NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Ark Shvetsky
Date: 2009 Mar 26, 10:05 -0700
From: Peter Fogg <piterr11@gmail.com>
To: NavList@fer3.com
Sent: Thursday, March 26, 2009 9:15:10 AM
Subject: [NavList 7793] Re: Interpolation of Meridional Part Table
Would anyone have a copy of The Journal of Navigation (British) Volume 49, published in 1963 or 1964? Apparently it contains an article entitled "Practical Rhumb Line Calculations on the Spheroid" by GG Bennett, including numerical examples. The author no longer has a copy and would be grateful for a scan of such.
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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