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    Re: A Lunar Example: Sun and Moon vertically aligned
    From: Paul Hirose
    Date: 2017 Dec 27, 14:00 -0800

    As an exercise I created my own example. At 2017-10-24, 0 h UTC,
    5.71132° S 154.93815° E, height zero, unrefracted Moon altitude is 20°
    and the Sun is directly above.
    
    Verify with the JPL Horizons online calculator:
    107.5471  19.9999  Moon az, unrefracted el
    107.5473  68.2811  Sun
    
    Azimuths are not exactly equal, but a better figure of merit is the
    inclination of the Moon / Sun / observer plane with respect to the
    zenith. I get .00009°, about as good as we can expect at the precision
    of Horizons.
    
    (I omitted my procedure for computing the place where the desired
    conditions occur, but if anyone is interested I'll explain.)
    
    Next, create simulated observations (1013 millibars and 15 C):
    20°17' Moon alt (upper limb, zero dip)
    68°01' Sun (lower limb)
    47°43.8' lunar distance, near limbs
    
    Solve with the Lunar4 program. To make the problem more difficult,
    inject errors of 30 minutes in time and 5° in lat and lon. Initial
    settings are 00h30m UTC, zero latitude, 160 E. Enter the angles in the
    input boxes, select .1' accuracy, check the "solve for time" box, and
    click OK.
    
    Oops, the program throws an exception. The reason is that Lunar4
    iteratively seeks the time and place where all three angles occur. But
    with both bodies on the same azimuth, the algorithm fails to converge.
    So first let's solve for time only. Delete the altitudes in the input
    boxes but not the lunar distance. Now, with "solve for time" still
    checked, when you click OK the program assumes its position is correct,
    and solves for time only. (I call this a "lunar time sight.")
    
    The result is 00:03:16. Enter that in Lunar4, enter Moon and Sun
    altitudes, UNCHECK the "solve for time" box, click OK, read the
    intercepts (observed - computed) at the new time:
    Moon -3°42.9′ at azimuth 110.3°
    Sun -2°33.9′ at azimuth 128.3°
    
    With those intercepts and azimuths, correct the position with the
    numerical method in the Nautical Almanac. Result is 1.6 S 155.4 E. (The
    Almanac says to iterate until the correction is less than 20 miles, but
    I didn't do that.) Enter the corrected position in Lunar4 and solve the
    lunar time sight. Result is 00:00:00.6 UTC. That's virtually perfect,
    but of course in the real world we wouldn't know that. We must verify
    the solution has converged, so compute intercepts at the new time:
    
    Moon 0°50.9′ at 109°01.5′
    Sun 1°08.3′ at 117°14.3′
    
    Calculate an improved position: 3.9 S 155.5 E. Solve the lunar time
    sight at the new position. Result is virtually unchanged: 00:00:01 UTC.
    And, the intercepts are only a few minutes of arc, so the lunar distance
    corrections for refraction and parallax will be correct. The solution is
    accurate to one second of time. (The lunar distance rate is +0.37′ per
    minute time and the simulated observation was rounded to .1′, so I got
    lucky.)
    
    I may adopt the method above in the next release of my program.
    

       
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