|
|
|
|
|||||
|
||||||
Subject: Re: Spherical Law of Cosines
From: Trevor J. Kenchington (Gadus@XXX.XXX)
Date: Sat Oct 26 2002 - 10:10:22 EDT
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@XXX.XXX Gadus Associates, Office(902) 889-9250 R.R.#1, Musquodoboit Harbour, Fax (902) 889-9251 Nova Scotia B0J 2L0, CANADA Home (902) 889-3555Science Serving the Fisheries http://home.istar.ca/~gadus
|