~ 大约太阳正午
lw = 88.743 # my longitude
jdate = Date.ordinal_to_jd(Time.now.year, Time.now.yday)
n = (jdate - 2451545 - 0.0009 - lw / 360).round # lw is users longitude west of 0.
j_noon = 2451545 + 0.0009 + lw / 360 + n
puts j_noon
=> 2455616.24740833
作为更新，部分混淆是太阳正午是自公元前 4713 年 1 月 1 日格林威治中午以来所有计算开始的地方。
Date.ordinal_to_jd 的正确使用并没有弥补这一事实。因此，通过像这样添加或减去 12 小时：
jdn = Date.ordinal_to_jd(Time.now.year, Time.now.yday) - 0.5
我们应该得到更少的错误。既然我们的计算是从昨天中午开始的，那么我们使用哪个呢？
该代码来自此页面Sunrise_equation的两个方程式。
我从这里的一位用户那里得到的第一个答案是，我们不了解 0.0009 和 lw / 360 的使用。lw / 360 似乎是本初子午线的小数日。至于 0.0009，它必须是自公元前 4713 年 1 月 1 日格林威治中午以来以秒为单位的少量方差。有关更多信息，请参阅 IAU 标准
根据此页面，我计算为 0.007776 秒。
我有一些来自 Date 类的信息，不包括方法细节。
=begin
--------------------------------------------------------------------- Class: Date
Class representing a date.
See the documentation to the file date.rb for an overview.
Internally, the date is represented as an Astronomical Julian Day Number, ajd.
The Day of Calendar Reform, sg, is also stored, for conversions to other date formats.
(There is also an of field for a time zone offset,
but this is only for the use of the DateTime subclass.)
A new Date object is created using one of the object creation class methods named
after the corresponding date format, and the arguments appropriate to that date
format; for instance, Date::civil()
(aliased to Date::new()) with year, month, and day-of-month, or Date::ordinal() with
year and day-of-year.
All of these object creation class methods also take the Day of Calendar Reform as an
optional argument.
Date objects are immutable once created.
Once a Date has been created, date values can be retrieved for the different date
formats supported using instance methods. For instance, #mon() gives the Civil month,
#cwday() gives the Commercial day of the week, and #yday() gives the Ordinal day of
the year. Date values can be retrieved in any format, regardless of what format was
used to create the Date instance.
The Date class includes the Comparable module, allowing date objects to be compared
and sorted, ranges of dates to be created, and so forth.
---------------------------------------------------------------------------------
Includes:
Comparable(<, <=, ==, >, >=, between?)
Constants:
MONTHNAMES: [nil] + %w(January February March April May June July August
September October November December)
DAYNAMES: %w(Sunday Monday Tuesday Wednesday Thursday Friday Saturday)
ABBR_MONTHNAMES: [nil] + %w(Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec)
ABBR_DAYNAMES: %w(Sun Mon Tue Wed Thu Fri Sat)
ITALY: 2299161
ENGLAND: 2361222
JULIAN: Infinity.new
GREGORIAN: -Infinity.new
Class methods:
_load, _parse, _strptime, ajd_to_amjd, ajd_to_jd, amjd_to_ajd, civil, civil_to_jd,
commercial, commercial_to_jd, day_fraction_to_time, gregorian?, gregorian_leap?, jd,
jd_to_ajd, jd_to_civil, jd_to_commercial, jd_to_ld, jd_to_mjd, jd_to_ordinal,
jd_to_wday, julian?, julian_leap?, ld_to_jd, mjd_to_jd, new, now, ordinal,
ordinal_to_jd, parse, s3e, strptime, time_to_day_fraction, today, valid_civil?,
valid_commercial?, valid_jd?, valid_ordinal?, valid_time?
Instance methods:
+, -, <<, <=>, ===, >>, _dump, ajd, amjd, asctime, civil, commercial, ctime, cwday,
cweek, cwyear, day, day_fraction, downto, england, eql?, gregorian, gregorian?, hash,
hour, inspect, italy, jd, julian, julian?, ld, leap?, mday, min, mjd, mon, month,
new_offset, new_start, next, next_day, offset, ordinal, sec, sec_fraction, start,
step, strftime, succ, time, to_s, to_yaml, upto, wday, weeknum0, weeknum1, wnum0,
wnum1, yday, year, zone
=end
附带说明一下，Ruby 有一种计算朱利安日期的方法真是太好了。我正在研究NOAA的 Javascript 代码。
这是我受到链接启发而写的一堂课。
class JulianDayNumber
def initialize(year = 2000, month = 1, day = 1) #defaults to Jan. 01, 2000
@year = year
@month = month
@day = day
end
def calcJDN
if (@month <= 2) then
@year -= 1
@month += 12
end
varA = (@year/100).floor
varB = 2 - varA + (varA/4).floor
jdn = (365.25*(@year + 4716)).floor \
+ (30.6001*(@month+1)).floor \
+ @day + varB - 1524.5
return jdn
end
end
jd = JulianDayNumber.new(2011, 3, 2)
julianday = jd.calcJDN
puts julianday
=> 2455622.5
现在这让我到了那里，但我仍在研究如何返回一个数字，例如由最上面的方程计算的数字。试试这个，我们可以看到我们在 JDN 中确实得到了 0.5。谁是对的？红宝石还是 NOAA？
NOAA 使用从 jd 中减去的 2000 年 1 月 1 日值 2451545.0 来获得像这样的分数世纪时间
def calcTimeJulianCent(j)
t = (j - 2451545.0)/36525.0
return t
end