What is are the angles of view (relative to the level direction at the pilot's location)?
These values enable us to recover the azimuth as the inverse tangent of the cosine and sine.Ī pilot in an airplane flying west at 4000 meters, located at (lat, lon) = (39, -75), sees a jet far ahead located at (39, -76) and flying at 12000 meters. Such a vector is (-y, x, 0), because it clearly is perpendicular to (x,y,z) (the up direction) and the northern direction. We also need the sine of the azimuth, which is similarly obtained once we know a vector pointing due East (locally). These two calculations confirm that this is indeed the desired vector). Its projection onto the Equatorial plane is proportional to (-x,-y) which points directly inward, making it the projection of a north-pointing vector. (The inner product of these two vectors is zero, proving it is level. One such vector at the location (x,y,z) is (-z x, -zy, x^2+y^2). We need a level vector (u,v,w) that points due north. This is the "elevation." AzimuthĪ similar calculation obtains the local direction of view ("azimuth"). Subtract this from 90 degrees if you want the angle of view relative to a nominal horizon. Given two points (x,y,z) and (x',y',z') in an earth-centered coordinate system, the vector from the first to the second is (dx,dy,dz) = (x'-x, y'-y, z'-z), whence the cosine of the angle made to the normal at (x,y,z) is the inner product of the unit length versions of those vectors: Cos(elevation) = (x*dx + y*dy + z*dz) / Sqrt((x^2+y^2+z^2)*(dx^2+dy^2+dz^2)) (If you don't do this, you need to retain additional information about the local normal directions in order to compute angles accurately at nonzero elevations.) Elevation For accurate calculations, convert (lat, lon, elevation) directly to earth-centered (x,y,z).
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