Astronomy Answers
Julian Day Number

In astronomical formulas that contain a date, it is not convenient to write that date as a combination of years, months, and days, especially because not all years have the same number of days and not all months have the same number of days. It is much more convenient to measure the date as the number of days since some fixed day. The Julian Day Number (JDN) or Julian Date (JD) and its relatives are much used for this in astronomy. This page explains how you can translate a date from various calendars to the Julian Date, or the other way around.

There are many different algorithms in use for converting between calendar dates and day numbers. The only thing that counts is that they provide the correct answers for all day numbers and calendar dates. If there is only a limited number of input values and possible outcomes (for example for calculating the month number and day-within-the-month number from the day-within-the-year number), then you can usually find many different algorithms that all give just the right answers for those particular values (but perhaps very different answers for all other input values), even if there is no logical connection between that algorithm and the calendar. In such a case there is often also little or no logical connection between the algorithm that converts from date to day number and the algorithm that converts from day number to date, but the algorithms may be shorter then.

For every algorithm, you should know for which input values it was designed to work. Some calendar algorithms only work well with positive numbers and give wrong results for negative years or negative day numbers. The algorithms I give below are designed to work for all years (including negative ones), all months in each year, all days in each month, and for all Julian Day Numbers (including negative ones).

Some of the fomulas express a choice: If a certain condition is met, then a certain action should be taken (for example, apply a correction to something), and if that condition is not met, then a different action or no action at all should be taken. Such alternative paths are easy to use in manual calculations, in applications that process one date at a time, and in applications written in a programming language that requires compilation before the program can be executed (for example, written in C or Fortran) but are not very convenient for use in applications that can handle many dates at once (in an array) and that are written in a programming language that can be executed immediately (such as Basic or Perl or IDL), for which the execution of an if-then statement per date is much slower than the execution of a single fixed calculation for each date in an array.

If the calendrical calculations below here involve such choices, then where possible I give a formula that circumvents if-then constructions and is therefore convenient for application to arrays of dates in direct-execution programming languagues. The accompanying text explains which choice is being worked around.

1. Different Kinds of Day Numbers

For astronomical calculations that depend on time it is useful to have a continuous time scale that uses only a single unit, and not three like most calendars (days, months, years), and that in addition uses a fixed time zone so that there is no confusion about which precise instant of time is meant.

The IAU has adopted the Julian Date (JD) for this purpose. This should not be confused with "a date in the Julian Calendar". There are various other timescales that look like JD. The next table describes a few of them.

"Whole" means that that timescale uses whole numbers only and counts whole days only. "Fractional" means that that timescale uses fractional numbers (with a part after the decimal point) and indicates instants of time. The "Begins" column shows when each next calendar day begins. 12:00 UTC means that a new calendar day begins at 12:00 (noon) universal time, which is not 12:00 noon local time ― except if you happen to be in a time zone equal to UTC, such as in Great Britain in winter. 00:00 LT means that a new calendar day begins at midnight local time. The "From JD"-column shows how the other numbers can be calculated from JD. In that column, "TZ" stands for an adjustment for your local time zone; that adjustment is 0 for UTC.

In practice, "fractional" JDs and CJDs are usually written with at least one digit after the decimal marker (even if that digit is 0), and "whole" JDNs and CJDNs without digits after the decimal marker.

The zero point of JD (i.e., JD 0.0) corresponds to 12:00 UTC on 1 January −4712 in the Julian calendar. The zero point of CJD corresponds to 00:00 (midnight) local time on 1 January −4712. JDN 0 corresponds to the period from 12:00 UTC on 1 January −4712 to 12:00 UTC on 2 January −4712. CJDN 0 corresponds to 1 January −4712 (the whole day, in local time).

To avoid round-off errors it is best to do calendrical calculations with whole numbers only. In what follows we use CJDN for that.

2. Notation

2.1. Rounding

In the formulas given below, many numbers are rounded to a nearby whole number. It is important that those numbers get rounded in the right direction. Whole numbers are already rounded, so those don't change.

The types of rounding that we use here are downward rounding to the nearest whole number (for x this is indicated by ⌊x⌋) or upward rounding to the nearest whole number (⌈x⌉). The standard rounding function on calculators usually rounds to the nearest whole number (our notation for that is [x]) and the standard conversion from a fractional number to a whole number in a computer language is usually by rounding in the direction of zero (our notation for that is trunc(x)). We do not want those kinds of rounding here! The below table shows these three different kinds of rounding for some values.

2.2. Modular Arithmetic

If you divide the whole number y by the whole number x then you get a quotient q and a remainder r. Sometimes we're interested in the quotient, and sometimes in the remainder. We choose to have the remainder never be negative. Then 0 ≤ r < |x| and

(Eq. 1) q = ⌊y/x⌋

(Eq. 2) r = y mod x = y − x⌊y/x⌋

(Eq. 3) y = qx + r

The notation y mod x means: the non-negative remainder from the division of y by x.

We find

(Eq. 4) y/x = ⌊y/x⌋ + ((y/x) mod 1) = ⌊y/x⌋ + (y mod x)/x

and if x = 1 then

(Eq. 5) y = ⌊y⌋ + (y mod 1)

The notation x ≡ y (mod n) means that x and y differ by some multiple of n. One says that x and y are congruent modulo n, and n is called the modulus. The equation x ≡ y (mod n) is called a congruence.

Modular arithmetic is arithmetic in which all calculations are modulo a fixed n. That means that you can subtract arbitrary multiples of n from all terms and factors without changing the congruence. If x ≡ y (mod n) and p is a whole number, then also x + p×n ≡ y (mod n), and p×x ≡ p×y (mod n), and also p×x ≡ p×y (mod p×n).

