NOTE. The above TABLE includes all the regular sizes and qualities of tin plates, except "wasters." Other sizes, such as 10 X 10, 11 X 11, 13 X 13, &c., of the different brands, are often imported into the United States to order. Common English Sheet Iron, Nos. 10 to 28, Birmingham gauge, widths from 24 to 36 inches. R. G. Sheet Iron, Nos. 10 to 30, Birmingham gauge, widths from 24 to 36 inches. American Puddled Sheet Iron, Nos. 22 to 28, Birmingham gauge, widths from 24 to 36 inches. Russia Sheet Iron, Nos. 16 to 8 inclusive, Russia gauge, sheets 28 X 56 inches. Sheet Zinc, Nos. 16 to 8, Liege gauge, widths from 24 to 40 inches; length 84 inches. Copper Sheathing, 14 × 48 inches, 14 to 32 oz. (even numbers), per square foot. Yellow Metal, in sheets, 48 × 14 inches, 14 to 32 oz. (even numbers), per square foot. TABLE Showing the Capacity, in Wine Gallons, of Cylindrical Cans, of Diameter in Inches. different diameters, at One Inch depth. Diam'r. Gallons. Diam'r. Gallons. Diam'r. Gallons. Diam'r. Gallons. inches. Applications of the foregoing TABLE. EXAMPLE. ·A cylindrical can is 11 inches in diameter, and its depth is 183 inches; required its capacity. .4303 X 188 gallons. Ans. EXAMPLE. The diameter of a can containing oil is 26 inches, and the oil is 14 inches in depth. How many gallons are there of the oil? 2.38814434.6 gallons. Ans. EXAMPLE.A can is to be constructed that will hold just 36 gallons, and its diameter is to be 18 inches; what must be its depth? 361.102=323 inches. Ans. EXAMPLE.A cylindrical can is to be constructed that shall have a depth of 15 inches and a capacity of just 5 gallons; what must be its diameter ? 515.3333 = capacity of can in gallons for each inch of depth; and against .3333 gallon in the table, or the quantity in gallons nearest thereto, is 10 inches, the required, or nearest tabular diameter. Ans. NOTE. The table is not intended to meet demands of the nature of the one contained in the last example, with accuracy, unless the fractional part of the diameter, if there be a fractional part, is, or inch. As, however, the diameter opposite the tabular gallon nearest the one sought, even at its greatest possible remove, can be but about 4 inch from the diameter required, we can, by inspection, determine the diameter to be taken, or true answer to the inquiry, sufficiently near for practical purposes, be the fraction what it may. Or, to throw the demand into a mathematical formula: As the tabular gallon nearest the one sought is to the diameter opposite, so is the tabular gallon required to the required diameter, nearly. Thus, in answer to the last query, .3400 10 3333 9.8 inches, the required or true diameter, nearly. For a mathematical formula strictly applicable to this question, see Gauging Or, for a formula more strictly geometrical, we have WEIGHT OF PIPES. -the The weight of ONE FOOT IN LENGTH of a pipe, of any diameter and thickness, may be ascertained by multiplying the square of its exterior diameter, in inches, by the weight of 12 cylindrical inches of the material of which the pipe is composed, and by multiplying the square of its interior diameter, in inches, by the same factor and subtracting the product of the latter from that of the former, remainder or difference will be the weight. This is evident from the fact that the process obtains the weight of two solid cylinders of equal length, (one foot,) the diameter of one being that of the pipe, and the other that of the vacancy, or bore. For very large pipes, the dimensions may be taken in feet, and the weight of a cylindrical foot of the material used as the factor, or multiplier, if desired. The weight of 12 cylindrical inches (length 1 foot, diameter 1 inch) of Therefore-EXAMPLE. - Required the weight of a copper pipe whose length is 5 feet, exterior diameter 34 inches, and interior diameter 3 inches. 31 10.5625 X 3.0317 =32.022 + = 13 X 13 = 27.285 + Ans. 4.737 X 5=23.685 lbs. EXAMPLE. Required the weight of a cast iron pipe, whose length is 10 feet, exterior diameter 38 inches, and interior diameter 3 feet. 382 X 2.4573-362 X 2.4573=363.68 X 10=3636.8 lbs. Ans. Or, 382 362 148 X 2.4573 = 363.68 X 10 =3636.8 lbs. Ans. EXAMPLE. — Required the weight of a lead pipe, whose length is 1200 feet, exterior diameter of an inch, and of an inch. }}==.765625, and 1 × 1% : .765625 = interior diameter 2856 = .316406, and .316406.449219 X 3.8697 X 1200 =2086 lbs. Ans. EXAMPLE.-The length of a cast-iron cylinder is 1 foot, its exterior diameter is 12 inches, and its interior diameter 10 inches; required its weight. 122 — 102 = 44 X 2.4573 = 108.12 lbs. Ans. The following TABLE exhibits the coefficients of weight, in pounds, of one foot in length, of various thicknesses, of different kinds of pipe, of any diameter whatever. To obtain the weight of pipes by means of the above TABLE— RULE.-Multiply the diameter of the pipe, taken from the interior surface of the metal on the one side to the exterior surface on the opposite, (interior diameter + thickness,) in inches, by the number in the table under the respective metal's name, and opposite the thickness corresponding to that of the pipe the product will be the weight, in pounds, of ONE foot in length of the pipe, and that product multiplied by the length of the pipe, in feet, will give the weight for any length required. EXAMPLE. Required the weight of a copper pipe whose length is 5 feet, interior diameter and thickness 3 inches, and thickness of an inch. 3253.125 X 1.516 X 5 = 23.687 lbs. Ans. EXAMPLE.Required the weight of a cast iron pipe, 10 feet in length, whose interior diameter is 3 feet, and whose thickness is 1 inch. 36+1=37 × 9.829 × 10 = 3636.73 lbs. Ans. |