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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Peter Fogg
Date: 2009 Mar 27, 03:15 +1100
Yes because Bowditch uses the sphere for a, and the WGS72 for f
The different values for the Earth models gives:
Ellipsoid
a
f
BOWDITCH
3437.746771
nm
0.003352779
WGS84
3443.918467
nm
0.003352811
WGS72
3443.917387
nm
0.003352779
ESFERA
3437.746771
nm
0
NAD83
3443.918467
nm
0.003352811
EUROPE1950
3444.053996
nm
0.003367003
And the value for lat = 45º
WGS84 MP( 45.000000 ) = 3019.058271
WGS72 MP( 45.000000 ) = 3019.057477
NAD83 MP( 45.000000 ) = 3019.058271
EUROPE1950 MP( 45.000000 ) = 3019.108031
E7784 WGS84 MP( 45.000000 ) = 3019.058271357112100000 [NavList 7784] equation
E7784 BWD MP( 45.000000 ) = 3013.648100569424100000 [NavList 7784] equation for BOWDITCH model
BOWDITCH MP( 45.000000 ) = 3013.648101
ESFERA MP( 45.000000 ) = 3029.939203
-----Mensaje original-----
De: NavList@fer3.com [mailto:NavList@fer3.com] En nombre de Ark Shvetsky
Enviado el: jueves, 26 de marzo de 2009 16:35
Para: NavList@fer3.com
Asunto: [NavList 7791] Re: Interpolation of Meridional Part Table
Quite interesting. I got 3015.6145.
e=0.08181919
a=3437.7468
Bowditch 1968 gives a-3437.68, e=0.082271854
----- Original Message ----
From: George Huxtable <george@hux.me.uk>
To: NavList@fer3.com
Sent: Thursday, March 26, 2009 7:47:15 AM
Subject: [NavList 7788] Re: Interpolation of Meridional Part Table
The plot thickens-
Andres calculates, from the expression he gives in [7784]
WGS84
a = 6378137 [m] = = 6378137/1852 [nm]
f = 1.0/298.257223563
The result is:
MP( 45.000000 ) = 3019.058271357112100000
The expression Andres quoted used eccentricity (epsilon) rather than the
flattening f that he tells us is f = 1.0/298.257223563 for WGS84. However,
we can deduce one from the other using the expression given in Meeus, in
which
eccentricity = square root of ((2 * f) - (f * f)) for which I have taken
the result to be 0.08181919, exactly in keeping with Earle's figure.
===========================
On the other hand, I have tried to do exactly the same thing with my pocket
calculator and with the following values, taken from Earle-
E= 0.08181919
A = 3437.7468, the Earth's equatorial radius in geodetic miles.
Latitude L= 45
and here's a transcription of the equation I've used, now corrected, which
runs on a 20-year-old Casio programmable calculator-
M=A*LN(TAN(45+.5*ABSL)*((1-E*SINABSL)/(1+E*SINABSL))^(E/2))
and that gives me M(45º) = 3013.648, which conforms with Bowditch, and not
(quite) with Andres..
Why do Andres and I differ (slightly), and why do I agree with Bowditch,
when Andres and I are trying to calculate the same thing with the same
quantities, using what appears to be the same expression?
You can see that I've rejoined the nit-picking purists, as my natural home.
George.
contact George Huxtable, at george@hux.me.uk
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
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