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    Re: No solution yet
    From: Andrés Ruiz
    Date: 2011 Dec 27, 10:50 +0100
    Hello Fred,

    the method is by Gauss. It calculates the latitude of the two points of intersection of two simultaneous circles of equal altitude.
    the source code in ANSi C is:

    int LatitudeBySimultaneousDoubleAltitudes( double GHA1, double Dec1, double Ho1,                                        double GHA2, double Dec2, double Ho2,
                       double* Bi, double* Bj )
    {
    double D = 0;
    double alpha = 0;
    double alpha_mas_beta = 0;
    double beta = 0;

    *Bi = *Bj = 0.0;
    // Punto 1º de intersección
    D = ACOS( SIN( Dec1 )*SIN( Dec2 )+COS( Dec1 )*COS( Dec2 )*COS( GHA1-GHA2 ) );

    alpha = ACOS( (SIN( Ho2 )-SIN( Ho1 )*COS( D ))/(COS( Ho1 )*SIN( D )) );

    alpha_mas_beta = ACOS( (SIN( Dec2 )-SIN( Dec1 )*COS( D ))/(COS( Dec1 )*SIN( D )) );

    beta = alpha_mas_beta - alpha;

    *Bi = ASIN( SIN( Dec1 )*SIN( Ho1 )+COS( Dec1 )*COS( Ho1 )*COS( beta ) );

    // Punto 2º de intersección
    beta = alpha_mas_beta + alpha;

    *Bj = ASIN( SIN( Dec1 )*SIN( Ho1 )+COS( Dec1 )*COS( Ho1 )*COS( beta ) );

    return(0);
    }

    Regards,
    -- 
    Andrés Ruiz
    Navigational Algorithms
    http://sites.google.com/site/navigationalalgorithms/ 


    2011/12/27 Fred Stevens <fredstevens@yahoo.com>

    After spending much time on the "Latitude by two altitudes" problem detailed on page 6-3 of Umland's Short Guide to CN, I am totally lost.

    I used the following from the USNO almanac online:
    Sirius GHA 80* 0.7' dec s16* 44.1' H 37* 16.2' Zn 159.1
    Jupiter 152* 42.9' n10* 24.5' 34* 23.6' Zn 259.1

    I am assuming that the south dec is what is causing the pain but
    whatever it is, I am nowhere near the right answer, which should
    be close to 33.22*.

    Pointers, anyone?

    Thanks,
    Fred

       
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