Now that we are managing angles in degrees or radians, we have one of the elementary blocks of the library.
One other important notion is time. Time is complicated, we are going to talk about it a lot in this series. But first, we can tackle a time notion in astronomy that is going to follow us all the way.
Julian day number
Most of us now use the Gregorian calendar that was introduced on October 1582 and starts on the 15th of October, 1582. Before this day, the Julian Calendar was used. They are both very similar calendars, but they space leap years differently, which makes the Gregorian calendar more accurate regarding the real motion of the Earth around the Sun.
Other calendars still exist, like the Hebrew calendar, the Hijri calendar. Some existed but are not used anymore, like the French Revolutionary calendar or the Maya calendar.
We didn't talk about leap years, time zones or leap seconds but we can understand from here that time is a complicated, if not ambiguous, notion to grasp. While ambiguity is not something that we want to deal with in astronomy and science in general.
Therefore, the Julian day was introduced by Joseph Scaliger in 1583 to manipulate time as a universal entity. Its definition is the number of days that have elapsed since noon at Greenwich, England, on January 1, 4713 BC. Usually, the Julian day is written as a decimal number, with the decimal part being the fraction of the day that passed. The Julian day is the Julian day number plus the fractional day.
Using the Julian day is very useful in astronomical calculations, especially when comparing two different dates, in order to know the time that separates them.
Converting a date and time to a Julian day
In Astronomical Algorithms, Jean Meeus details a simple formula to calculate the Julian day associated to a date. The following method works for any positive Julian day, which means any date starting from January 1, 4713 BC.
Given a date, sliced in year (
Y), number of the month (
M), day of the month (
D), hour (
h), minute (
m) and second (
M > 2, then
M stays the same.
Y - 1 and
M + 12.
If the date is on the Gregorian calendar, meaning equal of greater than 15th October 1582, then:
A = INT(Y / 100) B = 2 - A + INT(A/4)
INT() being a truncation function so that it represents the whole part of the number.
If the date is on the Julian calendar, then
A is not necessary and
Then, the Julian day number is defined by the following formula:
INT(365.25 (Y + 4716)) + INT(30.6001 (M + 1)) + D + B - 1524.5
The Julian day, which is the Julian day number plus the fraction of the day that passed, is the same formula, replacing
D as such:
D = D + h / 24 + m / 1440 + s / 86400
One notion to keep in mind is that the Julian day dating started at noon, which means a time at noon will give a whole Julian day, while any other time will give a Julian day with decimals.
Also, the dating started at noon at Greenwich, so any time defined on another time zone than UTC will need to be converted into UTC before being converted into a Julian day. But we will talk more about time zones in future posts.
Implementation in Ruby
The Ruby standard library provides two methods to compute the Julian day number and the Julian day.
DateTime#jd returns the corresponding Julian day number of a
date_time. It is the "chronological" Julian day number, which means if suffers from half a day advance from the original Julian day number. Also, it doesn't include the fraction of time passed in the day. To have the original Julian day from
DateTime#jd, it is required to add a few instructions to it:
DateTime.new(2022, 4, 1, 13, 30, 0).jd + 13 / 24.0 + 30 / 1440.0 - 0.5 # => 2459671.0625
DateTime#ajd returns the "astronomical" Julian day, which is the original one. However, it returns it as a fractional number, so it needs to be converted to be displayed in a way that is understandable to the human eye:
ajd = DateTime.new(2022, 4, 1, 13, 30, 0).ajd # => (39354737/16) ajd.to_f # => 2459671.0625
Using the most performant method
Even if it is tempting to use something ready out of the box, what the Ruby standard library provides us is not perfect. It needs complementary work, or conversion, and this may sometimes reflect reduced performance.
I developed a simple benchmark to compare
DateTime#jd with adjustments,
DateTime#ajd with conversion, and the custom algorithm provided by Jean Meeus.
