EXAMP. 2. For the year 1791.

1791
I

10)1792(94

171

82
76

6 Golden Number, and 6×11=66, therefore 30)66(2

RULE. Subtract 11 from the Julian Epa&t: If the subtraction can. not be made, add 30 to the Julian Epact; then subtract, and the re mainder will be the Gregorian Epact; if nothing remain, the Epa& is 29.

Or, take 1 from the Golden Number, and divide the remainder by 3; if 1 remain, add 10 to the dividend, which sum will be the Epact; if 2 remain, add 20 to the dividend; but if nothing remain, the div. idend is the Epact.

Therefore, as 2 remains, add

20 to the dividend, and it gives the Epact 25, as before.

A general Rule for finding the Gregorian Epal forever.

Divide the centuries of any year of the Christian Era by 4, (rejecting the subsequent numbers ;) multiply the remainder by 17, and

to

to this product add the quotient multiplied by 43; divide this sum plus 86 by 25, multiplying the Golden Number by 11, from which subtract the last quotient, and rejecting the thirties, the remainder will be the Epact.

EXAMP. For the year 1786.

Rejecting the subsequent numbers 86, it will be 17.

4)17(4

ATABLE of the nineteen Epads for the Julian and Gregorian

Accounts, by the Golden Number.

Julian Greg.

Julian Greg.

Julian Greg.
G. N. Epact. | Epact. G. N. Epact. Epact. G. N. | Epact. | Epact

1

29

PROBLEM VIII.

To calculate the Moon's Age on any given day.

RULE. To the given day of the month, add the Epact and number of the month: If the sum be less than 30, it is the Moon's age, but if it exceed 30, then take 30 from it, and the remainder will be the Moon's age.

Note. The numbers to be added to the following months, are as fol

EXAMPLE. For January 25th, 1786.

Add Given day

Epact

= 25

=29

No. of the month == 00

To find the times of the New and Full Moon, and the first and last Quarters. RULE. Find the Moon's age on the given day, then, if it be 15, the Moon will be full on that day, and by counting 7 days backward and forward you will have the first and last quarters, and by counting backward and forward 15 days, you will have the times of the last and next change; but if the age of the Moon be greater than 15, take 15 from it, and the remainder will shew how many days have passed since the last full moon, and counting these backward, you will have the day the last full moon happened on, and by know. ing that, we can find the change, or either of the quarters, as before. of the moon, on the assumed day, be less than Again, if the age 15, then take that from 15, and the remainder will shew how many days are to run till the next full moon, which you will have by adding the remainder to the assumed day; and, proceeding as before, you will have the days of the change, and either quarter as above. EXAMP. For Jan. 25th, 1786.

Add

Assumed day
Epact
Number of the month

=25

=29

Take the days since the last full moon= 9
From the assumed day=25

To the day of the full moon=16th,

Add 15

PROBLEM X.

The time of the Moon's coming to the South, after the Sun, being given, to find the age of the Moon.

RULE. AS 24 hours, the whole difference of time, are to 30, the number of days from change to change, so is the difference of time, to the Moon's age.

EXAMPLE. I observed the Moon to be on the meridian, or due south, at 6 o'clock in the afternoon: What is the Moon's age?

PROBLEM XI.

24: 30 :: 6 : 74 days, Ans.

To find the time of the Moon's Southing.

RULE. Multiply the Moon's age, on the given day, by 48 minutes, and divide the product by 60, the minutes in an hour, (or multiply by 4, and divide by 5) and the quotient will show how many hours and minutes the moon is later, in coming on the meridian, than the sun, and counting so many hours and minutes forward from 12 o'clock, we have the time of the Moon's southing; if the hours and minutes, found as above, be less than 12, then, that will be the time of the Moon's southing after noon; but, if greater than 12, then, take 12 from them, and the remainder will be the time of the Moon's southing in the morning.

EXAMP. 1. Required the time of the Moon's southing on the 25th. day of January 1786?

Moon's age =24

Or, 24

h. m.

48

4

60)480(8 0 afternoon, is the time of the Moon's

Note. From the change to the full, the Moon comes to the south afternoon; but from the full to the change, before noon.

PROBLEM

PROBLEM XII.

To find on what day of the week, any given day in any month will fall. As one of the first seven letters of the alphabet is prefixed to every day in the year beginning with A, which is always prefixed to the first day of January: And as, in common years, the letter, annexed to the first Sunday in January, shews the Dominical Letter for that year; but every leap year having two Dominical Letters, the first of which serving to the twenty fourth of February, and the other for the rest of the year, consequently, in any common year, the Dominical Letter being known, the first of January may be easily found, reckoning from A according to the natural order of the letters: and in any leap year, the first of its two Dominical Letters will shew as above, counting from A 1, B 2, C 3, &c. and by counting backward, you may have the day of the week, on which the first of January will happen.

RULE. Find the day of the week answering to the first of January that year, then add together the days contained in each month from the beginning of the year to the proposed day of the month inclusively; divide this sum by 7, and if any thing remain, after the division, then, count so many forward, beginning with that day on which the first of January falls, and yon will have the day of the week, on which the proposed day will fall: but if nothing remain, then the day of the week, preceding that day on which the first of January falls, answers to the proposed day.

EXAMPLE.

On what day of the week will the 5th day of May 1786 fall?

PROBLEM XIII.

To Find the Cycle of the Sun.

RULE. Add 25* to the given year; divide the the remainder, after division, is the Cycle required; remain, the Cycle is 28.

6 from Jan. 1.

sum by 28, and but if nothing EXAMPLE.

* From the commencement of this century, 9+16=25 must be added to the given year. The leap year having been omitted in the year 1800, makes it neceffary to add 25 to the date of the year, and then dividing by 28, it will give the Cycle right during the prefent century. And this is a general rule to be obferved, that when a leap year has been abated, add 16 to the number which was before added to the year, rejecting 28, when it exceeds it, and this number beingadded to the year, and the fum divided by 28, the remainder after divifion, will be the Cycle for finding the Dominical Letter. Thus in the nineteenth century, it will be 9+16=25, and in the twentieth century 25+16-28-13, which number will ferve two centuries, for the year 2000 is a leap year.