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 or Julian Date (JD) is much used for this in astronomy. This page explains how you can translate a date from the Gregorian calendar 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.

About 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 between 1 and 12, all monthly day numbers between 1 and 31, and for all Julian Day Numbers (including negative ones).

1. Derivation of the Algorithms

1.1. Use the Right Kind of 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.

1.2. Months

All months in the Julian and Gregorian calendars have the same length in each year, except for February: that month has 29 days in a leap year, instead of 28 days. It is easier to calculate if the deviating thing is at the very end of the period, so we work with periods of a year that begin with March 1st and that run until just before the next March 1st. Let's call that the calculation year. If there is a leap day, then it comes at the very end of the calculation year. For convenience we keep calling March month number 3 (of the calculation year), so then January becomes month number 13 and February becomes month number 14. We regard January and February of a particular calendar year as the last two months of the preceding calculation year.

The lengths of the months in the calculation year then are:

1.3. The Method with Decomposition into Periods

We see two successive sets of 31-30-31-30-31 days (March through July, and then August through December) and then part of a third such set (January and February), but that one stops after 28 or 29 days in the second month of the set. Such a period of 31-30-31-30-31 is 153 days long, so we can say that the year is divided into periods of 153 days, and each such a period has the same sequence of month lengths, until the year is full.

Within such a period of 153 days we see three repeats of the sequence 31-30 days (61 days), again with the last repeat truncated (because that one only contains one month). And within such a period of 61 days we see a repeat of periods (months) of 31 days, again with the last period truncated, because that one contains only 30 days.

All in all we can say that the day number n since the beginning of the calculation year (March 1 has n = 1), and the month number m in the calculation year (m = 3 through 12 for March through December, and 13 and 14 for the immediately following January and February) are equal to

(Eq. 1) n − 1 = Q₁ = 153*c₃ + 61*c₂ + 31*c₁ + d − 1

(Eq. 2) m − 3 = Q₂ = 5*c₃ + 2*c₂ + c₁

where all c numbers are whole numbers greater than or equal to 0, and c₁ is less than 2, c₂ is less than 5, and c₃is less than 3.

1.4. Algorithm for Decomposition into Periods

The trick is to find the right c₁, c₂, and c₃ that go with the given Q₁ (if you want to go from day number to date) or Q₂ (if you want to go from date to day number).

In a bit more mathematical language, the algorithm for "decomposing into periods" goes as follows. Suppose that

(Eq. 3) x = p₁*c₁ + p₂*c₂ + … + pn*cn + x₀

and that you know x and all pi and want to calculate ci and x₀. We assume that all pi are greater than 0. If a particular pi is negative, then remove its minus sign. We sort the pi so that p₁ < p₂ < p₃ < … < pn. Then set

(Eq. 4) xn = x

and then for every i from n through 1:

(Eq. 5) ci = ⌊xi/pi⌋

(Eq. 6) x(i−1) = xi − ci*pi = xi mod pi

Warning! Don't remove the − 1 from the left hand side and right hand side of equation 1, otherwise the cs aren't correct anymore in some cases, such as for n = 31. Equation 1 then yields n − 1 = 30 = Q₁ = 153*0 + 61*0 + 31*0 + d − 1 so d = 31 and (from equation 2) m − 3 = 5*0 + 2*0 + 0 = 0 so m = 3, which means March 31st. If the − 1 are removed then equation 1 yields

(Eq. 7) n = 153*c₃ + 61*c₂ + 31*c₁ + d (wrong!)

which in this case yields 31 = 153*0 + 61*0 + 31*1 + 0 and then equation 2 leads to m − 3 = 5*0 + 2*0 + 1 = 1 so m = 4, which means April 0th ― which is incorrect.

1.5. Algorithm with the Straight Line

Another method is to find a straight line that (with rounding) yields the right results. If there are only two different month lengths that differ by only one day (except that the very last month may be shorter), then you can try to find such a line.

