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Re: No solution yet
From: Andrés Ruiz
Date: 2011 Dec 27, 10:50 +0100
Andrés Ruiz
Navigational Algorithms
http://sites.google.com/site/navigationalalgorithms/
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 259.1I 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