Our Calendar Part 5

You’re reading novel Our Calendar Part 5 online at LightNovelFree.com. Please use the follow button to get notification about the latest chapter next time when you visit LightNovelFree.com. Use F11 button to read novel in full-screen(PC only). Drop by anytime you want to read free – fast – latest novel. It’s great if you could leave a comment, share your opinion about the new chapters, new novel with others on the internet. We’ll do our best to bring you the finest, latest novel everyday. Enjoy!

4th. Take the remainder from 3 or 10, from the commencement of the era to October 5th, 1582. From October 15th, 1582 to 1700, from 6 or 13. From 1700 to 1800, from 7, and so on. See table on 49th page.

We divide by 4 because the intercalary period is four years; and as every fourth year contains the divisor 4 once more than any of the three preceding years, so there is one more added to the fourth year than there is to any of the three preceding years; and as every year consists of 52 weeks and one day, this additional year gives an additional day to the remainder after dividing by 7. For example, the year

1 of the era consists of 52 w. 1 d.

2 years consist of 104 w. 2 d.

3 years consist of 156 w. 3 d.

(4 4) + 4 = 5 years consist of 260 w. 5 d.

Hence the numbers thus formed will be 1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 15, and so on.

We divide by 7, because there are seven days in the week, and the remainders show how many days more than an even number of weeks there are in the given year. Take, for example, the first twelve years of the era after being increased by one-fourth, and we have

1 7 = 0 remainder 1 Then 3 - 1 = 2 = B 2 7 = 0 " 2 " 3 - 2 = 1 = A 3 7 = 0 " 3 " 10 - 3 = 7 = G 5 7 = 0 " 5 " 10 - 5 = 5 = F E 6 7 = 0 " 6 " 10 - 6 = 4 = D 7 7 = 1 " 0 " 3 - 0 = 3 = C 8 7 = 1 " 1 " 3 - 1 = 2 = B 10 7 = 1 " 3 " 10 - 3 = 7 = A G 11 7 = 1 " 4 " 10 - 4 = 6 = F 12 7 = 1 " 5 " 10 - 5 = 5 = E 13 7 = 1 " 6 " 10 - 6 = 4 = D 15 7 = 2 " 1 " 3 - 1 = 2 = C B

From this table it may be seen that it is these remainders representing the number of days more than an even number of weeks in the given year, that we have to deal with in finding the dominical letter.

Did the year consist of 364 days, or 52 weeks, invariably, there would be no change in the dominical letter from year to year, but the letter that represents Sunday in any given year would represent Sunday in every year.

Did the year consist of only 363 days, thus wanting one day of an even number of weeks, then these remainders, instead of being taken from a given remainder, would be added to that number, thus removing the dominical letter forward one place, and the beginning of the year, instead of being one day later, would be one day earlier in the week than in the preceding year.

Thus, if the year 1 of the era be taken from 3, we would have 3 - 1 = 2; therefore, B being the second letter, is dominical letter for the year 1.

But if the year consist of only 363 days, then the 1 instead of being taken from 3 would be added to 3; then we would have 3 + 1 = 4; therefore, D being the fourth letter would be dominical letter for the year 1. The former going back from C to B, the latter forward from C to D; or which amounts to the same thing, make the year to consist of 51 weeks and 6 days; then 10 - 6 = 4, making D the dominical letter as before.

As seven is the number of days in the week, and the object of these subtractions is to remove the dominical letter back one place every common year, and two in leap-year, why not take these remainders from 7? We answer, all depends upon the day of the week on which the era commenced.

Had G, the seventh letter been dominical letter for the year preceding the era, then these remainders would be taken from 7; and 7 would be used until change of style in 1582. But we know from computation that C, the third letter, is dominical letter for the year preceding the era; so we commence with three, and take the smaller remainders, 1 and 2 from 3; that brings us to A. We take the larger remainders, from 3 to 6, from 3 + 7 = 10. We add the 7 because there are seven days in the week. We use the number 10 until we get back to C, the third letter, the place from whence we started. For example, we have

3 - 1 = 2 = B 3 - 2 = 1 = A 10 - 3 = 7 = G 10 - 4 = 6 = F 10 - 5 = 5 = E 10 - 6 = 4 = D 3 - 0 = 3 = C

The cycle of seven days being completed, we commence with the number three again, and so on until 1582, when on account of the errors of the Julian calendar, ten days were suppressed to restore the coincidence of the solar and civil year. Now every day suppressed removes the dominical letter forward one place; so counting from C to C again is seven, D is eight, E is nine, and F is ten. As F is the sixth letter, we take the remainders from 1 to 5, from 6; if the remainder be 6, take it from 6 + 7 = 13. Then 6 or 13 is used till 1700, when, another day being suppressed, the number is increased to 7. And again in 1800, for the same reason, a change is made to 1 or 8; in 1900 to 2 or 9, and so on. It will be seen by the table on the 49th page that the smaller numbers run from 1 to 7; the larger ones from 8 to 13.