First we prepare some formulas. Let

(Eq. 8) y = y₀ + z

(Eq. 9) y₀ = ⌊px⌋

for whole numbers x, y, z, with p ≥ 1 and 0 ≤ z ≤ ⌈p⌉. Then when x increases by one, then y₀ increases by either ⌊p⌋ or ⌈p⌉.

In the other direction we have

(Eq. 10) x = ⌈(y + 1)/p⌉ − 1

(Eq. 11) z = y − ⌊px⌋

where one or more values of y yield the same x. With these formulas, y = 0 corresponds to x = 0, z = 0.

That these pairs of formulas are each other's inverse can be seen as follows. First we rewrite equation 9 to

(Eq. 12) y₀ = px − δ

where δ is the difference that must be subtracted from px to find the nearest whole number, with 0 ≤ δ < 1. If we insert this rewritten formula into 10 with also z = 0 then we find

(Eq. 13) x = ⌈(px − δ + 1)/p⌉ − 1 = ⌈x + (1 − δ)/p⌉ − 1 = x − 1 + ⌈(1 − δ)/p⌉ = x

because 0 ≤ δ < 1 so 0 < 1 − δ ≤ 1 so 0 < (1 − δ)/p ≤ 1/p ≤ 1 so ⌈(1 − δ)/p⌉ = 1.

If y increases or decreases then at a certain point the corresponding x (from equation 8) changes, too. Suppose that a certain x = x' corresponds to a certain y₀ = y₀'. We want the change in x to occur just before y₀', i.e., that if y = y₀' − 1 then x = x' − 1.

If y = y₀' − 1 then we find from equations 10 and 12 that

(Eq. 14) x = ⌈(y₀' − 1 + 1)/p⌉ − 1 = ⌈y₀'/p⌉ − 1 = ⌈(px₀ − δ)/p⌉ − 1 = x₀ − 1 + ⌈−δ/p⌉ = x₀ − 1

because 0 ≤ δ < 1 so −1 < −δ ≤ 0 so ⌈−δ/p⌉ = 0.

We have now shown that the shift to another x occurs just before the y₀ that follows from equation 9.

It is inconvenient that two different kinds of rounding are used in these formulas. Rounding up (⌈⌉) is not often available on electronic calculators. ⌈q⌉ = ⌊q⌋ + 1 except if q is a whole number, because then ⌈q⌉ = ⌊q⌋. So, if we can arrange things such that q cannot be a whole number (for the input values that interest us), then we can replace equation 10 by

(Eq. 15) x = ⌊(y + 1)/p⌋   ((y + 1)/p not a whole number)

If p is a rational number, p = f/g with f, g whole numbers, then (y + 1)/p can be a whole number (if y + 1 is a multiple of f), and (y + 1)/p can differ from a whole number by 1/f or a multiple of that, but cannot differ less than that from a whole number. If we shift (y + 1)/p by an amount greater than 0 but less than 1/f, then it can never be a whole number anymore. We choose a shift of half of 1/f. The shift must be downward. Then equation 10 yields

(Eq. 16) x = ⌈(y + 1)/p − (1/(2f))⌉ − 1 = ⌈(2*g*y + 2*g − 1)/(2*f)⌉ − 1

and equation 15

(Eq. 17) x = ⌊(y + 1)/p − (1/(2*f))⌋ = ⌊(2*g*y + 2*g − 1)/(2*f)⌋

Warning! Equations 16 and 17 are only correct for all y if p is a whole number or ratio (i.e., a rational number). If p is not a whole number or a ratio (i.e., if it is transcendent) then these formulas are only valid for a limited range of y.

Now we are ready to apply these formulas to the calendar.

