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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{at}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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    File: 115493.hdiff.xls
       
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