In a double-decked vessel, the length is reckoned from the fore part of the main stem to the after side of the sternpost above the upper deck; the breadth is taken at the broadest part above the main wales, and half this breadth is taken for the depth. In a single-decked vessel the length and breadth are taken as for a double-decked vessel, and the distance between the ceiling of the hold and the under side of the deck plank is taken as the depth. EXAMPLE. -The length of a double-decked vessel is 260 feet, and the breadth is 60 feet; required the tonnage. 5 = 224×60×6=403200 ÷ 95=4244.2 tons. Ans. EXAMPLE. -The length of a single-decked vessel is 180 feet, the breadth 34 feet, and depth 18 feet; required the tonnage. 180 of 34 = 159.6 × 34 X 18 ÷ 95 = 1028.16 tons. Ans. CARPENTER'S MEASUREMENT. For a double-decked length of keel X breadth main beam breadth For a single-decked 95 length of keel X breadth main beam X depth of hold 95 tonnage. tonnage. OF CONDUITS OR PIPES. Pressure of Water in Vertical Pipes, &c. λ= height of column in inches; o circumference of column in inches; t = thickness of pipe in inches equal in strength to lateral pressure at base of column; wweight of a cubic inch of water in pounds; C cohesive strength in pounds per inch area of transverse section of the material of which the pipe is composed-TABLE, p. 72. ho area of interior of pipe in inches; hw pressure in pounds per square inch at the base of the column, or maximum lateral pressure in pounds per square inch on the pipe tending to burst it; how maximum lateral pressure in pounds on the pipe, tending to burst it at the bottom; and how 2 mean lateral pressure in pounds on the pipe, or pressure in pounds on the pipe tending to burst it at half the height of the column. how ÷ C = t; how÷t = C; Ct ÷ ow h; Cthw = = 0. NOTE. -The reliable cohesion of a material is not above its ultimate force, as given in the Table of Cohesive Forces. By experiment, it has been found that a cast iron pipe 15 inches in diameter and of an inch thick, will support a head of water of 600 feet; and that one of the same diameter made of oak, and two inches thick, will support a head of 180 feet: 12000 lbs. per square inch for cast iron, 1200 for cak, 750 for lead, are counted safe estimates. The ultimate cohesion of an alloy, composed of lead 8 parts and zinc I part, is 3000 pounds per square inch. Concerning the Discharge of Pipes, &c. Small pipes, whether vertical, horizontal, or inclined, under equal heads, discharge proportionally less water than large ones. That form of pipe, therefore, which presents the least perimeter to its area, other things being equal, will give the greatest discharge. A round pipe, consequently, will discharge more water in a given time than a pipe of any other form, of equal area. The greater the length of a pipe discharging vertically, the greater the discharge. Because the friction of the particles against its sides, and consequent retardation, is more than overcome by the gravity of the fluid. The greater the length of a pipe discharging horizontally, the less proportionally will be the discharge. The proportion compared with a less length is in the inverse ratio of the square root of the two lengths, nearly. Other things being equal, rectilinear pipes give a greater discharge than curvilinear, and curvilinear greater than angular. The head, the diameters and the lengths being the same, the time occupied in passing an equal quantity of water through a straight pipe is 9, through one curved to a semicircle 10, and through one having one right angle, otherwise straight, 14. All interior inequalities and roughness should be avoided. It has been ascertained that a velocity of 60 feet a minute (1 foot a second) through a horizontal pipe, 4 inches in diameter and 100 feet in length, is produced by a head 24 inches, only of an inch above the upper surface of the orifice; and that, to maintain an equal velocity through a pipe similarly situated, of equal length, having a diameter of inch only, a head of 12 feet is required. To increase the velocity through the last mentioned pipe to 2 feet a second, requires a head 419 feet; to 3 feet, a head of 10; to 4 feet, a head of 1719, &c. From the foregoing, the following, it is believed, reliable rules, are deduced. To find the velocity of water passing through a straight horizontal pipe of any length and diameter, the head, or height of the fluid above the centre of the orifice, being known. RULE. Multiply the head, in feet, by 2500, and divide the product by the length of the pipe, in feet, multiplied by 13.9, divided by the interior diameter of the pipe in inches; the square root of the quotient will be the velocity in feet per second. EXAMPLE. -The head is 6 feet, the length of the pipe 1340 feet, and its diameter 5 inches; required the velocity of the water passing through it. 2500 X 6 = 15000 ÷ (1340X13.9) second. 