Editing Julian day (section)

== Julian day number calculation ==<!-- This section is linked from [[Gregorian calendar]] -->
The Julian day number can be calculated using the following formulas ([[integer division]] rounding towards zero is used exclusively, that is, positive values are rounded down and negative values are rounded up):{{efn|Doggett in Seidenmann 1992, p. 603, indicates the algorithms are inspired by Fliegel & Van Flanderen 1968. That paper gives algorithms in [[Fortran]]. The Fortran computer language performs integer division by truncating, which is functionally equivalent to rounding toward zero.}}

The months January to December are numbered 1 to 12. For the year, [[astronomical year numbering]] is used, thus 1&nbsp;BC is 0, 2&nbsp;BC is −1, and 4713&nbsp;BC is −4712. ''JDN'' is the Julian Day Number. Use the previous day of the month if trying to find the JDN of an instant before midday UT.

===Converting Gregorian calendar date to Julian Day Number===
The algorithm is valid for all (possibly [[Proleptic Gregorian calendar|proleptic]]) Gregorian calendar dates after November 23, &minus;4713. '''Divisions are integer divisions towards zero'''; fractional parts are ignored.<ref>L. E. Doggett, Ch. 12, "Calendars", p. 604, in Seidelmann 1992. "These algorithms are valid for all Gregorian calendar dates corresponding to JD >= 0, i.e, dates after &minus;4713 November 23."</ref>

{{block indent|1=<math>\text{JDN}  =  \frac{1461 \cdot \left( \text{Y} + 4800 + \frac{\text{M}-14}{12} \right)}{4}  + \frac{367 \cdot \left( \text{M} - 2 - 12 \cdot \frac{\text{M}-14}{12} \right)}{12} - \frac{3 \cdot \frac{\text{Y} + 4900 + \frac{\text{M} - 14}{12}}{100}}{4}  + \text{D} - 32075</math>}}

===Converting Julian calendar date to Julian Day Number===
The algorithm<ref>L. E. Doggett, Ch. 12, "Calendars", p. 606, in Seidelmann 1992</ref> is valid for all (possibly [[Proleptic Julian calendar|proleptic]]) Julian calendar years ≥&nbsp;−4712, that is, for all JDN&nbsp;≥&nbsp;0. Divisions are integer divisions, fractional parts are ignored.

{{block indent|1=<math>\text{JDN} = 367 \cdot \text{Y} - \frac{7 \cdot \left(\text{Y} + 5001 + \frac{\text{M} - 9}{7}\right)}{4} + \frac{275 \cdot \text{M}}{9} + D + 1729777</math>}}

=== Finding Julian date given Julian day number and time of day ===
For the full Julian Date of a moment after 12:00&nbsp;UT one can use the following. Divisions are [[real number]]s.

{{block indent|1=<math>\begin{matrix}J\!D & = & J\!D\!N + \frac{\text{hour} - 12}{24} + \frac{\text{minute}}{1440} + \frac{\text{second}}{86400}\end{matrix}</math>}}

So, for example, January 1, 2000, at 18:00:00&nbsp;UT corresponds to ''JD''&nbsp;=&nbsp;2451545.25 and January 1, 2000, at 6:00:00&nbsp;UT corresponds to ''JD''&nbsp;=&nbsp;2451544.75.

=== Finding day of week given Julian day number ===
Because a Julian day starts at noon while a civil day starts at midnight, the Julian day number needs to be adjusted to find the day of week: for a point in time in a given Julian day after midnight UT and before 12:00&nbsp;UT, add 1 or use the JDN of the next afternoon.

The US day of the [[week]] '''W1''' (for an afternoon or evening UT) can be determined from the Julian Day Number '''J''' with the expression:
{{block indent|1='''W1''' = [[Modular arithmetic|mod]](''J'' + 1, 7)<ref>Richards 2013, pp. 592, 618.</ref>}}

{| class="wikitable"
|-style="text-align:center;"
!W1
| 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
!Day of the week
|Sun||Mon||Tue||Wed||Thu||Fri||Sat
|}

If the moment in time is after midnight UT (and before 12:00&nbsp;UT), then one is already in the next day of the week.

