NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Andrés Ruiz
Date: 2009 Oct 23, 12:41 +0200
I answer
in bleu.
-----Mensaje original-----
De:
Enviado el: viernes, 16 de octubre de 2009 21:27
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
Antoine M. "Kermit" Couette
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