PROB. XXX. To find the area of a space bounded on one side by a curve line. RULE. Let perpendiculars be erected upon the base, so numerous, that the part of the curve between any two nearest to one another shall differ very little from a straight line. Then add the perpendiculars at the extremities of the base, if there are any, and to half their sum add the rest of the perpendiculars. Multiply the sum by the base, and divide the product by the number of parts into which the base is divided by the perpendiculars: the quotient will be the area nearly. 1. Suppose the perpendiculars at the extremities of the base to be 10 and 16, and the other perpendiculars to be 11, 14, 16, and the base to be 20 feet. (10+16) × 13 B a A a b C 11 14 16 54 20 4) 1080 Ans. 270 square feet the area. 2. A curve-lined space meets the base at one of its extremities, and the perpendicular at the other extremity is 96, the other perpendiculars are 83, 70, 64, 51, 38, 25, and the base 325 links. What is the area? Ans. 17596 square links. 3. An offset meets the base at both extremities, the base is 252 links, and the perpendiculars are 24, 36, 42, 54, 67, 76, 58, 49, 33, and 19. Required the area. Ans. 10492 square links. 4. Perpendiculars were raised from the base to a curve; those at the ends were 364 and 578, the others were 396, 418, 453, 512, and 554 links, the base 1260 links. Ans. 5 acres 3 roods 22 perches 4 yards 3.2 feet. 5. A curve meets the base at one extremity, the base is 2364, the perpendicular at the other extremity 758, and the others are 642, 587, 524, 432, 417, and 335 links. Ans. 11198604 links, 11 acres 31 perches 23.5 yards. NOTE 1. This rule supposes the figure to be divided into trapeziods, and would be exact if the breadths of the trapeziods were all equal. But the common rule is to add all the perpendiculars, and to multiply by the base, and divide by the number of perpendiculars; which is not much easier, and gives the answer sometimes considerably erroneous. Thus the third example would come to 11541·6. NOTE 2. If the distances between the perpendiculars be equal, the curvature, if single, may be considered as parabolical. And taking care to have an odd number of perpendiculars, add the first and last perpendiculars into one sum, the second, fourth, &c. into another, and all the rest into a third sum; then add the first sum, twice the third, and four times the second sum together, multiply this by the base, and divide by three times the number of parts into which the base is divided. The quotient is the area. Thus, in the first example, the first sum is 26, the second 27, and the third 14; therefore (26+4×27+2×14) × × 20=270. MENSURATION OF SOLIDS. THE SOLID CONTENT of a body is the number of cubical inches, feet, &c. which the body contains. A CUBICAL INCH is a solid contained by six square inches ; or it is a solid, of which the length, breadth, and thickness are each of them an inch. And the same is to be understood respecting a cubical foot, yard, &c. NOTE. 231 cubical inches make a wine gallon, 282 cubical inches make an ale gallon, 2150-42 cubical inches make a malt bushel, and 104-2 such inches make a Scotch pint. All these measures are now laid aside by act of parliament, and the only legal standard for measuring both liquid and dry goods is declared to be the imperial gallon, containing 10 pounds avoirdupois weight of distilled water weighed in air at the temperature of 62 degrees of Fahrenheit's thermometer, the barometer being at 30 inches; each avoirdupois pound containing 7000 troy grains. It is declared that this gallon is to contain 277-274 cubic inches of rain water. A pint is the eighth part of a gallon, 8 gallons make a bushel of 4 pecks, and 8 bushels make a quarter. Hence a wine gallon is 0-8331109 imperial gallon, an ale gallon 1017045 imperial gallon, a Winchester bushel 0-969448 imperial bushel, a Scotch wheat firlot 0.998256 imperial bushel, a Scotch barley firlot 1-4562794 imperial bushel, and a Scotch pint 0.375814 imperial gallon. PROB. I. To find the surface of a prism. A PRISM is a solid of which the ends are equal, similar, and parallel rectilineals, and the other sides are parallelograms. NOTE. If the ends be parallelograms, the prism is called a Parallelopiped; and if all its sides be squares, it is called a Cube. RULE. Find the area of one of its ends, and to its double add the sum of the areas of the parallelograms. 1. Required the surface of a cube, upon a line of 37 inches. Ans. Surface 8214 square inches. 2. Required the surface of a rectangular parallelopiped, of which the length is 11 feet, and each side of the base 27 inches. Ans. 109 feet 18 inches. 3. Required the surface of a pentagonal prism, the length 14 feet, and each side of the base 33 inches. Ans. 218-52 feet. TO FORM A PRISM WITH PASTEBOARD. Let ABCD be one of the parallelograms of which the sides are compounded, AB the length, and AD a side of the base. Extend AD and BC, and make the parallelograms DK, AL, FM, &c. each equal to AC, and upon AD and BC make figures equal to the bases. Then if the figure thus formed be cut out of the pasteboard, and folded at the sides of the parallelograms till they meet, T HGF A NM L B the prism will be formed, and its surface is the figure cut out. 4. Required the surface of a chest, of which the length is 7 feet 8 inches, the breadth 4 feet 7 inches, and the depth 2 feet 9 inches. Ans. 137 feet 7 inches 10 parts. 5. Required the surface of a triangular prism, of which the length is 13 feet, and the sides of the base 23, 34, and 19 inches. Ans. 85.2241 square feet. PROB. II. To find the solid content of a prism. RULE. Find the area of one of the ends, and multiply it by the length or perpendicular height. NOTE. If the height be one foot, the solid will contain as many cubical feet as there are square feet in the base; if the height be two feet, the solid will contain twice as many cubical feet; if the height be three feet, it will contain three times as many, and so on. 1. Required the solid content of a triangular prism, of which the height is 9 feet, and each side E of the base 34 inches. Tabular number 0.4330127 34 X 34 = 1156 square 500.5626812 9 feet. 144) 4505-0641308 inches. Ans. Content 31.2851676 cubic feet. 2. Required the solid content of a rectangular cistern, of which the length is 3 feet 2 inches, the A breadth 2 feet 8 inches, and the depth 2 feet 6 inches. Ans. 21 feet 1 inch 4 parts. 3. Required the solid content of a heptagonal prism, of which the length is 21 feet, and each side of the base 43 inches. D B C H E F F Ans. 979-86934 cubic feet. 4. Required the solid content of a pentagonal prism, the length 23 feet, and each side of the base 54 inches. Ans. 801 cubic feet 539-739 cubic inches. 5. Required the solid content of a quadrilateral prism, the length 19 feet, the sides of the base 43, 54, 62, and 38, and the diagonal between the first and second 70 inches. Ans. 306 cubic feet 81.976 inches. PROB. III. To find the surface of a cylinder. A CYLINDER is a round solid of uniform thickness, of which the bases are equal and parallel circles. RULE. Multiply the circumference of the base by the height: the product is the curve surface, to which add the areas of the two bases. See Appendix, Prop. 79. 1. What is the curve surface of a cylinder, of which the length is 16 feet, and the diameter of the base 27 inches? 3.1416 7.0686 16 Ans. Surface 113-0976 square feet. 2. Required the whole surface of a cylinder 13 feet long, and having the circumference of its base 57 inches. Ans. 65-3409 square feet. |