EXERCISES. 1. What is the content of a cask, whose bung diameter is 32 inches, end diameter 18, and length 38 inches? Ans. 69.4. 2. Find the content of a cask, whose diameters are 40 and 20, and length 50 inches. Ans. 132.2. MEAN DIAMETERS OF CASKS. 167. The mean diameter of a cask is the diameter of a cylinder of the same length, whose capacity is equal to that of the cask. The mean diameter may be found by means of the following table, the construction of which is this:-If the bung diameter be denoted by 1, the contents of the four varieties of casks will be expressed by 15 }(2+H2), (8+4 H+3 H2), 3 (1+H2), 3(1+H+H2), multiplied by 7854 L; but if D = the mean diameters, the contents are also expressed by 7854 D2L; hence, as each of the above expressions × 7854 L is equal to the last, therefore these expressions themselves are = D2 for the four varieties, or D for these varieties is equal to the square root of each of them. For example, when H = 5, B being = 1, then, for the first variety, D=√(2+H2)= {√3 =866; which is the number under the first variety in the following table opposite to 5 or H', which is just the value of H when B=1. In the same manner, all the other numbers in th ble are found, for H, or rather H′ = ·51, ·52, ·53,... up to 1. The numbers marked H' are just the ratio of the bung to the head diameter, and the numbers under the different varieties are the mean diameters when the bung diameter is = 1, and the ratio H' is its head diameter. For when B is assumed 1, that is, as the unit of measure, the following proportion will exist, namely, B:1-H:H' or H'H ÷ B. = TABLE OF MEAN DIAMETERS, WHEN THE BUNG DIAMETER (The numbers are all decimal fractions, except those in the last line, 50 8660 8465 7905 7637 51 8680 8493 7937 7681 8520 7970 7725 52 8700 8002 7769 54 8740 8576 8036 7813 8070 7858 8104 7902 7947 55 8760 8605 8662 8140 8210 8282 8320 63 8938 8835 8357 64 8962 8865 8395 8433 65 8986 8894 66 9010 8924 67 9034 8954 8472 8511 8551 68 9060 8983 69 9084 9013 70 9110 8590 9044 8631 71 9136 9074 8672 72 9162 9104 8713 73 9188 9135 8754 74 9215 9166 8796 8732 75 9442 9196 8838 8780 H' 76 9270 9227 8881 8827 9296 9258 8944 8874 77 78 9324 9290 8967 8922 79 9352 9320 9011 8970 80 9380 9352 9055 9018 81 9409 9383 9100 9066 82 9438 9415 9144 9114 83 9467 9446 9189 9163 84 9496 9478 9234 9211 85 9526 86 9556 9542 9326 9308 87 9586 9574 9372 9357 88 9616 9606 9419 9406 89 9647 9638 9466 9455 90 9678 9504 91 9710 92 9740 93 9772 94 9804 8497 9671 9513 9703 9560 9553 9736 9608 9602 9768 9656 9652 9801 9704 9701 9834 9753 9751 9867 9802 9800 9900 9851 9850 9933 9900 9900 9966 9950 9950 100 10000 10000 10000 10000 95 9836 8544 96 9868 8590 97 9901 8637 98 9933 7992 8037 8082 8128 8173 8220 8265 8311 8357 8404 8450 First Second Third Fourth Variety. Variety. Variety. Variety. 168. PROBLEM XIII.-To find the capacity of a cask of any of the four varieties, by means of their mean diameters, found by the table. 'Divide the head by the bung diameter, and find the quotient in the column marked H in the table, and opposite to it and under the proper variety is the mean diameter of a similar cask, whose bung diameter is 1. Multiply this tabular mean diameter by the given bung diameter, and the product is the required mean diameter, the square of which, multiplied by the length, and divided by 353-036, is the content in imperial gallons." If H' ratio of H to D, and D' = tabular mean diameter, then D' is found opposite to H', and then D = BD', and CD2L ÷ 353·036, or C =·0028325 D2L. For, if D = the mean diameter, the content is C= -7854 277.274 ·D2LD2L ÷ 353.036. Instead of this divisor, the corresponding multiplier may be used, and then C0028325 D2L. By the Sliding Rule. 'Set the gauge point 18-79 on D to the length L on C, then opposite to the mean diameter D on D is the content on C.' EXAMPLE. Find the content of a cask of the first variety, whose diameters are 30 and 24, and length 36. H 24 H' = = = 8, and opposite to 8 is 938 D'; B 30 then D=BD'30 × 93828.14, and C0028325 D2L = ·0028325 × 28·142 × 36 = 80·7. Or, set 18-79 on D to 36 on C, then at 28.14 on D is 80-7 on C. EXERCISE.