Sea Navigation: Great Circles
Date: Fri, 4 Nov 1994 09:53:00 -0700 From: Paul Kairis Subject: let's do math The problem is found in sea navigation. Given two points A,B with LON and LAT coordinates, find the travel distance from A to B. The distance is actually the great circle passing A and B. The solution is needed in the form of a double integral where the first int covers the LONs and the second the LATs. I guess the diff. part is to form the vector. My calculus books are not with me, can you help? Thanks -Paul
From: Dr. Ken Date: Fri, 4 Nov 1994 12:15:29 -0500 (EST) Paul! I wanted to let you know that we're working on your problem. The first thing my colleague Ethan noticed, however, is that you probably don't want to look at a double integral at all, but instead a single integral. The reason for this is that when you find the length of a curve in three space, you integrate the number one over the region in question, i.e. the curve you want the area of. I think that there's an easier way to handle your problem, though. Use the distance formula to find the real distance (through the crust of the earth) between the two points. Then you've reduced the problem to a two-dimensional Euclidean Geometry problem, namely finding the length of an arc of a circle cut off by a segment with a certain length. I hope this helps, and let us know if it doesn't! -Ken
Date: Fri, 4 Nov 1994 10:40:39 -0700 From: Paul Kairis Ken, thanks for your input. What you have suggested is quite true. I am interested in the symbolic representation though, e.g. Here're my notes: ----------------- Let dS = sqrt[ dx^2 + dy^2 + dz^2 ] where x = r cosf cosg y = r cosf sing z = r sinf then, latB lonB / / S = Radius | | [1+cos^2(f)] df dg / / latA lonA This is what I came up with but it is not right! -Paul
From: Dr. Ken Date: Mon, 7 Nov 1994 12:02:16 -0500 (EST) Hello there! I think the main problem is that you're trying to use a double integral to find the length of a curve, but the double integral will give you the AREA of the curve. And that's Zero. I'm not quite sure how you got that particular integral, but that's why it doesn't work. -Ken "Dr." Math Note: There's a complete solution of this type of problem, available for your viewing pleasure, at http://mathforum.org/dr.math/problems/longandlat.html .
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