Bram Dorreman wrote: > > Who can tell me the formula to calculate the stellar > > magnitude from a self defined absolute magnitude (phase 90° > > and distance 1000 km), range and sun-target-observer angle? > > > > Another question: what might be a reasonable value for > > Rosetta's standard magnitude? Ted Molczan replied: > I have derived the following formula, for use with data provided by the > ephemeris: > > mag = 32.17 + 5 * log10("delta") + 0.01 * ("S-T-O" - 90), +/- 2 > > where: "delta" = Target apparent range relative to observer. Units: AU > > "S-T-O" = Sun-Target-Observer angle; target's apparent PHASE ANGLE as > seen at observer's location at print time. Units: DEGREES > > The above formula is based on my estimated standard magnitude of 6.3 +/- 2 mag > (1000 km, phase angle 90 deg), based upon the ESA's description: "a large > aluminium box with dimensions 2.8 x 2.1 x 2.0 metres" Thanks Ted and other repliers too. Before I received these replies I recalculated my topocentric predictions so that distance unit is km. What would be the formula when "delta" unit is km? Meanwhile I can start with the original output in which "delta" is in AU. ESA's statement that Rosetta would be mag +8 or +9 at 10000 km range makes it mag +3 or +4 at 1000 km (100 times as bright -> mag 5 difference). I now know what standard magnitude range I can apply in my further calculations. It may all be in vain because we now have a real winterperiode with clouded skies, snow and temperatures below zero (Celsius) during the whole days. This will cause we cannot "hunt happily". Best regards..Bram Dorreman ------------------------------------------------------------------------- Subscribe/Unsubscribe info, Frequently Asked Questions, SeeSat-L archive: http://www.satobs.org/seesat/seesatindex.html
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