-What are the contents of each of four casks of the four varieties, their diameters being 32 and 24, and length 40? Ans. The contents will be nearly the same as those in the former examples for the four varieties, and also the same as those in the first exercise of the next problem. CONTENTS OF CASKS WHOSE BUNG DIAMETERS AND LENGTHS ARE UNITY. 169. The contents of casks may be more readily computed by means of a table of the contents of casks, whose bung diameters and lengths are = 1. Let D' have the same meaning as in the preceding problem, that is, let it denote the numbers in the preceding table under the different varieties, which are just the mean diameters of casks, whose bung diameters are 1 and head diameters H'; also let C' be the content of a cask, whose mean diameter is D' and length 1, and which may be called the standard cask, then C' = D'2L' X -7854 277-274 = D'2÷353·036, for L' = 1. Hence, if the numbers in the preceding table are squared, and the square divided by 353-036, or multiplied by 0028326, the results will be the contents C' required. The following table can thus be constructed from the preceding one. Thus, for example, when HB, or H' is 75, then, by the preceding table, D' for the second variety is 9196, and = C′ = D'2 ÷ 353·036 — ·9196 ÷ 353·036 — ·0023959, which is just the capacity opposite to 75 in the following table. TABLE OF CONTENTS IN IMPERIAL GALLONS OF STANDARD CASKS C'. H' *50 51 *52 *53 •54 *55 *56 ⚫57 •58 *59 *60 61 *62 *63 '64 *65 '66 *67 *68 *69 •70 71 *72 *73 ⚫74 ⚫75 First Variety. Second Variety. Third Variety. Fourth Variety. *0021244 0021340 *0,021437 *0021536 *0021637 ⚫0021740 *0021845 ⚫0021951 ⚫0022060 *0022170 ⚫0022283 ⚫0022397 ⚫0022513 ⚫0022631 ⚫0022751 *0022873 *0022997 ⚫0023122 ⚫0023250 ⚫0023379 ⚫0023510 '0023643 *0023778 ⚫0023915 ⚫0024054 *0024195 '0020300 *0020433 *0020567 ⚫0020702 ⚫0020838 *0020975 ⚫0021114 ⚫0021253 ⚫0021394 *0021536 *0021679 ⚫0021823 *0021968 ⚫0022114 ⚫0022262 ⚫0022410 *0022560 *0022711 '0022863 ⚫0023016 ⚫0023170 ⚫0023326 ⚫0023482 ⚫0023640 ⚫0023799 *0023959 *0017704 *0017847 ⚫0017993 0018141 0018293 *0018447 *0018604 *0018764 *0018927 ⚫0019093 ⚫0019261 ⚫0019433 *0019607 *0019784 ⚫0019964 ⚫0020147 ⚫0020332 ⚫0020521 *0020712 ⚫0020906 ⚫0021103 ⚫0021302 *0021505 0021710 *0021918 *0022129 ⚫0016523 *0016713 *0016905 *0017098 ⚫0017294 ⚫0017491 ⚫0017690 ⚫0017891 *0018094 0018299 *0018506 *0018715 ⚫0018925 ⚫0019138 ⚫0019352 *0019568 *0019786 *0020006 ⚫0020228 0020452 ⚫0020678 ⚫0020905 *0021135 *0021366 *0021599 ⚫0021834 H' *76 ⚫77 -78 *79 ⚫80 81 ⚫82 ⚫83 ⚫84 ⚫85 •86 ⚫87 .88 *89 '90 ⚫91 *92 *93 ⚫94 *95 ⚫96 ⚫97 *98 *99 1.00 *0024337 *0024482 ⚫0024628 *0024777 *0024927 *0025079 ⚫0025233 *0025388 *0025546 *0025706 *0025867 *0026030 0026196 *0026363 ⚫0026532 *0026703 *0026875 ⚫0027050 *0027227 *0027405 *0027585 *0027768 *0027952 *0028138 ⚫0028326 First Variety. Second Variety. Third Variety. Fourth Variety. *0024120 ⚫0024282 *0024445 ⚫0024610 *0024776 *0024942 *0025110 *8025279 *0025449 ⚫0025621 *0025793 *0025967 ⚫0026141 *0026317 *0026494 *0026672 ⚫0026851 ⚫0027032 *0027213 ⚫0027396 *0027579 ⚫0027764 *0027950 ⚫0028137 ⚫0028326 *0022343 ⚫0022560 ⚫0022780 *0023002 ⚫0023227 ⚫0023455 *0023686 ⚫0023920 *0024156 *0024396 ⚫0024638 ⚫0024883 ⚫0025131 *0025381 *0025635 *0025891 *0026150 *0026412 *0026677 *0026945 ⚫0027215 ⚫0027489 *0027765 ⚫0028044 *0028326 *0022071 ⚫0022310 *0022551 *0022794 ⚫0023038 ⚫0023285 *0023533 *0023783 '0024035 *0024289 *0024545 ⚫0024803 ⚫0025063 *0025324 *0025588 *0025853 ⚫0026120 *0026389 *0026660 *0026933 *0027208 *0027484 *0027763 0028043 ⚫0028326 170. PROBLEM XIV. To find the content of a cask by means of the table of contents of standard casks. 'Divide the head by the bung diameter, and find the quotient in the column H', and opposite to it and under the proper variety is the content C' of the standard cask; multiply this tabular content by the square of the bung diameter of the given cask, and this product by the length, and the result will be the required content in imperial gallons.' H' H÷B, opposite to H' is C', then C = C'B2L. = and D2 B2D'2, also (169) C'D'2353-036; EXAMPLE. Find the content of a cask of the first variety, whose diameters are 30 and 24, and length 36. |
