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    Re: Time Sight and Ex-Meridian by Bygrave
    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) · t/ 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 ·  t  with π/6 = 0.52, Greg's formula.

    Lars

       
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