# Error when calculating Alt/Az from Ra/Dec

I wrote a program based on this tutorial: http://www.stjarnhimlen.se/comp/ppcomp.html to calculate altitude and azimuth of a celestial object. When I compare the calculated RA/Dec values to the ones from Stellarium, they are very accurate. But when I compare the Alt/Az values to the ones from Stellarium, there is an error of about 2 degrees! I use the following method to calculate Alt, Az:

GMST0 = Ls + 180_degrees # Ls = Sun's longitude
GMST = GMST0 + UT
LST  = GMST + local_longitude
HA = LST - RA

x = cos(HA) * cos(Decl)
y = sin(HA) * cos(Decl)
z = sin(Decl)

xhor = x * sin(lat) - z * cos(lat)
yhor = y
zhor = x * cos(lat) + z * sin(lat)

az  = atan2( yhor, xhor ) + 180_degrees
alt = asin( zhor ) = atan2( zhor, sqrt(xhor*xhor+yhor*yhor) )


Some specific example:

Test date 15.09.2018, time 15:00 UT Planet: Mercury

Coordinates: +47.55777777° +8.89888888

What stellarium says:

RA = 11h 18m 13.26s

Dec = +6°25'08.5"

Az = +250°21'13.2"

Alt = +25°25'00.1"

What my program says:

RA = 11h 18m 14s

Dec = 6° 25' 6.59"

Az = +248° 49' 6.9"

Alt = +26° 33' 16.43"

• Could you give us some specific examples? Is the 2 degree consistent with different latitudes/longitudes/ra/dec, or does it vary based on the input parameters? My vague first suspicion is that, because the Earth is an ellipsoid and not a sphere, your zenith direction ("surface normal") isn't exactly opposite the direction to the center of the Earth. – barrycarter Sep 15 '18 at 14:34
• The 2 degree error is not consistent, it's just the maximum error, but a lot of calculations (especially azimuth) were off by about 2°. I'll edit the question with some specific examples. – Michael Siebenmann Sep 15 '18 at 14:42
• Stellarium usually gives both precessed and unprecessed coordinates. Are you using J2000 or current precession. The difference is small, but might be enough to explain the error. I may look into this deeper, but this was just a thought I had. – barrycarter Sep 15 '18 at 15:09
• It says "Precession is computed in a simplified way, by a simple addition to the ecliptic longitude" on the website. – Michael Siebenmann Sep 19 '18 at 15:43
• An interesting thing is that the azimuth always is about 1.5° under the expected value, except for the moon. – Michael Siebenmann Sep 19 '18 at 18:08