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))

I was all ready to write my own module in exactly the same fashion! Thanks so much for doing this :-)

ReplyDeleteThere is an error in the Western Hemisphere.

ReplyDeleteline 80 should read:

self.timezone=offset.seconds/3600.0 + (offset.days * 24)

because time zones west of UTC have an offset value of -1 days plus a number of seconds.

@VernonDCole You're right, thanks for the comment

ReplyDeleteWhy does this return UTC instead of local time? I'm getting sunset times of 24 hours, which sunrise.py chokes on. Why?

ReplyDelete@Anonymous: I don't understand your question. Can you give an example?

ReplyDeleteThanks for publishing this code. I took the liberty of reworking it to give just a day/night boolean return, and would appreciate your approval before releasing this revised code. Please see www.flightdatacommunity.com where we plan to use this for determining whether aircraft take off or land in day or night conditions.

ReplyDeleteMany Thanks

Dave Jesse

Hi Michel, thanks for code and...anonymous, strangely I am just in process of doing the same thing :-)

ReplyDeleteThanks!

ReplyDeleteEphem has now an update for python 3.Xxx

Hey Michel,

ReplyDeletecool script, thx...

I used it in a pet project for my raspi.

Is it ok for you to put it on GitHub?

https://github.com/ChristianKniep/raspiGpioCtrl

Cheers

Christian

Hello,

ReplyDeleteThanks for the algorithm. When I try:

sunrise.sun(49, -123).sunset(when=datetime.datetime(2013, 6, 1))

I get:

ValueError: hour must be in 0..23

The error seems to occur for any longitude under -61. Any idea why this would occur?

I'm also getting this error:

Deletes = sun(lat=-41.3,long=174.8)

I get that error when the time zone does not match the longitude. Also, as a check I simply copied and pasted the formulas from the NOAA Excel spread sheet into Python (see below) and everything seems to work OK. Changes to the functions case (capitals to lower case) , removing the dollar sign from the absolute references (eg, $B$5 >> B5), and changing the MOD function from Excel format to Python format were the only changes required. The code then looked like this:

Delete#!/mnt/us/python/bin/python2.7

# Calculate sunrise and sunset based on equations from NOAA

# http://www.srrb.noaa.gov/highlights/sunrise/calcdetails.html

# tested on IDLE

from math import sin, asin, cos, acos, tan, atan2, radians, degrees

def __timefromdecimalday(day):

hours = 24.0*day

h = int(hours)

minutes= (hours-h)*60

m = int(minutes)

seconds= (minutes-m)*60

s = int(seconds)

return h,m,s

time_zone = 10 # non DST

lat = -38 # Melbourne, Australia

lon = 145

date = 41645 # number format from excel spreadsheet

time = 0.1/24

# setup parameters

B3 = lat

B4 = lon

B5 = time_zone

B7 = date

# cut and pasted equations fron NOAA spreadsheet

# may have to change case (capitals to lower case, etc) of functions

D2 = B7

E2 = 0.1/24 # day is a decimal fraction of 24 hours between 0.0 and 1.0 (e.g. noon = 0.5)

F2 = D2 + 2415018.5 + E2 - B5 / 24

G2 = (F2 - 2451545) / 36525

I2 = (280.46646+G2*(36000.76983 + G2*0.0003032)) % 360

J2 = 357.52911+G2*(35999.05029 - 0.0001537*G2)

K2 = 0.016708634-G2*(0.000042037+0.0000001267*G2)

L2 = sin(radians(J2))*(1.914602-G2*(0.004817+0.000014*G2))+sin(radians(2*J2))*(0.019993-0.000101*G2)+sin(radians(3*J2))*0.000289

M2 = I2 + L2

P2 = M2-0.00569-0.00478*sin(radians(125.04-1934.136*G2))

Q2 = 23+(26+((21.448-G2*(46.815+G2*(0.00059-G2*0.001813))))/60)/60

R2 = Q2+0.00256*cos(radians(125.04-1934.136*G2))

T2 = degrees(asin(sin(radians(R2))*sin(radians(P2))))

U2 = tan(radians(R2/2))*tan(radians(R2/2))

V2 = 4*degrees(U2*sin(2*radians(I2))-2*K2*sin(radians(J2))+4*K2*U2*sin(radians(J2))*cos(2*radians(I2))-0.5*U2*U2*sin(4*radians(I2))-1.25*K2*K2*sin(2*radians(J2)))

W2 = degrees(acos(cos(radians(90.833))/(cos(radians(B3))*cos(radians(T2)))-tan(radians(B3))*tan(radians(T2))))

X2 = (720-4*B4-V2+B5*60)/1440

Y2 = X2-W2*4/1440 # sunrise

Z2 = X2+W2*4/1440 # sunset

hrs,mins,secs = __timefromdecimalday(Y2)

print "sunrise {:02}:{:02}:{:02}".format(hrs,mins,secs)

hrs,mins,secs = __timefromdecimalday(Z2)

print "sunset {:02}:{:02}:{:02}".format(hrs,mins,secs)

Sorry, i´m a absolute beginner in Python and programming. The sunrise.py works fine, but when i try the typical use, i get an syntax error in the line "print('sunrise at ',s.sunrise(when=datetime.datetime.now()) "

ReplyDeleteif i comment this line out, i get the error " name sun is not defined"...

whats wrong?

