If the magnitude of any body be multiplied by its specific gravity, the product will be its absolute weight. EXAMPLE. What weight of rose copper will be necessary to cover a man of war’s bottom, measuring 4000 feet, and the thickness of the copper ido of a foot: ido of 4000 equal 40 cubic feet, and its specific gravity is 9000 oances Avoirdupois per cubic foot: therefore 40 X 9000 = 360000 oz. = 10 tons O cwt. 3 qrs, 16 lbs. Ans. CHRONOLOGICAL CYCLES. To find the dominical letter for any year, according to the Julian (or old stile) method of calculation. RULE. Add to the given year its fourth part, and 4, and divide that sum by 7, and if nothing remains, the dominical letter is G; but if there be a remainder, it shews the letter in a retrograde order from G, beginning the reckoning with F, E, &c. Or if it be subtracted from 7, you will have the index of the letter, beginning with A, counting as follows: Required the dominical letter for 1827? 1827 4 2287--7=326 and 5 over: then the dominical letter will be B. Thus 7-5=2, which count, A 1, B 2. Or begin at F, and count backwards, and 5 will come to B; therefore B is the dominical letter for 1827. To find the dominical letter for any year, according to the new stile, or Gregorian method, RULE. Divide the year and its fourth part by 7, subtract the remainder, after the division, from 7, and this remainder will be the index of the dominical letter, as before. If nothing remains, it will be G. EXAMPLE Required the dominical letter for 1827? 1827 Fourth part 456 -2283 -7= 326 and 1 over; then 7-16. Therefore F is the dominical letter. NOTE. Every leap-year has two dominical letters: that found last is the dominical letter from the 25th Feb. to the end of the year; and the next in the order of the alphabet serves from the 1st Jan. to Feb. 94th. To find the golden number. Rule. Add 1 to the given year, divide that sum by 19, - and the remainder, after the division, will be the golden number : if nothing remains, then 19 will be the golden number, EXAMPLE Required the golden number for 1827? 1827+1=1828--19–96 and 4 over; therefore 4 is the golden number, To find the Julian epact, old stile. RULE. First find the golden number, which multiply by 11, and the product, if less than 30, will be the number required; but if the product exceeds 30, divide it by 30, and the remainder is the epact. EXAMPLE Required the Julian (or old stile) epact for 18277 Golden number 4 X11=430=1 and 14 over; therefore 14 is the épacto To find the Gregorian (or new stile) epact. RULE. Subtract 11 from the Julian epact, and if the subtraction cannot be made, add 30, and then subtract, and the remainder will be the Gregorian epact: but if nothing remains, 29 will be the epact. EXAMPLE. Required the Gregorian (or new stile) epact for 1827? Julian epact as above 14-11=3, and 3+30=33-30=3 Greg. epact. To find the solar or cycle of the sun, RULE. Add 9 to the given year, divide that sum by 28, and the remainder, after division, is the cycle required: but if nothing remains, the cycle will be 28. EXAMPLE. Let it be required to find the solar cycle for 1827? 1827 +9=1836-28-65 and 16 over ; therefore 16 is the solar eyele: To find the Roman indiction. Rule. Add 3 to the given year, divide that sum by 15, and the remainder, after division, will be the indiction: but if nothing remains, it will be 15. EXAMPLE. Required the year of indiction for 1827? 1827+3=1830*15=122 and no remainder: 15 is the Roman indiction. To find the Julian period. 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 1827 7 To find the year of the christian æra, by having the cycle of the sun, the golden number, and Roman indiction. Rule. Multiply 4845 by the cycle of the sun, 4200 by the golden number, 6916 by the Roman indiction, add the several products together, and divide the sum by 7980, and the remainder, after division, will be the Julian period, from which subtract 4713, and the remainder will be the year of the christian æra. EXAMPLE. The cycle of the sun being 16, the golden number 4, and the Roman indiction 15, what year of the christian æra is it? 4845x16= 77520 -198060; 7980=24, and 6540 over, which 6540 is the Julian period before found. Now 6540 — 4713 - 1827, the year of the christian æra. To find the year of the Dionysian period. RULE. Add to the given year 457, divide that sum by 532, and the remainder will be the number required. EXAMPLE. Required the year of the Dionysian period for 1827? 1827 + 457=2284-532=4, and 156 over; therefore 156 is the Dionysian period for 1827. To find the age of the moon 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 exceeds 30, take 30 from it, and the remainder will be the moon's age. A Table for the number of the months. Jan Feb Mar Apr " Deci 10 2 EXAMPLE. Required the age of the moon this day, 28th Sept. 1827? 28 day of the month.-. To find the time of the new and full moon, and the first and last quarters. RULE." Find the moon's age on the given day, and if it be 15, the moon will be full on that day. By counting 71 days backwards and forwards, you will have the first and last quarters; and by counting backwards and forwards 15 days, you will have the time 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. By counting these backwards, you will have the day the last full moon happened on, by knowing which, you can find the change, or either of the quarters, as before. Again, if the age of the moon, on the given day, be less than 15, take that from 15, and the remainder will shew how many days are to come, 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 quarters, as above. EXAMPLE. For September the 28th, 1827. 28 4-3+-8339—30=9 days, the age of the moon, as before, being less than 15 : then 15-9=6 days the moon has to rojo till the next full moon, Proved thus: The assured day 28 6=34 days, from which take the days in the month of September, and you will find the day we will have full moon in the next month. Thus, 34–30=4, bringing the full moon on the 4th October. By adding 7 days to the last 4 days, (with the odd hours of the 4th Oct.) you will have the last quarter to be on the 12th of Oct. and by taking 7 davs from the 4th Oct. you will have the first quarter on the 27th of Sept. By taking 7 days from the 27th Sept. you will have new moon on the 20th September, &c. NOTE. From the change to the full, the moon comes to the south in the afternoon; but from the full to the change, it comes before noon. A PERPETUAL ALMANACK.. Jan. * Feb. Jan. Sept. June Dec. July 5 6 7 12 13 14 21 28 Days of the Years. Years. Days of the week. Friday 1847 Monday The use of the perpetual almanack. To find on what day of the week any given day in any month will fall, EXAMPLE. On what day of the week is the 28th Sept. 1827? Observe the day of the week annexed to the year 1827, in the column, which you will find to be Thursday; and in the table ander the given month, (Sept.) in the sixth row of figures, you will find Thursday to be the 6th, 13th, 20th, and 27th of that month : then count forwards to the 28th, and you will have Friday to be the day required. Again, what day of the week will next Christmas day fall on? Look as before to the column of 1827, and you have Thursrlay annexed to it, and then look in the sixth row of figures for December, and you will find tbat Thurs. day is the 6th, 13th, 20th, and 27th days of that month: then count back. wards, thus, 'Thursday the 27th, Wednesday the 26th, Tuesday the 25th, which is Christmas day. Therefore Tuesday is the 25th Dec. 1827, and so on, NOTE. In leap-years, January and February must be taken in the colama marked * Leap-years are marked |