Converting between Helioprojective and AltAz Coordinate

How to find the Sun in the sky as viewed from a particular location.

from astropy.coordinates import EarthLocation, AltAz, SkyCoord
from astropy.time import Time
from sunpy.coordinates import frames, sun
import astropy.units as u

We use SkyCoord to define the center of the Sun

obstime = "2013-09-21 16:00:00"
c = SkyCoord(0 * u.arcsec, 0 * u.arcsec, obstime=obstime, frame=frames.Helioprojective)

Now we establish our location on the Earth, in this case let’s consider a high altitude balloon launched from Fort Sumner, NM.

Fort_Sumner = EarthLocation(lat=34.4900*u.deg, lon=-104.221800*u.deg, height=40*u.km)

Now lets convert this to a local measurement of Altitude and Azimuth.

frame_altaz = AltAz(obstime=Time(obstime), location=Fort_Sumner)
sun_altaz = c.transform_to(frame_altaz)
print(f'Altitude is {sun_altaz.T.alt} and Azimuth is {sun_altaz.T.az}')

Out:

Altitude is 37.782959956075395 deg and Azimuth is 121.342173388297 deg

Next let’s check this calculation by converting it back to helioprojective. We should get our original input which was the center of the Sun. To go from Altitude/Azimuth to Helioprojective, you will need the distance to the Sun. solar distance. Define distance with SunPy’s almanac.

distance = sun.earth_distance(obstime)
b = SkyCoord(az=sun_altaz.T.az, alt=sun_altaz.T.alt, distance=distance, frame=frame_altaz)
sun_helio = b.transform_to(frames.Helioprojective)
print(f'The helioprojective point is {sun_helio.T.Tx}, {sun_helio.T.Ty}')

Out:

The helioprojective point is -0.021860375301912427 arcsec, -0.00795802552508306 arcsec

Total running time of the script: ( 0 minutes 2.001 seconds)

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