Wolfram language 如何计算'；适合观看夜景'；当它是'；足够暗'；对于天文学，使用Wolfram语言？

我正试图用Wolfram语言（我对Mathematica没有经验）

我注意到“它”有“天文学知识”。因此，我想知道我是否可以计算如下：

tell me the dates and start/finish times of 'dark, socially friendly viewing opportunities during school holidays' where
* 'socially friendly' means between after 10am and before 10:30pm in local time at my vantage point
* it is 'dark' for at least an hour
* where 'dark' means:
  * it's between astronomical sunset and astronomical dawn
  * the moon is either not risen, or less than 15% of its face is lit
* 'school holidays' means school holidays in NSW, Australia, in 2019 (which Wolfram Alpha doesn't 'implicitly know')

put such dates and times as events into my Google Calendar somehow :-)

解决这个玩具问题将帮助我理解在这个平台上解决问题的“方法”，而且它似乎迎合Wolfram系统的优势。另外，这样每年我就可以省下一两个小时查找资料：-）

---

在Wolfram语言的“程序”（等式）中，我将如何处理这个问题？我可以在Wolfram Alpha中执行此操作，还是需要获取引擎？

不是您问题的完整答案。这只是一个起点

tz = Entity["TimeZone", "Australia/Sydney"];
date = DateObject[{2019, 07, 14}, TimeZone -> tz];

Sunrise[Entity["City", {"Sydney", "NewSouthWales", "Australia"}], date, TimeZone -> tz]

类似地，对于日落

sunset = Sunset[
  Entity["City", {"Sydney", "NewSouthWales", "Australia"}], 
  DateObject[{2019, 07, 14}, TimeZone -> tz], TimeZone -> tz]

你可以从结果中加/减2或3个小时，得到“天文”值

月相

MoonPhase[date]
(* 0.8913 *)

MoonPhase[date, "Name"]["Name"]
(* waxing gibbous moon *)

日落时月亮的位置

MoonPosition[Entity["City", {"Sydney", "NewSouthWales", "Australia"}], sunset]
(* {98.67\[Degree],25.03\[Degree]} *)

---

请看视频，了解WL中天文和空间相关实体数据的概述。

这不是对您问题的完整答案。这只是一个起点

tz = Entity["TimeZone", "Australia/Sydney"];
date = DateObject[{2019, 07, 14}, TimeZone -> tz];

Sunrise[Entity["City", {"Sydney", "NewSouthWales", "Australia"}], date, TimeZone -> tz]

类似地，对于日落

sunset = Sunset[
  Entity["City", {"Sydney", "NewSouthWales", "Australia"}], 
  DateObject[{2019, 07, 14}, TimeZone -> tz], TimeZone -> tz]

你可以从结果中加/减2或3个小时，得到“天文”值

月相

MoonPhase[date]
(* 0.8913 *)

MoonPhase[date, "Name"]["Name"]
(* waxing gibbous moon *)

日落时月亮的位置

MoonPosition[Entity["City", {"Sydney", "NewSouthWales", "Australia"}], sunset]
(* {98.67\[Degree],25.03\[Degree]} *)

请看视频，了解WL中天文和空间相关实体数据的概述