Note the difference in notation: z = x mod y means that z is equal to the remainder of dividing x by y, and z ≡ x (mod y) means that z is equal to x except for an arbitrary multiple of y. If z = x mod y then also z ≡ x (mod y), but the opposite need not be true,

3. The Gregorian Calendar

3.1. From Gregorian Date to CJDN

One algorithm to calculate a CJDN J from a Gregorian date (calendar year j, calendar month m, calendar day d) is:

(Eq. 6) c₀ = ⌊(m − 3)/12⌋

(Eq. 7) x₄ = j + c₀

(Eq. 8) x₃ = ⌊x₄/100⌋

(Eq. 9) x₂ = x₄ mod 100

(Eq. 10) x₁ = m − 12c₀ − 3

(Eq. 11) J = ⌊146097x₃/4⌋ + ⌊36525x₂/100⌋ + ⌊(153x₁ + 2)/5⌋ + d + 1721119

Here J₁ is the number of days from the beginning of calculation year 0 until the beginning of the current 100-calculation-year-period, J₂ the number of days since the beginning of the current 100-calculation-year-period, and J₃ the number of days minus one from the beginning of the current calculation year until the beginning of the current month. Calculation years run from 1 March to 1 March.

3.2. From CJDN to Gregorian Date

The algorithm to calculate a Gregorian date (calendar year j, calendar month m, calendar day d) from a CJDN J is:

(Eq. 12) k₃ = 4×(J − 1721120) + 3

(Eq. 13) x₃ = ⌊k₃/146097⌋

(Eq. 14) k₂ = 100⌊(k₃ mod 146097)/4⌋ + 99

(Eq. 15) x₂ = ⌊k₂/36525⌋

(Eq. 16) x₁ = ⌊(5⌊(k₂ mod 36525)/100⌋ + 2)/153⌋

(Eq. 17) c₀ = ⌊(x₁ + 2)/12⌋

(Eq. 18) j = 100×x₃ + x₂ + c₀

(Eq. 19) m = x₁ − 12c₀ + 3

(Eq. 20) d = ⌊(k₁ mod 153)/5⌋ + 1

Here x₃ is the number of calculation centuries since that time, x₂ the number of calculation years since that time, and x₁ the number of months since the beginning of the current calculation year.

3.3. From Milanković Date to CJDN

Some Eastern Orthodox churches use a calendar invented by Milutin Milanković, that differs from the Gregorian calendar only in the rules for which century years are leap years. One algorithm to calculate a CJDN J from a Milanković date (calendar year j, calendar month m, calendar day d) is:

(Eq. 21) c₀ = ⌊(m − 3)/12⌋

(Eq. 22) x₄ = j + c₀

(Eq. 23) x₃ = ⌊x₄/100⌋

(Eq. 24) x₂ = x₄ mod 100

(Eq. 25) x₁ = m − 12c₀ − 3

(Eq. 26) J = ⌊(328718x₃ + 6)/9⌋ + ⌊36525x₂/100⌋ + ⌊(153x₁ + 2)/5⌋ + d + 1721119

3.4. From CJDN to Milanković Date

An algorithm to convert CJDN J to year j, month m, day d in the Milanković calendar is:

(Eq. 27) k₃ = 9×(J − 1721120) + 2

(Eq. 28) x₃ = ⌊k₃/328718⌋

(Eq. 29) k₂ = 100⌊(k₃ mod 328718)/9⌋ + 99

(Eq. 30) x₂ = ⌊k₂/36525⌋

(Eq. 31) x₁ = ⌊(5⌊(k₂ mod 36525)/100⌋ + 2)/153⌋

(Eq. 32) c₀ = ⌊(x₁ + 2)/12⌋

(Eq. 33) j = 100x₃ + x₂ + c₀

(Eq. 34) m = x₁ − 12c₀ + 3

(Eq. 35) d = ⌊(k₁ mod 153)/5⌋ + 1

4. The Julian Calendar

4.1. From Julian Date to CJDN

The algorithm to calculate a CJDN J from a Julian date (calendar year j, calendar month m, calendar day d) is:

(Eq. 36) J₀ = 1721117

(Eq. 37) c₀ = ⌊(m − 3)/12⌋

(Eq. 38) J₁ = ⌊1461×(j + c₀)/4⌋

(Eq. 39) J₂ = ⌊(153m − 1836c₀ − 457)/5⌋

(Eq. 40) J = J₁ + J₂ + d + J₀ = ⌊1461×(j + c₀)/4⌋ + ⌊(153m − 1836c₀ − 457)/5⌋ + d + 1721117

Here J₁ is the number of days between the beginning of calculation year 0 and the beginning of the current calculation year, and J₂ the number of days between the beginning of the current calculation year and the beginning of the current month.

4.2. From CJDN to Julian Date

The algorithm to calculate a Julian date (calendar year j, calendar month m, calendar day d) from a CJDN J is:

(Eq. 41) y₂ = J − 1721118

(Eq. 42) k₂ = 4y₂ + 3

(Eq. 43) k₁ = 5⌊(k₂ mod 1461)/4⌋ + 2

(Eq. 44) x₁ = ⌊k₁/153⌋

(Eq. 45) c₀ = ⌊(x₁ + 2)/12⌋

(Eq. 46) j = ⌊k₂/1461⌋ + c₀

(Eq. 47) m = x₁ − 12c₀ + 3

(Eq. 48) d = ⌊(k₁ mod 153)/5⌋ + 1

4.3. From Islamic Date to CJDN

The religious Islamic calendar depends on observations and so cannot be caught in formulas. The administrative calendar has fixed rules so it can be caught in formulas. The difference between the religious and administrative calendars should usually be no more than 1 day.