require "benchmark" require "bigdecimal" require "date" year = 1957 month = 10 day = 4 hour = 19 minute = 28 second = 34 iterations = 10_000_000 Benchmark.bm do |benchmark| benchmark.report("DateTime#jd") do iterations.times do DateTime.new(year, month, day, hour, minute, second).jd + hour / 24.0 + minute / 1440.0 + second / 86400.0 - 0.5 end end benchmark.report("DateTime#ajd") do iterations.times do DateTime .new(year, month, day, hour, minute, second) .ajd .to_f end end benchmark.report("Meeus") do iterations.times do if month > 2 m = month y = year else m = month + 12 y = year - 1 end if ( year > 1582 || (year == 1582 && month > 10) || (year == 1582 && month == 10 && day >= 15) ) a = y / 100 b = 2 - a + a / 4 else b = 0 end julian_day_number = (365.25 * (y + 4716)).floor + (30.6001 * (m + 1)).floor + day + b - 1524.5 time_of_day = hour / 24.0 + minute / 1440.0 + second / 86400.0 julian_day_number + time_of_day end end end # user system total real # DateTime#jd 2.766334 0.003965 2.770299 (2.770522) # DateTime#ajd 5.054064 0.000000 5.054064 (5.054338) # Meeus 1.978400 0.000000 1.978400 (1.978430)
I also ran this benchmark on multiple computers and Ruby versions, with very similar results. Thanks to my teammates at Getaround for their help.
The custom algorithm from Jean Meeus is the most performant, it is the one I am going to implement in astronoby.
Converting a Julian day to a date and time
Still from Jean Meeus' book, given
J the Julian day augmented by
Z = INT(J) and
F = DEC(J)
DEC() being a truncation function so that it represents the decimal part of the number.
Z < 2299161, then we define
A as the same value as
Z is greater or equal to
2299161, then we define
A as such:
α = INT((Z - 1867216.25) / 36524.25) A = Z + 1 + α - INT(α / 4)
We can now define
E as such:
B = A + 1524 C = INT((B - 122.1) / 365.25) D = INT(265.25 * C) E = INT((B - D) / 30.6001)
Finally, we have the day of the month, as a decimal number, equal to
B - D - INT(30.6001 * E) + F.
The number of the month (
M) is equal to
E - 1 if
E < 14, and equal to
E - 13 if
E is equal to
The year number is equal to
C - 4716 if
M > 2, and equal to
C - 4715 if
M is equal to
Implementation in Ruby
DateTime provides a class method
jd to instantiate
Datetime from a chronological Julian day. Adding
0.5 makes it astronomical, and will result into the accurate date and time associated.
DateTime.jd(2436116.31 + 0.5) # => #<DateTime: 1957-10-04T19:26:24+00:00 ((2436116j,69984s,4828n),+0s,2299161j)>
Using the most performant method
I developed a second benchmark to compare
DateTime#jd with a custom implementation.
require "benchmark" require "bigdecimal" require "date" j = BigDecimal("2436116.31") iterations = 1_000_000 Benchmark.bm do |benchmark| benchmark.report("DateTime::jd") do iterations.times do DateTime.jd(j + 0.5) end end benchmark.report("Meeus") do iterations.times do julian_day = j + 0.5 z = julian_day.floor f = julian_day.abs - julian_day.floor if z < 2299161 a = z else aa = ((z - 1867216.25) / 36524.25 ).floor a = z + 1 + aa - aa / 4 end b = a + 1524 c = ((b - 122.1) / 365.25).floor d = (365.25 * c).floor e = ((b - d) / 30.6001).floor day = b - d - (30.6001 * e).floor month = e < 14 ? e - 1 : e - 13 year = month > 2 ? c - 4716 : c - 4715 hour = f * 24 minute = (hour - hour.floor) * 60 second = ((minute - minute.floor) * 60).floor DateTime.new( year, month, day, hour.floor, minute.floor, second.floor, ) end end end # user system total real # DateTime::jd 6.562874 0.000000 6.562874 (6.563703) # Meeus 6.316613 0.000000 6.316613 (6.317376)
This time, the custom implementation is only slightly more performant than the standard one. I would normally use the standard implementation for such a small performance difference, however the custom one will let me have access to each part of the date time (year, month, ...) without having to extract it afterwards, and it will probably be useful in the future.
I am really satisfied with this research as one of my goal with this series is to both implement and explain astronomical calculations. By having custom algorithms being as performant or more performant than those implemented in the standard library, it is a logical choice to implement the custom ones and therefore to have an explanatory source code.
However, it takes a lot of time to translate my research into code, and then write about the theory and the Ruby implementation. It already takes me quite some time to understand these notions, try them and add them to astronoby. Duplicating the Ruby code into articles is going to take too much time, and will probably be a bit difficult to read.
In future articles, I will try to find something in the middle, with Ruby code only when it's appropriate and necessary.
-  Gregorian calendar
-  Julian day
-  Astronomical Algorithms by Jean Meeus, Chapter 7, page 59
-  Astronomical Algorithms by Jean Meeus, Chapter 7, page 63
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