If m is the month number (m = 3 for the first month of the calculation year) and d is the day number within the month (d = 1 for the first day of the month) and n is the day number within the calculation year (n = 1 for the first day of the calculation year), then

(Eq. 18) n = ⌊p*(m + a)⌋ − ⌊p*(3 + a)⌋ + d

(Eq. 19) m = ⌈(n + ⌊p*(3 + a)⌋)/p⌉ − a − 1

(Eq. 20) d = n − ⌊p*(m + a)⌋ + ⌊p*(3 + a)⌋

These formulas follow from equation 8ff if we set x = m + a, y = n + ⌊p*(3 + a)⌋ − 1, and z = d − 1.

The trick is now to find p and a that yield the correct month lengths. The p determines the succession of month lengths (and the average month length, for an infinite number of months), and a determines which of those months gets to be number 1.

If you allow m to run through all whole numbers, and if p is a rational number, then the sequence of month lengths keeps repeating itself. If p = f/g with whole numbers f, g, then the sequence of month lengths repeats every g months. If p cannot be written as a ratio (i.e., it is a transcendental number), then the sequence of month lengths does not repeat itself in the long run.

For the Julian and Gregorian calendars, the month lengths are 30 or 31 days (except for February, which is shorter), so p must be between 30 and 31. We need two successive months with 31 days (e.g., July and August), so p must be greater than 30.5. It turns out that p = 30.6 = 153⁄5 and a = 1 yield the right sequence of month lengths for m between 3 and 14 (inclusive). This means

(Eq. 21) n = ⌊30.6*(m + 1)⌋ + d − 122 = ⌊153*(m + 1)/5⌋ + d − 122 = ⌊(153*m − 457)/5⌋ + d

(Eq. 22) m = ⌊(n + 122)/30.6001⌋ − 1

(Eq. 23) m = ⌊(10*(n + 121) + 10 − 1)/306⌋ − 1 = ⌊(10*n + 913)/306⌋

(Eq. 24) d = n + 122 − ⌊30.6*(m + 1)⌋ = n + 122 − ⌊153*(m + 1)/5⌋ = n − ⌊(153*m − 457)/5⌋

In equation 22 the divisor is 30.6001 rather than 30.6, to ensure that the number to be rounded is never already a whole number. Other values greater than 30.6 but not greater than 30.615 also yield the right results (for m between 3 and 14). If you used 30.6 in equation 22, then you'd get incorrect results for example when n = 31. In equations 21 and 24 you can also use 30.6001 instead of 30.6.

1.6. Julian Years, with Decomposition into Periods

In the Julian calendar, every three regular years of 365 days are followed by a leap year of 366 days. This period of four years contains 3*365 + 366 = 1461 days. Each period of four years has the same sequence of months with their month lengths. Now we have the opposite problem to the one we faced for the months: During the earlier calculations with months the last period was truncated, but now that we calculate with years the last period (the 4th year) gets longer. If we're not careful, then the extra bit (the leap day) is regarded as a very short extra period, rather than belonging at the end of the previous period.

If we follow the same ideas as for the months, then we'd have for the calculation year number j and day number N since the previous leap day:

(Eq. 25) N − 1 = 1461*c₅ + 365*c₄ + n − 1

(Eq. 26) j = 4*c₅ + c₄

where c₄ is less than 4. But if we apply this to N = 1461 (the leap day at the very end of the period) then we find c₅ = 0, c₄ = 4, n = 1, j = 4 (which corresponds to the first day of year 4) instead of c₅ = 0, c₄ = 3, n = 366, j = 3 (the last day of year 3).

We want to keep the results from equation 25 for the first 1460 days, but to have c₄ for day number 1461 equal to 3 instead of 4. We manage that by applying a correction to c₄ and n. If you calculate c₄ and n first and only then apply the correction, then that correction is

(Eq. 27) n → n + 365*⌊c₄/4⌋

(Eq. 28) c₄ → c₄ − ⌊c₄/4⌋

If you first calculate c₄, then apply the correction to c₄, and only then calculate n (based on the corrected c₄), then you don't have to correct n any further.

These corrections should be applied only when we go from day number to calendar date.

