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    Re: A celestial navigation problem
    From: Antoine Couëtte
    Date: 2011 Nov 17, 19:01 -0800

    RE : [NavList] A celestial navigation problem [17427]
    From: slk1000---com
    Date: 16 Nov 2011 21:44


    Dear Stan,


    Since you did not receive my privately sent answer - I suspect it might have been treated as a "spam" - let me simply indicate it here.


    Assuming that there is no index error and no watch error, and treating this Nav Problem a a "local apparent noon" problem (instead of dealing with it through 12 individual LOP's) I find that the time of South transit is 22h07m45.7s, hence an Observer's Longitude equal to W 151d50'0 , and I also compute Observer's Latitude as being S 11d24'9. I also find that the goodness of fit is very good : dispersion of observations inferior to - i.e. better than - 0.1' .

    See enclosed document which gives the computation results.

    Correcting for index error ( indicated by Gary as being -1.3' ) I get corrected Latitude S 11d26'2.

    Correcting for chronometer error (5 seconds fast), we need to shift Longitude by
    1.25' to the East, hence a final Longitude of W 151d48'7

    To recap, observed position is :

    S 11d26'2 W 151d48'7


    A few remarks : if we treat this example through LOP's, we - again - find that the LOP's all cross together almost at the very same point (dispersion inferior to 0.1 NM)

    Observed position is quite close from actual position since actual position has to be the southern tip of Island, i.e. the only point on Island where observer could see the Sun over the sea at all times.


    Thanks again for a great exercise


    Kermit

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    File: 117452.111118-stans-lan-log.doc
       
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