Our Calendar: The Julian calendar and its errors. How corrected by the Gregorian. Rules for finding the dominical letter, and the day of the week of any event from the days of Julius Caesar 46 B.C. to the year of our Lord four thousand; a new and easy method of fixing the date of Easter. Hebrew calendar; showing the correspondence in the date of events recorded in the Bible with our present Gregorian calendar. Illustrated by valuable tables and charts.Packer, George Nichols
History
Our Calendar: The Julian calendar and its errors. How corrected by the Gregorian. Rules for finding the dominical letter, and the day of the week of any event from the days of Julius Caesar 46 B.C. to the year of our Lord four thousand; a new and easy method of fixing the date of Easter. Hebrew calendar; showing the correspondence in the date of events recorded in the Bible with our present Gregorian calendar. Illustrated by valuable tables and charts.
Packer, George Nichols
Calendar; Jewish calendar
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
Public-domain text, read in full here on John Shaqi.
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