4.03 =2 feet per Ans. To find the head necessary to produce a required velocity through a pipe of given length and diameter. RULE. Multiply the square of the required velocity, in feet, per second, by the length of the pipe multiplied by the quotient obtained by dividing 13.9 by the diameter of the pipe in inches, and divide the product thus obtained by 2500; the quotient will be the head in feet. EXAMPLE. The length of a pipe lying horizontal and straight is 1340 feet, and its diameter is 5 inches; what head is necessary to cause the water to flow through it at the rate of 2 feet a second? 22 X 1340 X 13.9 ÷ 25006 feet. Ans. To find the quantity of water flowing through a pipe of any length and diameter. RULE. Multiply the velocity in feet per second by the area of the discharging orifice, in feet, and the product is the quantity in cubic feet discharged per second. EXAMPLE. The velocity is 2 feet a second, and the diameter of the pipe 5 inches; what quantity of water is discharged in each second of time? 512.4166, and .41662 X .7854 X 2 = .273 cubic foot. Ans. A MISCELLANEOUS PROBLEMS. To find the specific gravity of a body heavier than water. RULE.Weigh the body in water and out of water, and divide the weight out of water by the difference of the two weights. EXAMPLE. - A piece of metal weighs 10 lbs. in atmosphere, and but 84 in water; required its specific gravity. 10 8.25 1.75, and 101.75 = = 5.714. Ans. To find the specific gravity of a body lighter than water. RULE.Weigh the body in air; then connect it with a piece of metal whose weight, both in and out of water, is known, and of sufficient weight that the two will sink in water; and find their combined weight in water; then divide the weight of the body in air by the weight of the two substances in air, less the sum of the difference of the weight of the metal in air and water and the combined weight of the two substances in water, and the quotient will be the specific gravity sought. EXAMPLE. The combined weight, in water, of a piece of wood, and piece of metal, is 4 lbs. ; the wood weighs in atmosphere 10 lbs. ; and the metal in atmosphere 12, and in water 11 lbs.; required the specific gravity of the wood. 10 ÷ (10+ 12 12 11+4) .588. Ans. To find the specific gravity of a fluid. RULE.Multiply the known specific gravity of a body by the difference of its weight in and out of the fluid, and divide the product by its weight out of the fluid; the quotient will be the specific gravity of the fluid in which the body is weighed. EXAMPLE. The specific gravity of a brass ball is 8.6; its weight in atmosphere is 8 oz., and in a certain fluid 7 oz.; required the specific gravity of the fluid. 8 - 7.25 .75, and 8.6 X .75 = 6.45, and 6.45 ÷ 8.806. Ans. To find the proportion of one to the other of two simples forming a compound, or the extent to which a metal is debased, (the metal and the alloy used being known.) The Rule strictly bears upon that of Alligation Alternate, which see. EXAMPLE.The specific gravity of gold is 19.258, and that of copper, 8.788; an article composed of the two metals, has a specific gravity of 18; in what proportion are the metals mixed? 18 18 19.258 X 8.788 = 11.055 Ans. 11.055 +177.4 11.055 :: 18 = 1.056 copper, Or, 18 To find the lifting power of a balloon. RULE.-Multiply the capacity of the balloon, in feet, by the difference of weight between a cubic foot of atmosphere and a cubic foot of the gas used to inflate the balloon, and the product is the weight the balloon will raise. EXAMPLE.A balloon, whose diameter is 24 feet, is inflated with hydrogen; what weight will it raise? Specific gravity of air is 1, weight of a cubic foot 527.04 grains; specific gravity of hydrogen is .0689. 527.04.0689: - 36.31 grains weight of 1 cubic foot of hydrogen. 527.04 36.31 490.73 grs. dif. of weight of air and hydrogen. 243 X .5236 = 7238.24 capacity in cubic feet of balloon. = = Then, 7238.24 X 490.73 = 3552021 grs.: 3552021-507 lbs. = 7000 Ans. To find the diameter of a balloon that shall be equal to the raising of a given weight. The weight to be raised is 507 507.4 X 7000490.73: 7238.24, and 7238.24÷.5236 = 13824 =24 feet. Ans. To find the thickness of a concave or hollow metallic ball or globe, that shall have a given buoyancy in a given liquid. EXAMPLE. A concave globe is to be made of brass, specific gravity 8.6, and its diameter is to be 12 inches; what must be its thickness that it may sink exactly to its centre in pure water? Weight of a cubic inch of water .036169 lb.; of the brass .3112 lb. Then, 123 X.5236 X .036169 ÷ 2 = = 16.3625 cubic inches of water to be displaced. 16.3625.3112 = 52.5787 cubic inches of metal in the ball. 122 X 3.1416 452.39 square inches of surface of the ball. And, 52.5787452.39 = .1162+ = inch thick, full. Ans. To cut a square sheet of copper, tin, etc., so as to form a vessel of the greatest cubical capacity the sheet admits of. RULE. From each corner of the sheet, at right angles to the side, cut part of the length of the side, and turn up the sides till the corners meet. |