This is a relatively simple problem to solve. A noon sight just to keep us on our celestial toes.
The day in question is March 25. We will use the 2005 Nautical Almanac for this solution. The Height of Eye is 6 feet. The Index Error is 2 minutes Off the arc. Viracocha is at a DR position of 25 degrees 30 minutes S by 92 degrees 45 minutes W. The Hs of a Lower Limb shot of the Sun is 62 degrees 10.8 minutes. We want to calculate the time of noon in GMT and then find our latitude.
As usual, the procedure is to go to the Sun pages in the Nautical Almanac for March 25. We want to find the time of Local Apparent Noon. According to the Sun column at the bottom right hand of the page, the time for LAN on the 25th is 12 hours 6 minutes Local Time. To find the exact time the Sun crosses our meridian, we convert our longitude of 92 degrees 45 minutes to time.
To do this, we go to the rear of the NA to the Conversion of Arc to Time table. We find by searching the columns that the Sun takes 6 hours 8 minutes. For the 45 minutes of the longitude, we see it takes the Sun 3 minutes to travel that distance. Thus, we have the following:
Mer Pass 12:06
+ 92 degrees 6:08
+ 45 minutes 0:03
Time of Local Apparent Noon in GMT = 18 hours 17 minutes GMT.
Next we need to reduce the Hs of the Sun to Ho:
Hc 62 degrees 10.8 minutes
– dip 2.4 minutes
+ IC 2.0 minutes
Ha 62 degrees 10.4 minutes
+ 3rd cor 15.7 minutes
Ho 62 degrees 26.1 minutes
Next we need to find the Declination at the time of the sight. Our time of sight was 18 hours 17 minutes 15 seconds GMT, and the dec of the Sun at that time when calculated is 02 degrees 03.6 minutes N. The d corr is +1 minute and is reduced to zero for the sake of this problem in terms of whether there is any additional declination correction.
We are now ready to find latitude using the formula Latitude = 90 degrees – Ho = ZD +/- Dec:
90 degrees – 62 degrees 26.1 minutes (Ho)
27 degrees 33.9 minutes (ZD)
– 2 degrees 03.6 minutes (Dec)
25 degrees 30.3 minutes (Latitude)