我尝试使用下面提供的链接,使用python计算日落/上升时间。
通过Excel和Python完成的结果与实际值不匹配。对我可能做错的事情有什么看法吗?
我的Excel工作表可以在下找到。http://transpotools.com/sun_time.xls

# 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()

最佳答案

为什么所有的呼叫都指向radiansdegrees?我以为输入的数据已经是十进制的。
如果我:
去掉所有对弧度和度数的调用
将lat/long分别更正为:45.9-66.6
以UTC为单位的正确时间为0到24。
编辑:
正如J.F.Sebastian指出的,根据问题中链接的电子表格,该位置的日出时间答案和使用ephem的观察者类提供的答案在07:01-07:02区域。
当我在正确的大概范围内得到一个数字(在实施中的注释中的07:34)后,我就不再寻找Dassouki执行美国海军天文台算法的错误。
从中可以看出,该算法对“日出”的构成做了一些简化,并对其中的一些内容进行了讨论。然而,根据我最近对这个问题的了解,我认为这些变化只会导致日出时间的几分钟差异,而不是超过半小时。

10-02 11:09