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    Re: A celestial navigation problem
    From: Gary LaPook
    Date: 2011 Nov 30, 09:14 -0800
    But only sights 2 and 11 were an equal pair, there were no other pairs except for the four sights right around LAN.

    The reason that you use the pair that has the greatest separation in time is due to the rate of change in altitude near LAN. it changes very slowly at LAN and then more and more rapidly the farther away from LAN. This allows a more accurate determination of LAN by averaging those two times. When the change is slow, the accuracy is lost in the noise of the observation.

    gl

    --- On Wed, 11/30/11, 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: Wednesday, November 30, 2011, 8:12 AM

    Hi GL thanks for reply the 45 in site 2 was a typo error and my average was correct at 18-07-46. Also I wonder why after the round of 12 sites equal altitudes were not used as per some nav texts. I thought the more equal alt pairs you have,especially the ones that result in close LANS, the more accurate your LAN. I realize 5-8 do not help but 9-12 could have been treated as equal alts,i.e., set the sextant to the AM alts. Sorry to be-labor the point and thanks for input.
    Mike Burkes  
     

    Date: Mon, 28 Nov 2011 19:23:17 -0800
    From: garylapook@pacbell.net
    Subject: [NavList] Re: A celestial navigation problem
    To: NavList@fer3.com

    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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