I expected to find dozens of readily available implementations of sunrise and sunset calculations in Python on the web but this turned out to be a disappointment. Therefore I resolved to write my own straightforward implementation.
The sunrise equation is pretty easy to find on Wikipedia but actual implementations are not, certainly not in Python 3.x. (If you are willing to stick to Python 2.x there is of course the excellent PyEphem package, see the end of this article) Fortunately NOAA provides annotated equations in the form of an OpenOffice spreadsheet. This allows for simple re-engineering in Python and gives us a way to verify the results of a new implementation against those in the spreadsheet.
If you save the code below as sunrise.py then calculating the time of sunrise today would be straightforward as shown in this example:
import datetime
import sunrise
s = sun(lat=49,long=3)
print('sunrise at ',s.sunrise(when=datetime.datetime.now())
The sun class also provides a sunset() method and solarnoon() method. All three methods take a when parameter that should be a datetime.datetime object. If this object contains timezone information or daylight saving time information, this information is used when calculating the times of sunrise, sunset and the solar noon.Note that if no when parameter is given, a default datetime is used that is initialized with a LocalTimezone object from the timezone module. I have not provided that module here but you can implement one simple enough by copying the example in Python's documentation or you can comment out the import statement below and always supply a when parameter.
from math import cos,sin,acos,asin,tan
from math import degrees as deg, radians as rad
from datetime import date,datetime,time
# this module is not provided here. See text.
from timezone import LocalTimezone
class sun:
"""
Calculate sunrise and sunset based on equations from NOAA
http://www.srrb.noaa.gov/highlights/sunrise/calcdetails.html
typical use, calculating the sunrise at the present day:
import datetime
import sunrise
s = sun(lat=49,long=3)
print('sunrise at ',s.sunrise(when=datetime.datetime.now())
"""
def __init__(self,lat=52.37,long=4.90): # default Amsterdam
self.lat=lat
self.long=long
def sunrise(self,when=None):
"""
return the time of sunrise as a datetime.time object
when is a datetime.datetime object. If none is given
a local time zone is assumed (including daylight saving
if present)
"""
if when is None : when = datetime.now(tz=LocalTimezone())
self.__preptime(when)
self.__calc()
return sun.__timefromdecimalday(self.sunrise_t)
def sunset(self,when=None):
if when is None : when = datetime.now(tz=LocalTimezone())
self.__preptime(when)
self.__calc()
return sun.__timefromdecimalday(self.sunset_t)
def solarnoon(self,when=None):
if when is None : when = datetime.now(tz=LocalTimezone())
self.__preptime(when)
self.__calc()
return sun.__timefromdecimalday(self.solarnoon_t)
@staticmethod
def __timefromdecimalday(day):
"""
returns a datetime.time object.
day is a decimal day between 0.0 and 1.0, e.g. noon = 0.5
"""
hours = 24.0*day
h = int(hours)
minutes= (hours-h)*60
m = int(minutes)
seconds= (minutes-m)*60
s = int(seconds)
return time(hour=h,minute=m,second=s)
def __preptime(self,when):
"""
Extract information in a suitable format from when,
a datetime.datetime object.
"""
# datetime days are numbered in the Gregorian calendar
# while the calculations from NOAA are distibuted as
# OpenOffice spreadsheets with days numbered from
# 1/1/1900. The difference are those numbers taken for
# 18/12/2010
self.day = when.toordinal()-(734124-40529)
t=when.time()
self.time= (t.hour + t.minute/60.0 + t.second/3600.0)/24.0
self.timezone=0
offset=when.utcoffset()
if not offset is None:
self.timezone=offset.seconds/3600.0
def __calc(self):
"""
Perform the actual calculations for sunrise, sunset and
a number of related quantities.
The results are stored in the instance variables
sunrise_t, sunset_t and solarnoon_t
"""
timezone = self.timezone # in hours, east is positive
longitude= self.long # in decimal degrees, east is positive
latitude = self.lat # in decimal degrees, north is positive
time = self.time # percentage past midnight, i.e. noon is 0.5
day = self.day # daynumber 1=1/1/1900
Jday =day+2415018.5+time-timezone/24 # Julian day
Jcent =(Jday-2451545)/36525 # Julian century
Manom = 357.52911+Jcent*(35999.05029-0.0001537*Jcent)
Mlong = 280.46646+Jcent*(36000.76983+Jcent*0.0003032)%360
Eccent = 0.016708634-Jcent*(0.000042037+0.0001537*Jcent)
Mobliq = 23+(26+((21.448-Jcent*(46.815+Jcent*(0.00059-Jcent*0.001813))))/60)/60
obliq = Mobliq+0.00256*cos(rad(125.04-1934.136*Jcent))
vary = tan(rad(obliq/2))*tan(rad(obliq/2))
Seqcent = sin(rad(Manom))*(1.914602-Jcent*(0.004817+0.000014*Jcent))+sin(rad(2*Manom))*(0.019993-0.000101*Jcent)+sin(rad(3*Manom))*0.000289
Struelong= Mlong+Seqcent
Sapplong = Struelong-0.00569-0.00478*sin(rad(125.04-1934.136*Jcent))
declination = deg(asin(sin(rad(obliq))*sin(rad(Sapplong))))
eqtime = 4*deg(vary*sin(2*rad(Mlong))-2*Eccent*sin(rad(Manom))+4*Eccent*vary*sin(rad(Manom))*cos(2*rad(Mlong))-0.5*vary*vary*sin(4*rad(Mlong))-1.25*Eccent*Eccent*sin(2*rad(Manom)))
hourangle= deg(acos(cos(rad(90.833))/(cos(rad(latitude))*cos(rad(declination)))-tan(rad(latitude))*tan(rad(declination))))
self.solarnoon_t=(720-4*longitude-eqtime+timezone*60)/1440
self.sunrise_t =self.solarnoon_t-hourangle*4/1440
self.sunset_t =self.solarnoon_t+hourangle*4/1440
if __name__ == "__main__":
s=sun(lat=52.37,long=4.90)
print(datetime.today())
print(s.sunrise(),s.solarnoon(),s.sunset())
For people willing to stick to Python 2.x there is a simple and good alternative in the form of the PyEphem package. It can do a lot more than just calculating sunsise. An example is shown below.
import ephem o=ephem.Observer() o.lat='49' o.long='3' s=ephem.Sun() s.compute() print ephem.localtime(o.next_rising(s))