ASTRONOMICAL PROBLEMS. Art. 332.-PROB. I. To find the dominical letter for any year in the present century, and also to find on what day of the week January will begin. RULE. To the given year add its fourth part, rejecting the fractions; divide this sum by 7; if nothing remains, the dominical letter is A; but, if there be a remainder, subtract it from 8, and the residue will show the dominical letter, reckoning 0=A, 2=B, 3=C, 4=D, 5=E, 6=F, 7=G. These letters will also show on what day of the week January commences. For, when A is the dominical letter, January begins on Sunday; when B is the dominical letter, January begins on Saturday; C begins it on Friday; D begins it on Thursday; E on Wednesday; F on Tuesday; G on Monday. 1. Required the dominical letter for 1825. 4)1825 8-6=2=B, dominical letter. 456 7)2281 325-6 As B is the dominical letter, January will begin on Saturday, and the second day will be the Sabbath. 2. Required the dominical letter for 1842. 3. Required the dominical letter for 1837. Ans. B. Ans. A. 4. What is the dominical letter for 1801 ? Ans. D. 5. What is the dominical letter for 1845 ? Ans. E. Art. 333.--PROB. II. To find on what day of the week any given day of the month will happen. RULE. Find by the last problem the dominical letter for the given year, and on what day in January will be the first Sabbath; and the corresponding days in the succeeding months will be as follows: Wednesday for February; Wednesday for March; Saturday for April; Monday for May; Thursday for June; Saturday for July; Tuesday for August; Friday for September; Sunday for October; Wednesday for November; Friday for December. Having found the day of the week for any day in the month, any other day may be easily obtained, as may be seen in the following example. 1. Let it be required to ascertain on what day of the week will be the 25th day of September, 1842. The dominical letter for 1842 is B; therefore, the 2d of January will be the Sabbath; and, by the above rule, the 2d of February will be Wednesday; the 2d of March will be Wednesday; the 2d of April will be Saturday; the 2d of May will be Monday; the 2d of June will be Thursday; the 2d of July will be Saturday; the 2d of August will be Tuesday; the 2d of September will be Friday. If the 2d be Friday, the 9th, 16th, and 23d will be Fridays. And if the 23d be Friday, the 24th will be Saturday, the 25th will be the Sabbath, the day required. 2. On what day of the week will be December 8, 1849 ? Ans. Saturday. 3. On what day of the week will happen July 4, 1857 ? Ans. Saturday. 4. On what day of the week were you born? OF BALLS AND SHELLS. Art. 334.-An iron ball, 4 inches in diameter, weighs 9 lbs., nearly; and a leaden one, 41, weighs about 17 lbs., and a pound of gunpowder measures about 30 cubic inches. Given the diameter of an iron ball, to find the weight, and the converse. RULE. Divide the cube of the diameter by 73; the quotient will be the weight in pounds. Multiply the weight by 7. The cube root of the product will be the diameter. 1. What is the weight of an iron ball, of which the diameter is 3 inches? Ans. 3.5 76.0293 lbs. 2. What is the diameter of an iron ball which weighs 24 lbs. ? Ans. √/24×73=170.6 5.547 inches diameter. 3. What does an iron ball weigh, whose diameter is 5.5 inches? Ans. 23.3965 lbs. 4. What is the diameter of an iron ball weighing 48 lbs? Ans. 6.988 inches. 5. What does an iron ball weigh whose diameter is 4.6 inches? Ans. 13.688 lbs. 6. What is the diameter of an iron ball which weighs 36 lbs? Ans. 6.349 inches. Art. 335.-Given the diameter of a leaden ball, to find its weight, and the converse. RULE. Divide the cube of the diameter by 41: the quotient will be the weight in lbs. Multiply the weight by 4: the of the product will be the diameter in inches. 1. What is the weight of a leaden ball, whose diameter is 4.25 inches? Ans. 4.253-4=17.059 lbs. 2. What is the diameter of a leaden ball which weighs 36 lbs. ? Ans. 5.45 inches. 