Making a Simple Calendar - PHP Science labs

The BASICS OF MAKING A STANDARD CALENDAR FOR ANY DATE

To construct a calendar for any given date, we can break down the process into 4 essential steps.

Compute the standard calendrical JD number for the given date.
Determine the day of the week on which the given date falls.
Determine the day of the week on which the month containing the given date begins.
Determine the number of days in the month containing the given date.

Once we know on what day of the week any given month of any given year begins and the number of days in that month, then we have all the information we need to construct a standard calendar for that month.

Step 1:

Compute The Standard Calendrical JD (Julian Day) Number For the Given Date

NOTE:
Here, the calendrical JD number is being distinguished from the astronomical JD number in that the time of day is ignored. The calendrical JD number simply serves its original purpose as a unique, linear, integer serial number, one for each date on the standard calendar. There are some rare historical deviations from the standard calendar formula, but those special cases are not being considered here.

The JD number is the single most important number we will use for a variety of general Julian and Gregorian calendar computations. It contains all the information we need to create a complete standard calendar for any month of any year over a very long time period extending thousands of years into the past, prior to the actual dates either calendar officially existed, as well as thousands of years into the theoretical future.

To construct a calendar, we begin by computing the simple calendrical JD (Julian Day) number for the given calendar date and calendar mode. This simple calendrical JD number will always equate to a positive integer value, measured forward from origin JD=0 in the distant past.

The prehistoric mathematical origin of the Julian calendar model is:
JD=0 on 4713 BC Jan 01 = Monday

The prehistoric mathematical origin of the Gregorian calendar model is:
JD=0 on 4714 BC Nov 24 = Monday

Since each calendar system begins with JD=0 being on a Monday and every 7th day thereafter is a Monday, this leads to a simple formula to derive the day of the week (DoW) directly from the JD number. This formula will follow later. We have to compute the JD number first of all.

In the simplest terms, the JD number for any given date is the serial count of the number of days on the calendar that have passed to date since the mathematical origin of the respective calendar system model.

For example, the current Universal Gregorian Date is:
2024 Apr 20, for which JD=2460421, which means that today is sequential calendar day number 2460421, as reckoned from the mathematical origin (JD=0) of the Gregorian calendar system that began on Monday, November 24th, 4714 BC.

Every date on the calendar has such a unique JD serial number reckoned from the origin. Since it works like a linear yardstick, the difference between the JD numbers of any two dates equates to the exact number of calendar days between those two dates, which includes accounting for any leap years in between as well.

Compute the simple calendrical JD (Julian Day) number for any given integer-encoded date and calendar mode. The numerical value of this JD number will always equate to a positive integer value.

ISOymd = ISO integer-encoded date value (Negative = BC date)
CalMode = Calendar mode (0=Julian or 1=Gregorian)

 YearSignVal = (ISOymd < 0)? -1 : 1
 u = abs(ISOymd)
 v = floor(u / 10000)

 y = YearSignVal * v
 m = floor((u - 10000*v) / 100)
 d = u - 10000*v - 100*m

 w = (y < 0)? y+1 : y

 a = floor((14-m) / 12)
 b = w - a
 c = floor(b/100)

 JD = floor(30.6001*(12*a + m + 1))
    + floor(365.25*(b + 4716)) - 1524
    + CalMode*(floor(c/4) - c + 2) + d

 JD EXAMPLE: For the interface date entered above.

 ISOymd  = 20240420 = 2024 April 20
 CalMode = 1 (Gregorian)

 -----------------------------------------------------------
 YearSignVal = ((2024 < 0)? -1 : 1) = 1

 u = abs(20240420) = 20240420
 v = floor(20240420 / 10000) = 2024

 y = 1 * 2024 = 2024
 m = floor((20240420 - 10000*2024) / 100) = 4
 d = 20240420 - 10000*2024 - 100*4 = 20

 w = ((2024 < 0)? 2024+1 : 2024) = 2024

 a = floor((14-4) / 12) = 0
 b = 2024 - 0 = 2024
 c = floor(2024/100) = 20

 JD = floor(30.6001*(12*0 + 4 + 1)) = 153
    + floor(365.25*(2024 + 4716)) - 1524 = 2460261
    + 1*(floor(20/4) - 20 + 2) + 20 = 7

    = 153 + 2460261 + 7 = 2460421

Step 2:

Determine the Day of the Week (DoW) On Which the Given Date Falls

From the calendrical JD serial number for the given date, we can determine the corresponding day of the week (DoW) index. This is simply a number code from 0 to 6, interpreted to represent the days of the week from Sunday to Saturday respectively. The formula is based on the fact that both calendar system models begin (JD=0) on a Monday (DoW=1), by definition, and reckoned cyclically forward from there, every 7th day is a Monday.

Compute the day-of-the-week (DoW) index from the standard Julian Day (JD) number. The same formula applies to both the Julian and Gregorian calendars.

JD = Calendrical JD number for given date
DoW = Day of the week index code corresponding to JD

DoW = (JD + 1) mod 7

Where the DoW indices are: 0=Sun, 1=Mon, 2=Tue, 3=Wed, 4=Thu, 5=Fri and 6=Sat


DoW EXAMPLE: For the interface date entered above.

 ISOymd = 20240420 = 2024 April 20 (Gregorian Calendar)

     JD = 2460421

DoW = (JD + 1) mod 7
    = 2460422 mod 7 = 6 = Saturday

Get the number of days in any given month of any given year. The number of days in the months are held in an indexed array, from which the values can easily be extracted according to the numerical value of the month. For leap years, a special adjustment is made for February.

DaysInMonth = Indexed array holding the number of days in each month
          y = Year (Negative = BC)
          m = Month number (1 to 12)
       leap = Leap year flag (1=Leap or 0=Common)
      mDays = Number of days in given month of given year

DaysInMonth = array (31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)

mDays = DaysInMonth[m-1]

w = (y < 0)? y+1 : y

if (CalMode == 0)
   {
    leap = (w mod 4 == 0)? 1:0
   }
else
   {
    leap = ((!(w mod 4) && (w mod 100)) || !(w mod 400))? 1:0
   }

if (m == 2 && leap == 1)  {mDays = 29}

This does the same as the above, but using a more direct algorithmic approach, without using any array.

w = (y < 0 )? y+1 : y

if (CalMode == 0)
   {
    leap = (w mod 4 == 0)? 1:0
   }
else
   {
    leap = ((!(w mod 4) && (w mod 100)) || !(w mod 400))? 1:0
   }

mDays = 30 - ((m < 8)? -1:1)*(m mod 2) + floor(m/8) - ((m == 2)? 2:0)
      + ((m == 2 && leap == 1)? 1:0)

Making a Simple Calendar
Date: ±Y M D (Neg=BC)
Calendar Mode: Julian Gregorian