ART. 3. To find the solid contents of the frustrum of a cone, or ta pering stick of round timber. RULE. Multiply the diameters of the bases or ends together, and to the pro. duct add one third of the square of the difference of the diameters; then multiplying this sum by,7854 gives the mean area between the two bases or ends, which multiplied by the length gives the solid contents.* EXAMPLE. If the diameter of one end of a tapering stick of round timber be 21 inches, and the diameter of the other end 12 inches, and the length 30 feet, what is the solid contents ? Then, 279,7854×30÷144-45,65+ solid feet, Ans. PROB. VI.-To find how many solid feet a round stick of timber, of equal bigness from end to end, will contain when hewn square. RULE. Multiply twice the square of the semi-diameter in inches by the length in feet, and divide the product by 144, and the quotient will be the answer. EXAMPLE. If the diameter of a round stick of timber be 21 inches, and its length 20ft., how many solid feet will it contain when hewn square? Diameter 21in÷÷2=10,5in, semi-diameter. Then 10,5×10,5×2×20÷144=30,625 feet, Ans. PROB. VII. To find the solid contents of a globe or sphere. . RULE. Multiplying the circumference and diameter together gives the area, which multiplied by one-sixth of the diameter gives the solid contents. *This is a correct way of measuring round logs, or timber, which taper gradually from end to end. But the old English method of measuring round timber was, to girt the stick in the middle, and call one-fourth of this girt the side of a square equal to the circumference. This one-fourth part squared, and the square multiplied by the length, they called the solid contents, which was an erroneous method; for, the girt in the middle is not the mean between the ends, nor one-fourth of the girt equal to the side of a square of equal area with circumference; and from this old erroneous practice of measuring timber was introduced the custom of calling 40 feet of round timber and 50 feet of hewn timber a ton, for 40 feet of round timber, measured by this method, will actually make about 50 feet of hewn timber. We suppose, that when timber is accurately measured, 40 feet of every kind should make a ton. EXAMPLE. What is the solid contents of a globe or ball whose diameter is 18 inches? Thus, the diameter being 18 inches, the circumference is found to be 56,57 inches. (See Rule, page 235.) Then the circumference, 56,57 inches, multiplied by the diameter, 18 inches, 1018,+ which multiplied by one-sixth of the diameter, (=3 inches) gives 3054 solid inches, Ans. 1 solid foot 1326in. PROB. VIII. To find how many bushels, or gallons, will be contained in a vessel of given dimensions, (whether it be cubic, cylindric, or globular.) RULE. If it be a cubic vessel, find its contents or capacity in inches, by Prob I. or II. p. 236. If it be cylindric, find its contents by Prob. IV If it be globular, by Prob. VII. Then, as 2150,4 cubic inches make a bushel, therefore, dividing by 2150,4 will give the bushels; and divide by 231, the number of inches in a wine gallon, gives the number of wine gallons; or, divide by 282, and you will have the beer gallons. EXAMPLES. 1. How many bushels, and how many wine gallons, will a cistern hold that is 3 feet long, 2 feet wide, and 2 feet deep? Thus, 3 feet 36 inches, and 2 feet 24 inches. Then 36×24×24=20736 cubic inches. And 20736+2150,4 9,64 bushels, Answer. And 20736÷231-Ans. 89 wine gallons, 177in. rem. How many wine gallons are contained in a cylindrical vessel whose diameter is 18 inches, and depth 12 inches? Thus, 18X18X,7854×12, (by Prob. 4, page 237,)=3053,6352 cubic inches, and 3053,6352÷231—13,2+ gallons, Ans. PROB. IX. The dimensions of the walls of a brick building given, to find how many bricks are sufficient to build it. RULE. From the whole extent of the wall, measured round on the outside, subtract four times its thickness, and multiply the remainder by the height, and that product by the thickness of the wall, gives the solid contents of the whole, which multiplied by the number of bricks in a solid foot, gives the answer. Vote. To find the number of bricks in a solid foot, multiply the tength in inches by the breadth in inches, and that product by the thickness; then divide 1728 by this product. EXAMPLES. How many bricks, 8 inches long, 4 inches wide, and 2 inches thick, will it take to build the walls of a house 40 feet long, 30 feet wide, and 20 feet high, the walls to be one foot thick ? · X4X280 solid inches in a brick; then 1728+80=21,6 bricks number of bricks in a solid foot X21,6 Answer 58752 bricks. PROB. X.-To find the tonnage of a Ship. RULE. Multiply the length of the keel by the breadth of the beam, and that product by the depth of the hold, and divide the product by 95, the quotient will be the tonnage. NOTE.-If the vessel be doubledecked, half the breadth of the mainbeam is accounted the depth of the hold. EXAMPLES. What is the tonnage of a vessel 65 feet in length by the keel, the breadth of her beam 22 feet, and the depth of the hold 10ft. 6in,= 10,5ft.? Thus, 65×22×10,5÷95—Ans. 158 tons, 5 rem. PROB. XI. To find the number of gallons, &c., that are contained in a vessel in the form of the frustrum of a cone, or a tub, whose top and bottom diameters are unequal. Find the cubic contents of the given vessel in inches, (by the rule, Prob. V. Art. 3, for finding the contents of the frustrum of a cone,) which divided by 231 will give the wine gallons, &c. Ex. How many wine gallons are contained in a tub, whose bottom diameter is 27 inches, top diameter 36 inches, and depth 50 inches? Thus, 36-27-9 diff. x9=81, sq. of diff.