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Re: FW: Mag. Variation
From: Herbert Prinz
Date: 2004 Feb 24, 15:54 -0500
From: Herbert Prinz
Date: 2004 Feb 24, 15:54 -0500
George Huxtable wrote: > hav (180-Z) = sec h sec L cos S cos (S - p) > > I have quite a collection of navigational texts, but in none of them can I > find any referece to that same expression for azimuth. Although Bowditch is > usually rather good at explaining or deriving the expressions he uses, in > this case he doesn't seem to do so (or I can't find it). It's always the same boring old trick: 2 cos((a+b)/2)*cos((a-b)/2) = cos(a) + cos(b) and hav(a) = (1-cos(x))/2 as the name suggests. So, from [1 - cos(180-z)] * cos(h) * cos(L) = 2 * cos((h+L+p)/2) * cos((h+L-p)/2) we get [1+ cos(z)] * cos(h) * cos(L) = cos(h+L) + cos(p) = cos(h)*cos(L) - sin(h)sin(L) + cos(p) Eliminate bracket, move terms around and obtain the plain cosine formula for the PZX triangle: sin(h)*sin(L) + cos(h)*cos(L)* cos(z) = cos(p) Best regards Herbert Prinz