I have a sometimes rude habit of asking pointed questions without giving context. When I worked at ComputerVision, I implemented an algorithm for polygonal area. For each edge of the polygon, I computed the area of the quadrilateral between the edge and its shadow on the x-axis via orthogonal projection. The edges were directed counterclockwise around the polygon, allowing computation of directed normals, the areas were added or subtracted according to whether the normal pointed toward or away from the x-axis. Adding up all these areas gave the area of the polygon. Later, I generalized the algorithm to polyhedra. Here faces were projected onto the xy-plane to produce cylindrical volumes. Face vertices were oriented in a counterclockwise direction around the face, again allowing computation of directed face normals, and volumes were added or subtracted based on the whether the normal pointed toward or away from the xy-plane. Adding the volumes gave the volume of the polyhedron. Just recently, my new boss told me he had worked at a mapping company, and one of his last tasks, which he did not finish, was to compute the area of a map region (spherical polygon) with vertices given in lat-long coordinates. I was trying to apply the same technique, projecting the edges down to the equator and adding the areas of the quadrilaterals thus formed, hoping that this area might have some tractable formula in terms of the lat-longs. 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. ----- Original Message ----- From: "Fred lunnon" <fred.lunnon@gmail.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Tuesday, February 08, 2011 9:07 PM Subject: Re: [math-fun] Spherical trig question
On 2/7/11, Bill Gosper <billgosper@gmail.com> wrote:
Ah, from 3.5 yrs ago, ...
If there is a connection with the original thread, I'm afraid it has escaped me. However ...
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