Hello:
As a former teacher I really appreciate a simple explanation with realistic but simplified conditions. FER's recent Galapagos sight nudged me to review this method using some sight data from 2013. In brief review, here is FER's results using the sextant as a sun dial in the Galapagos sight. The Galapagos are near the equator, at 90-91 degrees west(zone 6), sight recorded on the equinox so sun's declination is about zero. Sight at local time 8:00:00 . Corrected altitude =30 degrees , zenith distance = 60 degrees or 4 hours. For the sight UT = 6 + 8:00:00 =14:00:00, or sun GHA = 30 degrees. LHA = 60 degrees. Longitude = 30 + 60 = 90 degrees W.
For typical conditions, the LHA must be calculated using one of several available methods and the latitude must be known(or approximated). For my April 2013 sight, I've provided more data than is needed. Date : 4/15/2013; zone 5,latitude about 42* N at a place where Thoreau liked to walk. Sun is to the south east.
UT = 14:18:56 (GHA = 34* 44.8'); (Declination N 9* 57.5')
Hs = 45* 22.8' , IC =0, dip = - 5.7, HO = 45* 32.4'
I calculate LHA using cosLHA =(sin Ho - [sinLat]*[sindec])/([coslat]*[cosdec])
where North is plus sign.
LHA + GHA = Longitude
Location is ...................
Best regards
Bruce
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