjecture of Schwartz and Hooper that no neighborhood of the Veech triangles $V_{2^k}$ can be covered by a finite union of orbit tiles.
