Local solar time function from UTC and longitude

Question

I'm wondering if there is a python function/module that calculates the local time after midnight (or local solar time) given the UTC time and longitude? It doesn't need to take into account daylight saving time.

Thanks in advance.

Answer 1

For a very simple function & very very approximated local time: Time variation goes from -12h to +12h and longitude goes from -180 to 180. Then:

---

import datetime as dt

def localTimeApprox(myDateTime, longitude):
   """Returns local hour approximation"""
   return myDateTime+dt.timedelta(hours=(longitude*12/180))

---

Sample calling: localTimeApprox(dt.datetime(2014, 7, 9, 20, 00, 00), -75)

Returns: datetime.datetime(2014, 7, 9, 15, 0)

Answer 2

Using ephem's sidereal_time() method:

import ephem # pip install pyephem (on Python 2)
             # pip install ephem   (on Python 3)

def solartime(observer, sun=ephem.Sun()):
    sun.compute(observer)
    # sidereal time == ra (right ascension) is the highest point (noon)
    hour_angle = observer.sidereal_time() - sun.ra
    return ephem.hours(hour_angle + ephem.hours('12:00')).norm  # norm for 24h

Note: ephem.hours is a float number that represents an angle in radians and converts to/from a string as "hh:mm:ss.ff".

For comparison, here's the "utc + longitude" formula:

import math
from datetime import timedelta

def ul_time(observer):
    utc_dt = observer.date.datetime()
    longitude = observer.long
    return utc_dt + timedelta(hours=longitude / math.pi * 12)

Example

from datetime import datetime

# "solar time" for some other cities
for name in ['Los Angeles', 'New York', 'London',
             'Paris', 'Moscow', 'Beijing', 'Tokyo']:
    city = ephem.city(name)
    print("%-11s %11s %s" % (name, solartime(city),
                             ul_time(city).strftime('%T')))

# set date, longitude manually
o = ephem.Observer()
o.date = datetime(2012, 4, 15, 1, 0, 2) # some utc time
o.long = '00:00:00.0' # longitude (you could also use a float (radians) here)
print("%s %s" % (solartime(o), ul_time(o).strftime('%T')))

Output

Los Angeles 14:59:34.11 14:44:30
New York    17:56:31.27 17:41:27
London      22:52:02.04 22:36:58
Paris       23:01:56.56 22:46:53
Moscow       1:23:00.33 01:07:57
Beijing      6:38:09.53 06:23:06
Tokyo        8:11:17.81 07:56:15
1:00:00.10 01:00:01

Answer 3

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()

Answer 4

Trying again, with ephem. I included latitude and elevation as arguments, but they are not needed of course. You can just call them 0 for your purposes.

#!/usr/local/bin/python

import sys
from datetime import datetime, time, timedelta
import ephem

def hour_angle(dt, longit, latit, elev):
    obs = ephem.Observer()
    obs.date = dt.strftime('%Y/%m/%d %H:%M:%S')
    obs.lon = longit
    obs.lat = latit
    obs.elevation = elev
    sun = ephem.Sun()
    sun.compute(obs)
    # get right ascention
    ra = ephem.degrees(sun.g_ra) - 2 * ephem.pi

    # get sidereal time at greenwich (AA ch12)
    jd = ephem.julian_date(dt)
    t = (jd - 2451545.0) / 36525
    theta = 280.46061837 + 360.98564736629 * (jd - 2451545) \
            + .000387933 * t**2 - t**3 / 38710000

    # hour angle (AA ch13)
    ha = (theta + longit - ra * 180 / ephem.pi) % 360
    return ha

def main():
    if len(sys.argv) != 6:
        print 'Usage: hour_angle.py [YYYY/MM/DD] [HH:MM:SS] [longitude] [latitude] [elev]'
        sys.exit()
    else:
        dt = datetime.strptime(sys.argv[1] + ' ' + sys.argv[2], '%Y/%m/%d %H:%M:%S')
        longit = float(sys.argv[3])
        latit = float(sys.argv[4])
        elev = float(sys.argv[5])

    # get hour angle
    ha = hour_angle(dt, longit, latit, elev)

    # convert hour angle to timedelta from noon
    days = ha / 360
    if days > 0.5:
        days -= 0.5
    td = timedelta(days=days)

    # make solar time
    solar_time = datetime.combine(dt.date(), time(12)) + td
    print solar_time

if __name__ == '__main__':
    main()

Answer 5

I took a look at Jean Meeus' Astronomical Algorithms. I think you might be asking for local hour angle, which can be expressed in time (0-24hr), degrees (0-360) or radians (0-2pi).

