DW>Fred Lunnon's analysis uncovered a simpler approach, to wit, project the edges not to the equator, but to the north pole. Instead of adding up quadrilateral areas below the edges, I could add up triangular areas above the edges. The sides of the triangle impinging on the pole are easy to obtain (complements of the vertex latitudes) as is angle between them (difference of the vertex longitudes). From there, standard spherical trigonometry should allow me to derive the remaining side (the original edge, via law of cosines), angles (via law of sines), and area (law of excess), the prize being the area in terms of edge vertex lat-longs. The sign of the area is determined by local order of the vertex latitudes. The trig, however, is too daunting for me, which is why I posted to brighter lights than myself. I was hoping that some of the computational ugliness might implode leaving a relatively tractable expression. (rwg)>Then can't we do the whole thing with just the law of excess, and the general formula for the intersection angle of two arcs, given the (lat,long)s of their endpoints? How bad can that be? --rwg