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    The purpose of the ex-meridian was to get a latitude line even if the
body was obscured at the actual time of meridian passage in the days
before accurate time pieces were the norm.  With an ex-meridian, you
have to double interpolate the tables to get the �a� factor, even if
you use the formula C=a*t^2/60 to get the actual correction.  I am not
fond of this sight, and will just as soon not shoot an LAN then do an
ex-meridian of one.  In my opinion, ex-meridians are next to useless
these days at sea and are really only a good academic exercise.
Frankly, in real life, I will just run a sunline calculation which
would take me far less time, even if I did it by tabular methods.
    All types of bodies can be shot at ex-meridian, and at either upper
or lower transit.  The most common is the sun near LAN, but stars,
planets, and the moon can also be observed.
    My methodology of sight reduction of the ex-meridian is by the
meridian angle (t) method.  This saves me from having to calculate the
time of LAN at the actual position of the sight (if it is even known)
or the DR, which would require a sailing and another iteration of
calculating the time of LAN.
    The first step is to take the sight as you would a sunline, and mark
the exact time.  Next you derive the declination and GHA of the sun at
the time of the sight.  Next find the difference between the GHA of
the sun and your longitude converted to GHA.  The number should be
fairly small as the ex-meridian tables do not allow for too much time
difference between LAN and the sight time.  The number can be either
positive or negative.  Remember meridian angle is not always LHA.
Meridian angle can be measured east or west of your longitude.  Use
the arc to time table or formula for the sun; but for other bodies,
the increments and corrections page in the Almanac should be used to
turn this angular measure into minutes of time.  This will be a big
source of error if it is done incorrectly.
    The �a� factor is found in the navigational tables (Table 24 in the
2003 Bowditch) and is entered using the declination of the body and
the latitude of the observer.   This table must be double interpolated
in order to obtain an accurate value, especially as the declination
and latitude values get  closer together.   This number is then
entered into the formula C= a*t^2/60; where C is the correction to Ho
that is added for upper transits and subtracted for lower transits,
�a� is from the table, and �t� is meridian angle in minutes of time.
As an alternative, you can enter and interpolate another table (table
25 in 2003 Bowditch) to get the Ho or �C� correction.  Tabular values
must be interpolated.
    Once Ho is corrected with the �C� correction, the problem is solved
like an ordinary meridian transit problem.
    The tricky part about ex-meridians is that the table and the formula
from which it is derived fails at high altitude sights.  If there is a
blank spot where the declination and latitude meet, the formula can
give error that may be too great for general navigation which usually
occurs when the sun�s declination is the same name as, and very close
numerically, to the latitude of the observer.  This can also occur
when you are near the equator near the equinoxes.   In my example we
have less than 5 degrees of difference between the declination of the
sun and the latitude, so the �a� value will be very large and hard to
calculate accurately.
    In the given example my solution is as follows:
GHA hr      210-53.6
t-corr        2-11.5
GHA         213-05.1
GHA-o       214-20.0 (360 minus East Longitude is the �GHA� of the
observer)
t             1-15.0 = 5 minutes (convert arc to time)
�a� is interpolated from the tables, since it is borderline, it was
only interpolated for Declination which will add a bit of error$, but
since the latitude is fairly close to 15 degrees, the error should not
be unreasonable for general navigation.  I got 22.5 for �a.�  Be sure
to use the table that states Declination and Latitude are the same
name.
C= a*t^2/60 = 22.5*25/60 = 9.4�

Hs           85-22.0
IC             + 1.0
Dip             -9.9
T/P              0.0
HA           85-13.1
Body           +15.8
Ho           85-28.9
C               +9.4
Ho�          85-38.3
Z-dis (90deg minus Ho�) = 4 deg 21.7�

Dec hr            19 deg 36.3� N
Tcorr(+0.6)             + 0.1�
Dec               19 deg 36.4� N
Z-dis            -04 deg 21.7�  (sun dec is same name and > Lat, so
Dec-Z-dis = Lat.)
Latitude          15 deg 14.7�N
Error 0.7 nm north

When I ran a sunline using the given DR as the assumed position I
computed an intercept of 0.8 towards an Az of 015 deg.  Very close to
the ex-meridian latitude so the �a� correction and observation must be
fairly accurate.

Jeremy

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