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.
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5 回答
14
使用ephem'ssidereal_time()方法：
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
注意：ephem.hours是一个浮点数，表示以弧度表示的角度，并转换为/从字符串转换为“hh:mm:ss.ff”。
为了比较，这里是“UTC + 经度”公式：
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)
例子
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')))
输出
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
于 2012-11-16T22:41:35.387 回答
8
或者，如果您想更短，您可以使用NOAA 的低精度方程：
#!/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()
于 2012-11-16T21:17:03.633 回答
4
我看了一下 Jean Meeus 的天文算法。我想你可能会问当地的小时角，它可以用时间（0-24hr）、度（0-360）或弧度（0-2pi）来表示。
我猜你可以用ehem做到这一点。但只是为了它，这里有一些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()
于 2012-11-15T04:44:36.380 回答
3
对于一个非常简单的函数和非常近似的本地时间：时间变化从 -12h 到 +12h，经度从 -180 到 180。然后：
import datetime as dt
def localTimeApprox(myDateTime, longitude):
"""Returns local hour approximation"""
return myDateTime+dt.timedelta(hours=(longitude*12/180))
示例调用：localTimeApprox(dt.datetime(2014, 7, 9, 20, 00, 00), -75)
回报：datetime.datetime(2014, 7, 9, 15, 0)
于 2014-07-09T23:02:23.123 回答
2
再试一次，用ehem。我将纬度和海拔高度作为参数，但当然不需要它们。您可以0出于您的目的调用它们。
#!/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()
于 2012-11-16T20:29:22.477 回答