Page:Catholic Encyclopedia, volume 3.djvu/825

 CHRONOLOGY

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CHRONOLOGY

first datum all the rest follows. The succeeding days of March increase their figures each by 1, on account of the increased day number. When 7 is passed it is only the figures which remain, after division by that number, which are to be considered; thus 11 may be treated as 4 (7 + 4) and 30 as 2 (28 + 2). In gen- eral, any exact multiple of 7 (14. 21, 28) may be added or subtracted when convenient without affect- ing the result. Instead of adding any number (e. g. 1 or 4) we may subtract its difference from 7 or a mul- tiple of 7 (e. g. 6 or 3). The remainder in a division is equivalent to 7. and thus in calculating for the day of the week it signifies Saturday.

As the days of the leading month, so those of the months preceding it follow naturally. As March contains 31 days (i. e. 28 + 3), April necessarily begins with a day 3 places later in the weekly sequence, and its month number instead of 1 is 4. So of other months, according to the number of days in that which preceded. The following are the month num- bers throughout the year which never change: — March 1; April 4; May 6; June 2; July 4; August 0; September 3; October 5; November 1 ; December 3; January 6; February 2. a. d. 1, being a common year of 365 days (or .52 weeks +1 day), ends with the same day of the week — Tuesday — with which it commenced. Consequently the next year, A. d. 2, commences a day later, with Wednesday for 1 .March, and as its year number is increased to 2, we get 2+1 + 1=4. So in A. D. 3, the year number b.n imes 3, and 1 March is Thursday. But on account of 29 February preceding 1 March, A. d. 4. this day falls 366 days (or 52 weeks + 2 days) after 1 March, A. D. 3, or on Saturday, and its year number must be increased to 5; 5+1 + 1 = 7. Thus, to find the number belonging to any year within its own century, we must find how many days beyond an exact number of weeks there have been since that cen- tury commenced. As every common year contains one day more than fifty-two weeks, and every leap year two days more, by adding at any period the number of leap years which there have been in the century to the total number of years in the same, we obtain the number of days required. To obtain the number of leap years, we divide the last two figures of the date (i. e. those in the tens and units place) by four. The quotient (neglecting any remainder) shows the number of leap years; which, added to the same two figures, rives the number of days over and above the sets of fifty-two weeks which the years con- tain. Thus, for example, the year '39 of any century 53 I, 1839, 1939) will have 6 for its year num- ber, for in such year is extra days will have accumu- lated since the corresponding day of the centurial year (00), viz. 1 day for each of the 30 common years, and 18 days for the 9 leap years.

The Century. — One more element of calculation remains to be considered, viz. the Century. We begin with the Julian system, or O. S. — according to which all centuries contain 75 common years of 365 days, and 2"> leap years .>f 366, and accordingly 125 days in all, over and above 5200 weeks. But 125 days=17 weeks +6 days. Therefore a Julian century ends with the day of the week tiro days previous to that with whicb it began, and the succeeding century will begin with the day of the week, one day earlier than its predecessor. Thus, a. d. 1 March, 11500, being Tuesday, in 1400 it would be Monday, in 1500 Sun- day, in 1600 Saturday. Having obtained the cen- turial number for any century, we add to it the year numbers of the years which follow to the close of thai century. Centurial numbers O. S. are obtained by subtracting the centurial figure or figures (viz. those preceding 110 1 from the multiple of 7 next above, the remainder being the number required. Thus for A. D. 1100 the centurial number is 3 (14-11), for 1500, 6 (21-15), for 1900, 2 (21-19).

Under the N. S. three centuries in every four con- tain 76 common years and 24 leap years, and thus have only 124 days over 5200 weeks, or 17 weeks and 5 days, and end with the day of the week three earlier than they began. The following century, beginning two days earlier than that which it follows, has its cen- turial number less by 2. Thus 1 March, a. d. 1700, was Monday, and the centurial number (or 7). 1 March, 1800, was Saturday, and the centurial number 5. Every fourth centurial year N. S., being a leap year (1600, 2000, 2400, etc.), has 366 days; and the century to which it belongs, like those of the O. S., diminishes its centurial number only by 1 from the preceding. N. S. having been introduced in the six- teenth century, it is only for dates 15 — and upwards that N. S. centurial numbers are required. They are asfollows: for 1500=3; 1600=2; 1700 = 7; 1800=5; 1900=3; 2000=2. It will be seen that the same figures constantly recur. Leap year centuries (with the first two figures exactly divisible by 4) having the centurial number 2, and the three centuries fol- lowing having 7 (or 0), 5, and 3 respectively, after which 2 comes round again. The centurial number N. S. can be obtained from that of O. S. if the difference of days between O. S. and N. S. be allowed for. This is done by subtract ing the said difference from the O. S. centurial number, increased by as many times 7 as the subtraction requires. As we have seen, for the sixteenth and seventeenth cen- turies, the difference was 10 days; for the eighteenth, 11; for the nineteenth. 12; for the twentieth and twenty-first, 13. Thus:—

a.d. 1500 etc.

C.N.

(O.S.)=6(N.S.)=3(6+ 7-10).

A.D. 1600

do. =5 do. =2(5+7-10).

A.D. 1700

do. =4 do. =0(7) (4+ 7-11).

A.D. 1800

do. =3 do. =5(3+ 14-12).

A.D. 1900

do. =2 do. =3(2+ 14-13).

a.d. 2000

do. =1 do. =2(1+14-13).

+ 4 + ]

Rule to find tint/ of week for an;/ date: — Take the sum of the centurial number + year number + month number + day number; divide this by 7; the re- mainder gives day of week, O. S. or N. S., according to centurial number used.

Examples. — (1) King John was crowned 27 May, 1199. What day? Century (O. S.) Year Month Day)

6or )_l =40=7x5 + 5. ■j 6(27=21 + 6) )+ 2 ')

Therefore the day was Thursday.

(2) Waterloo was fought 18 June, 1815. What day?

Century (N. S.) Year Month Day 1

5 + )4asL 8 i r + 4+2 + 18}= 43 = 7 >< 6 + 1 -

Therefore the day was Sunday, as leaders of "Vanity Fair" will recollect.

(3) Columbus discovered the New World 12 Octo- ber, 1492. What day.'

Century (O. S.) Year Month Day

20 +3+5+12 =^; remainder 6.

Therefore the day was Friday.

(4) If St. Patrick died 17 March, 4o3, required the day of the week.

Century (O. S.) Year Month Day

22 3 + 1 + 1 + 17 =~ ; remainder 1.

Therefore the day was Sunday.

(5) Mary Queen of Scots was executed S February,

1587 (158=), which was a Wednesday. Was this

O. S. orN. S.?

Century (O. S.) Year 1586 Month Day

6 + 2 + 2+8 =18= Wednesday.