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Re: Spherical Law of Cosines
From: Trevor Kenchington
Date: 2002 Oct 26, 11:10 -0300
From: Trevor Kenchington
Date: 2002 Oct 26, 11:10 -0300
Dan Allen wrote: [snip] > However, I went back and found support in Smart's book for the > form that I had written, i.e., > > cos(c) = sin(a)*sin(b) + cos(a)*cos(b)*cos(ab) > > in determining the length of twilight and other such calculations. > > In thinking about things I realized that both versions are > right, but it simply is a matter of origin. Are the angles > measured down from the pole (co-latitudes and such) or are > they measured from the equator up (latitudes)? > They are equivalent. > > The mental picture that I work from is the canonical version, > > cos(c) = cos(a)*cos(b) + sin(a)*sin(b)*cos(ab) If it were true that: cos(c) = sin(a)*sin(b) + cos(a)*cos(b)*cos(ab) and: cos(c) = cos(a)*cos(b) + sin(a)*sin(b)*cos(ab) then it would necessarily be true that: sin(a)*sin(b) + cos(a)*cos(b)*cos(ab) = cos(a)*cos(b) + sin(a)*sin(b)*cos(ab) since both are equal to cos(c). And so we would have to suppose that sin(a)=cos(a), which is obviously absurd. The correct equation of this pair is Dan's "cononical version" (unless I am wildly off base). His alternate should, I suspect, be written: cos(c) = sin(A)*sin(B) + cos(A)*cos(B)*cos(ab) where A=90-a and B=90-b. Stretching memory back to high-school triginometry, I think it is true that sin(90-a)=cos(a), making this form ofthe alternate identical to the canonical version. Of course, provided one is careful over using (e.g.) latitudes rather than co-latitudes with the alternate, you could forget about explicitly subtracting anything from 90. But remembering the alternate version as a solution to the spherical triangle could get you into serious confusion. Trevor Kenchington -- Trevor J. Kenchington PhD Gadus@iStar.ca Gadus Associates, Office(902) 889-9250 R.R.#1, Musquodoboit Harbour, Fax (902) 889-9251 Nova Scotia B0J 2L0, CANADA Home (902) 889-3555 Science Serving the Fisheries http://home.istar.ca/~gadus