The CJDN can be derived from the year number j, month number m, and day number d in the most commonly used administrative Islamic calendar (used by al-Fazārī, al-Khwārizmī, al-Battānī, and in the Toledan and Alfonsine Tables) as follows:

(Eq. 49) J = ⌊(10631j − 10617)/30⌋ + ⌊(325m − 320)/11⌋ + d + 1948439

4.4. From CJDN to Islamic Date

The following formulas calculate the year number j, month number m, and day number d in the most commonly used administrative Islamic calendar from the CJDN J:

(Eq. 50) k₂ = 30×(J − 1948440) + 15

(Eq. 51) k₁ = 11⌊(k₂ mod 10631)/30⌋ + 5

(Eq. 52) j = ⌊k₂/10631⌋ + 1

(Eq. 53) m = ⌊k₁/325⌋ + 1

(Eq. 54) d = ⌊(k₁ mod 325)/11⌋ + 1

4.5. From Babylonian Date to CJDN

The Babylonians had a lunisolar calendar in which the beginning of each month was determined from observations of the Moon, but the number of months in each year followed a fixed 19-year pattern. The below formulas for an administrative Babylonian calendar uses that same 19-year pattern, and should usually deviate by at most one day from the calendar that the Babylonians used that depended on observations.

If j is the year number in the era of Seleukos, and m is the month number (beginning at 1) in the current year, and d is the day number (beginning at 1) in the current month, then the CJDN J can be found as follows:

(Eq. 55) y₁ = ⌊(235j − 241)/19⌋ + m

(Eq. 56) J = ⌊6940y₁/235⌋ + d + 1607557

4.6. From CJDN to Babylonian Date

From CJDN J, we find the Babylonian year number j, month number m, and day number d as follows:

(Eq. 57) k₂ = 235×(J − 1607558) + 234

(Eq. 58) k₁ = 19⌊k₂/6940⌋ + 5

(Eq. 59) j = ⌊k₁/235⌋ + 1

(Eq. 60) m = ⌊(k₁ mod 235)/19⌋ + 1

(Eq. 61) d = ⌊(k₂ mod 6940)/235⌋ + 1

4.7. From Jewish Date to CJDN

The Jewish calendar is a lunisolar calendar with complicated rules, and repeats itself only after 251,827,457 days = 35,975,351 weeks = 8,527,680 months = 689,472 years. The algorithm to transform a Jewish date into CJDN is long, and it features very large intermediate results. The algorithm that I provide here uses only whole numbers (to avoid round-off error), and assumes that all input values, intermediate results, and output values cannot be greater than 2³¹ = 2,147,483,648.

In the Jewish calendar, a year has 12 or 13 months, and a month has 29 or 30 days. New Year is the first day of the 7th month, Tishri. The months in a year of 12 months are: 1 = Nisan (נִיסָן‎), 2 = Iyar (אִיָּר / אייר‎), 3 = Sivan (סִיוָן / סיוון‎), 4 = Tammuz (תַּמּוּז‎), 5 = Av (אָב‎), 6 = Elul (אֱלוּל‎), 7 = Tishri (תִּשׁרִי‎), 8 = Ḥeshvan (מַרְחֶשְׁוָן / מרחשוון‎), 9 = Kislev (כִּסְלֵו / כסליו‎), 10 = Tevet (טֵבֵת‎), 11 = Shevat (שְׁבָט‎), 12 = Adar (אֲדָר‎). In a year of 13 months, embolismic month Adar Ⅰ (‎אֲדָר א׳) is inserted between Shevat and regular Adar, and that regular Adar is then temporarily renamed to Adar Ⅱ (אֲדָר ב׳‎). In a year of 13 months, we count Adar Ⅰ as month number 12, and Adar Ⅱ as month number 13.

One algorithm to transform calendar year j, calendar month m (from 1 through 12 or 13), and calendar day d (from 1 through 29 or 30) into CJDN is then:

(Eq. 62) c₀ = ⌊(13 − m)/7⌋

(Eq. 63) x₁ = j − 1 + c₀

(Eq. 64) x₃ = m − 1

(Eq. 65) z₄ = d − 1

(Eq. 66) c₁(x₁) = ⌊(235x₁ + 1)/19⌋

(Eq. 67) q(x₁) = ⌊c₁(x₁)/1095⌋

(Eq. 68) r(x₁) = c₁(x₁) mod 1095

(Eq. 69) υ₁(x₁) = 32336q(x₁) + ⌊(15q(x₁) + 765433r(x₁) + 12084)/25920⌋

(Eq. 70) υ₂(x₁) = υ₁(x₁) + (⌊6×(υ₁(x₁) mod 7)/7⌋ mod 2)

Calculate in the same way also υ₂(x₁ + 1). Then

(Eq. 71) L₂(x₁) = υ₂(x₁ + 1) − υ₂(x₁)

(Eq. 72) L₂(x₁ − 1) = υ₂(x₁) − υ₂(x₁ − 1)

(Eq. 73) v₃ = 2(⌊(L₂(x₁) + 19)/15⌋ mod 2)

(Eq. 74) v₄ = ⌊(L₂(x₁ − 1) + 7)/15⌋ mod 2

(Eq. 75) c₂(x₁) = υ₂ + v₃ + v₄

Calculate in the same way also c₂(x₁ + 1) (which means you need to calculate υ₂(x₁ + 2), too). Then

(Eq. 76) L = c₂(x₁ + 1) − c₂(x₁)

(Eq. 77) c₈ = ⌊(L + 7)/2⌋ mod 15

(Eq. 78) c₉ = −(⌊(385 − L)/2⌋ mod 15)

(Eq. 79) c₃ = ⌊(384×x₃ + 7)/13⌋ + c₈×⌊(x₃ + 4)/12⌋ + c₉×⌊(x₃ + 3)/12⌋

(Eq. 80) c₄(x₁,x₃) = c₂(x₁) + c₃(x₃)

(Eq. 81) J = J₀ − 177 + c₄(x₁,x₃) + z₄ = 347821 + c₂(x₁) + c₃ + z₄

4.8. From CJDN to Jewish Calendar

From CJDN J we go to Jewish year j, month m, and day d:

(Eq. 82) y₄ = J − 347821

(Eq. 83) q = ⌊y₄/1447⌋

(Eq. 84) r = y₄ mod 1447

(Eq. 85) y₁′ = 49q + ⌊(23q + 25920r + 13835)/765433⌋

(Eq. 86) γ₁ = y₁′ + 1

(Eq. 87) ξ₁ = ⌊(19γ₁ + 17)/235⌋

(Eq. 88) μ₁ = γ₁ − ⌊(235ξ₁ + 1)/19⌋

Calculate c₄′ = c₄(ξ₁, μ₁) in the way described in the previous section. Then

(Eq. 89) ζ₁ = y₄ − c₄′

(Eq. 90) γ₂ = γ₁ + ⌊ζ₁/33⌋

(Eq. 91) ξ₂ = ⌊(19γ₂ + 17)/235⌋

(Eq. 92) μ₂ = γ₂ − ⌊(235ξ₂ + 1)/19⌋

Calculate c₄′ = c₄(ξ₂, μ₂) in the way described in the previous section. Then

(Eq. 93) ζ₂ = y₄ − c₄′

(Eq. 94) γ₃ = γ₂ + ⌊ζ₁/33⌋

(Eq. 95) x₁ = ξ₃ = ⌊(19γ₃ + 17)/235⌋

(Eq. 96) x₃ = μ₃ = γ₃ − ⌊(235ξ₃ + 1)/19⌋

Calculate c₄′ = c₄(ξ₃, μ₃) in the way described in the previous section. Then

(Eq. 97) z₄ = ζ₃ = y₁ − c₄′

(Eq. 98) c = ⌊(12 − x₃)/7⌋

(Eq. 99) j = x₁ + 1 − c

(Eq. 100) m = x₃ + 1

(Eq. 101) d = z₄ + 1

5. Derivation of the General Algorithms

We can derive many calendar formulas by finding straight lines that, with rounding, yield the correct results.

5.1. Large Intermediate Results

Many of the calendar calculations that are described below are of the type

(Eq. 102) y = ⌊(fx + t)/d⌋

(Eq. 103) e = (fx + t) mod d

where f, t, d are fixed whole numbers with 1 < f, d and |t| < d, and y, x, e are variable but whole numbers. The intermediate result fx + t is then usually much greater in magnitude than x and than the end result. This can give trouble on calculators and in computer programs where whole numbers cannot exceed a certain size that we'll refer to as w. (Floating-point numbers like 1.234 × 1020 can be much greater, but have limited accuracy, and then you have to worry about round-off errors.)

Intermediate results greater than w give trouble, so to avoid trouble we'd need roughly |fx| < w, so |x| < w/f, which is much less than w itself.

If |t| ≥ d, then we can rewrite equation 102ff to

(Eq. 104) y = ⌊(fx + ⌊t/d⌋d + (t mod d))/d⌋ = ⌊t/d⌋ + ⌊(fx + (t mod d))/d⌋

(Eq. 105) e = (fx + t) mod d = (fx + (t mod d)) mod d

where the second term of both equations has the same form as equation 102, if you rename t mod d to a new t. Below, we assume that that has been done, which means that |t| < d.

If f > d, then we can rewrite equation 102ff to

(Eq. 106) y = ⌊((⌊f/d⌋d + (f mod d))x + t)/d⌋ = ⌊f/d⌋x + ⌊((f mod d)x + t)/d⌋

(Eq. 107) e = ((⌊f/d⌋d + (f mod d))x + t) mod d = ((f mod d)x + t) mod d

Then |x| may be nearly as great as w/(f mod d) before we run into trouble.

Equations 106ff again have the form of equations 102ff, if you rename f mod d to a new f. Below, we assume that this has been done, so that f < d.

If we could first do the division by d and then the multiplication by f, then the intermediate results would stay smaller than x itself. We'd like to calculate something like f⌊x/d⌋ which is roughly equal to y, and then apply a correction to get the real y.

We can rewrite equation 102ff to

(Eq. 108) y = ⌊(fx + t)/d⌋ = ⌊(f(⌊x/d⌋d + (x mod d)) + t)/d⌋ = f⌊x/d⌋ + ⌊(f(x mod d) + t)/d⌋

(Eq. 109) e = (fx + t) mod d = (f(⌊x/d⌋d + (x mod d)) + t) mod d = (f(x mod d) + t) mod d

Now the greatest intermediate result is no longer fx + t, which can get infinitely large, but f×(x mod d) + t, which cannot exceed fd.

Sometimes an intermediate result of (nearly) fd is still too great, because fd can be much greater than w even if d and f are each much less than w. If fd must remain less than w, then d and f cannot both be greater than √w, which is a lot less than w itself. For example, for 32-bit numbers, w = 231 = 2,147,483,648 and √w = 46341, so even if d and f are both just a bit greater than 46431, which isn't really very large, then fd is already too great to fit into a 32-bit number.

Suppose we pick a P and then calculate once

(Eq. 110) Q = ⌊fP/d⌋

(Eq. 111) R = fP mod d

so

(Eq. 112) fP = dQ + R

Then, for each desired x, calculate

(Eq. 113) q = ⌊x/P⌋

(Eq. 114) r = x mod P

so

(Eq. 115) x = qP + r

and then

(Eq. 116) y = ⌊(fx + t)/d⌋ = ⌊(fqP + fr + t)/d⌋ = ⌊(qdQ + qR + fr + t)/d⌋ = qQ + ⌊(qR + fr + t)/d⌋

(Eq. 117) e = (fx + t) mod d = (fqP + fr + t) mod d = (qdQ + qR + fr + t) mod d = (qR + fr + t) mod d

Now the greatest intermediate result is qR + fr + t. We know that r < P, so fr < fP. To stay out of trouble we need |qR + fr + t| ≤ |q|R + fP + |t| < w, so |⌊x/P⌋|R + fP + |t| ≤ |x|R/P + fP + |t| < w so

(Eq. 118) |x| < X ≡ (w − fP − |t|)×P/R

We want to make the range of x as large as possible, so we seek P and R such that X becomes as great as possible, given w, f, t. This means that R should be small, and P must certainly be less than w/f, because otherwise X ≤ 0.

|x| cannot exceed w, so all P, R for which X ≥ w give the same greatest possible range for x. Within that group, we prefer small P, Q, R, because that makes for easier calculations.

