We are aboard Eagle on October 26, 2006. Our DR for the time of the sight is 30° 31’N by 76° 47’W. The Height of Eye is 30 feet. There is no sextant error. The Hs at the time of the sight is 47° 01.6′ This is a noon sight so the first thing we do is calculate the time of LAN at our meridian. We will calculate our time in GMT. This is a Lower Limb shot of the sun.
The steps for finding LAN are the following:
On the daily sun page find time for Meridian Passage for the day in question. In this case the time of Meridian Passage is 11 hours 44 minutes
Go to the Conversion of Arc to Time Table at the rear of the Almanac. Look up longitude of 76°. The table tells us that it takes the sun 5 hours and four minutes to travel that distance. Next look up 47′ of longitude. It takes the sun an additional 3 minutes and 8 seconds to travel this distance. Add these two numbers to the time of Meridian passage and the resulting number is the time of LAN at GMT:
+ 03min 8sec
LAN at 16 hours 51 minutes (forget the 8 seconds)
Next we need to reduce the Hs to Ho.
Hs 47° 01.6′
app alt 46° 56.3′
3rd corr +15.3
Unfortunately this answer differs from the problem. That is because a mistake was made in the calculations that were submitted. Assume that the answer given here is correct. That will mean of course that the printed answers in the problem are incorrect. I’m sure many of you have already found this out.
Next we have to find the declination at the time of the sight. For 1600 hours the declination of the sun is at S 12° 32.8′. The d correction is 0.9 and it is increasing. Thus at the time of predicted LAN the declination of the sun is:
S 12° 32.8′
+ 0.8′ (the d correction found in increment table for 51 minutes)
S 12° 33.6′ Declination
Now we will find latitude using the formula Latitude= 90°-Ho=Zenith Distance+/- declination. Thus we have the following:
ZD 42° 48.4′
Lat. 30° 14.8′
At the time of the sight the longitude was 76° 43.4’W. This was found by inspection in the NA for the time of the sight when LHA equals 0°.