Coordinates transformations: how to do it* Claudio Bogazzi NIKHEF Antares AWG Videoconference 06/05/2010 * Or how I think we should do it… Outlines • The ANTARES reference system • Slalib library • Comparison with ConvertCoordinates • Some checks Antares reference system N (x, y, z ) = (E, N, up) UTM grid y φ x E The azimuth angle is measured positively from East to North Do we all agree about it? I hope yes… Slalib • “…a library of routines intended to make accurate and reliable positional-astronomy applications easier to write…” ( from the manual, 218 pages!!! ) • What do we need for the transformations? Basically only 4 functions ① SLA_CALDJ (from the Gregorian Calendar to Modified Julian Date) ② SLA_OAP (from local horizontal coordinates to APPARENT [δ,α] )* ③ SLA_AMP (from apparent [δ,α] to mean place) ④ SLA_EQGAL (from equatorial to galactic coordinates) * Here we take into account annual aberration, nutation and precession SLA_OAP • http://starwww.rl.ac.uk/star/docs/sun67.htx/node138.html • INPUT: azimuth, ZENITH*, UTC date/time (MJD), ΔUT=UT1-UTC=-0.2, longitude, latitude and a lot of values that we set with the default values (pressure, humidity, effective wavelength) • OUTPUT: geocentric apparent [α,δ] * Here we used the zenith angle while in the function SLA_dh2e we need the elevation (zenith – 90) SLA_OAP: which zenith? • Two different way to define the zenith angle: acos( dz ) (as defined in RECO) zenith of the track acos(- dz ) zenith of the source Since acos (-x) = pi – acos (x) we have a difference of 180° between these two zeniths • Which of these two zeniths is the good one for slalib? I think the zenith of the source…any other opinion? Slalib reference system N E y y φant x φsla E (x, y, z ) = (E, N, up) ANTARES φsla = 90° – φant – 1.93° x (x, y, z ) = (N, E, up) Slalib N φsla = 90° – φant – 1.93° • Now remember that slalib wants the zenith and the azimuth of the source! The previous one was the azimuth of the track! φsla source = φslatrack + 180° • Now we have φsla source = 90° -φant -1.93° +180° = = 270° -φant -1.93° φsla = 270° – φant – 1.93° • How can we be sure about this transformation? • Indirect way: do you remember AstroCoordinates? It’s the program written by Gilles Maurin that permits to compute equatorial and galactic coordinates of an event. • It uses the ConvertCoordinates.hh library (it’s in the Physics package) in which “the azimuth is measured positively westwards from the South” • Here (x,y,z) = (S,W,up) i.e. 180° rotated to the slalib one! ConvertCoordinates • In AstroCoordinates you can find this transformation for the azimuth angle (from the detector one to the one used as input for the functions) φcc = 270° -φant – 180° – 1.93° If we consider this transformation correct, then also the previous one should be! Test • Only for dec and R.A. for the first 10 values suggested by Thomas

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