For example, let it be required to find the solidity of an irregular solid, where the equidistant sections, taken at 9 ft. apart, are in order as follows: 5, 6, 14, 17, 20, 12, and 9 square feet. Here sum of the extreme areas= 5+9=14, sum of the even areas= 6+17+12=35, sum of the odd areas = 14+20=34; .. Solidity={14 + 4 × 35 + 2 x 34} = 666 c. ft. 30. To find the content of a hay-stack. Let the annexed figure represent a hay-stack, whose horizontal sections are rectangles. Between ABC and NOP the dimensions of the equidistant sections, taken at 6 ft. apart, are as follows: AB=40, BC = 14, DE 42, EF = 18, GH=46, HI=23, KL= 45, LM = 20, NO 44, OP= 14 feet. The perpendicular height of SR from NOP is 5 ft., and SR 40 ft. Required the number of cubic feet in the stack. = Here we find the areas of the equidistant sections to be in their order, 560, 756, 1058, 900, and 616. .. Content ANPB = (560+616+4(756+900)+2 × 1058} = 19832 c. ft. = The top part NOPS, calculated by the prismoidal formula, will be found to be 1493 c. ft.; therefore we find the total content to be 21325 c. ft. 31. To gauge a cask. Let the annexed figure represent a vessel whose horizontal sections are circles. The equidistant diameters, taken at 12 inches apart, are as follows; AB = 50, CD=60, EF=70, GH = 60, and KL=54 inches. What is its content in imperial gallons? Content in c. in. -= E G H F C D B = 12 × ·7854 {502 + 542 + 4(602 + 602) + 2 × 702}. Dividing this result by 277-274, we find the number of gallons to be 499 nearly. 32. The general formula of Art. 26. will hold true for the area of a curvilineal space ABN M, by merely changing the equidistant areas into equidistant ordinates. (See fig. to Art. 29.) Let the face QVRS be parallel to ABNM, and AQSм perpendicular to it. Put b for the uniform breadth AQ, and p1, P2, P3, &c., for the equidistant ordinates AB, CD, EF, &c.; then the content of this solid will be equal to the area of ABDNM multiplied by b. But a1=p1b, m1=p2b, ɑ2=p3 b, m2=P4 b, a3 = p5 b, and so on; hence we have by equality and substitution, area ABNM× b d 2 =z {P1b+p+b+4(p2b+P4b+P6b)+2(p3b+p5b)} . Dividing by b, we have, d area ABNM={P1+Pr+4(P2+P4 + P6) + 2(P3+P5)} From which is the usual form of Thomas Simpson's rule. the demonstration here given, it appears that this formula is merely a particular case of the one given in Art. 26. Let it be required, for example, to find the area of a curvilineal space, whose extreme ordinates are 8 and 9 ft. respectively, the intermediate equidistant ordinates, taken in order, are 9, 11, 15, 16, and 14 ft.; and the common distance between them 5 ft. Here the sum of the extreme ordinates =8+9=17. .. Area(17+ 4 x 38 + 2 x 27)=3713 sq. ft. SPECIFIC GRAVITY. 33. The specific gravity of a body is its weight as compared with an equal bulk of pure water. As the weight of a cubic foot of water, at the temperature of 40°, is 1000 oz., it is customary to consider the specific gravity of a body as the weight of a cubic foot of it. The following table contains the weight of a cubic foot of the various kinds of material in ounces. 34. To find the weight of a body from its magnitude. 1. What is the weight of a block of marble 12 feet long, 3 ft. broad, and 1 ft. 6 in. thick? Content: = 12 × 3 × 11 = 54 c. ft. Wt. of 1 c. ft. = 2700 oz. .. Wt. of 54 c. ft. = 54 times 2700 oz. = =9112.5 lbs. 2. Required the weight of a beam of fir 20 ft. long, 9 in. deep, and 3 in. thick. Ans. 129.6 lbs. 3. A cast-iron pipe is 3 inches diameter in the bore, and inch in thickness; required the weight of a running foot. Here, by Problem 11., we find the section of the metal to be 5.4978 sq. in. .. Content of 1 ft. of pipe = 5·4978 × 1 ÷ 144 4. What will be the weight of a running foot of leaden. pipe, 2 inches diameter in the bore, and inch in thickness? Ans. 8.7 lbs. 35. To find the magnitude of a body from its weight. 1. A piece of beech weighs 300 lbs.; required its solidity. Weight of 1 c. ft. = 690 oz. .. No. c. ft. 300 × 16÷690=7 c. ft. nearly. 2. Required the number of cubic feet in 1 ton of iron. Ans. 4.655. 3. If a horse can draw 1 tons, how many cubic feet of oak will he draw? Ans. 57.55. 4. How many cart-loads of clay will there be in a drain 90 ft. long, 3 ft. broad, and 2 ft. deep, allowing 14 tons for each load? Ans. 26.03. FLOATING BODIES - TONNAGE OF VESSELS. 36. When a body floats in water, the weight of the fluid displaced is equal to the weight of the body. Thus, for example, if the weight of the floating body be 4000 oz., it would exactly displace 4 cubic feet of water. The tonnage of a vessel is the weight, in tons, necessary to sink it to a certain depth, or line of floatation. To find the tonnage of a vessel, therefore, we must calculate the volume of that portion of the vessel which is immersed; then the weight of water, in tons, equal to this volume, minus the weight of the vessel itself, will be the tonnage. The chief difficulty in this calculation consists in finding the volume of the solid immersed. The tonnage may be found in all cases, with a sufficient degree of precision, by the following method:1. Divide the length of the vessel into any convenient even number of equal parts, and make exact drawings of these sections. 2. Calculate the areas of these sections below the water line, by measuring off equidistant ordinates after the method explained in Art. 32. 3. Calculate the volume of the solid of immersion in cubic feet by Art. 29. 4. Multiply this volume by 62.5, the weight of a cubic foot of water in lbs., and divide this product by 2240, and it will give the number of tons in the whole floating mass; from which subtract the weight of the vessel in tons, and it will give the tonnage required. For example, let the areas of the equidistant sections, taken in order, at every 9 ft. apart, be 0, 20, 60, 30, and O square ft. ; and the weight of the vessel 5 tons. Required the tonnage. Vol. displaced fluid=} {4 (20 + 30) + 2 × 60} =960 c. ft. .. Wt. displaced fluid=960 × 62·5÷2240 = 26'7 tons. .. The tonnage 267-5=21.7. The weight of the vessel may be ascertained, by adding together the weight of the material in the different parts of |
