320. To find the solidity of a sphere.* RULE.-Multiply the cube of the diameter by .5236, or multiply the square of the diameter by one 6th of the circumference. 1. What is the content of a sphere whose diameter is 12 inches? 12×12×12×.5236 904.7808, Ans. 2. What is the solid content of the earth, its circumference being 25000 miles? Ans. 26385814912 miles. Guaging. 321. Guaging teaches to measure all kinds of vessels, as pipes, hogsheads, barrels, &c. RULE. To the square of the bung diameter add the square of the head diameter; multiply the sum by the length, and the product by .0014 for ale gallons, or by .0017 for wine gallons. 1. What is the content of a cask, whose length is 40 inches, and its diameters 24 and 32 inches? 32x32+24x24x40-64000, Ans. 64000X.0014-89.6 a. gal., Áns. 64000.0017-108.8 w. gal., Ans. 2. What is the content of a cask whose length is 20 inches, and diameters 12 and 16? 11.2 a. gal. 13.6 w. gal. Ans.{ SECTION III. PHILOSOPHICAL MATTERS. 1. Of the Fall of Heavy Bodies. 322. Heavy Bodies near the surface of the earth, fall one foot the first quarter of a second, three feet the second quarter, five feet the third quarter, and seven feet the fourth quarter, equal to 16 feet the first second. The velocities acquired by falling bodies, are in proportion to the squares of the times in which they fall; that is, if 3 bullets be dropped at the same time, and the first be stopped at the end of the first second, the second at the end of the second, and the third at the end of the third, the first will have fallen 16 feet, the second (2X2=4) four times 16, equal to 64; and the third (3×3=9) nine times 16, equal to 144 feet, and so on. Or, if 16 The surface of a sphere is found by multiplying its diameter by its circumference feet be multiplied by so many of the odd numbers, beginning at 1, as there are seconds in the given time, these several products will be the spaces passed through in each of the several seconds, and their sum will be the whole distance fallen. 323. The velocity given to find the space fallen through. RULE.-Divide the velocity in feet by 8, and the square of the quotient will be the space fallen through to acquire that velocity. 1. From what height must a body fall to acquire the velocity of a cannon ball, which is about 660 feet per second? 6608-82.5, and 82.5X 82.5-806.25ft. 137 miles, Ans. 2. From what height must a body fall to acquire a velocity of 1200 feet per second? Ans 22500 feet. 324. The time given to find the space fallen through. RULE.-Multiply the time in seconds by 4, and the square of the product will be the space fallen through in the given time. 1. How many feet will a body fall in five seconds? 5×4=20, and 20×20=400 feet, Ans. | 3. Ascending bodies are retarded in the same ratio that descending bodies are accelerated; therefore, if a ball, fired upwards, return to the earth 2. A stone, dropped into a in 16 seconds, how high did it well, reached the bottom in 3 ascend? The ball being half seconds; what was its depth? | the time, or 8 seconds, in its 3×4-12, and 12×12=144 | ascent: therefore 8X4-32, feet, Ans. and 32x32 1024ft., Ans. 325. The velocity per second given to find the time. RULE.-Divide the given velocity by 8, and one fourth part of the quotient will be the answer. 1. How long must a body be falling to acquire a velocity of 160 feet per second? 160-8-20, and 20-4-5 seconds, Ans. 2. How long must a body be falling to acquire a velocity of 400 feet per second? Ans. 123 seconds. 326. The space given to find the time the body has been falling. RULE.-Divide the square root of the space fallen through by 4, and the quotient will be the time. 1. In how many seconds will a body fall 400 feet? 400-20, and 2045 seconds, Ans. 2. In how many seconds will a bullet fall through a space of 11025 feet? Ans. 26 seconds. 327. To find the velocity per second, with which a body will begin to descend at any distance from the earth's surface. RULE.-As the square of the earth's semi-diameter is to 16 feet, so is the square of any other distance from the earth's centre, inversely, to the velocity with which it begins to descend per second. 1. Admitting the semi-diameter of the earth to be 4000 miles, with what velocity per second will a body begin to descend, if raised 4000 miles above the earth's surface? As 4000x4000: 16: : 8000 X8000: 4 feet, Ans. 2. How high above the earth's surface must a ball be raised, to begin to descend with a velocity of 4 feet per second? Ans. 4000 miles. 328. To find the velocity acquired by a falling body, per second, at the end of any given period of time. RULE.