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    Re: Time of meridian passage accuracy
    From: Andrés Ruiz
    Date: 2009 Oct 23, 12:41 +0200

    I answer in bleu.


    -----Mensaje original-----

    De: navlist@fer3.com [mailto:navlist@fer3.com] En nombre de antoine.m.couette@club-internet.fr

    Enviado el: viernes, 16 de octubre de 2009 21:27

    Para: NavList@fer3.com

    Asunto: [NavList 10166] Re: Time of meridian passage accuracy



    Dear Andres,


    In [Navlist 10132] , you did requested feedback on your results and comments in [Navlist 10065].


    Here are just a few , (although unsolicited) :


    -     The results you published in [Navlist 10065] as regards both Culmination Height / Time are perfect. Your indicated values are 18h14m18s / 46d15'68196, while to "higher" (what does it really mean anyway ??? ) accuracy ([Navlist 10041] ) they are :  18h14m18.05s / 46d15'715


    -     Your interpretation of the difficulty to get adequate values for both LAN Time and Height looks very true to me. A Formula like the one published by Jim cannot sensibly be used since current "exotic" environment is too far from the "usual environment" which permits/authorizes its approximations.


    -     Your remark on the "always good results" obtained by either "brute force" Marcq Saint Hilaire method, or the more elegant/sophisticated Kaplan algorithm are very true : such methods will "almost always" work very well (only exception occurs when azimuths are too narrow, which is most often the case for most of the LAN's),


    No exception, good results. See [NavList 5168] Series of Sun sights in relatively rapid succession. [NavList 5178], …

    It is used in the star-trakers [NavList 10145].



    -     It brings back an interesting point: whatever the algorithm you use to deal with LAN data, these data are BY NATURE somewhat inaccurate since the azimuths remain (far) too narrow most of the time. NO WAY – that I know of – to go around that hard fact, although (and again) : "better one LAN than no LAN at all" !


    -     One query from my side. What do you exactly mean in your second conclusion of [Navlist 10065] when you wrote :


    "For obtain the Hmax, the Least squares fitting H = a0+a1*t+a2*t2 is OK. The maximum gives:  t max = -a1/(2a2) and  Hmax = a0-a2t2. It retains the asymmetrical nature of the curve."


    I am just curious here about the meaning of the very last sentence ( it "retains" ). Would not it seem that, by nature, and even with a non zero term in t**1, a second degree curve (a parabola) is always fully symmetric ?


    Yes, is symmetric by a vertical axis through the maximum/min  .

    What I mean is that for a little time around the maximum a 2nd fit is good. Graphs in [NavList 10099]


    Thank you for your Kind Attention and

    Best Regards





    Antoine M. "Kermit" Couette

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