Here are the calculations for March 1th of the Julian calculation years 2007−2012. Warning: Because we regard January and February from the current calendar year as being at the end of the previous calculation year, the leap day from calendar year 2008 falls at the end of calculation year 2007 (= j).

Here are the calculations from day number to calendar date for three days around the leap day 29 February 2008:

1.7. Gregoriaanse jaren, met ontbinding in perioden

De Gregoriaanse kalender is als de Juliaanse, maar heeft twee extra regels voor welke jaren schrikkeljaren zijn, en daarmee twee grotere perioden in de berekeningen. De eerste extra regel betekent dat in elke periode van 100 jaar één schrikkeldag wordt overgeslagen. Zo'n periode van 100 jaar bevat dan 24*1461 + 1460 = 25*1461 − 1 = 36524 dagen. De twee extra regel betekent dat elke 400 jaar toch weer een extra schrikkeldag wordt ingelast. Zo'n periode van 400 jaar bevat dan 3*36524 + 36525 = 146097 dagen.

1.8. Gregorian Years, with Decomposition into Periods

The Gregorian calendar is like the Julian one, but has two extra rules for which years are leap years, which means it has two greater periods in its calculations. The first extra rule means that from every period of 100 years one leap day gets omitted. Such a period of 100 days then contains 24*1461 + 1460 = 25*1461 − 1 = 36524 days. The second extra rule means that every period of 400 years gets one extra leap day again. Such a period of 400 years then contains 3*36524 + 36525 = 146097 days.

All in all we find for the calculation year number j and the day number N of March 0 of the calculation year:

(Eq. 29) N = 146097*c₇ + 36524*c₆ + 1461*c₅ + 365*c₄

(Eq. 30) j = 400*c₇ + 100*c₆ + 4*c₅ + c₄

but when we calculate from day number to calendar date then we have to apply leap day corrections again. In this case we have to apply two corrections, because both the periods of 400 years (relative to that of 100 years) and the period of 4 years (relative to that of 1 year) contain an extra (leap) day. In that case it is better to rewrite the formulas to

(Eq. 31) N = 146097*c₇ + 36524*c₆ + n₁

(Eq. 32) n₁ → n₁ + 36524*⌊c₆/4⌋

(Eq. 33) c₆ → c₆ − ⌊c₆/4⌋

(Eq. 34) n₁ = 1461*c₅ + 365*c₄ + n₂

(Eq. 35) n₂ → n₂ + 365*⌊c₄/4⌋

(Eq. 36) c₄ → c₄ − ⌊c₄/4⌋

(Eq. 37) j = 400*c₇ + 100*c₆ + 4*c₅ + c₄

Here c₄ is eventually less than 4, c₅ less than 25 and c₆ less than 4.

1.9. All Together, with Decomposition into Periods

All together we now find:

(Eq. 38) J − c₀ = 146097*c₇ + 36524*c₆ + 1461*c₅ + 365*c₄ + 153*c₃ + 61*c₂ + 31*c₁ + d − 1

(Eq. 39) m − 3 = 5*c₃ + 2*c₂ + c₁

(Eq. 40) j = 400*c₇ + 100*c₆ + 4*c₅ + c₄

but we have to add in the corrections for leap days, if we are calculating calendar dates from day numbers. In that case it is better to rewrite the previous formulas to

(Eq. 41) J − c₀ = 146097*c₇ + 36524*c₆ + n₁

(Eq. 42) n₁ → n₁ + 36524*⌊c₆/4⌋

(Eq. 43) c₆ → c₆ − ⌊c₆/4⌋

(Eq. 44) n₁ = 1461*c₅ + 365*c₄ + n₂

(Eq. 45) n₂ → n₂ + 365*⌊c₄/4⌋

(Eq. 46) c₄ → c₄ − ⌊c₄/4⌋

(Eq. 47) n₂ = 153*c₃ + 61*c₂ + 31*c₁ + d − 1

(Eq. 48) m − 3 = 5*c₃ + 2*c₂ + c₁

(Eq. 49) j = 400*c₇ + 100*c₆ + 4*c₅ + c₄

Here c₀ is a number that ties the formulas to the beginning of the Julian day count. It is the Julian day number of March 1st of the year 0, in the Julian and Gregorian proleptic calendars.

