1. Required the content of an irregular block of common stone, which weighs 1 cwt. or 112 pounds. Thus, 2520 oz.: 1722 oz. :: 1728 cubic in.: 1228 cubic inches. Ans. 2. How many cubic inches of gunpowder are there in one pound? Ans. 14.91. 3. How many cubic feet are there in a ton of dry oak, its specific gravity being 925? Ans. 38188 feet. The linear dimensions or magnitude of a body being given, and its specific gravity, to find its weight. RULE. As one cubic foot, or 1728 cubic inches, is to the solidity of the body, so is the tabular specific gravity of the body to the weight in avoirdupois ounces. 1. Required the weight of a block of marble whose length is 63 feet, and its breadth and thickness each 12 feet, and specific gravity 27 ounces. Thus, 122 × 60 = = 9072 solid feet. Then 19072 :: 2700: 244944 oz. = 6941 tons. 2. What is the weight of a block of dry oak, which measures 10 feet in length, and 3 feet in breadth, and 2.5 feet deep; specific gravity 925? Ans. 43351 lbs. To find the specific gravity of a body, when the body is heavier than water. RULE.-Weigh it both in and out of water, and the difference will be the weight lost in the water; then as the weight lost in the water is to the whole weight, so is the specific gravity of water to the specific gravity of the body. 1. A piece of platina weighed 83-1886 pounds out of water, and in water only 79.5717 pounds; required its specific gravity, that of water being 1000. Thus, 83-1886 — 79·5717 — 3-6169 pounds, which is the weight lost in water. Then 3-6169 83-1886 1000 23000, the specific gravity, or the weight of a cubic foot of metal in ounces. 2. A piece of stone weighed 10 pounds in the air, but in water only 6.75; what is the specific gravity? Having the magnitude and weight of any body given, to find the specific gravity. RULE. Divide the weight by the magnitude, and the quotient will be the specific gravity. 1. A piece of marble contains 8 cubic feet, and weighs 1353-5 pounds, or 21656 ounces; required the specific gravity. Ans. 2707. To find the quantity of pressure against the sluice or bank which contains water. RULE.-Multiply the area of the sluice under water by the depth of the centre of gravity (which is equal to half the depth of water) in feet, and that product again by 62.5, the number of pounds avoirdupois in a cubic foot of fresh water, or by 64·4, the avoirdupois weight of a cubic foot of salt water, and the product will be the number of pounds required. 1. The length of a sluice or floom being 30 feet, the width at the bottom 4 feet, and the depth of water 4 feet; what is the pressure of water against the sluice? Thus, 30 × 4 = 120 feet, area of bottom; and 120 × 2 (the depth of the centre of gravity) = 240 cubic feet; and 240 × 623 = 15000 lb... = 6 tons, 13 cwt. 3 qrs. 20 lb. Ans. GAUGING, OR MEASURING CASKS. There are four different forms of casks, which are frustums of different kinds of solids, named from the greater or less curvature of their sides. 1. The middle frustum of a prolate spheroid. To find the content of a cask of the first variety. RULE. To the square of the head diameter add double the square of the bung diameter, and multiply the sum by the length of the cask; then let the product be multiplied by 00091 for ale gallons, and multiplied by 0011 for wine gallons. 1. Required the content of a spheri- 100cal cask whose length is 40 inches, 118.9547 wine gallons. 2. What is the content of a spheroidal cask whose length is 45 inches, bung diameter 34 inches, and head diameters each 25 inches, in ale gallons? Ans. 122 716 ale gallons. 3. A spheroidal cask is 42 inches long, diameter at the bung 32 inches, and the head diameters each 23 inches; what is the content in wine gallons? Ans. 122-7147 wine gallons. To find the content of a cask of the second form. RULE. To the square of the head diameter add double the square of the bung diameter, and from the sum take of the square of the difference of the diameters; then multiply the remainder by the length, and the product again by 0009 for ale gallons, or by 0011 for wine gallons. 