NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: James N Wilson
Date: 2011 Nov 28, 20:44 -0800
For figuring the time of LAN you only look at the two sights that have the exactly the same altitude, sights 2 and 11. All the sights that have the same altitude around LAN, sights 5-8 don't help. If you are faster with the sextant you could have had a lot more sights with that altitude. For figuring LAN you take the two sights that have the exact same altitude and that are spaced as far apart as possible.
It looks like you misread the time for sight 2, you put 18-02-45 and the correct time is 18-02-05, the average is 18-07-46. Plus ZD minus WE makes the GMT of LAN 22-07-41. Since your time is 4 seconds too soon your longitude is one minute to far to the east so add one minute to your answer and you have it.
gl
--- On Mon, 11/28/11, Mike Burkes <m_burkes@msn.com> wrote:
From: Mike Burkes <m_burkes@msn.com>
Subject: [NavList] Re: A celestial navigation problem
To: NavList@fer3.com
Date: Monday, November 28, 2011, 6:55 PM
Hi folks,after deciphering all the solutions,unless I am missing something really dumb and discussing watch error how is the latter figured into the solution? Here is my solution and I interpolated for LAN using equal altitudes.
Sight AM Hs PM AM+PM LAN=Total/2
2 18-02-45 77 44.1 18-13-27 36-15-32 18-07-46
3 18-04-05 77 46.8 18-11-19 36-15-24 18-07-42
4 18-06-05 77 48.3 18-09-14 36-15-19 18-07-40
5 18-07-25 77 48.7 18-07-50 36-14-75 18-07-38
9 18-09-05 77 48.4 18-06-25 36-15-30 18-07-45
--------
Total 90-35-211
Average 18-07-42= LAN
WE -5 fast
ZT 18-07-37
ZD +4
UT LAN 22-07-37
GHA 22 hr 149 53.6
increments 7m 37s 1 54.3
Longitude 151 47.9 W
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