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    Re: Meridian Transit of the Sun and Daily Pages
    From: Paul Hirose
    Date: 2015 Oct 11, 15:04 -0700

    I don't have a 2015 Almanac, but here's an example from 2013. At
    longitude W097 36.0, what time is the Sun's meridian passage on Sep 27
    local date?
    
    1. In the daily pages for Sep 27, look for the greatest Sun GHA that's
    less than the west longitude. In this case, it's 92 17.8 at 1800 UT.
    
    2. That GHA is 5 18.2 less than longitude. In the increments and
    corrections table, sun and planets column, look up that angle and find
    21m 13s.
    
    3. Thus meridian passage occurs at 18h 21m 13s UT. Again, that's *2013*
    Sep 27.
    
    The USNO MICA program says I missed by .07s.
    
    
    If the longitude is east, convert to the equivalent west longitude by
    taking the explement (subtract from 360). For instance, E097 36.0 is the
    same as w262 24.0. Following the procedure I outlined above, opposite 5h
    in the Sun GHA column find 257 15.0. That's 5 09 short of the desired
    angle, or 20m 36s. So transit time is 5h 20m 36s UT.
    
    
    In some cases the conversion to local time will put you in the wrong
    date. The remedy is fairly intuitive. For instance, if longitude is w179
    00.0 and local time is 12h behind UT, when is meridian passage on 2013
    Feb 10 local date?
    
    In the daily pages Feb 10 tabulation, the greatest Sun GHA less than
    longitude is 176 26.9, which is opposite 0h UT. Oops. Obviously, after
    you subtract 12h from UT to convert to local time, you're in the
    previous day.
    
    So go to Feb 11. Opposite 0h find 176 26.8, which is 2 33.2 less than
    the longitude. That corresponds to an increment of 10m 13s. So meridian
    passage is Feb 11 0h 10m 13s UT, or Feb 10 12h 10m 13s local time.
    
    
    Well, that's how *I* would calculate meridian passage with the Almanac.
    But the original poster said, "specifically referencing the data located
    on the right hand bottom of the far right hand page," and, "The reality
    for our class is that our sights require time to the nearest second."
    
    Assuming that means transit time to 1 second (far in excess of
    navigational requirements), we have to utilize the equation of time.
     From the 2013 Almanac:
    
          0h UT  12h UT
    27  8m 55s  9m 06s
    28  9m 16s  9m 26s
    
    Those times are on the Greenwich meridian. The table is not shaded,
    which means apparent time is ahead of mean time.
    
    At W097 36.0 meridian passage is about 18h UT on the 27th. Thus we have
    to interpolate between 27th 12h and 28th 0h. The desired time is (97.6 /
    180) of the way from the former to the latter. Note the EoT increases
    only 10s in that period, so no great accuracy is necessary. With a slide
    rule I get an increment of 5.4s, so EoT is 9m 11.4s. I.e., the Sun is 9m
    11.4s past the meridian at noon mean time on the observer's meridian. So
    meridian passage is at 11h 50m 48.6s
    
    But civil time is based on the 90° W meridian (? - observer's longitude
    is near a zone boundary), which is 7°36' to the east. The "conversion of
    arc to time" table on page i gives the increments as 28m (for 7°) and 2m
    24s (for 36'). So the zone (standard) time is 11h 50m 48.6s + 28m + 2m
    24s = 12h 21m 13s. That agrees with my calculation via GHA, but it's a
    lot more trouble.
    
    

       
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