From the commencement of the Christian era to October 5th, 1582, take the remainders, after dividing by 7, from 3 or 10; from October 15th,

1582 to 1700 from 6 or 13 1700 to 1800 " 7 1800 to 1900 " 1 or 8 1900 to 2100 " 2 or 9 2100 to 2200 " 3 or 10 2200 to 2300 " 4 or 11 2300 to 2500 " 5 or 12 2500 to 2600 " 6 or 13 2600 to 2700 " 7 2700 to 2900 " 1 or 8 2900 to 3000 " 2 or 9 3000 to 3100 " 3 or 10 3100 to 3300 " 4 or 11 3300 to 3400 " 5 or 12 3400 to 3500 " 6 or 13 3500 to 3700 " 7 3700 to 3800 " 1 or 8 3800 to 3900 " 2 or 9 3900 to 4000 " 3 or 10 4000 to 4100 " 4 or 11 4100 to 4200 " 5 or 12 4200 to 4300 " 6 or 13 4300 to 4500 " 7 4500 to 4600 " 1 or 8 4600 to 4700 " 2 or 9 4700 to 4900 " 3 or 10 4900 to 5000 " 4 or 11 5000 to 5100 " 5 or 12

CHAPTER V.

RULE FOR FINDING THE DAY OF THE WEEK OF ANY GIVEN DATE, FOR BOTH OLD AND NEW STYLES.

By arranging the dominical letters in the order in which the different months commence, the day of the week on which any month of any year, or day of the month has fallen or will fall, from the commencement of the Christian era to the year of our Lord 4000, may be calculated. (Appendix G.) They have been arranged thus in the following couplet, in which At stands for January, Dover for February, Dwells for March, etc.

At Dover Dwells George Brown, Esquire, Good Carlos Finch, and David Fryer.

Now if A be dominical or Sunday letter for a given year, then January and October being represented by the same letter, begin on Sunday; February, March and November, for the same reason, begin on Wednesday; April and July on Sat.u.r.day; May on Monday, June on Thursday, August on Tuesday, September and December on Friday. It is evident that every month in the year must commence on some one day of the week represented by one of the first seven letters of the alphabet. Now let

January 1st be represented by A, Sun.

Feb. 1st (4 w. 3 d. from the preceding date) by D, Wed.

Mar. 1st 4 w. 0 d. " " " by D, Wed.

Apr. 1st 4 w. 3 d. " " " by G, Sat.

May 1st 4 w. 2 d. " " " by B, Mon.

June 1st 4 w. 3 d. " " " by E, Thur.

July 1st 4 w. 2 d. " " " by G, Sat.

Aug. 1st 4 w. 3 d. " " " by C, Tues.

Sept. 1st 4 w. 3 d. " " " by F, Fri.

Oct. 1st 4 w. 2 d. " " " by A, Sun.

Nov. 1st 4 w. 3 d. " " " by D, Wed.

Dec. 1st 4 w. 2 d. " " " by F, Fri.

Now each of these letters placed opposite the months respectively represents the day of the week on which the month commences, and they are the first letters of each word in the preceding couplet.

To find the day of the week on which a given day of any year will occur, we have the following

RULE.

Find the dominical letter for the year. Read from this to the letter which begins the given month, always reading from A toward G, calling the dominical letter Sunday, the next Monday, etc. This will show on what day of the week the month commenced; then reckoning the number of days from this will give the day required.

EXAMPLES.

History records the fall of Constantinople on May 29th, 1453. On what day of the week did it occur? We have then 1453 4 = 363 +; 1453 + 363 = 1816; 1816 7 = 259, remainder 3. Then 10 - 3 = 7; therefore, G being the seventh letter is dominical letter for 1453. Now reading from G to B, the letter for May, we have G Sunday, A Monday, and B Tuesday; hence May commenced on Tuesday and the 29th was Tuesday.

