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    Thank you for those who tried the problem.  I will take heed of the
advice Peter offered and try to remember to give date in a more
universal format.
    I hesitate to work problems out for two major reasons: 1) there are
many methods of sight reduction and I really don�t have the time to
demonstrate more than one, and 2) the books I use are quite
�American� (i.e. HO 249, HO 229, American Practical Navigator) and I
am not sure what publications other members may use.  Of course all
information can be calculated in short order with computers and
calculators.
    My reductions at sea are usually performed with a navigation software
package called �SkyMate Pro,� which is available as shareware online.
It is incredibly easy to use and can reduce sights from GPS positions
without using assumed position reductions as is typical with tables
and paper.  I am, however, well versed in nearly every reduction using
the standard US government tables.  My reduction paradigms are similar
to the US Coast Guard methods as these were the ones I was taught, and
was tested on in my Merchant Marine Officer examinations.  I will be
using these methods in my solutions to the problems I offer up to the
list.
The first problem was a LAN latitude line from a ship drifting at sea
(no sailings are needed).  Part �A� asked for a time of LAN,
approximate sextant altitude, and AZ of the sun.  This information is
useful to the navigator at sea so that he is looking in the right
place at the right time for the body.
    The time of LAN is usually calculated by either the GHA method or the
Equation of Time method.  I prefer the GHA method as it has always
been easier for me to work.  Our longitude at noon was 145 deg 40�
East.  This equates to a GHA of 214 deg 20� (180+ (180-145 d 40�)).
Looking in the nautical almanac, the GHA of the sun that was the
closest, and less then,  214 deg was 02h00m with a GHA of 210 deg
54.5�.  When you find the difference of these two angles you get 3 deg
25.5�.  Enter the �arc to time� table in the nautical almanac and you
get 13min 42 seconds.  Since we are keeping ZD -10, add 10 hours to
0200 and then add the 13-42 seconds.  This will give us a local zone
time of 12h 13m 42s which is the time of LAN.
    The declination of the sun at 1200 hrs local zone time on 16 May 08
(0200 UTC) is 19 deg 09.5� North.  Since this is the same name, but a
greater number then our assumed Latitude, the sun will bear North (no
need to calculate this to the minute, you are just looking to see what
general bearing the sun will be, this is especially important with
high altitude sights where the azimuth of the body can be hard to tell
from inspection).  Given the location of the sun in relationship to
the ship and the equator, we can see that the sun�s declination minus
the zenith distance (90-Ho) will give us Latitude.  Using a bit of
algebra we can see that the approximate Hs will be 90-(Dec-Lat).  So
90-(19.1-15.2)= 86.1 degrees or 86 deg 6 minutes on the sextant.
    I shot the sun at 12 hr 13m 40 sec as I observed start to descend
from its highest altitude.  My Hs was 85 deg 59.0 minutes.  I then
plugged this number into my standard Hs to Ho format (you can
disregard temp and pressure in this case since the correction is 0).
Hs  85-59.0
IC 00-00.8 (off the arc so it is positive)
Dip -00-09.9 (106 feet is nearly 10 minutes of correction!  Found in
the nautical almanac)
Ha 85-05.7
Body 00-15.8 (lower limb sight, remember to use the correct column for
May)
Ho 86-05.7

We now determine the declination at the time of sight.  Since 13min 40
seconds past since the time of the sight, we need to go to the back of
the almanac and find that time in the increments and corrections
section.  We find a correction of +0.1 since the declination is
increasing by the hour at a rate of 0.6� per hour.  This gives our
declination at the time of the sight of 19 deg 09.6� North
    Again since we have the sun in the same hemisphere and its
declination is greater then our latitude, then we need to subtract
zenith distance from declination to find latitude.  Simply put Dec-(90-
Ho)= Latitude North.  19d 09.6�-(90-86 deg 05.7�)= 15 deg 15.3�
North.  This sight was 1.3 miles north of the actual latitude of 15
deg 14.0� North.
JCA



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