SECTION V. PRACTICAL RULES AND TABLES. 342. MEASURES OF CAPACITY. The English Winchester bushel, containing 2150.4 cubic inches, or 77.6274 lb. avoirdupois, of pure water, at its maximum density, is established, at the custom-houses in the United States, as the standard of dry measure; and the wine gallon of 231 cubic inches, or 8.339 lb. of water, as above, is established as the standard of liquid measure. The above are also the measures established by law in Vermont and some other states. But in New York, according to their revised laws, the legal bushel contains 2211,84 cubic inches, and the liquid gallon 221.18 cubic inches. In measuring coal, lime, ashes, and some other articles, it is customary to use a larger measure, In Vermont the bushel for these articles is established by law at 38 quarts, of which the common bushel holds 32, but in most places the bushel for coal, &c. contains 40 quarts. 1 cubic foot-0.80356 bush. Winchester measure. 1 cubic foot 0.67669 bush. Vermont coal, &c. measure. 1 cubic foot 0.64285 bush. com. coal, &c. measure. 343. To find how many bushels any bin, box, or coal-house will contain. RULE. Find the content in feet, and multiply it by the decimal of a bushel standing against 1 cubic foot in the above table. EXAMPLES. 1. The dimensions of a coal-box were length 12.5 ft., height 3.4 ft., width at the top 3.94 ft., width at the bottom 2.7 ft.; how many bushels of each of the above measures will it hold? 3.94+2.7-6.64, and 6.64-2-3.32, and 3.32×3.4×12.5—141.1 cubic feet. Then 141.1X0.8 112.88sh. Win. 141.1X0.677 95.52 bush. Vt. coal meas. 141.1X0.64 90.30 bush. com. coal meas. 2. If a coal-house be 50 feet long, 40 feet wide, and 20 feet high, how many bushels will it hold? 50X40X20 40000 cu. ft., and 40000×0.67669-27067 b.Vt. m. 50×40×20=40000 cu. ft., and 40000×0.64285-25714 b. c. m. 344. Having two dimensions in feet of a bin, box, or coal-house, to find what the other must be in order to hold a given quantity. RULE.-Multiply the given dimensions together for a divisor, and multiply the given quantity by the cubic feet in a bushel, as expressed in the above table; the quotient will be the other dimension. 1. A coal-box is 25 feet wide and 4 feet long; how high must it be to hold 10 bushels? 2.5×4=10 divisor, 10×1.4777=14.777 & 14.777÷10=1.4777 ft.=1ft. 5gin. 4. How high must the above garner be to hold 1000 bushels of wheat? Ans. 20×8=160 for a divisor, and 1000×1.2444—1244.4 for a dividend. Then 1244.4-160-7.77 feet, for the height of the garner. 1210.7854||19|1.9689||26|3.6863|33|5.9395||401 8.7179||47|12.0482] 13 0.9218 20 2.1817 27 3.9753 34 6.3050 41 9.1684 48 12.5664 14 1.0691 21 2.4048 28 4.2760||35|6.6813 42 9.6211 49 13.0954 151.2272 22 2.6393 29 4.5869 36 7.0686 43 10.0847 50 13.6354 16 1.3963 23 2.8847 30 4.9087 37 7.4667 44 10.5592 51 14.1861 171.5762 24 3.1416 31 5.2414 38 7.8758 45 11.0447 52 14.7479 181.7671253.408232 5.5851||39|8.2957||46|11.5410||53 15.3201 The column marked diameter is the diameter in inches, and the column marked area is the area of a section of the cylinder in feet and decimal parts. To illustrate the use of this table, I will give a few examples, viz. 1. How many cubic feet in a round stick of timber, 20 feet long, and 18 inches diameter ? Look in the table under the head of diameter, and against 18 in the column of areas is 1.7671, which multiplied into the length in feet, will give the number of cubic feet such stick containsthat is, 1.7671×20=35.342 cubic feet. 2. How many cubic feet in a round log 24 inches diameter and 16 feet long? Ans. 3.1416x16 50.2656 cubic feet. 3. Suppose the mean diameter of a cask to be 3 feet, and its length 5 feet, how many cubic feet will it contain, and how many bushels of wheat will it hold Ans. 7.0686X5-35.343 cubic ft., which X0.8 28.2744 bush. EXPLANATION OF THE TABLE OF SQUARE TIMBER MEASURE. The two first columns contain the size of the timber in inches, and the third column contains the area of a section of such stick in feet; so that if you find the size of the stick in the two first columns, and multiply its length in feet into the number in the third column, marked "areas of sections," the product will be the cubic feet and decimal parts which such stick of timber contains. One example will be sufficient : What number of cubic feet in a stick of timber 18 by 15 inches, and 25 feet long? Ans. 1.875X25-46.875 cubic feet. 347. To determine how big a stick you can hew square ou of a round log (317), and how big a round log is required to be, to make a square stick of given dimensions. In the first case, multiply the diameter of the log by 0.7071, the natural sine of 45°; and in the second case, multiply the side of the stick required by 1.4142, the natural secant of 45°. EXAMPLES. 1. How big will a log square that is 2.5 feet diameter? Ans. 0.7071X2.5 1.76775 feet for one side of the square. 2. A stick of timber is required 1.5 feet square; how large a round log is required to make it? Ans. 1.4142X1.5-2.1213 feet diameter. 348. To take off the corners of a square so as to form an octagon. Multiply the side of the square by 0.2929, and the product will be the distance to measure from the corners to form the octagon. Deduct twice the product from the side of the square, and it will leave one side of the octagon required. ABCD is a tower, 20 feet square, on which an octagon is to be erected; what will be its side, and what distance from the corner to the octagon post? Ans. AB=20X E 0.29295.858 = AF and AB-AF-GB= FG 8.284 for one side. of the octagon. If a diagonal square, as HIKL, is required to be formed on the above said square tower, then multiply one side by 0.7071 (360), and the product will be one side of the inscribed diagonal square. That is, AB=20X0.7071= 14.142=HI, HL, KL or KI. H If the side of a square tower be 16 feet, what will be the side of an octagon erected upon it? Ans. 6.6272 feet. fourth of the length of the beam, as KĎ. Braces are gener ally placed equidis tant each way from the corner, as FG, but sometimes farther one way than the other, as HI. To find the length of rafters when they rise one third of the length of the beam, multiply one half the length of the beam or the base of the rafter by 1.20185; and to get the length of studs under the rafters, multiply so much of the base as is contained between the foot of the rafter and the foot of stud by 0.6666. Consequently the half length of the beam, 12x1.2 (omitting the other figures), is 14.4 for the length from A to B; and if a stud is placed 9 feet from the foot of the rafter, its length will be 0.6666×9—6 feet. If the roof is raised 30 degrees to C, then 12×1.15468= 13.856 for the length of the rafter; and the length of studs under the rafter will be obtained by multiplying as above by 0.57735. If the roof rises one fourth of the length of the beam, then 12X1.118034 13.416 for the length of the rafter; and the length of the studs in this case will be half the distance from the foot of the rafter to the foot of the stud. For the length of braces subtending a right angle, and extending equidistant each way, multiply the length of one of the sides containing the right angle by 1.4142; or if you have the brace, and wish to know how far from the corners to make the mortices for it, multiply the length of the brace by 0.7071. The brace FG is 6 feet each way from the corner, and 6X1.4142-8.485 its length. The brace HI is found by the last case of rafters, thus 8X1.118-8.944 its length. They may also be found by the square root. (268) |