NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Antoine Couëtte
Date: 2013 Jun 27, 13:18 -0700
RE : http://fer3.com/arc/m2.aspx/Longitude-SeanC-jun-2013-g24511
Hello Sean,
Your averaged observations time is UT = 14h24m57.0S
Your averaged observations heights is 104°40'48. Assuming an estimate of -3'12 for your index error yields an observed averaged value of 104°44'0 to be divided by 2 into a "conventional/standard height" of 52°22'.
Since you observed overlapped images, treat the Sun as a 0'0 Semi-diameter body and apply corrections for only refraction (-0'7) and parallax (0'1), i.e. use a geocentric observed height close to 52°21'3 (probably accurate to a few arc minutes as a result of your environment).
At Latitude N37°03'3 and BEFORE Noon, I am deriving Longitude W076°25'5 ... which puts you somewhere near the Chesapeake Bay ....
Hampton ? Newport News ?
Kermit
Antoine M. "Kermit" Couëtte
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