therefore, in this problem, the couch or cistern gauge is to be taken in preference, according as it is the greater. EXAMPLE. If the cistern and couch gauges, after being more than 72 hours out of the cistern, are respectively 131-2 and 132, and the floor gauge 205-2 bushels, which will afford the greatest or duty gauge? The couch gauge exceeds the cistern gauge; hence 132 × 1.6 = 211.2. So that the couch bushels produce 6 bushels more than the floor gauge. EXERCISES. 1. Whether will 120 cistern bushels or 146 floor bushels, which have been less than 72 hours out of the cistern, produce the greatest gauge? Ans. The floor gauge by 2 bushels. 2. Whether will 145 couch bushels or 230 floor bushels, after the malt has been more than 72 hours out of the cistern, produce the greatest gauge? Ans. The couch gauge by 2 bushels. 177. PROBLEM XX.-To find the content of a cistern, couch, or floor of malt. 'Make the malt of as nearly a uniform depth as possible, then measure the length and breadth, and take a number of equidistant depths, the sum of which, divided by their number, will give the mean depth; find the content at one inch deep as formerly, and multiply it by the mean depth.' By the Sliding Rule. When the body of malt is of a rectangular form, set any of the three dimensions, namely, the length, breadth, or mean depth, on the line MD to either of the other two on B, then opposite to the third on A is the content in bushels on B. Or, find a mean proportional between the length and breadth, and set the proper gauge point for square vessels on D to the depth on C, and at the mean proportional on D will be the content on C.' Let 1, b, d, be the length, breadth, and depth of a rectangular floor of malt, and c its content; then lbd lb с 2218 = if m = 2218; m d' = c, or and hence, Ll+ Lb LmLc-Ld [1] When the slider for the line MD is in its place, let b' = the distance between 1 on the slider and b on MD; then, since the slider and line MD are equal logarithmic lines, the latter being inverted, and the number 2218 on it opposite to 10 on the slider, it follows that L10-L=Lb-Lm. Or, LbL+Lm+ L 10, and substituting this value for Lb in [1], LI Lb Lc Ld L 10. But since subtracting the logarithm of 10 implies that the number connected with it is to be divided by 10, the value of c found on the proper line is to be divided by 10, so that the figures found on the rule will be the content, which therefore will be obtained by the rule prescribed above. EXAMPLE. Find the quantity of malt in a rectangular floor, its length being 48 inches, its breadth 32, and depth, at six different places, 6·1, 5·8, 6·3, 5·9, 6·4, and 5·5. (6·1+5·8+6·3+5·9+6·4+5·5) = 3 × 36 = 6, and c 48 x 32 x = 48 × 32 × 6 ×·0004508 6 Or, set 48 on MD to 32 on B, then at 6 on A is 4.15 on B. EXERCISES. 1. What is the content in imperial bushels of a cistern of malt, whose length and breadth are 160 and 108 inches, and mean depth 4.68 inches? Ans. 36.46. 2. What is the content of a floor of malt, the length of which is 280 inches, the breadth 144 inches, and the depths at 5 places are 21.6, 22-3, 22.9, 23·4, and 23·55? Ans. 413-51. 3. Find the content of a regular hexagonal cistern of malt, the length of its side being 69 inches, and its mean depth 5 inches. Ans. 446.6. THE DIAGONAL ROD. 178. The diagonal gauging rod is 4 feet long and 4 of an inch square. The four sides of it contain different lines; the principal one of which is a line for imperial gallons for gauging casks. The use of the principal line is to determine the content of a cask of the most common form, by merely measuring with the rod the diagonal extending from the bung-hole to the opposite side of the head (BH, fig. to art. 172), that is, to the part where the staff opposite to the bung-hole meets the head; then the number on the rod at the bung-hole on the principal or first side is the number of gallons in the content of the cask.' The principal line is constructed thus:-It is found that for a cask of the common form, whose diagonal d = 40 inches, the content c is = 144 imperial gallons nearly, and therefore at 40 inches from the end of the rod is placed 144. Hence, if D, C, are the diagonal and content of any other similar cask, then, since the contents of similar solids are proportional to the cubes of any two of their corresponding dimensions, hence, C = √ D3 = d3 d3: D3 = c: C; 144 From this formula the numbers showing the contents can easily be calculated. Thus, to find the content C for a diagonal D of 30 inches, 9 9 C= D3 = × 303 = 60.7. 