The smallest possible positive value of R is equal to the greatest common divisor of d and f, and the corresponding P, Q > 0 can be found using the Extended Algorithm of Euclid ― or by trying all d possibilities for P.

If that P = P₁, Q = Q₁, R = R₁ fit together, then

(Eq. 119) fP₁ − dQ₁ = R₁

Then also P = R₁P₁, Q = R₁Q₁, R₁ belong together.

It is not useful to allow P ≥ d, because otherwise the greatest intermediate result can be as great or greater than fP = fd which can become greater than w, because if it couldn't exceed w then we could have used equation 108 and would not need the current more complicated method at all. We seek P < d.

If we must have kP₁ = P < d, then k < d/P₁ and that can be quite small, and then there are only few corresponding combinations of P, Q, R. Fortunately, we can find additional good combinations.

If P₂ = P + md, then Q₂ = ⌊fP₂/d⌋ = ⌊(fP + fmd)/d⌋ = ⌊fP/d⌋ + mf = Q + mf and R₂ = fP₂ − dQ₂ = fP + fmd − (dQ + dmf) = fP − dQ = R, so if we add or subtract a multiple of d from P, then Q shifts by a corresponding multiple of f, and R remains the same. In particular, we can take

(Eq. 120) P = kP₁ mod d

(Eq. 121) Q = kQ₁ mod f

(Eq. 122) R = kR₁

for k < ⌊d/R₁⌋ because kR < d.

5.2. The Simplest Case

For convenience, we'll work with a calendar that has only "days" and "months", where the months are longer than the days, but the formulas also work for other units of time. Later, we'll discuss calendars with more than two time periods.

In the simplest case, the average length of the month is equal to p days, all months are either ⌊p⌋ or ⌈p⌉ = ⌊p⌋ + 1 days long, and the running day number y depends on the month number x and the day number z within the current month as follows:

(Eq. 123) y = c + z = ⌊px⌋ + z

where x, y, c, z are whole numbers, and 1 ≤ p < ∞. The first month has x = 0 and the first day of the month has z = 0, for easier calculation.

We define

(Eq. 124) ψ = p − ⌊p⌋ = p mod 1

(Eq. 125) p = q + ψ

The fraction of all months that are long (contain q + 1 days) is equal to ψ. Then the length L of month x is equal to

(Eq. 126) L = ⌊p×(x + 1)⌋ − ⌊px⌋ = q + ⌊ψx + ψ⌋ − ⌊ψx⌋

and month x is a long month if

(Eq. 127) G = ⌊ψx + ψ⌋ − ⌊ψx⌋

is equal to 1, and is a short month (with length q days) if G = 0. The number of long months between month 0 and the beginning of month x is equal to

(Eq. 128) N = ⌊ψx⌋

There is another long month whenever ⌊ψx⌋ has grown by one; that is on average after

(Eq. 129) Q = 1/ψ

months, but Q is not always a whole number, so in practice the number of months between two long months is equal to either ⌊Q⌋ or ⌈Q⌉.

Using equation 123, we can calculate running day number y from calendar date x, z, but we also want to go in the other direction. We derive the reverse formula. From equation 123 and

(Eq. 130) px − 1 < ⌊px⌋ ≤ px

follows

(Eq. 131) px + z − 1 < y ≤ px + z

(Eq. 132) px + z < y + 1 ≤ px + 1 + z

(Eq. 133) x + z/p < (y + 1)/p ≤ x + (z + 1)/p

If z = 0 then

(Eq. 134) x < (y + 1)/p ≤ x + (1/p)

x is a whole number, and p is greater than 1 but finite, so

(Eq. 135) x < ⌈(y + 1)/p⌉ ≤ ⌈x + (1/p)⌉ = x + 1

(Eq. 136) x − 1 < x₀ ≝ ⌈(y + 1)/p⌉ − 1 ≤ x

The only value for x₀ that satisfies these inequalities is x₀ = x, so the formulas to calculate x, z from y are

(Eq. 137) x = ⌈(y + 1)/p⌉ − 1

(Eq. 138) z = y − c = y − ⌊px⌋

The first day of month x' (with z = 0) has y = y' ≝ ⌊px'⌋. The day before that one has day number y = y' − 1 = ⌊px'⌋ − 1. Equation 137 then yields

(Eq. 139) x = ⌈⌊px'⌋/p⌉ − 1 = ⌈(px' − δ)/p⌉ − 1 = ⌈x' − δ/p⌉ − 1 = x' − 1

with

(Eq. 140) δ = px' − ⌊px'⌋, 0 ≤ δ < 1

so equation 137 puts the day that precedes day z = 0 of month x = x' in month x = x' − 1 where it belongs. The transition from one month to the next month lies just before z = 0, where it should be. This shows that equation 137 is correct for all days.

Calculating machines often work with a limited number of decimals behind the decimal separator, so you can get into trouble with round-off errors. If (y + 1)/p = 6.99999999998 but because of a very small round-off error your calculating machine thinks that (y + 1)/p = 7.00000000001, then your calculating machine thinks that ⌈(y + 1)/p⌉ − 1 = 7 rather than 6, and then you find the wrong month.

If the average length p is a ratio (of whole numbers), then we can avoid such round-off errors by rewriting the formulas based on ratios. We define

(Eq. 141) p ≝ f/g

(Eq. 142) ψ ≝ h/g

with f, g, h whole numbers (f, g greater than zero, h greater or equal to zero). Then we find

(Eq. 143) y = c + z = ⌊px⌋ + z = ⌊fx/g⌋ + z

(Eq. 144) G = ⌊ψx + ψ⌋ − ⌊ψx⌋ = ⌊h×(x + 1)/g⌋ − ⌊hx/g⌋

(Eq. 145) x = ⌈(y + 1)/p⌉ − 1 = ⌈(y + 1)×g/f⌉ − 1 = ⌈(yg + g − f)/f⌉ = ⌊(yg + g − 1)/f⌋

(Eq. 146) z = y − c = y − ⌊px⌋ = y − ⌊fx/g⌋

because

(Eq. 147) ⌈n/m⌉ = ⌊(n + m − 1)/m⌋

for whole numbers n, m with m not equal to 0. Now you can do all calculations with ratios, which have no numbers after the decimal separator and hence no round-off errors.