The ISO day of the week '''W0''' can be determined from the Julian Day Number '''J''' with the expression:
{{block indent|1='''W0''' = mod (''J'', 7) + 1}}

{| class="wikitable"
|- style="text-align:center;"
!W0
| 1 || 2 || 3 || 4 || 5 || 6 || 7
|-
!Day of the week
|Mon||Tue||Wed||Thu||Fri||Sat||Sun
|}

=== Julian or Gregorian calendar from Julian day number ===
This is an algorithm by Edward Graham Richards to convert a Julian Day Number, '''J''', to a date in the Gregorian calendar (proleptic, when applicable). Richards states the algorithm is valid for Julian day numbers greater than or equal to 0.<ref>Richards 2013, 617–619</ref><ref>Richards 1998, 316</ref> All variables are integer values, and the notation "''a''&nbsp;div&nbsp;''b''" indicates [[integer division]], and "mod(''a'',''b'')" denotes the [[Modular arithmetic|modulus operator]].

{| class="wikitable"
|+Algorithm parameters for Gregorian calendar
|-
! variable
! value
! variable
! value
|-
| ''y'' || 4716 || ''v'' || 3
|-
| ''j'' || 1401 || ''u'' || 5
|-
| ''m'' || 2 || ''s'' || 153
|-
| ''n'' || 12 || ''w'' || 2
|-
| ''r'' || 4 || ''B'' || 274277
|-
| ''p'' || 1461 || ''C'' || &minus;38
|}

For Julian calendar:
# ''f&nbsp;''=&nbsp;'''J'''&nbsp;+&nbsp;''j''
For Gregorian calendar:
# ''f&nbsp;''=&nbsp;'''J'''&nbsp;+&nbsp;''j''&nbsp;+&nbsp;(((4 ×&nbsp;'''J'''&nbsp;+&nbsp;''B'')&nbsp;div&nbsp;146097)&nbsp;×&nbsp;3)&nbsp;div&nbsp;4&nbsp;+&nbsp;''C''
For Julian or Gregorian, continue:
{{ordered list|start=2
|1=''e'' = ''r'' × ''f'' + ''v''
|2=''g'' = mod(''e'', ''p'') div ''r''
|3=''h'' = ''u'' × ''g'' + ''w''
|4='''D''' = (mod(''h, s'')) div ''u'' + 1
|5='''M''' = mod(''h'' div ''s'' + ''m'', ''n'') + 1
|6='''Y''' = (''e'' div ''p'') - ''y'' + (''n'' + ''m'' - '''M''') div ''n''}}

'''D''', '''M''', and '''Y''' are the numbers of the day, month, and year respectively for the afternoon at the beginning of the given Julian day.

=== Julian Period from indiction, Metonic and solar cycles ===

Let Y be the year BC or AD and i, m, and s respectively its positions in the indiction, Metonic and solar cycles. Divide 6916i&nbsp;+&nbsp;4200m&nbsp;+&nbsp;4845s by 7980 and call the remainder r.
{{block indent|1=If r>4713, Y = (r − 4713) and is a year AD.}}
{{block indent|1=If r<4714, Y = (4714 − r) and is a year BC.}}

Example

i&nbsp;=&nbsp;8, m&nbsp;=&nbsp;2, s&nbsp;=&nbsp;8. What is the year?

{{block indent|1=(6916 × 8) = 55328; (4200 × 2) = 8400: (4845 × 8) = 38760.   55328 + 8400 + 38760 = 102488.}}
{{block indent|1=102488/7980 = 12 remainder 6728.}}
{{block indent|1=Y = (6728 − 4713) = AD 2015.<ref>Heath 1760, p. 160.</ref>}}