PERPETUAL CALENDARS................ [Se preferir, veja a página em Português com mais detalhes]

See any year calendar (from 1 AD to infinity). We follow the catholic world rules. Therefore, you'll see in the year 1582, the transition year from Julian to Gregorian Calendar, we didn't have the days 5 to 14 in October month:

Year (1 to infinity)→

See, below, the Easter date, Carnival and Corpus Christi for the same year (see details about this, if you wish):

The EASTER Sunday CARNIVAL Tuesday C.CHRISTI Thursday

Note: If you compare this calendar with the one in Microsoft/Office/Excel, you will find a difference in year 1900. This is a problem in Excel, that incorrectly considers 1900 a leap year. Any attempt to calculate former dates (before 1900 March 1st), starting from Excel calendar, will result 1 day mistake.

You can see more about calendars at my page Julian Days.

The world smallest perpetual calendar!

Look at the new world smallest (and the unique) pocket perpetual calendar from 1 AD to infinity. Print it, cut out and fix the three disks with an eyelet.

This gadget comprises all thousand/hundred from 00 to 99, allowing, this way, the direct calculation of any past or future date, without any limit.

Example: Which day of the week will be August 6, 516763? (516763 - thousand/hundred=67 ten/unit=63). Place arch (A) under thousand/hundred 67 (B). Place AUG (C) under ten/unit 63 (D). You'll see that August 6 (E), 516763 will be Tuesday (F). Note that the outer and the inner disks show the whole wished month calendar.

Thousand/hundred 15J (Julian Calendar) is used only until October 4, 1582 and 15G (Gregorian calendar) is used only from October 15, 1582.
Match leap YEAR and JAN or FEB months marked with an asterisk (*). Note that 1700, 1800, 1900, 2100 and 2200 are not leap years, although divisible by 4.
Note that APR, JUN, SEP and NOV have only 30 days and FEB 28/29 days.
Everything repeats each 4 centuries, on Gregorian Calendar, while repeated each 7 centuries, on Julian Calendar.

Ask for a total free sample of my calendar, made in PVC. Tell me your postal address and you'll have it, any place in the world.

You can use my pocket perpetual calendar in other situations, like in the United Kingdom, where the "missing" days were September 3 to 13, 1752. Just take in account that 16J thousand/hundred would be on the "02 09 18 22" sector and 17J thousand/hundred (until September 2, 1752) would be on the "03 10" sector. You'll see that September 2, 1752 (julian calendar) was Wednesday and September 14, 1752 (gregorian calendar) was Thursday.

Also see other samples of pocket perpetual calendars, if you are an inquisitive person.

A bit of History...
The Julian Calendar began on the 46th year b.C. and ended theoretically on October 4, 1582. And the Gregorian Calendar began on 15 of the same month, "erasing" from history the days from October 5 to 14. Portugal and its colonies adopted the Gregorian reform at its beginning, but many nations, including England, did not. Thus, in the English-speaking world, the "missing" days were September 3 to 13, 1752.

Leap years
Although in the Julian Calendar every year that is divisible by 4 is a leap year, the Gregorian Calendar, for greater astronomical precision, provides that years divisible by 4 are leap years, except those ending in "00" and not divisible by 400. That is, the years 1584, 1588... 1600, 2000, 2400 etc. are leap years, but not 1700, 1800, 1900, 2100 etc.. Just because tropic year isn't exactly 365 days and a quarter lasting.

A timely reminder
The 21st century and the 3rd millenium began on January 1, 2001, not 2000.

Mental calculation of the weekday
Surprise your friends, making mental calculation of the weekday for dates from 1901Jan1 to 2099Dec31, using the routine below:

We know that 1901Jan1 was a Tuesday. Let's consider 1=Sun 2=Mon 3=Tue 4=Wed 5=Thu 6=Fri 0=Sat.

a) choose a date from 1901Jan1 to 2099Dec31
b) start from number 3
c) add 1 for each elapsed year (sevens out*)
d) add 1 for each 4 entire elapsed years (sevens out)
e) add the elapsed days of the chosen year (sevens out)
f) the amount (sevens out) will show the weekday

See a practical example:

a) 2013Mar27
b) starting point=3
c) 2013-1901=112 sevens out=0
d) 112÷4=28 sevens out=0
e) 31+28+26=85 sevens out=1
f) 3+0+0+1=4 sevens out=4 (Wednesday)

See another practical example:

a) 1940Apr17
b) starting point=3
c) 1940-1901=39 sevens out=4
d) 39÷4=9,75 and 9 sevens out=2
e) 31+29+31+16=107 sevens out=2
f) 3+4+2+2=11 sevens out=4 (Wednesday)

*sevens out=the remainder of a number divided by seven

The routine above, from 1901 to 2099, is the most searched one. However, you can calculate the weekday from the 1st year of the Christian Era to infinite, if you understand about secular non-leap years (1700, 1800, 1900, 2100, 2200, 2300, 2500 a.s.o.) and about the suppressed 10 days between Julian and Gregorian calendars (1582 October 5 to 14).

We know that 0001Jan1 was, theoretically, a Saturday. Let's consider 1=Sun 2=Mon 3=Tue 4=Wed 5=Thu 6=Fri 0=Sat:

a) choose a date, from 0001Jan1 to infinite
b) get 1 for each elapsed year (sevens out)
c) add 1 for each 4 entire elapsed years (sevens out)
d) add the elapsed days of the chosen year (sevens out)
e) add 6* for each secular non-leap elapsed year (sevens out)
f) add 4** for any date from 1582Nov1 to infinite
g) the amount (sevens out) will show the weekday

*indeed we want to "subtract 1" but it's better to "add 6" (it's the same thing on "sevens out" operations)
**indeed we want to "subtract 10" but it's better to "add 4" (it's the same thing on "sevens out" operations)

See a practical example:
a) 2918Jun3
b) 2918-1=2917 sevens out=5
c) 2917÷4=729,25 and 729 sevens out=1
d) 31+28+31+30+31+2=153 sevens out=6
e) 6x10=60 sevens out=4
f) 4
g) 5+1+6+4+4=20 sevens out=6 (Friday)

See another practical example:
a) 1500Apr21
b) 1500-1=1499 sevens out=1
c) 1499÷4=374,75 e 374 sevens out=3
d) 31+29+31+20=111 sevens out=6
e) 6x0=0
f) 0
g) 1+3+6+0+0=10 sevens out=3 (Tuesday)

See a third practical example, about 21-days-month (1582Oct):
a) 1582Oct22
b) 1582-1=1581 sevens out=6
c) 1581÷4=395,25 e 395 sevens out=3
d) 31+28+31+30+31+30+31+31+30+11=284 sevens out=4
e) 6x0=0
f) 0
g) 6+3+4+0+0=13 sevens out=6 (Friday)

© 20/05/1996 Atualizada em 13/08/2016