It's that time of the year again, Saint Patrick's day. That is St. Patrick, patron of Ireland, Nigeria, Montserrat and engineers

We all remember how Christianity was brought to the island of Ireland, and who provided the inspiration for Downpatrick's glass-and-stone visitor centre. Patrick was of course the inspiration for the name of the town. I can only thank local sensibility for keeping the Enler brown (the picture is the Chicago River in 2005). Unusually, St. Patrick's day also lies in holy week this year, which last occured in 1940, when the feast was celebrated on 3rd April rather than interfere with Palm Sunday. It won't happen again until 2160.  This is a result of the unusually early Easter, which is  calculated from a 5,700,000-year cycle. This year Easter Sunday falls on the 23rd March, the second-earliest possible, the earliest being the 22nd. 0.97% of all Easters within the aforementioned cycle lie on this day, compared to 3.9% on 19th April, the most common day.  The calculation of the date of Easter is in itself a complicated process, and is performed by use of the computus, named such because at the time of its discovery it was the most important function of mathematics.

What happens, however, when you don’t know when the Full Moon will be, or when the Vernal equinox occurs? To do this we need some complicated mathematics (arguably more so than the last paragraph).

Theory

We must assume that the solar year has 365 days, excluding a small remainder (.24 days). A Lunar year is assumed to have 354 days, 12 months of 29.5 days (again we’ve neglected the .03 of a day in each month). Thus the solar year is 11 days longer than the Lunar year. Suppose both years start on 1 January one year. The Lunar Year will finish on 20^th December, and the Solar Year will end on 31^st December. The solar year (we’ll call it sy) will start on 11^th January ly (lunar year). Next year 1^st January sy is the 22^nd January ly. Complicated? Not yet. This extra 11 days is known as an epact, and we must add them to the solar year to get the day in the lunar year. Once there are 30 epacts are reached an embolistic month is added into the lunar year, a bit like a leap year in normal life, and then 30 is subtracted from the epact. Leap days are not included in this calculation, which related the calendar year to the tropical year. The nineteen-year cycle (the Metonic cycle) assumes that 19 tropical years are equal to 235 synodic months. (A tropical year is the time between two Summer solstices, and a synodic month the average period between full moons). Next comes some complicated mathematics: After 19 years the lunations should fall the same way. 19 * 11 = 209 = 30 r 29, not 31. Thus an extra day must be added to the epactin order for the cycle to repeat. The extra 209 days fill seven embolistic months, giving a total of 235 lunations (19 x 12 +7) lunations. The sequence number of a year is called a Golden Number.

Method 1

By 1582 somebody decided that this was far to complicated, and devised a system using tables which could be pre-calculated. This can be found in the Book of Common  Prayer. For the latter half of the current Metonic cycle the table looks like this:

We can use the epacts to find the first day of a lunar month (new moon) by writing out a table of the 365 days of the year (excluding a leap day where applicable), and labelling each day with the numerals * , xxix, right down to i beginning on 1^st January, and repeat each time you reach i. For the next month only count 29 days, and so on. Eventually you will produce a calendarium. Then label each day A-G in cycles. The letter assigned to the year’s first Sunday becomes the dominical letter. In

practice we only need to do all of this for 8^th March to the 5^th April.  This gives rise to the table on the right. Knowing the epact of the year one can look it up against the chart, and find the ecclesiastical full moon. This year the epact is xi, and the full moon the 20^th. Hence the next Sunday is Easter, or the 23^rd March.

Method 2

The Meeus algorithm

a = Y mod 4

b = Y mod 7

c = Y mod 19

d = (19 × c + 15) mod 30

e = (2 × a + 4 × b - d + 34) mod 7

month = (d + e + 114) / 31

day = ((d + e + 114) mod 31) + 1

Where Y = Year.

Written in 1991 this method has no exceptions and requires no table. Simpler I say.

Much of my research was on Wikipedia, and their page on the computus is enlightening!

http://en.wikipedia.org/wiki/Computus