[amsat-bb] Unitec-1 prediction data at T+ 00.48.19
G0MRF@aol.com
G0MRF at aol.com
Sat May 15 09:11:24 PDT 2010
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
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