[AstroPy] Hour Angle from location and alt/az

[Josh Walawender]

Hi John,


I solved a similar problem using the RA, DEC, and time from the header (and knowing the telescopes location) then just using the RA-LST method.  Here’s some (pseudo)code which calculates the HA for the mid point of the exposure (adjust as needed for start time if you like):


# Location of the observation
loc = EarthLocation(lat=cfg_loc['latitude’], lon=cfg_loc['longitude’], height=cfg_loc['elevation’])
# Read the start time and exposure time of the image from the header
starttime = Time(time.strptime(header['DATE-OBS’], '%Y-%m-%dT%H:%M:%S'), location=loc)
exptime = TimeDelta(float(header['EXPTIME']), format='sec')
# estimate the exposure mid time
midtime = starttime + exptime/2.0
# determine the sidereal time at the mid point of the exposure
sidereal = midtime.sidereal_time('apparent')
# read the position of the field from the header (assumes J2000)
header_pointing = SkyCoord(ra=float(header['RA'])*u.degree, dec=float(header['DEC'])*u.degree)
## To calculate HA, we need to precess to the current equinox
## otherwise the RA - LST method will be off.
FK5_Jnow = FK5(equinox=midtime)
HA = header_pointing.transform_to(FK5_Jnow).ra.to(u.hourangle) - sidereal

This was good enough for my purposes, but did seem a bit cumbersome.

-Josh




> On Nov 4, 2016, at 2:04 PM, John K. Parejko <parejkoj at uw.edu> wrote:
> 
> Hello astropythians,
> 
> I’m trying to work around a distinct lack of UT1 and LST in some telescope file headers, and was wondering if there’s a straight-forward astropy way to get a source’s Hour Angle from the alt/az, ra/dec, and telescope location. I can build an altaz SkyCoord, but I’m not sure what to do with it from there:
> 
> loc = coord.EarthLocation.from_geodetic(long, lat, elevation)
> sky = coord.SkyCoord(az, alt, location=loc, frame='altaz', unit='deg’)
> 
> I can’t transform this to ICRS:
> 
> ipdb> sky.transform_to(coord.ICRS)
> *** TypeError: unsupported operand type(s) for -: 'NoneType' and ‘float’
> 
> and there’s not an “hour_angle” method that I can see in either of those.
> 
> Per the internet[1], one should be able to get HA from alt/az and dec directly (modulo refraction corrections):
> 
> HA = asin(- sin(az)*cos(alt) / cos(dec))
> 
> I can do that, but I was hoping for a nice and tidy astropy method.
> 
> Thanks,
> John
> 
> 1: http://star-www.st-and.ac.uk/~fv/webnotes/chapter7.htm
> 
> --
> *************************
> John Parejko
> parejkoj at uw.edu
> http://staff.washington.edu/parejkoj/
> Department of Physics and Astronomy
> University of Washington
> Seattle, WA
> **************************
> 
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