I ran into the same issue using Python 3 on a Win7 box.

DeleteSolution:

import datetime

import sunrise

# Latitude N is positive, Latitude S is negative

# Longitude E is positive, Longitude W is negative

s = sunrise.sun(lat=25,long=-80)

here is my solution:

ReplyDeleteimport calendar

from math import cos, sin, acos as arccos, asin as arcsin, tan as tg, degrees, radians

def mod(a,b):

return a % b

def isLeapYear(year):

return (year % 4 == 0 and year % 100 != 0) or year % 400 == 0

def getDayNumber(year, month, day):

cnt = 0

for t in range(1900,year):

if isLeapYear(t):

cnt += 366

else:

cnt += 365

for t in range(1,month):

cnt += calendar.monthrange(2014, 2)[1]

return cnt + day + 1

def getHHMMSS(h):

hh = int(h)

mm = (h - hh) * 60

ss = int(0.5 + (mm - int(mm)) * 60)

return "{0:2d}:{1:02d}:{2:02d}" . format(hh, int(mm), ss)

# based on: http://www.srrb.noaa.gov/highlights/sunrise/calcdetails.html

def getSunriseAndSunset(lat, lon, dst, year, month, day):

localtime = 12.00

b2 = lat

b3 = lon

b4 = dst

b5 = localtime / 24

b6 = year

d30 = getDayNumber(year, month, day)

e30 = b5

f30 = d30 + 2415018.5 + e30 - b4 / 24

g30 = (f30 - 2451545) / 36525

q30 = 23 + (26 + ((21.448 - g30 * (46.815 + g30 * (0.00059 - g30 * 0.001813)))) / 60) / 60

r30 = q30 + 0.00256 * cos(radians(125.04 - 1934.136 * g30))

j30 = 357.52911 + g30 * (35999.05029 - 0.0001537 * g30)

k30 = 0.016708634 - g30 * (0.000042037 + 0.0000001267 * g30)

l30 = sin(radians(j30)) * (1.914602 - g30 * (0.004817 + 0.000014 * g30)) + sin(radians(2 * j30)) * (0.019993 - 0.000101 * g30) + sin(radians(3 * j30)) * 0.000289

i30 = mod(280.46646 + g30 * (36000.76983 + g30 * 0.0003032), 360)

m30 = i30 + l30

p30 = m30 - 0.00569 - 0.00478 * sin(radians(125.04 - 1934.136 * g30))

t30 = degrees(arcsin(sin(radians(r30)) * sin(radians(p30))))

u30 = tg(radians(r30 / 2)) * tg(radians(r30 / 2))

v30 = 4 * degrees(u30 * sin(2 * radians(i30)) - 2 * k30 * sin(radians(j30)) + 4 * k30 * u30 * sin(radians(j30)) * cos(2 * radians(i30)) - 0.5 * u30 * u30 * sin(4 * radians(i30)) - 1.25 * k30 * k30 * sin(2 * radians(j30)))

w30 = degrees(arccos(cos(radians(90.833)) / (cos(radians(b2)) * cos(radians(t30))) - tg(radians(b2)) * tg(radians(t30))))

x30 = (720 - 4 * b3 - v30 + b4 * 60) / 1440

x30 = (720 - 4 * b3 - v30 + b4 * 60) / 1440

y30 = (x30 * 1440 - w30 * 4) / 1440

z30 = (x30 * 1440 + w30 * 4) / 1440

sunrise = y30 * 24

sunset = z30 * 24

return (sunrise, sunset)

# latitude (+N -S):

lat = 50.0877777777777

# longitude (+E -W):

lon = 14.4205555555555

# (+E -W)

dst = 1

(sunrise, sunset) = getSunriseAndSunset(lat, lon, dst, 2014, 1, 29)

print("sunrise=", getHHMMSS(sunrise))

print("sunset =", getHHMMSS(sunset))

please change line

Deletecnt += calendar.monthrange(2014, 2)[1]

to this

cnt += calendar.monthrange(year, t)[1]

As far as I see Your x30-line is double . . .

DeleteThanks a lot for sharing this module. I will be going to use it in my selfmade home automation system to control the electric shutters.

ReplyDeleteThats AWESOME!!!!! :D

DeleteThanks for sharing. This made my day easier.

ReplyDeleteUnfortunately you have not run into Python Linters yet. These are not annoying rules that critic our code but ways to help us make it more readable. From my experience you have some functions with way too many lines of code. Python likes at least three letter variables and less than 25 lines of code in a function. But that's these days not Py2.7 2010

ReplyDeleteno timezone module :-(

ReplyDeletesame here.

Deletetimezone or timezones? But there is no LocalTimezone object in the timezones

You have to provide an object yourself for that.

DeleteIf you use Python 3.7 I have made modified a version that does that:

https://gist.github.com/jacopofar/ca2397944f56412e81a8882e565038af

Hello, there is a mistake in the Eccent formula. In the python code, the coef for J^2 is 0.0001537 (same as Manom).

ReplyDeleteThe correct coef is 0.0000001267 (XLS file from NOAA, column K).

Thanks, this is amazing!

ReplyDeleteI'd like to use it to create a CLI tool, of course mentioning the source of the original code. Is it OK?