Determine orientation of spherical polygon without trig functions

Question

Is there a way of testing the orientation of a spherical polygon given an ordered list of its vertices that doesn’t involve computing (inverse) trigonometric functions? The polygon is not necessarily convex but it partitions the sphere in a way that both the north and south poles are on the same side of the boundary. The interior of the polygon (so that we have a fixed clockwise and counterclockwise direction) is the region that contains neither pole.

Answer 1

Let $n = (0,0,1)$ and $s = (0,0,-1)$ be the north and south poles.

Since the spherical polygon $P$ doesn't intersect $s$, you can stereographic project $P$ with respect to $s$ and map it to a polygon-like region $P'$ (with curved edges) on the tangent plane $T_n$ of unit sphere at $n$.

$$S^3 \setminus \{ s \} \supset P \ni (x,y,z) \quad\mapsto\quad \left(\frac{2x}{1+z},\frac{2y}{1+z},1\right) \in P' \subset T_n$$

The orientation of $P$ will be the same as its image $P'$. One can determine the later by looking at the sign of its signed area.

Let's say the vertices of $P$ are $p_i = (x_i,y_i,z_i)$ for $0 \le i < m$ and let $p_m = p_0$. When the vertices are find enough and $P'$ can be approximated by an ordinary polygon with straight edges, we have

$$\begin{align}\verb/Orientation/(P) & = \verb/Orientaion/(P') = \verb/sign/(\verb/signed-area/(P'))\\ & = \verb/sign/\left(\sum_{i=0}^{m-1} \frac{x_i y_{i+1} - x_{i+1} y_i}{(1+z_i)(1+z_{i+1})} \right) \end{align} $$ This approximation can fail in two cases, either some of the edges in $P$ is very long or very close to south pole $s$. In both cases, subdivide corresponding edges in a fine enough manner will solve the problem.

Answer 2

Project the vertices on a unit cube centered on the sphere to get a polyhedron. This takes Euclidean norms of vectors, but no trigonometry.

Then you compute the signed volume by the shoelace formula generalized to 3D.