# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: lat/long from meridian passage
From: Peter Hakel
Date: 2011 Jan 27, 00:28 -0800
Hello Antoine,

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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