All cs are whole numbers. Further restrictions are listed in the following table.

2. From Gregorian Date to Julian Day Number

The algorithm to calculate a Julian Day Number J from a Gregorian date (calendar year j, calendar month m (1 ≤ m ≤ 12), calendar day d) is:

1. If m < 3, then replace m by m + 12 and j by j − 1.

2. (Eq. 50) c₇ = ⌊j/400⌋

3. (Eq. 51) x₆ = j − 400*c₇ = j mod 400

4. (Eq. 52) c₆ = ⌊x₆/100⌋

5. (Eq. 53) x₅ = x₆ − 100*c₆ = x₆ mod 100

6. (Eq. 54) c₅ = ⌊x₅/4⌋

7. (Eq. 55) c₄ = x₅ − 4*c₅ = x₅ mod 4

8. (Eq. 56) c₃ = ⌊(m − 3)/5⌋

9. (Eq. 57) x₂ = m − 3 − 5*c₃ = (m − 3) mod 5

10. (Eq. 58) c₂ = ⌊x₂/2⌋

11. (Eq. 59) c₁ = x₂ − 2*c₂ = x₂ mod 2

12. (Eq. 60) J = 146097*c₇ + 36524*c₆ + 1461*c₅ + 365*c₄ + 153*c₃ + 61*c₂ + 31*c₁ + d + 1721118.5

3. From Julian Day Number to Gregorian Date

The algorithm to calculate a Gregorian date (calendar year j, calendar month m (1 ≤ m ≤ 12), calendar day d) from a Julian Day Number J is:

1. (Eq. 61) c₇ = ⌊(J − 1721119.5)/146097⌋

2. (Eq. 62) x₆ = J − 1721119.5 − 146097*c₇ = (J − 1721119.5) mod 146097

3. (Eq. 63) c₆ = ⌊x₆/36524⌋

4. (Eq. 64) c₆ → c₆ − ⌊c₆/4⌋

5. (Eq. 65) x₅ = x₆ − 36524*c₆

6. (Eq. 66) c₅ = ⌊x₅/1461⌋

7. (Eq. 67) x₄ = x₅ − 1461*c₅ = x₅ mod 1461

8. (Eq. 68) c₄ = ⌊x₄/365⌋

9. (Eq. 69) c₄ → c₄ − ⌊c₄/4⌋

10. (Eq. 70) x₃ = x₄ − 365*c₄

11. (Eq. 71) c₃ = ⌊x₃/153⌋

12. (Eq. 72) x₂ = x₃ − 153*c₃ = x₃ mod 153

13. (Eq. 73) c₂ = ⌊x₂/61⌋

14. (Eq. 74) x₁ = x₂ − 61*c₂ = x₂ mod 61

15. (Eq. 75) c₁ = ⌊x₁/31⌋

16. (Eq. 76) x₀ = x₁ − 31*c₁ = x₁ mod 31

17. (Eq. 77) j = 400*c₇ + 100*c₆ + 4*c₅ + c₄

18. (Eq. 78) m = 5*c₃ + 2*c₂ + c₁ + 3

19. (Eq. 79) d = x₀ + 1

20. If m > 12, then replace m by m − 12 and j by j + 1.

4. From Julian Date to Julian Day Number

The algorithm to calculate a Julian Day Number J from a Julian date (calendar year j, calendar month m (1 ≤ m ≤ 12), calendar day d is:

