March 1 September 8 To April 2 8 October EXAMPLE. For January 25th, 1786. Given day Add Epact =25 =29 No. of the month =00 54 Subtract 30 24 Moon's age. PROBLEM IX. 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 thesc backward, you will have the day the last full moon happened on, and by knowing that, we can find the change, or either of the quarters, as before. Again, if the age of the moon, on the assumed day, be less than 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 January 25th, 1786. Assumed day Add Epact =25 =29 Number of the month=00 54 Subtract 30 Moon's age=24 Subtract 15 Take the days since the last full moon= 9 To the day of the full moon=16th The time of the Moon's coming to the South, after the Sun, being given, tc 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? 24 306 71 days, Ans. PROBLEM XI. 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, h. m. h. m. Hence the Moon 60)1152(19 12 19 19 12 as before. 60)480(8 0 afternoon, is the time of the Moon's 48 [southing. Note. From the change to the full, the Moon comes to the south afternoon; but from the full to the change, before noon. 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 you 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? By the preceding observations, and by Prob. 4th, the first of January is found to fall on Sunday. Now, counting forward six days from Sunday, the first of January (inclusively) the 5th of May falls on Friday, Jan. 31 Feb. 28 May March 31 7)125(17 7 55 49 6 from Jan 1. RULE. Add 25* to the given year; divide the sum by 28, and the remainder, after division, is the Cycle required; but if nothing remain, the Cycle is 28. A TABLE OF THE DOMINICAL LETTERS FOR THE NEW STYLE, Cycle. Letter.||Cycle. | Letter.|| Cycle.|Letter. || Cycle. Letter. RULE. PROBLEM XIV. To find the year of the Dionysian Period. Add to the given year 457; divide the sum by 532, and the remainder will be the number required. EXAMPLE. Required the year of the Dionysian Period for the year 1786 ? To 1786 Add 457 532)2243(4 2128 115 Dionysian Period. 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 necessary to add 25 to the date of the year, and then dividing by 28, it will give the Cycle right during the present century. And this is a general rule to be observed, that when a leap year has been abated, add 16 to the number which was PROBLEM XV. To find the year of Indiction. RULE. Add 3 to the given year; divide the sum by 15, and the remainder, after division, will be the Indiction; if nothing remain, it will be 15. RULE. Add 4713 to the given year, and the sum will be the Julian Period. EXAMPLE. What year of the Julian Period will answer to the year 1786 ? To 1786 6499 Ans. PROBLEM XVII. To find the Cycle of the Sun, Golden Number, and Indiction, for any current year. RULE. To the current year add 4729 ;* divide the sum by 28, 19 and 15, respectively, and the several remainders will be the numbers required; when nothing remains, the divisor is the number required. EXAMPLE. What are the Cycle of the Sun, Golden Number, and Indiction, for the year 1807? before added to the year, rejecting 28 when it exceeds it, and this number being added to the year, and the sum divided by 28, the remainder after division, 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 serve two centuries, for the year 2000 is a leap year. *For any year in the nineteenth century add 47134-16-4729. |