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From: Andrés Ruiz
Date: 2006 Jun 12, 11:17 +0200
Another way, using spherical
triangles:
Position from two circles of equal altitude
Direct computation. Spherical trigonometry
D = ACOS( SIN( Dec1 )*SIN( Dec2 )+COS( Dec1 )*COS( Dec2
)*COS( GHA1-GHA2 ) );
ro = ACOS( (SIN( Ho2 )-SIN( Ho1 )*COS( D ))/(COS( Ho1 )*SIN(
D )) );
omega = ACOS( (SIN( Dec2 )-SIN( Dec1 )*COS( D ))/(COS( Dec1
)*SIN( D )) );
Intersecction Point 1
psi1 = fabs( omega - ro );
B1 = ASIN( SIN( Dec1 )*SIN( Ho1 )+COS( Dec1 )*COS( Ho1
)*COS( psi1 ) );
L11 = ACOS( (SIN( Ho1 )-SIN( Dec1 )*SIN( B1 ))/(COS( Dec1
)*COS( B1 )) ) - GHA1;
L12 = 360.0-ACOS( (SIN( Ho1 )-SIN( Dec1 )*SIN( B1 ))/(COS(
Dec1 )*COS( B1 )) ) - GHA1;
Intersecction Point 2
psi2 = omega + ro;
B2 = ASIN( SIN( Dec1 )*SIN( Ho1 )+COS( Dec1 )*COS( Ho1
)*COS( psi2 ) );
L21 = ACOS( (SIN( Ho1 )-SIN( Dec1 )*SIN( B2 ))/(COS( Dec1
)*COS( B2 )) ) - GHA1;
L22 = 360.0-ACOS( (SIN( Ho1 )-SIN( Dec1 )*SIN( B2 ))/(COS(
Dec1 )*COS( B2 )) ) - GHA1;
Andrés
http://www.geocities.com/CapeCanaveral/Runway/3568/index.html
Andrés
