third be taken from the fum, the remainder will be the content. EXAMPLE. Required the content of a cask of the second variety, whofe bung and head diameters are 32 and 24, and length 40 inches. By the Pen. Here (2 × 322 + 24-2 × 82) X 40 X 96.49 ale {*0009} = { 117.79 wine } gallons, the content required. By the Sliding Rule. Having fet 40 on c to 32.82 on D, against 32, 24, and 8 on D, ftand 38, 213, and 2.4; then 2, X 38 +213 21309 96-4 ale gallons. × 2·4 = 76 + And having fet 40 on c to 29'7 on D, against 32, 24, and 8 on D, stand 46°5, 26.1, and 2.9; then 2 × 46.5+ 26·1 × 2·993 + 26'1 1179 wine gallons. 1.2 To find the Content of a Cafk of the Third To the fquare of the bung diameter add the fquare of the head diameter, and multiply the fum by the length; then if the product be mult. f0014 or div. or div. S 718 by 0017 by 588 for { ale wine} the product or quotient will be the content required. BY THE SLIDING RULE. Set the length on c to then find the bung and 24:25} } head diameters on D, noting the two oppofite numbers on c, whofe fum will be the content required *. EXAMPLE. Required the content of a cask of the third variety whofe bung and head diameters are 32 and 24, and length 40 inches. By the Pen. J00141 0017 = { = Here (322 + 242) X 40 X 89.1 ale, 108.8 wine gallons, the content. By the Sliding Rule. Having fet 40 on c to 26.8 on D, against 32 and 24 on D, ftand 57.3 and 32; whofe fum is 89.3 ale gallons. And having fet 40 on c to 24.25 on D, against 32 and 24 on D, ftand 69.8 and 391, whofe fum is 108.9 wine gallons. PROBLEM IV. To find the Content of a Cak of the Fourth To three times the fquare of the fum of the di * For, by prob. 12 fect. 6 part 3, the content in inches is B2 + H2 2 2 X 231 X Ln; and I 588.233 I = .0017. 00139255, and Allo the numbers 26.8 and 24°25 are the roots of the numbers 718-105 and 588.233. ameters, add the fquare of the difference of the diameters; then if the fum be mult. 000233 or div. by *00028 { S 4308 by 1 3529 } for { ale wine the product or quotient will be the content required. BY THE SLIDING RULE. Set the length on c to ale {6564} for { wine } on D, and on D find the fum and the difference of the diameters, noting their oppofite numbers on c; then if the second be added to three times the first, the sum will be the content *. EXAMPLE. Required the content of a cask of the fourth variety, whose bung and head diameters are 32 and 24, and length 40 inches. By the Pen. Here (3 × 56 + 82) × 40 × S.000233 $87.9342 ale •Oc0285 107.348 wine By the Sliding Rule. Having = gallons, the content. fet 40 on c to 65.64 on D, against 56 and 8 on D, stand 29.1 and 0·6; then 3 X 29°10'6 = 87.3 + 0·6 87.9 ale gallons. For, by prob. 8 fect. I part 3, the content in inches is 12 X 282 4308.628 I = 3529'42 00028333. Alfo 65-64 and 5941 are the roots of the numbers 4308·628 and 3529′42. And, having fet 40 on c to 59'41 on D, against 56 and 8 on D, ftand 35.6 and 0.7; then 3 x 35.6 + 0.7 107.5 wine gallons. CHAPTER IV. Of Gauging Cafks by their Mean Diameters. PROBLEM 1. To find the Mean Diameter of a Cask of Any of the Four Varieties, having Given the Bung and Head Diameters. *Divide the head diameter by the bung diameter, and find the quotient in the firft column of the following table, marked Qu. Then if the bung diameter be multiplied by the number upon the fame line with it, and in the column anfwering to the proper variety, the product will be the true mean diameter, or the diameter of a cylinder of the fame content with the cafk propofed, cutting off four figures for decimals. *The Investigation of the numbers of this table. By the 3d part it appears that if the bung diameter be reprefented by 1, and the head diameter by h, then the content for the ift, 2d, 3d, and 4th varieties will be refpectively In drawn into but the content must alfo be equal to In 2+bb 3 8 + 4b + 3hb drawn into the fquare 15 2 1 + b + bb ; 3 of the mean diameter; Sand confequently the faid mean diameter, or multiplier in the table, will be refpectively equal to 2+bb 3 8 + 4b + 3b b 15 2 + b + bb 3 Then by writing, in thefe forms, the feveral values of h, viz. •50, 51, 52, &c, there will refult the correfponding numbers of the table. Qu Var Qu) Var 2 Var|3 Var[4 Var | Qu Var 2 Var 3 EXAMPLE. Suppofing the diameters to be 32 and 24, it is required to find the mean diameter for each variety. Dividing 24 by 32 we obtain 75, which being found in the column of quotients, oppofite thereto ftand the numbers 9242 which being each 9196 multiplied by 32, 8838 produce refpec•8780, tively (29.5744] for the corref 29-4272 ponding mean 28.2816 diameters re[28.0960) quired. |