NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Navigation exercise
From: Jeremy C
Date: 2008 May 17, 01:35 -0700
From: Jeremy C
Date: 2008 May 17, 01:35 -0700
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 --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To , email NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---