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Re: Time Sight and Ex-Meridian by Bygrave
From: Lars Bergman
Date: 2021 Apr 7, 14:27 -0700
From: Lars Bergman
Date: 2021 Apr 7, 14:27 -0700
Tony,
with latitude φ, declination δ, hour angle t, altitude h and reduction to the meridian x, you have
sin(h+x) = sin φ · sin δ + cos φ · cos δ
sin h = sin φ · sin δ + cos φ · cos δ · cos t
Subtraction gives
sin(h+x) - sin h = cos φ · cos δ · (1 - cos t)
sin h · cos x + cos h · sin x - sin h = cos φ · cos δ · 2·sin2(t/2)
With x and t small quantities you can approximate to
x·cos h = cos φ · cos δ · 2·(t/2)2 with x and t expressed in radians. Then
x = (cos φ · cos δ / cos h) · t2 / 2
For x expressed in minutes of arc and t in degrees, you must multiply the right hand side by 60·180/π · (π/180)2 which equals π/3. Thus
x = (cos φ · cos δ / cos h) · π/6 · t2 with π/6 = 0.52, Greg's formula.
Lars