The change from Old to New Style was made by Pope Gregory XIII, October 5th, 1582. On what day of the week did it occur? We have then 1582 4 = 395+; 1582 + 395 = 1977; 1977 7 = 282, remainder 3. Then 10 - 3 = 7; therefore, G being the seventh letter, is dominical letter for 1582. Now reading from G to A, the letter for October, we have G Sunday, A Monday, etc. Hence October commenced on Monday, and the 5th was Friday.

On what day of the week did the 15th of the same month fall in 1582? We have then 1582 4 = 395+; 1582 + 395 = 1977; 1977 7 = 282, remainder 3.

Then 6 - 3 = 3; therefore, C being the third letter, is the dominical letter for 1582. Now reading from C to A, the letter for October, we have C Sunday, D Monday, E Tuesday, etc. Hence October commenced on Friday, and the 15th was Friday.

How is this, says one? You have just shown by computation that October, 1582, commenced on Monday, you now say that it occurred on Friday. You also stated that the 5th was Friday; you now say that the 15th was Friday.

This is absurd; ten is not a multiple of seven. There is nothing absurd about it. The former computation was Old Style, the latter New Style, the Old being ten days behind the new.

As regards an interval of ten days between the two Fridays, there was none; Friday, the 5th, and Friday, the 15th, was one and the same day; there was no interval, nothing ever occurred, there was no time for anything to occur; the edict of the Pope decided it; he said the 5th should be called the 15th, and it was so.

Hence to October the 5th, 1582, the computation should be Old Style; from the 15th to the end of the year New Style.

On what day of the week did the years 1, 2 and 3, of the era commence?

None of these numbers can be divided by 4; neither are they divisible by 7; but they may be treated as remainders after dividing by 7. Now each of these numbers of years consists of an even number of weeks with remainders of 1, 2 and 3 days respectively. Hence we have then for the year 1, 3 - 1 = 2; therefore, B being the second letter, is the dominical letter for the year 1. Now reading from B to A, the letter for January, we have B Sunday, C Monday, D Tuesday, etc. Hence January commenced on Sat.u.r.day.

Then we have for the year 2, 3 - 2 = 1; therefore A being the first letter, is dominical letter for the year 2; hence it is evident that January commenced on Sunday. Again we have for the year 3, 10 - 3 = 7; therefore, G being the seventh letter, is dominical letter for the year 3.

Now reading from G to A, the letter for January, we have G Sunday, A Monday; hence January commenced on Monday.

On what day of the week did the year 4 commence? Now we have a number that is divisible by 4, it being the first leap-year in the era, so we have 4 4 = 1; 4 + 1 = 5; 5 7 = 0, remainder 5. Then 10 - 5 = 5; therefore, E being the 5th letter, is dominical letter for that part of the year which follows the 29th of February, while F, the letter that follows it, is dominical letter for January and February. Now reading from F to A, the letter for January, we have F Sunday, G Monday, A Tuesday; hence January commenced on Tuesday.

Now we have disposed of the first four years of the era; the dominical letters being B, A, G, and F, E. Hence it is evident, while one year consists of an even number of weeks and one day, two years of an even number of weeks and two days, three years of an even number of weeks and three days, that every fourth year, by intercalation, is made to consist of 366 days; so that four years consist of an even number of weeks and five days; for we have (4 4) + 4 = 5, the dominical letter going back from G in the year 3, to F, for January and February in the year 4, and from F to E for the rest of the year, causing the following year to commence two days later in the week than the year preceding.

The year 1 had 53 Sat.u.r.days; the year 2, 53 Sundays; the year 3, 53 Mondays, and the year 4, 53 Tuesdays and 53 Wednesdays, causing the year 5 to commence on Thursday, two days later in the week than the preceding year. Now what is true concerning the first four years of the era, is true concerning all the future years, and the reason for the divisions, additions and subtractions in finding the dominical letter is evident.

Our Calendar Part 5

You're reading novel Our Calendar Part 5 online at LightNovelFree.com. You can use the follow function to bookmark your favorite novel ( Only for registered users ). If you find any errors ( broken links, can't load photos, etc.. ), Please let us know so we can fix it as soon as possible. And when you start a conversation or debate about a certain topic with other people, please do not offend them just because you don't like their opinions.


Our Calendar Part 5 summary

You're reading Our Calendar Part 5. This novel has been translated by Updating. Author: George Nichols Packer already has 687 views.

It's great if you read and follow any novel on our website. We promise you that we'll bring you the latest, hottest novel everyday and FREE.

LightNovelFree.com is a most smartest website for reading novel online, it can automatic resize images to fit your pc screen, even on your mobile. Experience now by using your smartphone and access to LightNovelFree.com