-Multiply the perpendicular space fallen through by 64, and the square root of the product is the velocity required. 1. What velocity per second does a ball acquire by falling 225 feet? 225×64 14400, and 14400 120, Ans. 2. If a ball fall 484 feet in 5 seconds, with what velocity will it strike? Ans. 176. 329. The velocity with which a body strikes given to find the space fallen through. RULE.-Divide the square of the velocity by 64, and the quotient will be the space required. 1. If a ball strike the ground with a velocity of 56 feet per second, from what height did it fall? 56×56 64-49 feet, Ans. 2. If a stream move with a velocity of 12.649 feet per second, what is its perpendicular fall? Ans. 24 feet. 330. To find the force with which a falling body will strike. RULE. Multiply its weight by its velocity, and the product will be the force. 1. If a rammer for driving piles, weighing 4500 pounds, fall through the space of 10 feet, with what force will it strike? 10X64-25.3 velocity, and 25.3X4500-113850lb. Ans. 2. With what force will a 42lb. cannon ball strike, dropped from a height of 225 feet? Ans. 5040lb. 2. Of Pendulums. 331. The time of a vibration, a cycloid, is to the time of a heavy body's descent through half its length as the circumference of a circle to its diameter; therefore to find the length of a pendulum vibrating seconds, since a falling body descends 193.5 inches in the first second, say, as 3.1416X3.1416: 1x1:: 193.5,19.6 inches the length of the pendulum, and 19.6X2=39.2 inches, the length. 332. To find the length of a pendulum that will swing any given time. RULE.-Multiply the square of the time in seconds, by 39.2, and the product will be the length required in inches. 1. What are the lengths of three pendulums, which will swing respectively seconds, seconds, and two seconds? .5X.5X39.2 9.8 in. for seconds. Ans. 2. What is the length of a pendulum, which vibrates 4 times in a second? .25.25×39.2 2.42 inches, Ans. 3. Required the lengths of 2 pendulums, which will respectively swing minutes and hours? 60X60X39.2-141120in.-2m. 1200 feet. 3600X3600×39.2508032000-8018m. 960 feet. Ans. 333. To find the time which a pendulum of a given length will swing. RULE.-Divide the given length by 39.2, and the square root of the quotient will be the time in seconds. 1. In what time will a pendulum 9.8 inches in length vibrate? ✔/9.839.2—.5, or second, Ans. 2. I observed that while a ball was falling from the top of a steeple, a pendulum 2.45 inches long, made 10 vibrations; what was the height of the steeple? 2.45-39.2=25s. and .25X 10-2.5s.; then 2.5×4=10, and 10×10 100 feet, Ans. 334. To find the depth of a well by dropping a stone into it. RULE.-Find the time in seconds to the hearing of the stone strike, by a pendulum; multiply 73088 (=16×4×1142; 1142 feet being the distance sound moves in a second), by the time in seconds; to this product add 1304164 (the square of 1142), and from the square root of the sum take 1142; divide the square of the remainder by 64 (=16x4), and the quotient will be the depth of the well in feet; and if the depth be divided by 1142, the quotient will be the time of the sound's ascent, which, taken from the whole time, will leave the time of the stone's descent. 1. Suppose a stone, dropped into a well, is heard to strike the bottom in 4 seconds, what is the depth of the well? 73088X4+1304164-1142-121.53, and 121.53x121.53 64-230.77 feet, Ans. Then 230.77-1142—.2 of a second, the sound's ascent, and 4-2-3.8 seconds, stone's descent. B. Of the Lever. 335. It is a principle in mechanics that the power is to the weight as the velocity of the weight is to the velocity of the power. 336. To find what weight may be balanced by a given power. RULE. As the distance between the body to be raised or balanced, and the fulcrum, or prop, is to the distance between the prop and the point where the power is applied, so is the power to the weight which it will balance. 1. If a man weighing 160 lb. rest on a lever 12 feet long, what weight will he balance on the other end, supposing the prop to be 1 foot from the weight? 1:11:: 160: 1760 lb. Ans. 2. At what distance from a weight of 1440 lb. must a prop be placed, so that a power of 160 lb. applied 9 feet from the prop may balance it? 1440 160: 9:1 foot, Ans. 3. In giving directions for making a chaise, the length of the shafts between the axletree and back band being settled at 9 feet, a dispute arose whereabout on the shafts the centre of the body should be fixed; the chaise maker advised to place it 30 inches before the axletree; others supposed that 20 inches would be a sufficient incumbrance for the horse. Now suppos |