NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Andrés Ruiz
Date: 2009 Mar 25, 13:47 +0100
Three members of navlist have contacted with me off-line
asking information.
As I think this information might be of interest, here it
is.
The Meridional Parts can be calculated like this:
Fi = latitude
a = main axis
Flattening: f=1-b/a
Eccentricity: eoe = sqrt(2.0*f-SQ(f));
Put the values on this formula, and you obtain the values.
M1 = log(10)*log10(tan( 45.0+fi/2.0) );
M2 = SQ( eoe )*sin( fi );
M3 = pow( eoe, 4 )/3.0*pow( sin( fi ), 3);
M4 = pow( eoe, 6 )/5.0*pow( sin( fi ), 5);
M = a*(M1-M2-M3-M4)
Values for "a"
in nautical miles, and "f"
for various Earth models:
·
BOWDITCH 21600.0/(2.0*PI),
1.0/298.26
·
WGS84 6378137.0/1852.0,
1.0/298.257223563
·
WGS72 6378135.0/1852.0,
1.0/298.26
·
ESFERA 21600.0/(2.0*PI), 0
·
NAD83 6378137.0/1852.0,
1/298.257222100880
·
GRS80 NAD83
·
EUROPE1950 6378388.0/1852.0, 1/297.0
If a spherical-Earth model is used, then the to axis are
equal: a=b, and f=0, eoe = 0
More information at my web site: “The Sailings” paper.
You can see that BOWDITCH use "a" for
"f" spherical model and f for the WGS72. This is the error of
concept.
Andrés Ruiz
Navigational Algorithms
http://www.geocities.com/andresruizgonzalez
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