We can simplify the calculation of z. Let

(Eq. 148) k = yg + g − 1

and divide k by f, yielding quotient x and remainder r:

(Eq. 149) yg + g − 1 = k = fx + r

(Eq. 150) x = ⌊k/f⌋

(Eq. 151) r = k mod f

Then

(Eq. 152) z = y − ⌊fx/g⌋ = y − ⌊(yg + g − 1 − r)/g⌋ = y − ⌊y + 1 − (1 + r)/g⌋ = y − (y + 1) − ⌊−(1 + r)/g⌋ = −1 − ⌊−(r + 1)/g⌋ = −1 + ⌈(r + 1)/g⌉ = −1 + ⌊(r + 1 + g − 1)/g⌋ = ⌊r/g⌋ = ⌊(k mod f)/g⌋

5.3. With a Shift of the Pattern

We saw above that p = 30.6 yields a pattern of month lengths like in the Gregorian calendar, but the pattern (from formula 123) does not begin at the right month number. We want to shift the pattern so 31-30-31-30-31 begins at x = 0, y = 0 instead of at x = 4, y = 122. Then March is the first month (x = 0) of the calculation year, and March 1st (y = 0) is the first day of the calculation year.

If we shift the pattern by a months (a is a whole number), then we get

(Eq. 153) y = c + z = ⌊p×(x + a)⌋ − ⌊pa⌋ + z = ⌊px + pa⌋ − ⌊pa⌋ + z = ⌊px + pa − ⌊pa⌋⌋ + z = ⌊px + σ⌋ + z

(Eq. 154) σ ≝ pa − ⌊pa⌋ = pa mod 1

(Eq. 155) c = ⌊px + σ⌋

Equation 153 yields y = 0 for x = 0, z = 0, just like we want.

The derivation of the reverse formula (to calculate x, z from y) is analogous to that of equation 137. The results are

(Eq. 156) x = ⌈(y + 1 − σ)/p⌉ − 1

(Eq. 157) z = y − c = y − ⌊px + σ⌋

If p = f/g is a ratio of whole numbers, then we find

(Eq. 158) y = c + z = ⌊(fx + s)/g⌋ + z

(Eq. 159) s ≝ fa − g⌊fa/g⌋ = fa mod g

(Eq. 160) k = gy + g − 1 − s

(Eq. 161) x = ⌊k/f⌋

(Eq. 162) z = y − c = y − ⌊(fx + s)/g⌋ = ⌊(k mod f)/g⌋

5.4. Very Unequal Months

So far we have assumed that a long month was only one day longer than a short month, but that difference can be larger. Suppose that short months are q days long and that long months are q + d days long, with q, d whole numbers greater than 0, and that a fraction ψ (between 0 and 1) of all months is long. Then the average length of a month is equal to

(Eq. 163) p = q + dψ

days. The formula to calculate the running day number y from month number x and day number z in the current month is then

(Eq. 164) y = c + z = qx + d⌊ψ×(x + a)⌋ − d⌊ψa⌋ + z = qx + d⌊ψx + ψa − ⌊ψa⌋⌋ + z = qx + d⌊ψx + σ⌋ + z

(Eq. 165) σ ≝ ψa − ⌊ψa⌋ = ψa mod 1

(Eq. 166) c = qx + d⌊ψx + σ⌋

How do we go in the opposite direction? To figure that out we compare the running day number c that comes from equation 164 for the first day (z = 0) of month x with the running day number (for convenience renamed to cp) that comes from equation 123 for the same x and the same average month length p (but with d = 1 and a = 0).

(Eq. 167) cp ≝ ⌊px⌋ = ⌊qx + dψx⌋ = qx + ⌊dψx⌋

Then we find

(Eq. 168) c − cp = d⌊ψx + σ⌋ − ⌊dψx⌋ = d(ψx + σ − δ) − (dψx − ε) = dσ − dδ + ε

for certain δ, ε satisfying

(Eq. 169) 0 ≤ {δ,ε} < 1

so

(Eq. 170) dσ − d < c − cp < dσ + 1

(Eq. 171) dσ − d + 1 ≤ c − cp ≤ dσ

so the beginning of month x in the target calendar is between dσ + 1 − d and dσ days after the beginning of month x in the simpler calendar. Because 0 ≤ σ < 1, we also have −d < c − cp < d, so the difference between the beginning of a particular month in the target calendar and the same month in the simpler calendar is always less than d days (in either direction).

For that simpler calendar we have formula 156 to calculate the corresponding month number for any of the ⌊p⌋ or ⌈p⌉ days in that month, so if running day number y corresponds to one of the first ⌊p⌋ days of month x in the target calendar, then we'll have for that x

(Eq. 172) xlower ≤ x ≤ xupper

(Eq. 173) xlower ≝ ⌈(y − dσ + 1)/p⌉ − 1

(Eq. 174) xupper ≝ ⌈(y − dσ + d)/p⌉ − 1

In the target calendar, a month can be up to q + d days long, while in the corresponding simpler calendar a month cannot be longer than ⌈p⌉ days, which is less than q + d. A number (less than d) days beyond day ⌊p⌋ of a month can still belong to that same month in the target calendar, but already belong to a later month in the simpler calendar. For such days, formula 174 can yield a month number that is too great, but then formula 172 still holds.

