NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: A celestial navigation problem
From: Gary LaPook
Date: 2011 Nov 17, 01:18 -0800
From: Gary LaPook
Date: 2011 Nov 17, 01:18 -0800
You didn't include the sextant correction table, do you expect the students to figure them in their heads? Dip approx square root of 6, somewhere between 2 and 3 minutes, call it 2.5 (2.4 from table.) Refraction= zero above 63° (to the precision on 1') or 0.2' from table. Or do you use the S.D. of +16.3 minus the 0.2' refraction, total +16.1 or from the sun table of 16.0'. Increment for GHA for time of noon average of the times for both of the 77° 44.1' altitudes, 22:07:46 Z. Do they do the increment in their heads, 15' X 7 minutes plus 46/4' per second = 1° 56.5', same as in increments table. Approx longitude 150° W based on 2200Z time of noon. Sun moves from left to right so must be south. Approx declination 23.5° S, around winter solstice. Approx Ho = 78° so ZD approx 12° so approx lat is 11.5° S. Longitude at noon, 22:07:46 = GHA =149°53.4' plus 1°56.5' so longitude is 151°49.9' W. Hs = 77° 48.7' I.C. + 1.3' Dip - 2.4' S.D. + 16.3' Ref - 0.2' Ho = 78° 03.5' ZD = 89° 60.0' - 78° 03.5' 11° 56.3' Lat 23° 22.6' - 11° 56.3' 11° 26.3' S. So you are on Flint Island. It is easy to memorize the refraction corrections to a precision of 1'; 5' above 10° 4 12 3 16 2 21 1 33 0 63 It's a good test, student must be able to think through the problem. gl --- On Wed, 11/16/11, slk1000@aol.com <slk1000@aol.com> wrote:
|