I'm guessing you can do this with ephem. But just for the heck of it, here's some python:

#!/usr/local/bin/python

import sys
from datetime import datetime, time, timedelta
from math import pi, sin, cos, atan2, asin

# helpful constant
DEG_TO_RAD = pi / 180

# hardcode difference between Dynamical Time and Universal Time
# delta_T = TD - UT
# This comes from IERS Bulletin A
# ftp://maia.usno.navy.mil/ser7/ser7.dat
DELTA = 35.0

def coords(yr, mon, day):
    # @input year (int)
    # @input month (int)
    # @input day (float)
    # @output right ascention, in radians (float)
    # @output declination, in radians (float)

    # get julian day (AA ch7)
    day += DELTA / 60 / 60 / 24 # use dynamical time
    if mon <= 2:
        yr -= 1
        mon += 12
    a = yr / 100
    b = 2 - a + a / 4
    jd = int(365.25 * (yr + 4716)) + int(30.6 * (mon + 1)) + day + b - 1524.5

    # get sidereal time at greenwich (AA ch12)
    t = (jd - 2451545.0) / 36525

    # Calculate mean equinox of date (degrees)
    l = 280.46646 + 36000.76983 * t + 0.0003032 * t**2
    while (l > 360):
        l -= 360
    while (l < 0):
        l += 360

    # Calculate mean anomoly of sun (degrees)
    m = 357.52911 + 35999.05029 * t - 0.0001537 * t**2

    # Calculate eccentricity of Earth's orbit
    e = 0.016708634 - 0.000042037 * t - 0.0000001267 * t**2

    # Calculate sun's equation of center (degrees)
    c = (1.914602 - 0.004817 * t - .000014 * t**2) * sin(m * DEG_TO_RAD) \
        + (0.019993 - .000101 * t) * sin(2 * m * DEG_TO_RAD) \
        + 0.000289 * sin(3 * m * DEG_TO_RAD)

    # Calculate the sun's radius vector (AU)
    o = l + c # sun's true longitude (degrees)
    v = m + c # sun's true anomoly (degrees)

    r = (1.000001018 * (1 - e**2)) / (1 + e * cos(v * DEG_TO_RAD))

    # Calculate right ascension & declination
    seconds = 21.448 - t * (46.8150 + t * (0.00059 - t * 0.001813))
    e0 = 23 + (26 + (seconds / 60)) / 60

    ra = atan2(cos(e0 * DEG_TO_RAD) * sin(o * DEG_TO_RAD), cos(o * DEG_TO_RAD)) # (radians)
    decl = asin(sin(e0 * DEG_TO_RAD) * sin(o * DEG_TO_RAD)) # (radians)

    return ra, decl

def hour_angle(dt, longit):
    # @input UTC time (datetime)
    # @input longitude (float, negative west of Greenwich)
    # @output hour angle, in degrees (float)

    # get gregorian time including fractional day
    y = dt.year
    m = dt.month
    d = dt.day + ((dt.second / 60.0 + dt.minute) / 60 + dt.hour) / 24.0 

    # get right ascention
    ra, _ = coords(y, m, d)

    # get julian day (AA ch7)
    if m <= 2:
        y -= 1
        m += 12
    a = y / 100
    b = 2 - a + a / 4
    jd = int(365.25 * (y + 4716)) + int(30.6 * (m + 1)) + d + b - 1524.5

    # get sidereal time at greenwich (AA ch12)
    t = (jd - 2451545.0) / 36525
    theta = 280.46061837 + 360.98564736629 * (jd - 2451545) \
            + .000387933 * t**2 - t**3 / 38710000

    # hour angle (AA ch13)
    ha = (theta + longit - ra / DEG_TO_RAD) % 360

    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])
    ha = hour_angle(dt, longit)
    # convert hour angle to timedelta from noon
    days = ha / 360
    if days > 0.5:
        days -= 0.5
    td = timedelta(days=days)
    # make solar time
    solar_time = datetime.combine(dt.date(), time(12)) + td
    print solar_time

if __name__ == '__main__':
    main()