If d ≤ p then the difference between xupper and xlower is at most 1. In that case

(Eq. 175) cupper = qxupper + d⌊ψxupper + σ⌋

(Eq. 176) zupper = y − cupper

(Eq. 177) ζ = ⌊zupper/(q + d)⌋

(Eq. 178) x = xupper + ζ

(Eq. 179) z = y − c = y − qx − d⌊ψx + σ⌋

If p = f/g = q + dh/g is a ratio of whole numbers (f, g, q, d, h are whole numbers), then we find

(Eq. 180) y = c + z = qx + d⌊(hx + s)/g⌋ + z

(Eq. 181) s ≝ ha − g⌊ha/g⌋ = ha mod g

(Eq. 182) xupper = ⌊(gy + dg − ds − 1)/f⌋

(Eq. 183) cupper = qxupper + d⌊(hxupper + s)/g⌋

(Eq. 184) zupper = y − cupper

(Eq. 185) ζ = ⌊zupper/(q + d)⌋

(Eq. 186) x = xupper + ζ

(Eq. 187) z = y − c = y − qx − d⌊(hx + s)/g⌋

5.5. Many Kinds of Unequal Months

We now allow there to be more than two kinds of months with different month lengths. In equation 164, ⌊ψ×(x + a)⌋ gave the pattern of long months (with a shift as desired). We now allow an arbitrary number of such pattern, each with its own length difference di (which may now also be negative), relative frequency 0 < ψi < 1, and pattern shift ai. Then we find

(Eq. 188) y = c + z = qx + ∑idi⌊ψi×(x + ai)⌋ − ∑idi⌊ψiai⌋ + z = qx + ∑idi⌊ψix + ψiai − ⌊ψiai⌋⌋ + z = qx + ∑idi⌊ψix + σi⌋ + z

(Eq. 189) c = qx + ∑idi⌊ψix + σi⌋

(Eq. 190) σi ≝ ψiai − ⌊ψiai⌋ = ψiai mod 1

The derivation of the formulas for the opposite direction goes analogous to that of equation 174, but now we have to take into account possibly negative di. We find

(Eq. 191) p = q + ∑idiψi

(Eq. 192) cp = ⌊px⌋ = qx + ⌊∑idiψi⌋

(Eq. 193) c − cp = ∑idi⌊ψix + σi⌋ − ⌊∑idiψix⌋ = ∑idi×(ψix + σi − δi) − (∑idiψix − ε) = ∑idiσi − ∑idiδi + ε

for

(Eq. 194) 0 ≤ δi, ε < 1

If there is at least one di > 0, then

(Eq. 195) c − cp > ∑idiσi − ∑(>0)di

(where ∑(>0)di is the sum of all di that are greater than 0) and otherwise

(Eq. 196) c − cp ≥ ∑idiσi

Also

(Eq. 197) c − cp < ∑idiσi − ∑(<0)di + 1

Together this yields

(Eq. 198) ∑idiσi − ∑(>0)di + 1 ≤ y − yp ≤ ∑idiσi − ∑(<0)di

if there are positive di, and otherwise

(Eq. 199) ∑idiσi ≤ c − cp ≤ ∑idiσi − ∑idi

We can combine this to

(Eq. 200) ∑idiσi − ∑(>0)di ≤ c − cp ≤ ∑idiσi − ∑(<0)di

regardless of whether there are positive di. With that,

(Eq. 201) c − ∑idiσi + ∑(<0)di ≤ cp ≤ c − ∑idiσi + ∑(>0)di

and so

(Eq. 202) xlower = ⌈(y − ∑idiσi + ∑(<0)di)/p⌉ − 1

(Eq. 203) xupper = ⌈(y − ∑idiσi + ∑(>0)di)/p⌉ − 1

If ∑(>0)di − ∑(<0)di < p then xlower and xupper differ by at most 1. Then

(Eq. 204) cupper = qxupper + ∑idi⌊ψixupper + σi⌋

(Eq. 205) zupper = y − cupper

(Eq. 206) ζ = ⌊zupper/(q + ∑(>0)di)⌋

(Eq. 207) x = xupper + ζ

(Eq. 208) z = y − c − y − qx − ∑idi⌊ψix + σi⌋

If all ψi = hi/g are ratios of whole numbers with a common denominator g, then we find

(Eq. 209) f ≝ qg + ∑idihi

(Eq. 210) p = f/g

(Eq. 211) y = c + z = qx + ∑idi⌊(hix + si)/g⌋ + z

(Eq. 212) si ≝ hiai − g⌊hiai/g⌋ = hiai mod g

(Eq. 213) xupper = ⌊(gy + g∑(>0)di − ∑idisi − 1)/f⌋

(Eq. 214) cupper = qxupper + ∑idi⌊(hixupper + si)/g⌋

(Eq. 215) zupper = y − cupper

(Eq. 216) ζ = ⌊zupper/(q + ∑(>0)di)⌋

(Eq. 217) x = xupper + ζ

(Eq. 218) z = y − c = y − qx − ∑idi⌊(hix + si)/g⌋

g is the least common multiple of the denominators of all ψi, or is a multiple of that. You can take the product of the denominators of all ψi for g, but sometimes a smaller value can be used.

5.6. Very Unequal Month Lengths

The preceding methods for calculating the calendar date from the running day number for calendars with very unequal month lengths assumed that the variation in month lengths is less than the average month length: d ≤ p when there are two different month lengths, or ∑(>0)di − ∑(<0)di < p if there are more month lengths. What to do if that condition is not met?

In that case you'll have to find the correct month by searching for it. First try month number x = xupper and calculate the corresponding z = y − c and ζ. Is that ζ < 0? Then add ζ to x and try again, until ζ = 0: then you have found the correct x.

If you want to, then you can use this procedure also if d ≤ p.

The greatest number of months that you may have to try before you find the right one is equal to the greatest value that xupper − xlower can attain, plus 1. For calendars with two different month lengths we have

(Eq. 219) xupper − xlower = ⌈(y − dσ + d)/p⌉ − ⌈(y − dσ + 1)/p⌉

The greatest value that this can attain is ⌈(d − 1)/p⌉, so the number N of month that you have to try at most is equal to

(Eq. 220) N = ⌈(d − 1)/p⌉ + 1

For calendars with more than two different month lengths we have

(Eq. 221) xupper − xlower = ⌈(y − ∑idiσi + ∑(>0)di)/p⌉ − ⌈(y − ∑idiσi + ∑(<0)di)/p⌉

which has for its greatest possible value ⌈(∑(>0)di − ∑(<0)di)/p⌉ so for such a calendar

(Eq. 222) N = ⌈(∑(>0)di − ∑(<0)di)/p⌉ + 1

5.7. Month Lengths Without Internal Patterns

Previously we looked at calendars of which the month lengths can be expressed as the sum of various patterns, for example with an extra day every second month and two additional days every third month, so that each sixth month gets three extra days in total. But what if the months repeat after some time but show no other discernable pattern?

