日出/设定计算

Question

我正在尝试使用python根据下面提供的链接计算日落/上升时间.

我通过excel和python完成的结果与实际值不匹配.关于我可能做错的任何想法？

我的Excel表格可以在http://transpotools.com/sun_time.xls下找到

# Created on 2010-03-28

# @author: dassouki
# @source: [http://williams.best.vwh.net/sunrise_sunset_algorithm.htm][2]
# @summary: this is based on the Nautical Almanac Office, United States Naval
# Observatory.

import math, sys

class TimeOfDay(object):

  def calculate_time(self, in_day, in_month, in_year,
                     lat, long, is_rise, utc_time_zone):

    # is_rise is a bool when it's true it indicates rise,
    # and if it's false it indicates setting time

    #set Zenith
    zenith = 96
    # offical      = 90 degrees 50'
    # civil        = 96 degrees
    # nautical     = 102 degrees
    # astronomical = 108 degrees


    #1- calculate the day of year
    n1 = math.floor( 275 * in_month / 9 )
    n2 = math.floor( ( in_month + 9 ) / 12 )
    n3 = ( 1 + math.floor( in_year - 4 * math.floor( in_year / 4 ) + 2 ) / 3 )

    new_day = n1 - ( n2 * n3 ) + in_day - 30

    print "new_day ", new_day

    #2- calculate rising / setting time
    if is_rise:
      rise_or_set_time = new_day + ( ( 6 - ( long / 15 ) ) / 24 )
    else:
      rise_or_set_time = new_day + ( ( 18 - ( long/ 15 ) ) / 24 )

    print "rise / set", rise_or_set_time

    #3- calculate sun mean anamoly
    sun_mean_anomaly = ( 0.9856 * rise_or_set_time ) - 3.289
    print "sun mean anomaly", sun_mean_anomaly

    #4 calculate true longitude
    true_long = ( sun_mean_anomaly +
                  ( 1.916 * math.sin( math.radians( sun_mean_anomaly ) ) ) +
                  ( 0.020 * math.sin(  2 * math.radians( sun_mean_anomaly ) ) ) +
                  282.634 )
    print "true long ", true_long

    # make sure true_long is within 0, 360
    if true_long < 0:
      true_long = true_long + 360
    elif true_long > 360:
      true_long = true_long - 360
    else:
      true_long

    print "true long (360 if) ", true_long

    #5 calculate s_r_a (sun_right_ascenstion)
    s_r_a = math.degrees( math.atan( 0.91764 * math.tan( math.radians( true_long ) ) ) )
    print "s_r_a is ", s_r_a

    #make sure it's between 0 and 360
    if s_r_a < 0:
      s_r_a = s_r_a + 360
    elif true_long > 360:
      s_r_a = s_r_a - 360
    else:
      s_r_a

    print "s_r_a (modified) is ", s_r_a

    # s_r_a has to be in the same Quadrant as true_long
    true_long_quad = ( math.floor( true_long / 90 ) ) * 90
    s_r_a_quad = ( math.floor( s_r_a / 90 ) ) * 90
    s_r_a = s_r_a + ( true_long_quad - s_r_a_quad )

    print "s_r_a (quadrant) is ", s_r_a

    # convert s_r_a to hours
    s_r_a = s_r_a / 15

    print "s_r_a (to hours) is ", s_r_a


    #6- calculate sun diclanation in terms of cos and sin
    sin_declanation = 0.39782 * math.sin( math.radians ( true_long ) )
    cos_declanation = math.cos( math.asin( sin_declanation ) )

    print " sin/cos declanations ", sin_declanation, ", ", cos_declanation

    # sun local hour
    cos_hour = ( math.cos( math.radians( zenith ) ) -
                 ( sin_declanation * math.sin( math.radians ( lat ) ) ) /
                 ( cos_declanation * math.cos( math.radians ( lat ) ) ) )

    print "cos_hour ", cos_hour

    # extreme north / south
    if cos_hour > 1:
      print "Sun Never Rises at this location on this date, exiting"
      # sys.exit()
    elif cos_hour < -1:
      print "Sun Never Sets at this location on this date, exiting"
      # sys.exit()

    print "cos_hour (2)", cos_hour


    #7- sun/set local time calculations
    if is_rise:
      sun_local_hour =  ( 360 - math.degrees(math.acos( cos_hour ) ) ) / 15
    else:
      sun_local_hour = math.degrees( math.acos( cos_hour ) ) / 15


    print "sun local hour ", sun_local_hour

    sun_event_time = sun_local_hour + s_r_a - ( 0.06571 *
                                                rise_or_set_time ) - 6.622

    print "sun event time ", sun_event_time

    #final result
    time_in_utc =  sun_event_time - ( long / 15 ) + utc_time_zone

    return time_in_utc



#test through main
def main():
  print "Time of day App "
  # test: fredericton, NB
  # answer: 7:34 am
  long = 66.6
  lat = -45.9
  utc_time = -4
  d = 3
  m = 3
  y = 2010
  is_rise = True

  tod = TimeOfDay()
  print "TOD is ", tod.calculate_time(d, m, y, lat, long, is_rise, utc_time)


if __name__ == "__main__":
  main()

Answer 1

你可以使用ephempython模块:

#!/usr/bin/env python
import datetime
import ephem # to install, type$ pip install pyephem

def calculate_time(d, m, y, lat, long, is_rise, utc_time):
    o = ephem.Observer()
    o.lat, o.long, o.date = lat, long, datetime.date(y, m, d)
    sun = ephem.Sun(o)
    next_event = o.next_rising if is_rise else o.next_setting
    return ephem.Date(next_event(sun, start=o.date) + utc_time*ephem.hour
                      ).datetime().strftime('%H:%M')

例:

for town, kwarg in { "Fredericton": dict(d=3, m=3, y=2010,
                                         lat='45.959045', long='-66.640509',
                                         is_rise=True,
                                         utc_time=20),

                     "Beijing": dict(d=29, m=3, y=2010,
                                     lat='39:55', long='116:23',
                                     is_rise=True,
                                     utc_time=+8),

                     "Berlin": dict(d=4, m=4, y=2010,
                                    lat='52:30:2', long='13:23:56',
                                    is_rise=False,
                                    utc_time=+2) ,

                     "Moscow": dict(d=4, m=4, y=2010,
                                    lat='55.753975', long='37.625427',
                                    is_rise=True,
                                    utc_time=4) }.items():
    print town, calculate_time(**kwarg)

输出:

Beijing 06:02
Berlin 19:45
Moscow 06:53
Fredericton 07:01

Answer 2

为什么所有的电话都打给radians和degrees？我认为输入数据已经是十进制度了。

如果我满足以下条件，我会在上午 7:37 得到结果：

• 删除所有对弧度和度数的调用
• 将纬度/经度分别修正为：45.9和-66.6
• 将 time_in_utc 更正为 0 到 24 之间。

编辑：
正如 JF Sebastian 指出的那样，根据问题中链接的电子表格以及使用 Observer 类提供的答案，该位置日出时间的答案ephem位于 07:01-07:02 区域。

一旦我得到了正确的大致数字（实现中的评论中的 07:34），我就不再寻找 dassouki 实现美国海军天文台算法的错误。

仔细研究一下，该算法进行了一些简化，并且关于“日出”的构成存在变化，其中一些内容将在此处讨论。然而，根据我最近在这件事上了解到的情况来看，这些变化应该只会导致日出时间的差异几分钟，而不是超过半个小时。