NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Peter Hakel
Date: 2011 Jan 27, 00:28 -0800
I took the computed altitude formula for a given latitude L, declination D, and rate R, and did a Taylor expansion to second order around the time T of the upper transit. For simplicity I assumed that L > D. I also took L, D, and R to be constant; although this is not true if there is a declination change and/or the vessel is in motion, we still should be able to get a good general idea how to answer your question.
I calculated the difference:
dH = H_parabolic - H_exact
for the Sun (R = 15 degrees/hour) at Declination = 10 degrees, for nine combinations of latitudes L and times away from LAN: | t-T |. I attached the corresponding spreadsheet in case anyone else wants to try their own numbers.
| t - T |
Latitude | 10min | 20min | 30min |
20d | -0.25' | -3.65' | -16.46' |
40d | -0.01' | -0.11' | -0.55' |
60d | -0.00' | -0.01' | -0.07' |
Note the latitude dependence in that table; the transit curves get flatter for higher latitudes (lower transit altitudes), so you get a wider | t - T | "close-enough" (i.e. |dH| small) window for those. At the North Pole this time window becomes infinite, since H=const. The spreadsheet is not designed to handle cases in which Latitude <= Declination.
I have not studied by how much the latitude extracted from the least-squares quadratic fit would be thrown off by including data points that are too far away (whatever that means) from transit time T. However, this dH could be a guide in deciding what is still acceptably close to transit. Does |dH|<1' sound reasonable, or can we afford a bigger difference? I'd think that as long as |dH| is not larger than the error bar on your observations, then your shots are close enough to t=T for the resulting transit curve to yield a latitude that is within that error bar around the true value.
Peter Hakel
From: Antoine Couette <antoine.m.couette@club-internet.fr>
To: NavList@fer3.com
Sent: Mon, January 24, 2011 11:54:35 AM
Subject: [NavList] Re: lat/long from meridian passage
http://www.fer3.com/arc/m2.aspx?i=115445
[NavList] Re: lat/long from meridian passage
From: goold---edu
Date: 24 Jan 2011 13:12
Dear Patrick,
You are welcome !
Thanks to data processing techniques which were not widely available some 30 years ago, it is now possible to shoot Heights ANYTIME around LAN (no more requirement for "paired sights" - which of course - work very well as you have seen ... ) and still get the best from your observations. This works well as long as you do not observe "too far away" from LAN.
So if you want to give a try ...
Best Regards
Antoine M. "Kermit" Couëtte
PS : As a regards the "not too far away from LAN rule" as applicable to plain parabola regression, do anyone know any numbers ? Obviously the smaller the zenit culmination, and the smaller the "permitted time" around culmination time ... but according to which numerical criteria ? Does anyone know on NavList ? Peter ? Andres ? I have not seen anything published, as far as I can remember ...
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