# Coordinates transformations: how to do it* Claudio Bogazzi

```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°
• 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
```