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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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