NOTE. If the generating ellipse revolves about its major axis, the spheroid is pro late or oblong; if about its minor axis, the spheroid is oblate. OF PARABOLIC CONOIDS AND SPINDLES. PROBLEM XII. To find the Solidity of a Parabolic Conoid. Rule. Multiply the square of the diameter of the base by the altitude, and the product by 3927 (which is of 7854), and it will give the contents. To find the Solidity of a Frustrum of a Paraboloid Rule. Multiply the sum of the squares of the diameters of the two ends, a b and cd, by the height of the frustrum, e f, and the product by 3927 (which is of 7854), and it will give the contents. 542=2916 Then, 3700×18×3927=26153.82. Ans. 28= 784 3700 PROBLEM XIV. To find the Solidity of a Parabolic Spindle. Rule. Multiply the square of the middle diameter, c d, by the length of the spindle, I m, and the product by 41888 (which is fs of 7854), and it will give the solidity. To find the Solidity of the Middle Frustrum of a Parabolic Spindle. Rule. Add together 8 times the square of the greatest diameter, cd, 3 times the square of the least diameter, fe, and 4 times the product of these two diameters; multiply the sum by the length, a b, and the product by 05236 (which is 8 of 31416); this will give the solidity. Example -What is the solidity of the frustrum of a parabolic spindle, whose dimensions are as follows: a b, 60, cd, 40, fe, 30 inches? Rule. To the square of the radius of the base, a s, add the square of the middle diameter, mr; multiply this sum by the height, s f, and the product by 5236, and it will give the solidity. 1300 And 1300×30×5236-1728-11.817 cubic feet. PROBLEM XVII. To find the Solidity of the Frustrum of a Hyperbolic Conoid. (See the foregoing figure.) Rule. Add together the squares of the greatest and least semidiameters, a s and dr, and the square of the whole diameter, mr, in the middle of the two; multiply this sum by the height, rs, and the product by 5236, and it will give the solidity. Example. Required the solidity of the frustrum of a hyperbola, abd c, whose semidiameter, as, is 20 inches, and dr, 10 inches; the middle diameter, mr, 30 inches, and whose height is 20 inches? 20'=400 10=100 30=900 1400 Then, 1400×20×6236-1728=8.426 cubic feet. PROBLEM XVIII. To find the Convex Surface of a Cylindrical Ring. Rule. To the thickness of the ring, a c, add the inner diameter; then multiply this sum by the thickness, and the product by 9-8696 (which is the square of 3.1416), and it will give the convex surface required. Example. The thickness, a c, of a cylindrical ring is 4 inches, and the inner diameter, c d, is 14 inches; required the convex surface. ac+cd=4+14=18. Then, 18×4×9.8696=710.612 square inches convex surface. PROBLEM XIX. nC N a b To find the Solidity of a Cylindrical Ring. Rule. To the thickness of the ring, a c, add the inner diameter, cd; then multiply the sum by the square of the thickness, and the product by 2.4674 (which is of the square of 3.1416), and it will give the solidity. Example. Required the solidity of an anchor ring, whose inner diameter is 8 inches, and thickness in metal 3 inches. First, 3+8=11 3×3=9= square of thickness. 99×2-4674-244-2726= solidity in inches. GAUGING OF CASKS. Gauging is a practical art, which does not admit of being treated in a very scientific manner. Casks are not commonly constructed in exact conformity with any regular mathematical figure. By most writers on this subject, how-⚫ ever, they are considered as nearly coinciding with one of the following forms: and their contents in cubic inches may be found by the rules in men suration, for determining the solidity of these figures. To find the Contents of a Cask by four Dimensions. Rule.--Add together the squares of the bung and head diameter, and the square of double the diameter, taken in the middle between the bung and head; multiply the sum by the length of the cask, and the product by 1309. To find the Contents of a Cask in the form of the Middle Frustrum of a Spheroid. Rule. Add together the square of the head diameter and twice the square of the bung bung diameter; multiply the sum by of the length, and the product by 00355, for a wine gallon of New York standard measure, or 0034 for old English gallons. If Dand d = the two diameters, and 1 = the length, the capacity in inches = (2D2 × d2) × +l×7854. And by substituting 00355 for 7854, we have the capacity in wine gallons. Example. What is the capacity of a cask of the second form, whose length is 30 inches its head diameter 18 inches, and its bung diameter 24? 14760× 00355=52.39 wine gallons. Ans. To find the Contents of a Cask in the form of two equal Frustrums of a Cone. Rule. Add together the square of the head diameter, the square of the bung diameter, and the product of the two diameters; multiply the sum by of the length and the product by 00355 for New York wine gallons, or 0034 for old English gallons of 231 cubic inches. Example.--What is the capacity of a cask whose dimensions are as follows: 30 inches long, head diameter 18 inches, and bung diameter 24 inches? 18=324 24=576 Product of 2 diam. =432 1332×10=13320×00355=46.286. Or (D2+d2+Dd)×11×00355. OF ARTIFICERS' WORK. Artificers compute the contents of their works by several different measures, viz.: 1. Glazing and mason work, by the foot. 2 Painting, plastering, paving, &c., by the yard. 3. Flooring, partitioning, roofing, &c., by the foot. 4. Brickwork, by the 1000, or cubic foot. BRICKLAYERS' WORK. Brickwork is estimated at the rate of a brick and a half thick. So that, if a wall be more or less than this standard thickness, it must be reduced to it, as follows: Rule. Multiply the superficial contents of the wall by the number of half bricks in the thickness, and of that product will be the contents required. BRICKS AND LATHS. DIMENSIONS. 15 common bricks to a cubic foot of 8-inch wall when laid. Laths are to inches by 4 feet in length, are usually set of ar inch apart, and a bundle contains 100. Stourbridge fire-brick 9 by 4 by 2 inches. Example.-How many bricks will it require to build a house 30 feet square, 20 feet high, and 12 inches thick, above which is a triangular gable, rising 12 feet, and 8 inches thick? 30×6=180=1 gable end 30×6=180=1 360×15= 66 5400 52200 OF MASONRY. Masonry is the science of preparing and combining stones, so as to properly touch, indent, or lie on each other, and become masses of walling and arching for the purposes of building. In stone walling, the bedding joints ought each to be laid horizontally when the top of the wall is to terminate so. In building bridges and fence walls upon inclined surfaces, the bedding joints ought to follow the general direction of the work. |