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    Re: Time of meridian passage accuracy
    From: Gary LaPook
    Date: 2009 Oct 01, 18:15 -0700
    None of the solutions are real close.

    I had rounded the data to whole minutes. Here are the values to the one-tenth of a minute if that helps.



    17:00:00     42º 14.1'
        :05         42º  44.2'
        :10         43º  12.6'
        :15         43º  39.1'
        :20         44º  03.7'
        :25         44º  26.4'
        :30         44º  47.1'
        :35         45º  05.7'
        :40         45º  22.2'
        :45         45º  36.5'
        :50         45º  48.7'
        :55         45º  58.6'

    18:00:00    46º  06.3'
        :05         46º  11.7'
        :10         46º  14.8'
        :15         46º  15.7'
        :20         46º  14.2'
        :25         46º  10.4'
        :30         46º  04.4'
        :35         45º  56.0'
        :40         45º  45.5'
        :45         45º  32.7'
        :50         45º  17.7'
        :55         45º  00.6'

    19:00:00    44º  41.4'

    gl


    P H wrote:
    One question to ask is what latitude is to be entered into (any) equation, when it is significantly changing during the exercise.  Calculus probably cannot be avoided if one wanted to do this rigorously.

    In my method I needed the latitude to translate the E/W miles into change in longitude (and change in moment of meridian transit) in the E/W-DR portion of the algorithm.  I chose the latitude calculated from the fit.  This seems to be adequate for examples given by Jeremy and Antoine.  As you move faster, you cover larger distances, the curvature of the Earth becomes even more visible and this approximation of picking a fixed latitude will progressively break down.  In Gary's example, however, the course is along a meridian, so this ambiguity should not be an issue for the moment of transit determination - though at some speed the geometry will become more complicated.

    Perhaps Gary will tell us how he constructed this example and what therefore the "exact" solution is.  My longitude result is in the neighborhood of his DR value.  If one changes the speed, the Ho's themselves will change accordingly.  However, the moment of meridian transit should come out independent of speed, as long as the vessel moves along the same meridian.

    I do get the latitude from an Hmax; but this Hmax is calculated from a curve fit through N/S-DR/declination-change prorated altitudes (advanced to a common parallel of latitude), not directly from the given Ho's.


    Peter Hakel


    From: Andres Ruiz <aruiz{at}orona.es>
    To: navlist@fer3.com
    Sent: Thursday, October 1, 2009 2:53:01 AM
    Subject: [NavList 10014] Re: Time of meridian passage accuracy

    My solution:

     

     

    Peter,

    if I have not done anything wrong, the possible difference could come from this:

     

    Jim Wilson correction for time depends on latitude, B:

    Dt = 3600.0*180.0/PI * (dBdt - dDecdt) * (TAN( B ) - TAN( Dec )) / SQ( dLHAdt )

     

    In this case, high N/S component of the velocity, the difference between the time of maximum altitude and LAN is very important, and also the associated altitude.

    If you take estimation for latitude using the Hmax, de solution becomes erroneous.

     

    Andrés Ruiz

    Navigational Algorithms

    http://sites.google.com/site/navigationalalgorithms/

     


    De: navlist@fer3.com [mailto:navlist@fer3.com] En nombre de P H
    Enviado el: jueves, 01 de octubre de 2009 8:20
    Para: navlist@fer3.com
    Asunto: [NavList 10012] Re: Time of meridian passage accuracy

     

    UT of LAN:                     17:17:35

    Latitude at LAN:            N 22 deg 40.8'

    Longitude at LAN:        W 79 deg 42.4'

     

     

    Peter Hakel

     


    From: Gary LaPook <glapook---.net>
    To: navlist@fer3.com
    Sent: Wednesday, September 30, 2009 9:19:19 PM
    Subject: [NavList 10009] Re: Time of meridian passage accuracy

    Jim Wilson wrote on Sept. 26th:


    "I compiled a table in my Navigation paper showing differences between maximum alititude and meridian passage ranging from 10 seconds to 37 minutes. My trial which led me to write the paper had a difference of a minute and a half, and that was at a moderate latitude and sailboat speeds."

    I am curious how well it works at higher speeds. I have created a data set and I would like you to evaluate it and tell me the time of meridian passage, the longitude and the latitude at meridian passage.

    This hypothetical takes place on December 22, 2009 and the sights are taken at five minute intervals on the exact five minute mark and all sextant corrections have been applied so the values stated are the observed altitudes, Ho. They cover a two hour period from 1700 Z to 1900 Z.

    17:00:00     42º 14'
        :05         42º  44'
        :10         43º  13'
        :15         43º  39'
        :20         44º  04'
        :25         44º  26'
        :30         44º  47'
        :35         45º  06'
        :40         45º  22'
        :45         45º  36'
        :50         45º  49'
        :55         45º  59'

    18:00:00    46º  06'
        :05         46º  12'
        :10         46º  15'
        :15         46º  16'
        :20         46º  14'
        :25         46º  10'
        :30         46º  04'
        :35         45º  56'
        :40         45º  45'
        :45         45º  33'
        :50         45º  18'
        :55         45º  01'

    19:00:00    44º  41'

    The D.R. at 1700 Z is 24º  25' N, 79º  50' W.

    The course is 180º True.

    The ground speed is 300 knots.

    gl







    James N Wilson wrote:

    Douglas:

     

    You state that "the effect is very small as to be of no practical concern to practical navigators."

     

    I compiled a table in my Navigation paper showing differences between maximum alititude and meridian passage ranging from 10 seconds to 37 minutes. My trial which led me to write the paper had a difference of a minute and a half, and that was at a moderate latitude and sailboat speeds.

     

    Jim Wilson
     




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