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| This problem was a lot more difficult for the two men in Fox than it will be for us. The day in question is July 15, and we are using the 2005 Nautical Almanac. We are solving a Lower Limb shot of the sun at noon. The last DR position of Fox was 45 degrees 30 minutes N by 50 degrees 25 minutes W. The Height of Eye is 3 feet, and the Index Error is 3 minutes Off the Arc. The Hs at the time of the sight is 64 degrees 39.3 minutes. Based on the DR longitude, what is the time of Meridian Passage in GMT? Hours and minutes only. Solution: Go to the July 15 sun page in the NA and locate the time of meridian passage in LMT at bottom of page. For that day, the time is 12 hours 6 minutes. Next, go to the back of the book to Arc/Time conversion and look up the DR longitude of 50 degrees 25 minutes.
50 degrees = 3 hours 20 minutes 25 minutes = 1 minute 40 seconds + 12 hours 6 minutes GMT = 15 hours 27 minutes for where they thought they were at 50 degrees 25 minutes W.
The actual time of the meridian passage occurred at 15 hours 33 minutes. This means they were actually at 3 hours 27 minutes of longitude instead of the 3 hours 21 minutes 40 seconds they calculated. Back to the Arc/Time conversion table, we find that 3 hours 27 minutes converts to 52 degrees. This basically means it takes the sun 3 hours 27 minutes to move 52 degrees. So the men aboard Fox knew they were farther west than their calculated longitude. They actually were at about 52 degrees W instead of 50 degrees 25 minutes W. Next we want to reduce the Hs to Ho. This is standard operating procedure for any celestial sight. Hs = 64 degrees 39.3 minutes – Dip 1.7 minutes + IC 3.0 minutes Ha = 64 degrees 40.6 minutes + 3rd corr 15.5 minutes Ho = 64 degrees 56.1 minutes | |||
