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    Re: Nav exercise - ex-meridian
    From: John Cole
    Date: 2008 May 23, 15:12 -0700

    I used the ex-meridian tables in Burton's Nautical Tables (1963
    Edition)
    
    2008 May 18
    ZD -10
    Lat= 15d 14.0m N
    Long= 145d 40.0m E
    
    For the ex-meridian sight at 2h 8m 46s GMT
    Dec= 19d 36.4m N
    LHA= 23h 55.02m; 358.75d;  358d 45.0m
    Ho= 85d 29.0m (sextant altitude corrected=observed altitude)
    Hc= 85d 27.9m
    Az=15.00
    Intercept 1.1 toward
    
    From Burton:
    Table I (Dec, Lat): F=25.2
    Table II (LHA, F): 10.5 minutes of arc to be added to the observed
    altitude to determine the altitude at the time of observation
    Table III second correction to be subtracted from first correction
    (10.6m)=0.2m
    Total Correction= 10.3m to be added
    
    In Table I, owing to the closeness of Dec and Lat I had only 3 of the
    4 entries needed for accurate interpolation so there was some
    guesswork involved.
    
    Corrected Ho = 85d 29.0m + 10.3m =  85d 39.3m
    Corrected Obs ZD= 90 -  85d 39.3m = 4d 20.7m
    
    Lat= Dec-Corrected ZD= 19d 36.4m - 4d 20.7m = 15d  15.7m  N
    DR Lat= 15d 15.7m N
    
    No error found.
    
    John Cole
    
    
    
    On May 21, 2:54�am, Anabasis  wrote:
    > � � � � 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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