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    Re: determination of longitude and the prime vertical
    From: Paul Hirose
    Date: 2007 Apr 12, 16:14 -0700

    George Huxtable wrote:
    > | Date:     1 May, 2007
    > | UT:       20:41:45
    > | Position: N40 W60
    > | Hc:       24d 00!2
    > | Az:       270.00
    > | LHA:      71d 09!5
    > | dec:      15d 09!5
    I'll work the computation with the USNO MICA program, using the
    Terrestrial Time scale for best accuracy. IERS Bulletin A predicts UT1
    will be .11 second behind UTC on that date. In addition, UTC will be 33
    seconds behind TAI, and TAI 32.18 seconds behind TT.
    Delta T = .11 + 33 + 32.18 = 65.29 seconds. So 20:41:45 UT1 = 20:42:50.3 TT.
    For that time MICA gives RA 2h 34m 40.568s and dec. +15° 09′ 29.53″ with
    respect to the true equator and equinox of date.
    At 20:41:45 UT1, MICA says the Greenwich apparent sidereal time is 11h
    19m 20.1896s. (UT1, not TT, is the better time scale to use here.)
    Subtracting RA Sun gives GHA 131° 09′ 54″, and subtracting longitude
    gives LHA 71° 09′ 54″. That's only a tenth minute off what George's
    calculator gave.
    To be consistent with the TT time scale, the observer's longitude must
    be moved west by delta T times the solar / sidereal ratio, converted to
    arc. I.e., 65.29 * 1.002738 * 15″ = 16′ 22″. So the adjusted position is
    40° 00′ 00″ N 60° 16′ 22″ W. From that point on 2007 May 1 20:42:50.3
    TT, MICA gives altitude 24° 00′ 03″, azimuth 270° 00′ 02″.
    Everything gets much simpler if you input UT1 throughout and let MICA
    use its internal delta T table. MICA 2.0, the one I have, is a full 
    second of time in error, but that affects only the Sun's position 
    relative to the stars. To navigational accuracy, the end result is 
    Incidentally, the latest MICA version is 2.1, released in December 2006.
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