1. The length being 40 inches, and diameters 24 and 32 inches; required the content. Thus, 32- 24 = 25.6: 322 8 x8 = 64 x 1024 × 2 = Fig. 2. Then 2598-4 x40=10393.6 × 000996-1408 ale gallons. And 103936 x 0011117.9741 wine gallons. 2. What is the content in wine gallons, of a cask whose length is 40 inches, bung diameter 31 inches, and head diameter 24 inches? Ans. 122-7143 gallons. To find the solidity of a cask of the third variety. RULE. To the square of the bung diameter add the square of the head diameter; multiply the sum by the length, and the product again by 0014, for ale gallons, or by 0017, for wine gallons. 1. Required the content of a cask of the third form, whose length is 40, and the diameters 24 and 32 inches. 2. Required the content in ale gallons of a cask of the third form, whose length is 50 inches, bung diameter 30 inches, and head diameter 20 inches. Ans. 91 ale gallons. To find the content of a cask of the fourth variety. RULE. Add the square of the difference of the diameters to three times the square of their sum; then multiply the sum by the length, and the product again by 00023 for ale gallons, or by 000281 for wine gallons. 1. Required the content when the length is 40, and the diameters 24 and 32 inches. Thus, 32+24 = 56 x 563136 x 39408: 8 x 864 + 9408 = 9472 x 40378880 ×·00023 =87.90016 ale gallons. And 378880x-000283 = 107-34933 wine gallons. 2. The length of a cask of the fourth form is 50 inches, the bung diameter 30 inches, and head diameter 21 inches; required the number of ale gallons. Ans. 91.49+ ale gallons. 3. The head diameters of a cask of the fourth form are 18 inches, the bung diameter 30 inches, and the length 50 inches; required the content in wine gallons. Ans. 99.96. THE ULLAGE OF CASKS. The ullage of a cask is what it contains when only partly filled. The cask is considered either standing on the end, with its axis perpendicular to the horizon, or as lying down on its side with the axis parallel to the horizon. To ullage a standing cask. RULE.-Add all together, the square of the diameter at the surface of the liquor, the square of the diameter of the nearest end, and the square of double the diameter taken in the middle between the other two; then multiply the sum by the length between the surface and nearest end, and the product again by 00043 for ale gallons, or by 0005 for wine gallons in the less part of the cask, whether empty or filled. 1. The three diameters being 24, 27, and 29 inches; required the ullage for 10 wet inches. Thus, 24 x 24 576 =2916+ 841 + 20-2205 ale gallons; = 576: 292 841: 542 = 4333 × 10 43330 x .00043 43330 x .0005 = = = = 24.5535 wine gallons. 2. If the diameter at the surface of the liquor in a standing cask be 32 inches, the diameter of the nearest end 24 inches, the middle diameter 29 inches, and the distance between the surface of the liquor and the nearest end 12 inches; required the number of wine gallons in the cask. Ans. 33 gallons. To ullage a lying cask. RULE. Divide the wet inches by the bung diameter; find the quotient in the column of versed sines, in the table of circular segments, take out its corresponding segment; then multiply this segment by the whole content of the cask, and the product again by 14 for the ullage required, nearly. Thus, 8.00 32 25, whose tabular segment is 153546. 14.126232 x 11 17.65779 gallons. NOTE. The table of versed sines is omitted in this volume. = = To find the content of a cask by the mean diameter. RULE.-Multiply the difference of the head and bung diameters by 68 for the first variety, by 62 for the second variety, by -55 for the third, and by 5 for the fourth; when the difference between the head and bung diameters is less than 6 inches; but when the difference between these exceeds 6 inches, multiply that difference by 7 for the first variety, by 64 for the second, by 57 for the third, and by 52 for the fourth. Add this product to the head diameter, and the sum will be a mean diameter. Square this mean diameter, and multiply the square by the length of the cask; this product, divided by 359-05 for ale gallons, and by 294.1 for wine gallons. 11 |