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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Lars Bergman
Date: 2020 Nov 1, 07:49 -0800
Dave, you asked "Which antilog of 8.230947 is 59minutes 58 seconds?".
It is log hav 59m58s. The formula he use is
hav t = sec φ · csc p · cos s · sin(s-h)
where t is hour angle, φ is latitude, p is polar distance (or co-declination), h is altitude and s = (h+φ+p)/2. The "remainder" is (s-h). The formula could be derived from the ususal PZX triangle altitude formula:
sin h = sin φ · cos p + cos φ · sin p · cos t
Then cos t = (sin h - sin φ · cos p) / (cos φ · sin p) . Subtract each side of this equation from unity:
1 - cos t = (1 - (sin h - sin φ · cos p)) / (cos φ · sin p) = (cos φ · sin p - sin h + sin φ · cos p) / (cos φ · sin p) = (sin(φ+p) - sin h) · sec φ · csc p
Now,
sin(x+y) = sin x · cos y + cos x · sin y
sin(x-y) = sin x · cos y - cos x · sin y
From this follows that sin(x+y) - sin(x-y) = 2·cos x · sin y
We have
x+y = φ+p
x-y = h
Hence x = (φ+p+h)/2 and y = (φ+p-h)/2 = (φ+p+h)/2 - h. Thus we get sin(φ+p) - sin h = 2 · cos((φ+p+h)/2) · sin((φ+p+h)/2 - h) and the final formula becomes
(1 - cos t) / 2 = cos((φ+p+h)/2) · sin((φ+p+h)/2 - h)· sec φ · csc p, or
hav t = sec φ · csc p · cos s · sin(s-h), where s = (h+φ+p)/2. This form is suitable for logarithmic use.
Lars