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    Re: Time of meridian passage accuracy
    From: Antoine Cou�tte
    Date: 2009 Oct 3, 12:22 -0700

    Having some unexpected "free time" for this week-end, I was able to better 
    study the "exotic" LAN case which Gary published in [NavList 10009] . This 
    was made possible thanks to the "solution" he gave us in [NavList 10041] as 
    being LAN Time = 17h16m32s with LAN Position N22d47'7 W079d26'5 .
    
    I wish to share some interesting results.
    
    First, from Gary's solution I found that Culmination time is 18h14m18.05s with 
    a culmination height of 46d15'715. Hence the difference between LAN and 
    Culmination times is 57m29.80s .
    
    The only Culmination Time published as a answer to [NavList 10009] was made by 
    Andres in [NavList 10014]. Andres found a Culmination Time of 18h14m12.8s 
    which is pretty a impressive result given the difficulty of the case 
    submitted by Gary ("Exotic vessel" speeding at 300 kts due south).
    
    Two members published results for LAN :
    
    Andres in the same [NavList 10014] gave 17h12m45.9s , with LAN Position 
    N23d05'96 and W079d42'4 . At time 17h12m45.9s actual "ship" position computed 
    from Gary's results in [NavList 10014] was N23d05'84 and 79d26'5 . Therefore 
    the result computed by Andres, although not accurate as regards LAN Time was 
    EXCELLENT for the Latitude he finds at his indicated LAN Time.
    
    Peter Hakel in [NavList 10012] gave 17h13m35.0s , with LAN Position N22d40'8 
    and W079�42'4 . At time 17h13m35.0 s, actual airship position from Gary's 
    results in [NavList 10014] was 23d01'75. This result is off by som 14' , but 
    on the other hand Peter's result for LAN is fairly close from Gary's 
    position.
    
    The main cause of the differences in positions found by both Andres and Peter 
    -who, under "non exotic" cases, always publish accurate results - can 
    certainly be attributed to a fact which they both analyzed very well in 
    [NavList 10014] and [NavList 10012], i.e. results given by Formulae such as 
    the one published by Jim can no longer remain accurate under such huge 
    changes in Latitude during the time span of the Observations.
    
    
    And Last but not least, I discovered something I was not expecting at all 
    given the very high N/S speed which in my feeling would do more than 
    offset/compensate for a culmination hight quite far from zenith : the 
    Altitude curve of Gary's example is ALMOST PERFECTLY SYMMETRICAL when studied 
    from its Culmination time. Simple results as follows :
    
    Culmination Time / Height :  18h14m18.05s / 46d15'715
    
    15 minutes earlier and later :
    
             17h59m18.05s / 46d05'393    and    18h29m18.05s / 46d05'318
    
    
    30 minutes earlier and later :
    
             17h44m18.05s / 45d34'684    and    18h44m18.05s / 45d34'631
    
    
    60 minutes earlier and later :
    
             17h14m18.05s / 43d35'551    and    19h14m18.05s / 43d35'280
    
    
    I am quite surprised by these results because for culminations close to zenit 
    (say above 85d) the effect of a high N/S speed component and/or change in 
    declination significantly skews the heights vs times curve. Given the almost 
    perfect symmetry of this Altitude curve, the algorithm used by Andres - 
    apparently a second degree best fit as seems to be indicated on his site - 
    was very powerful to pick up almost the exact the Culmination time. Only the 
    presence of higher order terms (in t**4 and maybe t**6) prevented Andres to 
    do the perfect pick of the exact Culmination time.
    
    I would conclude by thanking Gary, Andres and Peter for the interesting 
    unusual/exotic case we had the opportunity to study together.
    
    Best Regards to all
    
    
    Antoine
    
    Antoine M. "Kermit" Couette
    
    
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