3. What is the weight of a leaden ball, of 4.6 inches i diameter ? Ans. 21.63 lbs. 4 What is the diameter of a leaden ball weighing 48 lbs. Ans. 6 inches. PILING OF BALLS. Art. 336.-BALLS and shells are piled up in horizontal courses, upon a base of the form of an equilateral triangle, or of a square, or of a rectangle. The number of balls in a row diminishes, till, in the two first forms, it ends in a single ball, and in the last in a single row. The number of rows is equal to the number of balls in the lesser side of the under row. The number in the top row of a rectangular pile is one more than the difference between the length and breadth of the bottom row. Art. 337.-PROB. I. To find the number of balls in a triangular pile. RULE. Multiply the number of balls in a side of the bottom row by that number increased by 1, and again by that number increased by 2: the product, divided by 6, will be the number of balls in the pile. Required the number of balls in a triangular pile, of which each side of the base contains 30 balls. Ans. 4960. Art. 338.-PROB. II. To find the number of balls in a square pile. RULE. To twice the number of balls in a side of the bottom, add 1, and multiply the sum by the number in that row, and by that number increased by 1: the product, divided by 6, will give the number of balls in the pile. Let the side of the bottom row of a square pile contain 20 balls. How many balls are in the pile ? Ans. 2870. Art. 339.—PROB. III. To find the number of balls in a rectangular pile. RULE. From 3 times the number in the length of the bottom row, increased by 1, subtract the number in the breadth, and multiply the remainder by the breadth, and by the breadth increased by 1: the product, divided by 6, will give the number of balls in the pile. Suppose the number of balls in the length of a rectangular pile to be 59, and in the breadth 20, what is the number in the pile? Ans. 11060. Art. 340.-PROB. IV. To find the number of balls in an incomplete pile. RULE. From the number of balls in the complete pile subtract the number in the pile that is wanting, both computed as before: the remainder is the number in the incomplete pile. Required the number of balls in a rectangular pile of 15 courses, the numbers in the bottom row being 60 and 25. Ans. 14590. TO FIND THE WEIGHT OF CATTLE. Art. 341.-TAKE the girt behind the shoulder, and the length from the fore part of the shoulder-blade to the buttock, both in feet. Multiply the square of the girt by 4 times the length, and divide by 21. Multiply this quotient by 16, and it will give the weight of the four quarters, nearly. OBS.-The four quarters are little more than the whole weight. The skin weighs nearly, and the tallow very nearly 2 What will the four quarters of an ox weigh, whose girt is 6 feet 6 inches, and length 5 feet 10 inches. Ans. 6.52×231÷21×16=751+lbs. MISCELLANEOUS QUESTIONS. Art. 342.—1. What is the product of 2s. 6d., multiplied by 2s. 6d. ? Ans. £ 2. Purchased a book for 15 cents, and sold it for 18. did I gain per cent. ? What Ans. 20 per cent. 3. Sold a book for 18 cents, and gained 20 per cent. did it cost me? What 4. Purchased a book for 15 cents; sold it so as to gain 20 per cent. What did I get for it? 5. If of 3 of 7 of 3 of 5 of a vessel be worth £378, how many dollars is of it worth? Ans. $9450. 6. A person owning of a ship, sold of his share for $3750. What was the whole ship worth? Ans. $15000. 7. What is the sum of the third and half third of 3s. 4d.? 8. How many solid feet in a stick of and 6 feet 5 inches long? 9. A man owning of a farm, sold What was the value of the farm? Ans. 1s. 8d. timber 17 inches square, Ans. 12 ft. 1517 in. of his share for $245. Ans. $1225. 10. A. holds B.'s note for $2000, dated June 1st, 1825, on which are the following endorsements, viz: Received Sept. 1st, 1825... Dec. 10th, 1825. Apr. 20th, 1826 July 1st, 1826.. How much remains due June 1st, 1827? 11. What principal, at 5 per cent., will 9 years? .$96. 15. 36. .200. 20. 90. Ans. $1767.228. amount to $725 in Ans. $500. |