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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.
>
>
>
>
>
>
> >
>
>   


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