Or if you want to go even shorter, you could use NOAA's low-accuracy equations:
#!/usr/local/bin/python
import sys
from datetime import datetime, time, timedelta
from math import pi, cos, sin
def solar_time(dt, longit):
return ha
def main():
if len(sys.argv) != 4:
print 'Usage: hour_angle.py [YYYY/MM/DD] [HH:MM:SS] [longitude]'
sys.exit()
else:
dt = datetime.strptime(sys.argv[1] + ' ' + sys.argv[2], '%Y/%m/%d %H:%M:%S')
longit = float(sys.argv[3])
gamma = 2 * pi / 365 * (dt.timetuple().tm_yday - 1 + float(dt.hour - 12) / 24)
eqtime = 229.18 * (0.000075 + 0.001868 * cos(gamma) - 0.032077 * sin(gamma) \
- 0.014615 * cos(2 * gamma) - 0.040849 * sin(2 * gamma))
decl = 0.006918 - 0.399912 * cos(gamma) + 0.070257 * sin(gamma) \
- 0.006758 * cos(2 * gamma) + 0.000907 * sin(2 * gamma) \
- 0.002697 * cos(3 * gamma) + 0.00148 * sin(3 * gamma)
time_offset = eqtime + 4 * longit
tst = dt.hour * 60 + dt.minute + dt.second / 60 + time_offset
solar_time = datetime.combine(dt.date(), time(0)) + timedelta(minutes=tst)
print solar_time
if __name__ == '__main__':
main()