400 400 So that at 30 inches from the end of the rod is placed 60-7 gallons, for the content of a cask whose diagonal is 30 inches. In a similar manner, the other numbers showing the content are calculated and marked on the rod. 9 D300225 D3 = D3. Another line on a different side of the rod is marked Seg. St. for ullaging a standing cask. Another side contains tables for ullaging lying casks. The remaining side contains lines for ullaging casks of known capacity, as firkins, barrels, &c., either lying or standing. WEIGHTS AND DIMENSIONS OF BALLS The following problems can be solved by means of the rules in articles 72 and 73; but they can be more concisely and easily solved by those given under this head. 179. PROBLEM I.-Given the diameter of an iron ball in inches, to find its weight in pounds, and conversely. Multiply the cube of the diameter by 9, and divide the product by 64, and the quotient will be the weight. Find the cube root of three times the weight in pounds, and of it will be the diameter in inches." Let d the diameter in inches, and w = the weight in lbs., then w= d3, and d = 3 w. The weight of an iron ball, whose diameter is 4 inches, is very nearly 9 lbs., and since similar solids are proportional to the cubes of their corresponding dimensions, therefore 43: d39:w, and w= d3. Hence also d3 = &* w, or d = 3/4 w = &/ (§ 4 × 3 w) = † †/ 3 w. EXERCISES. 1. Find the weight of a cast-iron ball, whose diameter is 3.5 inches. Ans. 6.028 lbs. whose diameter is 5 Ans. 17.58 lbs. lbs., what is its diaAns. 5.54 inches. 2. Find the weight of an iron ball, inches. 3. The weight of an iron ball is 24 meter? 4. The weight of a ball is 12 lbs., what is its diameter ? Ans. 4.4 inches. 180. PROBLEM II.-Given the diameter of a leaden ball in inches, to find its weight in pounds, and conversely. 'Multiply the cube of the diameter by 64, and divide this product by 289, and the quotient will be the weight. Or, take of the cube of the diameter, and from it subof itself, and the remainder will be the weight. tract WEIGHTS AND DIMENSIONS OF BALLS AND SHELLS. 113 Multiply the weight by 289, find the cube root of the product, and of it will be the diameter.' Or, w=d3, and d=289 w. 64 289 For a leaden ball 4 inches in diameter weighs 17 lbs. ; hence, as in the preceding problem, (44)3: d3 = 17: w, and w = 17 d3 × (4)3 = 84 d3. 64 64 = Or, nearly, and 3 — }=} hence the second rule. Again, } of ; and d3 = √(¥)3 w = (1)3 289 w; hence d=289 w. EXERCISES. 1. Find the weight of a leaden ball, whose diameter is 3 inches. Ans. 6 lbs. 2. What is the weight of a leaden ball, whose diameter is 5.24 inches? Ans. 32 lbs. 3. Find the diameter of a leaden ball, whose weight is 27 lbs. Ans. 4.96 inches. 4. What is the diameter of a leaden ball weighing 12 lbs. ? Ans. 3.78 inches. 81. PROBLEM III.-Given the internal and external diameters of an iron shell in inches, to find its weight in pounds. 'Find the difference between the cubes of the internal and external diameters, multiply it by 9, and divide the product by 64, and the quotient will be the weight.' Let D and d be the external and internal diameters, then the weights of two balls having these respectively as diameters, are (179) D3 and d3, and that of the shell = (D3-d3). EXERCISES. 1. Find the weight of a bomb-shell, whose external diameter is 10 inches, and the internal one 6 inches. Ans. 110-25 lbs. 2. Find the weight of a 9 inch bomb-shell, the metal being 1 inch thick; and also of a 13 inch shell, the metal being 2 inches thick. Ans. 72-14 lbs. and 206-4 lbs. |