NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Nav exercise - ex-meridian
From: Jeremy C
Date: 2008 May 21, 00:54 -0700
From: Jeremy C
Date: 2008 May 21, 00:54 -0700
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 --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To , email NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---