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Dear Frank,
using my clearing algorithm I found:
time 22:19:38 GMT, pos 14�N 35.3', 61�W 41.3'
LD 68� 13.13' (sextant LD 68� 19.40')
The position is roughly 5 miles from yours
Giuseppe
----- Original Message ----- 
From: 
To: 
Sent: Wednesday, July 16, 2008 4:03 AM
Subject: [NavList 5852] Re: My first Lunar


>
> Jeremy, you wrote:
> "Well I found my first lunar, and it will be tricky.  Here's the data that 
> I
> have.
>
> GMT Date is 26 January 1999.  GMT of the sight was about 2220.  Dip
> correction is -7.7' of arc.  The lunar was at evening twilight and a near
> limb observation between Jupiter and the Moon was taken.  The sextant LD 
> is
> 68deg 19.4'  IC is 0.0'.  An upper limb altitude of the moon was taken HS 
> is
> 66 deg 09.3'  The Hs of Jupiter is 45 deg 22.3.
>
> Here's the rub:  I have no idea where I was other then to say I was 
> probably
> somewhere in the Eastern Caribbean.  Best guess is about 20 deg North
> Latitude and 70 degrees West Longitude."
>
> Having a good DR position is convenient but not necessary when it comes to
> clearing a lunar. Of course if you want to assess the accuracy of the 
> sight,
> then you want the actual position and correct GMT as nearly as possible. 
> You
> can figure out where you are, more or less, by trial and error from your
> sight data. Go to the calculator on my web site, set the GMT of the sight 
> to
> 22:19:30 and set your DR Lat to 14d 31'N and your DR Lon to 61d 38.1W. 
> That
> nearly matches your sights, lunar and altitudes, too. So assuming your
> observations were good (and I would bet they were) you were probably about
> 30 miles west of Martinique. Does that fit your recollection?
>
> Now as it happens, this is yet another one of this miraculous lunar sights
> where you can do the clearing without using any spherical trig. If we take
> the pre-cleared altitudes and distance (the altitudes of the objects'
> centers and the center-to-center lunar distance) and add them up, we get
> nearly 180 degrees. So adjust the Moon's altitude higher by about 24 
> minutes
> of arc and then work it AS IF they were exactly opposite each other in the
> sky.
>
> -FER
>
>
>
> >
> 


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