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Re: No solution yet
From: Andrés Ruiz
Date: 2011 Dec 27, 18:36 +0100
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Andrés Ruiz
Navigational Algorithms
http://sites.google.com/site/navigationalalgorithms/
From: Andrés Ruiz
Date: 2011 Dec 27, 18:36 +0100
Robert, your solution is OK.
Here is it by some different methods:
TWO Circles of Position
CoP1 = 80.011667 -16.735000 37.270000
CoP2 = 152.715000 10.408333 34.393333
Vector Solution for the Intersection of two Circles of Equal Altitude
THE JOURNAL OF NAVIGATION (2008), 61, 355-365. The Royal Institute of Navigation
Andrés Ruiz González - Navigational Algorithms - http://sites.google.com/site/navigationalalgorithms/
iter Err Be Le B1 L1 B2 L2 GHA1f dec1f GHA2f dec2f
0 364.763403720619863 33.365742 -97.235712 -42.605056 -134.378443 33.365742 -97.235712 80.011667 -16.735000 152.715000 10.408333
1 0.000000000000000 33.365742 -97.235712 -42.605056 -134.378443 33.365742 -97.235712 80.011667 -16.735000 152.715000 10.408333
I1: -42.605056 -134.378443
I2: 33.365742 -97.235712
An analytical solution of the two star sight problem of celestial navigation
James A. Van Allen. NAVIGATION Vol. 28, No. 1, 1981
iter Err Be Le B1 L1 B2 L2 GHA1f dec1f GHA2f dec2f
0 364.763403720641691 33.365742 -97.235712 33.365742 -97.235712 -42.605056 -134.378443 80.011667 -16.735000 152.715000 10.408333
1 0.000000000000000 33.365742 -97.235712 33.365742 -97.235712 -42.605056 -134.378443 80.011667 -16.735000 152.715000 10.408333
I1: 33.365742 -97.235712
I2: -42.605056 -134.378443
Newton Raphson solution for the Intersection of two Circles of Equal Altitude
Andres Ruiz (c)2006
Iter = 4
I1: 33.365742 -97.235712
Applications of Complex Analysis to Celestial Navigation - Double Altitude Sights
Robin G. Stuart - NavList 10015
Zp1 = (0.128966, -0.732276 i)
ro1 = 0.495643
Zc1 = (0.185896, -1.055527 i)
r1 = 0.890627
Zp2 = (-1.066853, -0.550290 i)
ro2 = 0.527315
Zc2 = (-2.275100, -1.173512 i)
r2 = 2.147736
d = 2.463823
mu = -0.314604
nu = 0.310318
Z1 = (-0.233750, -1.841093 i)
Z2 = (-0.306976, -0.313709 i)
I1: 33.365742 -97.235712
I2: -42.605056 -134.378443
DeWit/USNO Nautical Almanac/Compac Data, Least squares algorithm for n LOPs
GHA DEC HO BO LO LHA HC Z p
80.0117 -16.7350 37.2700 33.3657 -97.2357 342.7760 37.2700 159.1247 0.0000
152.7150 10.4083 34.3933 33.3657 -97.2357 55.4793 34.3933 259.1265 -0.0000
Estimate position at time of fix:
Befix [deg] = 33.3657
Lefix [deg] = -97.2357
Least Squares information:
nObservations = 2
AA = 0.9086
BB = -0.1477
CC = 1.0914
DD = -0.0000
EE = 0.0000
FF = 0.0000
G = 0.9698
Error:
S = 0.0000
sigma = 0.0000 nm
sigmaB = 0.0000
sigmaL = 0.0000
Ellipse:
Prob = 0.9500
k = 0.0000
theta = 0.0000
a = 0.0000
b = 0.0000
Improved position at time of fix:
dB [deg] = -0.0000
dL [deg] = 0.0000
DO [deg] = 0.0000 nm
BI [deg] = 33.3657
LI [deg] = -97.2357
iteraciones = 4
--
Andrés Ruiz
Navigational Algorithms
http://sites.google.com/site/navigationalalgorithms/
2011/12/27 Robert Bernecky <bernecky@sbcglobal.net>
Here is the calculation done by calculator. It is a bit tedious, and so no guarantees I did not blunder somewhere...