COMMON METHOD OF MEASURING TIMBER. RULE. MULTIPLY the square of the mean quarter girt, or circumference, by the length for the solid contents. NOTE.-When all the dimensions are taken in feet, the contents will be in feet; but if the girt be taken in inches, and the length in feet, divide the product by 144. EXAMPLES. 1. What is the content of a piece of timber, the length of which is 9 feet 6 inches, and quarter girt 3 feet 6 inches? By Decimals. 3.5 3.5 175 105 12.25 9.5 6125 11025 Feet 116.375 Content. 2. If a piece of round timber be 96 inches in circumference, or the quarter girt be 24 inches, and the length 18 feet; how many feet of timber are contained therein? Ans. 72 feet. 3. If a piece of round timber be 86 inches in circumference, or the quarter girt be 21 inches, and length 20 feet, how many feet of timber does it contain? Ans. 64.2 feet. To find how much in length will make a foot of any square timber of equal thickness from end to end. RULE. Divide 1728 by the area of the end in inches, the quotient will be the length of a solid foot in inches. EXAMPLES. 1. If a piece of timber be 18 inches square, how much in length will make a solid foot? Ans. 5 inches. 2. If a piece of timber be 22 inches deep and 15 inches broad, how much in length will make a solid foot? Ans. 5.23 inches. GAUGING. GAUGING is the art of measuring the capacities of all kinds of casks, barrels, &c. used for liquor, and of ascertaining the quantity of liquor they will contain. To gauge a cask, you must measure the head diameters, and take the mean of them when they differ. Measure, also, the diameter at the bung, taking the measure within the cask; and the length of the cask, due allowance being made for the heads. With these dimensions we may calculate the contents in gallons or bushels, by the following. RULE. Take the difference between the head and bung diameters, multiply this by .62, and add the product to the head diameter; the sum will be the mean diameter; multiply the square of this by the length of the cask; and divide the product by 294.12 for wine gallons, by 359.05 for beer gallons, and by 2738 for bushels. NOTE. The quantity .62 is generally used by gaugers; but if the staves are nearly straight, it would be more accurate to take .55 or less. If, on the contrary, the cask is full on the quarter, it would be best to take .64 or .65. EXAMPLES. 1. What is the content in wine gallons of a cask, the dimensions being as follows: bung diameter=34.5 inches, the head diameter 30.7 inches, and length 59.3 inches? 34.5 bung. 3.8 diff. .62 common multiplier. 76 228 2.356 30.7 head diameter. 33.056 mean diameter, 33.056 218336 165280 991680 99168 1092.719136 square of mean diameter. 3278157408 9834472224 5463595680 294.12)64798.2447648 (220 wine gallons. 2. Suppose the bung diameter of a cask to be 26.5 inches, head diameter 23 inches, and length 28.3 inches; required the contents in ale or beer gallons. Ans. 50 gallons. A SHORT METHOD TO FIND THE CONTENTS OF A CASK. RULE. Measure the diagonal of the cask from the centre of the bung hole to the head in inches, and divide the cube of this diagonal by 368, the quotient will be the content in gallons. EXAMPLES. 1. Suppose the diagonal of a cask to be 36 inches, required the content in gallons. 36 × 36 × 36 ÷ 368 = 126.78 gallons. 2. Suppose the diagonal of a barrel to be 30 inches, how many gallons will it contain? Ans. 73.37. MEASURING GRAIN, &c. THE following is an easy way of ascertaining the number of bushels of grain or the like, lying on the barn floor, when the same is heaped up into the form of a cone. When the grain is not supported by the sides of the barn. RULE. Measure the perpendicular height of the heap, and also the slanting height from the top of the heap to the floor in inches, then multiply the difference of the squares of these two heights by the perpendicular height, and by .0005, the product will be the content in bushels. EXAMPLES. 1. Having threshed a parcel of wheat, I have collected it together on the barn floor in the form of a cone, and I find the perpendicular height to be 30 inches, and the slanting height to be 80 inches, required the number of bushels in the heap. The square of 80 = 6400 5500 difference of squares. 165000 .0005 82.5000 Ans. or 82 bushels. 2. Required the number of bushels of corn in a conical heap, the perpendicular height being 20 inches, and the slanting height 50 inches. Ans. 21 bushels. When the grain is heaped up against the side of the barn. RULE. Multiply the difference of the squares of the heights by one half of the perpendicular height, and this product by .0005; the result will be the content in bushels. EXAMPLES. 1. How many bushels of potatoes are in a heap piled up against the side of a loft, the perpendicular height being 40 inches, and the slanting height 60 inches? 60 × 60 = 3600 2. How many bushels of rye are in a heap, the perpendicular height being 50 inches, and the slanting height Ans. 93 bushels. When the grain is heaped up in the corner of the barn. 100 inches. RULE. " Multiply the difference of the squares of the heights by one fourth of the perpendicular height, and this product by .0005; the result will be the content in bushels. EXAMPLES. 1. Required the number of bushels of grain collected in the corner of the barn, the perpendicular height being 40 inches, and the slanting height 60 inches. Ans. 10 bushels. 2. Required the number of bushels of oats gathered into the corner of a bin, perpendicular height being 30 inches, and the slanting height 80 inches. Ans. 20.625 bushels. 3. Required the number of bushels of peas in the corner of a drying room, the perpendicular height being 40 inches, and the slanting height 70 inches. Ans. 16 bushels. |