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    Re: No solution yet
    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
    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



    image.png

    -- 
    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...

    File:


       
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