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Brad, regarding your comment on the least square 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."
It's actually surprising how well we can estimate a linear regression
line through data points with a reasonable correlation only by using a
ruler. In this case it would only require to adjust the scale of the
axis accordingly.

Marcel

On Thu, Apr 16, 2009 at 6:55 PM, Brad Morris  wrote:
>
> 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
> Brad
>
>
>
>
> -----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
> 
File: 
dwVERSION2.xls

>
>
>
>
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