1. Required the content of a cask of the first variety, in ale gallons, whose length is 40, bung and head diameters 32 and 24 inches. Thus, 32 — 24 — 8 × ·7 = 5·6 + 24 = 29·6, the mean di ameter. = 876.16 x 40: Then 29.62 × 40 2. Required the content of a cask of the second variety, in wine gallons, whose length is 20, bung and head diameter 16 and 12 inches. Ans. 14.25 wine gallons. SHIPS' TONNAGE.-CARPENTERS' TONNAGE. There are several methods by which to ascertain the measurement of ships, the following rule is probably the most correct. To find the tonnage by carpenters' measure. RULE. For single-decked vessels, multiply the length and breadth at the main beam, and depth of the hold together, and divide the product by 95. For double-decked vessels, take half the breadth of the main beam for the depth of the hold, and work as above. If the deck be bolted at any height above the main wale, it is customary to add half the difference to the former depth, for the depth used in calculating the tonnage. 1. What is the tonnage of a single-decked vessel whose length is 65 feet, breadth 20 feet 6 inches, and depth 8 feet 9 inches? Thus, 65 x 20.5 x 8.75 11659-37595 =3 122.73 tons. 2. What is the tonnage of a double-decked vessel whose length is 75 feet 9 inches, and breadth of main beam 24 feet 6 inches? Thus, 75-75 x 24·5 × 12·25 = 22734·46875 ÷÷ 95: = 239-31 tons. 3. What is the tonnage of a ship whose length is 97 feet, breadth 31 feet, depth 15.5 feet? = By another method.--Thus, the breadth 480-5 × 97 length = 46608·5 ÷ 94 15.5 x 31 breadth 495.83 tons. 4. Required the tonnage of a ship, of which the length is 75 feet, and the breadth 26 feet. Thus, 262 x 75 ÷ 188 = 270 tons, nearly. The content of Noah's ark is as follows, allowing the cubit to be 22 inches; length of keel 300 cubits, breadth of the midship beam 50 cubits, depth in the hold 30 cubits; its burthen as a man of war, 27729 tons; as a merchant ship, 29188-6 tons. FALLING BODIES. When a body descends freely by its own weight, the velocity is as the time, and space as the square of the time. The time and velocity then will be 1, 2, 3, &c. ; the space passed through as 4, 9, 16, &c.; and the spaces for each time as 1, 3, 5, 7, 9, &c. A falling body will descend through 16 feet in the first second of time, its velocity increasing so, that in the next second it will fall 32 feet. A Table, showing the time, space, and velocity. To find the velocity a falling body will acquire in a given time. RULE.-Multiply the time in seconds by 324, and it will give the velocity required in feet per second. 1. Required the velocity in 15 seconds. Thus, 15 x 321 482 feet. Ans. = 2. Required the velocity in 10 seconds. Ans. 321 ft. To find the velocity a body will acquire by falling from any given height. RULE.-Multiply the space in feet by 641, and the square root of the product will be the velocity acquired in feet per second; or, when the time is given, multiply the time in seconds by 321. 1. Required the velocity a ball has acquired in descending through 144 feet. Thus, 144 × 649312.2596.5 feet. 2. Required the velocity a ball has acquired in descending through 402, feet in 5 seconds. Ans. 160 feet. 12 To find the space through which a body will fall in any given time. RULE.-Multiply the square of the time in seconds by 16,12, and it will give the space in feet. 1. Required the space fallen through in 6 seconds. Thus, 6236 × 16,2 = 579 feet. Ans. 2. A bullet being dropped from the top of a building was 5 seconds in reaching the ground; required the height. Ans. 402 feet. To find the time that a body will be in falling through a given space. RULE.-Divide the space in feet by 162, and the square root of the quotient will give the required time in seconds. 1. Required the time a body will be in falling through 579 feet of space. (579162√366 seconds.) 2. How long will a body be in falling through 402 space? feet of Ans. 5 seconds. To find the space fallen through, the velocity being given. RULE.-Divide the velocity by 8·02, and the square of the quotient will be the distance fallen through to acquire that velocity. 1. If the velocity of a cannon ball be 579 feet per second, from what height must a body fall to acquire the same velocity? Thus, 5798.02 = 72.192 5211.3961 feet. 2. If the velocity of a cannon ball be 68918 8 feet per second, from what height must a body fall to acquire the same velocity? To find the time, the velocity per second being given. RULE. Divide the velocity by 321, and the quotient will be the time in seconds. 1. How long must a bullet be in falling to acquire a velocity of 772 feet per second? (7723224 seconds.) 2. How long must a bullet be in falling to acquire a velocity of 386 feet per second? Ans. 12 seconds. Ascending bodies are retarded in the same ratio that descending bodies are accelerated. To find the space moved through by a body projected upward with a given velocity. RULE.-Multiply the square of the time in seconds by 162, (see Table) the velocity of the projection in feet by the number of seconds the body is in motion, and the sum of these products will be the space in feet when projected downward; and the dif ference of the products will give the distance of the body from the point of projection, when projected upward. 1. If a rocket, projected upward, return to the earth in 12 seconds, how high did it ascend? The rocket is half the time in ascending 12 ÷ 2 × 6 = 1158; 62 Answer. = 36 × 1612 = 579; Ĭ158 579 6; 193 = 579. 2. A bullet is dropped from the top of a building, and found to reach the ground in 1.75 seconds; required the height. 1.757, and 7 x 749 feet. Ans. 3. In what time will a ball dropped from the top of a steeple 484 feet high, come to the ground? 484 =22 ÷ 4 = 5 seconds. Ans. To find the velocity per second with which a heavy 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 began to descend per second. 1. With what velocity per second will an iron ball begin to descend, if raised 3000 miles above the earth's surface? As 4000 x 4000: 16: 4000 + 3000 + 3000 : 5-22449 ft. Answer. 2. How high must a ball be raised above the earth's surface to begin to descend with a velocity of 5-22449 feet per second? Thus, 16: 4000 × 4000 :: 5-22449: 49000000. To find the velocity of a falling stream of water per second, at the end of any given time, the perpendicular distance being given. RULE. The velocity required at the end of every period is equal to twice the mean velocity with which it passes during that period. Or 2d, Multiply the perpendicular space fallen through by 64, and the square root of the product is the velocity required. 1. There is a sluice or flume, one end of which is 2.5 feet lower than the other; what is the velocity of the stream per second? Thus, 2.5 x 64 = 160; and 160 12-649 feet. Ans. The weight of a body, and the space fallen through given, to find the force with which it will strike. RULE. Find the velocity by the preceding rule, and multiply it by the weight, which will produce the weight required. 1. The rammer used for driving the piles of a bridge weighed 2 tons or 4500 pounds, and fell through a space of 10 feet, with what force did it strike the pile? Thus, 10 x 64 25.3= 113850 pounds. Ans. velocity; and 25.3 × 4500 Then reverse the question and rule; if the aforementioned rammer weighed 4500 pounds, and struck with a force of 113850 pounds, from what height did it fall? Thus, 113850 ÷ 4500 = 25.3; and 25.3 × 25·3 ÷ 64 = 10 feet. Ans. To ascertain with what momentum or force a fluid moving with a given velocity strikes upon a fixed obstacle. RULE.-Divide the square of the velocity by 64, and the quotient will be the height required. Multiply the height by 62 pounds avoirdupois for clear water, and by 63 for unclean water, and by 64 for salt water. 1. Admit a stream of clear water to move at the rate of 5 feet per second, and to meet with a fixed obstacle (or bulk head) 15 feet wide and 4 feet high, what is the momentary pressure of the stream? |