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    Re: Digital Camera Celestial Navigation
    From: Paul Hirose
    Date: 2008 Jul 09, 14:14 -0700

    Marcel Tschudin wrote:
    > What I tried to imagine is how the vertical refraction could possibly
    > affect the horizontal diameter.
    Imagine the center of the Sun and a point on the side limb are two ships 
    at, say, 20° north latitude (altitude). They are 16 miles apart. (The 
    Sun's unrefracted semidiameter is about 16'.) The difference in 
    longitude (azimuth) = 16' / cos 20° = 17.0268'. (Not rigorously correct, 
    but close enough.)
    At 20° altitude, refraction is 2.6' according to the Almanac table. OK, 
    the ships go 2.6' north from the initial positions. They are still 
    separated by 17.0268' of longitude (azimuth) because both ships stayed 
    on the same meridians. (Refraction moves objects along vertical circles, 
    i.e., exactly toward the zenith.) But due to the convergence of the 
    meridians, the distance between them (semidiameter) is less. The new 
    distance = 17.0268 * cos 20° 02.6' = 15.9956 miles. The decrease equals 
    .0044 miles, or .26 great circle seconds.
    Let's recalculate at 45°. Difference in azimuth = 16 / cos 45° = 
    22.6274'. Refraction = 1.0'. After applying refraction, semidiameter = 
    22.6274' * cos 45° 01' = 15.9953'. The decrease due to refraction = .28".
    These values are remarkably close to the output of my precise algorithm. 
    And as Meeus says, altitude seems to have little effect.
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