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    Re: Lat/Lon by "Noon Sun" & The Noon Fix PROVE IT
    From: Brad Morris
    Date: 2009 Apr 16, 11:55 -0400

    I agree with your result for a least squares fit of the parabola to the measured data. With the variation
    limited, there is no doubt that the least squares fit will provide a reasonable parabola and therefore a
    fairly good estimation of Lat and Lon.  If this is how the method is used in 
    practice, you will find little
    argument from me.
    However neither procedure, as I understand them to date, uses a least squares fit.  The
    navigator is urged to, in one case, fair a line thru a series of data points or in another, to visualize
    a parabola as we slide the paper.  That isn't a least squares fit by any stretch of the imagination.
    As such, we may find that the deviation in longitude is somewhat greater than your value of 5.68nm.
    Best Regards
    -----Original Message-----
    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of waldendand@YAHOO.COM
    Sent: Thursday, April 16, 2009 11:23 AM
    To: NavList@fer3.com
    Subject: [NavList 7970] Re: Lat/Lon by "Noon Sun" & The Noon Fix PROVE IT
    The .doc file works great. Thanks.
    I took George's spreadsheet and automated it with a macro.
    First, I used it to generate 1000 sets of 13 observations.  For each set, I 
    let excel fit the parabola, and find the time and altitude of its peak.  I 
    then find the difference between the resulting latitude and longitude thus 
    calculated and the true values.  Thus for each set of 1000, I get a mean and 
    standard deviation of the error.
    Using George's convenient "scatter switches", I tried various combinations.  
    He has four switches. In order, the first turns on/off the scatter in ship 
    speed (if one always uses the nominal 10 knots in the calculation), the 
    second changes the equation of time (as I understand it, changing the value 
    to be calculated as opposed to introducing scatter into the result), the 
    third changes the longitude (same comment), the fourth adds scatter to the 
    observed altitude. In the table below, the switches are listed in order, 
    1=on, 0=off.
    As can be seen for everything on, the standard deviation of longitude error in 
    nm is 5.68.  It can be seen the ship speed scatter seems to have a slightly 
    greater effect than the altitude reading scatter.
    The 1 mile lat and 5 mile long estimates look pretty good to me. YMMV.
    1,1,1,1 min lat min long        nm long
    mean    0.00    0.19    0.11
    stdev   0.40    10.16   5.68
    max     1.29    48.02   26.85
    0 1 1 1
    mean    0.01    0.19    0.10
    stdev   0.44    6.57    3.67
    max     1.55    22.64   12.66
    0 0 1 1
    mean    0.00    -1.00   -0.56
    stdev   0.42    6.73    3.77
    max     1.41    20.77   11.61
    0 0 0 1
    mean    -0.01   -0.38   -0.21
    stdev   0.44    6.72    3.76
    max     1.14    20.50   11.46
    1 1 1 0
    mean    0.00    -0.22   -0.12
    stdev   0.05    7.48    4.18
    max     0.25    24.18   13.52
    0 1 1 0
    mean    -0.01   -0.38   -0.21
    stdev   0.00    0.01    0.01
    max     0.00    -0.34   -0.19
    1 0 0 1
    mean    0.01    0.17    0.09
    stdev   0.42    9.97    5.58
    max     1.33    32.45   18.14
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