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Re: Latitude by Noon Sun for Beginners
From: Gary LaPook
Date: 2009 Apr 29, 05:47 +0200
From: Gary LaPook
Date: 2009 Apr 29, 05:47 +0200
Here is a link to the problem: http://www.oceannavigator.com/ME2/dirmod.asp?sid=&nm=&type=Publishing&mod=Publications%3A%3AArticle&mid=8F3A7027421841978F18BE895F87F791&tier=4&id=9C718558F98A4D429B392F5BC5CE4C4F The data for the problem is DR 59� 25' S 68� 25' W January 25, 2008 16:46:20 Z Hs 59� 09.3' Height of eye 8 feet Lower limb of the sun No index error or watch error. For a LAN latitude shot your DR plays no part, LAN is when it is, when the sun is highest in the sky. The GHA of the sun at the time given is 68� 31.2' W so your DR longitude is off by 6.2'. If you were actually at your DR, LAN should have occurred only 25 seconds earlier so I don't know why you mention a 12 minute difference, are you looking at the 2008 Almanac? The time of the sight is not critical for determining latitude (within reason) since you only need it to determine the declination of the sun which changes slowly and would have been the same 25 seconds earlier as at the stated time which was 18� 31.2' S. (Depending on the accuracy needed for latitude you might be able to get by with just a calendar for time as they did for centuries, before the invention of chronometers and the development of the lunar distance procedure.) First you correct the sextant altitude, subtract dip and refraction and add semi diameter for the lower limb if using the Air Almanac, or subtract dip and use the combined correction table in the Nautical Almanac which combines refraction, parallax and semi diameter for the lower limb and you should get the answer given in the magazine, Ho= 53� 22'. (Or, doing it in your head, the square root of 8 is almost 3, refraction between 33� and 63� is 1' and SD is 16' for a total correction of plus 12' giving an Ho of 53� 21'.) Subtract Ho from 90� to get zenith distance, the distance you are either north or south of the sun's GP. Instead of writing 90� is is easier to write 89� 60' (which is the same thing.) Subtracting Ho from 89� 60' you get a zenith distance of 36� 38' which is your distance south of the sun's GP since you are looking north at LAN. Since you must be closer to the pole than the sun you no you add the zenith distance to the declination of the sun and get a noon latitude of 55� 37.8' S, the same as in the magazine. The magazine goes on to plot an EP which does take into account the DR. To method used to find an EP is to plot a perpendicular from the DR to the LOP. Since in this case the LOP is a latitude line the perpendicular goes straight south from the DR to intersect the latitude LOP at the DR longitude. So which longitude is correct, the DR longitude or the longitude corresponding to the sun's GHA at the time of LAN given as 68� 31.2' W at 16:46:20 Z? If you have been following the discussions you should have noticed that due to the difficulty of accurately determining the exact second when the sun "hangs" that there is some uncertainty in that time which then causes uncertainty in a longitude based on the time of observed LAN so it is taken that the DR longitude is more accurate which is why the EP is plotted on the latitude line at the DR longitude. Another way to do these types of sights is to treat them exactly like any other sight, use a sight reduction table with the LHA of zero. Doing it this way you see why the exact time is not so important when determining latitude since LHA equals zero for a four minute period, you simply move the AP east or west but the Az stays the same, 360� in this case. Doing it this way you find an Hc of 53� 59.8' for an AP of 55� 00' S, 68� 31.2' W, resulting in an intercept of 37.8 NM Away from an AZ of 360� resulting is an LOP running east and west at latitude 55� 37.8' S, the same answer as before. Theoretically the LAN sight would produce a slightly more accurate latitude due to the slight change in altitude during the plus and minus two minute period when LHA = zero although nobody worries about this when doing normal sights on other bodies. To see how big a difference this might make I added two minutes to the time of the sight and used the same AP and the Hc came out to be 53� 59.6' a difference of only 0.2', a difference small enough to be lost in the "noise" of celestial navigation. gl JKP@obec.com wrote: > I've been following with great interest, if with less than complete comprehension, the recent threads on finding position at and around noon. > > I'd like to divert a tiny new rivulet from that stream-- a back-to-basics thread on noon sights for latitude. I don't like those titles for how-to books that end in "...for Dummies," so I'm using the kinder term "...for Beginners." > > I've been working the latest Nav Problem in Ocean Navigator magazine, and I'm finding I can't get the latitude the editor does from the noon sight data presented. My main stumbling block seems to be that the time given for the shot is not the time I consider Local Apparent Noon for the DR longitude used. Wouldn't that be essential? If the sun is observed at 12 minutes or more after LAN, how does one account for that difference? > > My main source of instruction is Dutton's (12th edition), which I find usually gives excellent and thorough explanations of procedures. The article on latitude by noon sun focuses on how to determine LAN and what to do with the Ho obtained then. I've found nothing so far about accounting for a time difference and the fact that the sun is not at its zenith when observed before or after LAN. Much is made of combining the vessel's change of position east or west with that of the sun to pinpoint LAN, but the magazine problem provides no data about course or speed. > > I'll bet one you can tell me somnething that will make me slap my forehead and exclaim "Of course!" Fire away! > > Thanks in advance, > John P. > > > > > > > > > > --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To , email NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---