Hi all. Does anyone know how to convert the spatial co-ordinates and 'velocity vectors' of Unitec-1 at separation into something that gives pointing data for elevation and azimuth. I can see it will be at an altitude of 6600km 48 mins and 19 seconds after its 21.44UTC launch, but converting the distances from the X Y and Z references is unfamiliar. Regards David G0MRF Table. UNITEC-1 Separation Prediction (as of 2010.5.12) Separation timing 2901.000 sec >>time after the launcher liftoff） >>17 May 2010, 22:32:35 UTC Inertial velocity 8957.4 m/s Elevation for inertial flight 47.627 deg Azimuth for inertial flight 96.499 deg Geocentric distance 12999.796 km Altitude 6627.232 km Inertial coordinate position: X 4630586 m Inertial coordinate position: Y -10174686 m Inertial coordinate position: Z -6635370 m Inertial coordinate velocity: X 7672.0 m/s Inertial coordinate velocity: Y -2377.4 m/s Inertial coordinate velocity: Z -3965.3 m/s Definition of position and velocity, Inertial coordinate system Origin：Earth centre X axis：Intersection of equator plane and meridian that contains the Greenwich at a nominal lift off time. Y axis：Right hand system (on equatorial plane) Z axis：North pole _______________________________________________ Sent via amsat...@amsat.org. Opinions expressed are those of the author. Not an AMSAT-NA member? Join now to support the amateur satellite program! Subscription settings: http://amsat.org/mailman/listinfo/amsat-bb
