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    Re: A celestial navigation problem
    From: Stan K
    Date: 2011 Nov 18, 10:13 -0500

    You are right - I found both of your previous e-mails in the spam folder.  Why?

    And your final position is correct, although my solution (and Gary's) used the average of the two equal-altitude sights to get a UT of LAN of 22-07-41.

    I would have liked to see your intermediate (approximate) answers.  These are the ones that I believe would force the students to think.

    So you like the low dispersion of the sights.  Of course they are all good sights - after all, he is SuperNavigator!  (And the values were all "hand-selected".)

    You make an excellent point that the actual position has to be at the southern tip of the island.  Well, maybe not the southern tip - it looks like the western shore might work, but not (near) the eastern shore, where I arbitrarily placed SuperNavigator without giving it appropriate consideration.  I am going to adjust index error by 0.6' to put SuperNavigator on the southern tip before I give the problem to the students, just to avoid this issue.

    Glad you enjoyed the exercise.  I certainly hope the class does.


    -----Original Message-----
    From: Antoine Couette <antoine.m.couette@club-internet.fr>
    To: NavList <NavList@fer3.com>
    Sent: Thu, Nov 17, 2011 10:02 pm
    Subject: [NavList] Re: A celestial navigation problem

    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

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