÷3=27, of sq. of diff. And top diam. 36 × bottom diam. 27=972,+ 27, sq. of diff.=999. Then 999,7854 × 50=39230,73 cubic inches, which by 231 gives 169,83 gallons, Ans. PROB. XII. -To guage a cask, or to find how many gal. lons it will hold. To guage a cask, you must measure the head diameter and the bung diameter, (taking the measure within the cask,) and the length of the cask, making allowance for the thickness of the heads; then take the difference between the head and bung diameter, and when the staves are about an ordinary curve, add about 6 tenths of the difference to the head diameter, which will reduce the cask to a cylinder; that is, it will give the mean diameter.* Then multiply the square of the mean diameter in inches by,7854, and that product by the length in inches, (Prob. 4.) gives the cubic *If the diameters of the heads are unequal, take their mean diameter, and if the staves are very much curved, take ,66 instead of six tenths; but if they be nearly straight, take,55, &c. contents in inches, which divided by 231, gives the wine gallons, and by 282-gives the beer gallons. But as the square of the mean diameter is always to be multiplied by,7854 and divided by 231 for wine gallons, we may contract the operation, and multiply by their quotient,,7854 thus, for wine gals. multiply by 34, pointing off 4 fig 231,0034;) 7854 282,0028) by 28, pointing ures for the decimals; and for ale or beer gals., 282 off four figures for decimals. Hence the following RULE. Multiply the square of the mean diameter by the lengtn, and mul tiply this product by ,0034 for wine, and by ,0028 for beer gallons. Ex. There is a cask whose head diameter is 25 inches, bung diameter 30 inches, and whose length is 38 inches. How many wine gal. Lons, and how many beer gallons does it contain? Thus, the diff. between the head and bung diameter is 5 inches; this multiplied by 6 tenths gives 3in. to be added to the head diameter, 25+3=28in. mean diam. x 28in.=784,sq. of mean diam. × 38in. length, 29792. Then 29792 x,0034-101,2928 wine gals.; and 29792×,0028=83,4176 beer gals. PROB. XIII.-Of Mechanical Powers. ART. 1. Of the Lever.-It is a fundamental principle in mechanics, that the power, and weight which will be raised, are to each other inversely as the spaces which they pass over. Hence to find what weight may be raised by a given power. 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. EXAMPLE. 1. There is a lever 10ft. long, and the fulcrum or prop, on which it turns, is 2ft. from one end; how many pounds weight at the end 2ft from the prop, will be balanced by a power of 56lbs. at the end 8ft from the prop? As 2ft.: 8ft. :: 56lbs. Ans. 224lbs. 2. What weight must be applied to the above lever eight feet from the prop, to balance one thousand pounds two feet from the prop? As 8ft. 2ft. :: 1000lbs. .... Ans. 250lbs. 3. If 1000lbs. be placed 2 feet from the prop, on the above lever, how far must 250lbs. be placed from the prop on the other side, to balance the 1000lbs. ? As 250lbs.: 1000lbs :: 2ft. Ans. 8ft. .... ART. 2. Of the Wheel and Axle.-The spaces passed over are as their diameters or circumferences. There is a windlass, the wheel of which is 60 inches in diameter, and the axis around which the rope coils is 6 inches in diameter; how many pounds on the axle will be balanced by 160lbs. at the wheel? 2. How many pounds at 1200lbs. on the axle ? ART. 3. Of the Screw. As 6in.: 60in. 160lbs. .... Ans. 1600lbs. raised, as the distance between the two threads of the screw, is to the circumference of a circle described by the end of the lever to which the power is applied. EXAMPLE. 1. There is a screw whose threads are 1 inch apart. If it be turned by a lever 7 feet long, what weight will be balanced by 130 pounds power? Thus, the lever is half the diameter, consequently the diameter is 14 feet. Then, as 7: 22 :: 14: 44, the circumference of the circle described 528 inches. Then, as lin. : 528in. :: 130lbs., the ratio being as 528 to 1, &c. without any allowance for friction for which it is common to add about one-third to the power. PROBLEM XIV.-To measure Loads of Wood. Loads of wood are generally estimated by cord-feet, 8 of which make a cord: 128 solid or cubic feet make a cord; hence every 16 solid feet make 1 cord-foot. RULE. The length, breadth, and height, multiplied together will give the solid contents, which divided by 16, will give the cord-feet. EXAMPLE. How many cord-feet of wood in a load 8 feet long, 3 feet 6 inches wide, and 2 feet 6 inches high? Thus, 3ft. 6in.=3,5ft. × 2ft. 6in.=2,5=8,75×8=70,00 +16=4 cord-feet. Or, by duodecimals, 3F 6'x2F 6'-8F 9'x8F 0'-70 solid feet. And 70÷16-43, Answer. But when the load is just 8 feet long, multiply the breadth and height together, and half the product will be the answer in cord-feet. Thus, 3,5ft. x 3,5ft.-8,75÷2-4,375 cord-feet, 4 ft. the same as above. |