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    Re: Time of meridian passage accuracy
    From: Antoine Cou�tte
    Date: 2009 Oct 16, 12:26 -0700

    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),
    
    -	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 ?
    
    Thank you for your Kind Attention and
    Best Regards
    
    Antoine
    
    
    Antoine M. "Kermit" Couette
    
    
    
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