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The calendar has f days in g months, for an average month length of p = f/g days. The relationship between the running day number y and the running month number x is a step function: after a (varying) number of days, the month number increases by one. Such a step of 1 that occurs just before day a and repeats itself every f days can be obtained through the formula ⌊(y + f − a)/f⌋ = ⌊(y − a)/f⌋ + 1. The + 1 ensures that the value for y = 0 is 0: that makes for easier calculations.

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When there are g months, each with its own ai (for i from 1 through g), then their combined effect is

(Eq. 223) x = g + ∑i⌊(y − ai)/f⌋

If the successive month lengths are mi, then the first step is at a₁ = 0, the second one is at a₂ = a₁ + m₁ = m₁, the third one is at a₃ = a₂ + m₂ = m₁ + m₂, and in general

(Eq. 224) ai = ∑(j<i)mj

Then

(Eq. 225) x = g + ∑i⌊(y − ∑(j<i)mj)/f⌋

If we go in the other direction then we also have a staircase, but now days and months exchange roles as length and height of the steps. With that, the formula for the running day number c of the first day of month x is:

(Eq. 226) c = ∑imi⌊(x + g − i)/g⌋

The formula to go from running month number x and day number z in the current month to running day number y (all beginning at 0) is then (with i running from 1 through g)

(Eq. 227) y = c + z = ∑imi⌊(x + g − i)/g⌋ + z

These formulas are, if you write out the summation, a lot longer than the formulas that we found earlier for calendars with internal patterns, so it is convenient if you recognize such patterns for a calendar, but not every calendar has such patterns.

The simple calendar from section 5.2 has y = ⌊px + σ⌋ + z. To have that calendar begin month x at day y we need y = ⌊px + σ⌋, from which follows y ≤ px + σ < y + 1, hence (for x > 0) (y − σ)/x ≤ p < (y + 1 − σ)/x. We know that 0 ≤ σ < 1, so (y − 1)/x < (y − σ)/x ≤ p < (y + 1 − σ)/x < (y + 1)/x, so

(Eq. 228) (y − 1)/x < p < (y + 1)/x

Not every p that meets these restrictions yields a simple calendar, but if a p does not meet these restrictions for at least one of its months, then there is certainly no simple calendar for these month lengths.

5.8. Combinations of straight lines

It is rare for a calendar to be fully defined by just one period, so usually you have to combine several periods. Let's assume that we have two straight lines for two periods:

(Eq. 229) y₁ = c₁(x₁) + z₁

(Eq. 230) y₂ = c₂(x₂) + z₂

and in the other direction

(Eq. 231) x₁ = c₁'(y₁)

(Eq. 232) z₁ = y₁ − c₁(x₁)

(Eq. 233) x₂ = c₂'(y₂)

(Eq. 234) z₂ = y₂ − c₂(x₂)

where c'(y) is the inverse of c(x): c'(c(x)) = x and c(c'(y)) = y.

We assume that the first line is for the smaller period and the second line is for the larger period. There are two ways to combine these: We can equate y₁ to z₂ or to x₂. In the first case, the two periods have the same leap unit, for example days. In the second case, the two periods have different leap units (for example, sometimes an extra day for the first period, and sometimes an extra month for the second period). If the larger period needs no leap rules at all, then you can choose which combination method to use.

Let's call the first case the "flat" combination, because the leap units remain the same. Let's call the second case the "stepped" combination, because the leap unit goes a step higher.

The combination of months and years in most (perhaps all) solar calendars (such as the Gregorian calendar, the Julian calendar, and the Egyptian calendar) is flat, i.e. of the first kind. A month can be a day shorter or longer than another month (and exactly one month can be a lot shorter), and a year can be a day shorter or longer than another year. The number of days in a year does not depend on the months, because the rules to calculate the length of the year depend on the year number but not on the month number.

The combination of months and years in a lunisolar calendar (such as the Hebrew and Babylonian calendars) can be flat with very unequal year lengths (d > 1) or stepped (then usually d = 1). A month can be a day longer or shorter than another month, and a year can be a month shorter or longer than another month. (For the flat combination, one month of the year can be much shorter.) In this way you can follow two separate (astronomical) cycles: the motion of the Sun (with the year) and the motion of the Moon (with the month). If the combination is stepped, then the length of the year depends on the length of the months, because the year is then defined in terms of a fixed number of months, not a fixed number of days.

Lunar calendars that are not lunisolar (such as the administrative Islamic calendar) usually do not have any leap rules, so then both methods can be used.

If a calendar has more than two important large periods with leap rules (for example, not just for the month and the year, but also for the century), then it is possible that some combinations are flat and others are stepped.

If we have to deal with more than one period, then it is important for translating a running day number into a calendar date to have the running numbers begin at 0 for each of those periods.

5.8.1. Flat combination

In this case z₂ = y₁, so

(Eq. 235) y₂ = c₂(x₂) + c₁(x₁) + z₁

and in the other direction

(Eq. 236) x₂ = c₂'(y₂)

(Eq. 237) y₁ = z₂ = y₂ − c₂(x₂)

(Eq. 238) x₁ = c₁'(y₁)

(Eq. 239) z₁ = y₁ − c₁(x₁)

Here (for example) y₂ is the running day number, x₂ is the year number, y₁ = z₂ is the day number in the current year, x₁ is the month number in the current year (with 0 for the first month), and z₁ is the day number in the current month (with 0 for the first day).

To calculate the calendar date from the running day number, we must first calculate the larger period (x₂, z₂), and then the smaller period (x₁, z₁).