1. If m < 3, then replace m by m + 12 and j by j − 1.

2. (Eq. 80) c₅ = ⌊j/4⌋

3. (Eq. 81) c₄ = j − 4*c₅ = x₅ mod 4

4. (Eq. 82) c₃ = ⌊(m − 3)/5⌋

5. (Eq. 83) x₂ = m − 3 − 5*c₃ = (m − 3) mod 5

6. (Eq. 84) c₂ = ⌊x₂/2⌋

7. (Eq. 85) c₁ = x₂ − 2*c₂ = c₂ mod 2

8. (Eq. 86) J = 1461*c₅ + 365*c₄ + 153*c₃ + 61*c₂ + 31*c₁ + d + 1721116.5

5. From Julian Day Number to Julian Date

The algorithm to convert from a Julian Day Number J to a Julian date (calendar year j, calendar month m (1 ≤ m ≤ 12), calendar day d) is:

1. (Eq. 87) c₅ = ⌊(J − 1721117.5)/1461⌋

2. (Eq. 88) x₄ = J − 1721117.5 − 1461*c₅ = (J − 1721117.5) mod 1461

3. (Eq. 89) c₄ = ⌊x₄/365⌋

4. (Eq. 90) c₄ → c₄ − ⌊c₄/4⌋

5. (Eq. 91) x₃ = x₄ − 365*c₄

6. (Eq. 92) c₃ = ⌊x₃/153⌋

7. (Eq. 93) x₂ = x₃ − 153*c₃ = x₃ mod 153

8. (Eq. 94) c₂ = ⌊x₂/61⌋

9. (Eq. 95) x₁ = x₂ − 61*c₂ = x₂ mod 61

10. (Eq. 96) c₁ = ⌊x₁/31⌋

11. (Eq. 97) x₀ = x₁ − 31*c₁ = x₁ mod 31

12. (Eq. 98) j = 4*c₅ + c₄

13. (Eq. 99) m = 5*c₃ + 2*c₂ + c₁ + 3

14. (Eq. 100) d = x₀ + 1

15. If m > 12, then replace m by m − 12 and j by j + 1.

6. Fast Estimate of the Number of Days Between Two Dates

If you want to know how many days there are between two dates, then you can calculate that exactly by subtracting the Julian Day numbers of the two dates. If you are satisfied with a possible error of a few days, then you can also estimate it from the difference measured in calendar years, calendar months, and calendar days.

Such information (the number of calendar years, months, and days) is not sufficient to calculate the exact number of days for all calendars, because not every year contains the same number of days as every other year, and not every month contains the same number of days as every other month (in all calendars).

Let's assume that the second date is j calendar years and m calendar months and d calendar days later than the first date (both in the same calendar, the Gregorian Calendar or the Jewish Calendar or the administrative Islamic Calendar), where you just subtract the year number, month number, and day number of the second date from those of the first date (which must be later). The variable j may be positive or zero, but not negative. The variables m and d may be positive or zero or negative, but m must not correspond to more than one calendar year, and d must not correspond to more than one calendar month. For the Jewish calendar we define Tishri to be month number 1 and define Adar II to belong to month number 6 together with Adar (so that month number 6 may contain up to 59 days). For the Egyptian calendar we count the last 5 days of the year as a 13th month. Then you can estimate the number n of days between those two dates using the formula

(Eq. 101) n = ⌊a j + b m + c d⌋

with a, b, and c from the following table:

The |Δ|max from the table shows the greatest error (measured in days) that the formula for that calendar yields (based on 10,000 random test dates) for dates that are not more than 400 years apart. The sigma shows the standard deviation (say, the average error, also measured in days). The p0 shows the probability that the estimated number n of days between the two dates is exactly right.

It is clear that the Egyptian calendar is the most predictable one, because it yields the least mistakes (namely zero). Then comes the administrative Islamic calendar, then the Gregorian calendar (with more variation in the month lengths than the Islamic calendar has), en lastly the Jewish calendar (with variation in the number of months in the year).

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Last updated: 2009-11-04