The Length and Breadth of a Parallelogram being given, to find its Area in Malt Bushels by the Line M D. Rule. Set either of the given dimensions on B, to the other on MD, then against i on A, is the required area on B. EXAMPLE. How many malt bushels can be con. tained on every inch of the depth of a cistern, whose length is 180, and breadth 72 inches? By setting 72 on B, to 180 on MD, against i on A, will appear nearly 6 bushels on B, which is the quantity required. To find, by the Line MD, the' Malt Bushels which may be contained in a Couch, Floor, or Cistern, whose length, breadth, and depth are given. Rule. Set one of the dimensions on B to another on mı, then against the third on A, will appear the content on B. EXAMPLE. Required the number of bushels in the cistern whose length is 230, breadth 58.2, and depth 5:4 inches. * Having set 230, on B to 5'4 on md, against 58•2 on A, I find 33•6 bushels on B, which is the content nearly. The use of the other parts or marks on the rule will appear in the examples farther on. Of the Gauging or Diagonal Rod. The diagonal rod is a square rule having four fides or faces, being generally four feet long, and folding together by means of joints. It takes its name from its use in measuring the diagonals of casks, and computing the contents from the said diagonal only. Where it may be noted, that by the diagonal of a cask, is meant the line from the bung to the intersection of the head with the stave opposite to it, and is commonly the longest line that can be drawn from the middle of the bung to any part within the calk. And, accordingly, upon one face of the rule is a scale of inches for taking the measure of the diagonal; to which is adapted the areas, in ale gallons, of circles to the corresponding diameters, like the lines on the under sides of the three fides in the Niding rule. Upon the opposite face are two scales of ale and wine gallons, expressing the contents of casks having the corresponding diagonals; and these are the lines which chiefly constitute the difference between this instrument and the sliding rule ; for all the other lines upon it are the same with those on that instrument, and are to be used in the same manner. E X A M P L E. Let it be required to find the content of a calk whose diagonal measures 34:4 inches, which agrees with the calk in the following chapter, whose head and bung diameters are 32 and 24, and length, 40 inches; for it to the square of 20, half the length, be added the square of 28, half the sum of the diameters, the square root of the sum will be 344 nearly. Now to this diagonal 34:4, corresponds, upon the rule, the content go ale or 111 wine gallons ; which differs from all the contents, in the next chapter, obtained by considering the cak as belonging to to each of the four proposed varieties; being indeed a kind of medium among them all, and falling in between the second and third variety; and so answering to the most common form of caíks. Of Casks considered as divided into several Varieties. According to the custom of most writers on this subject, I shall distinguish casks into four forms or varieties, viz. 1. The middle frustum of a spheroid, I omit here the middle frustums of circular, elliptic, and hyperbolic spindles, on account of the difficulty of their rules, which renders them unfit for the purpose of practical gauging. And indeed some of the above four forms are of very little real use ; for very few, if any, casks are to be met with which will hold so much as the first form, or so little as the third or fourth; so that the second is the most generally, if not the only useful one, of the four varieties. Note, 282 cubic inches make one ale gallon, 231 wine gallon, 2150042 a malt bushel. It is also to be noted that the dimensions are sup. posed to be inches, in the following rules. PRO To find the Content of a Cask of the First or Spheroidal Variety. *To the square of the head diameter add double the square of the bung diameter, and multiply the fum by the length of the cask. Then let the product be mult. by .00094, or divided by 1077 for ale gallons, or mult.by .0011, or divided by 882 for wine gallons BY THE SLIDING RULE. Set the length on c to 32.82, for ale, or to 29*7, for wine, on D, and on d find the bung and head diameters, noting the numbers opposite to them on c, then if the latter of these two nuinbers be added to the double of the former, the sum will be the measure in gallons. E X A M P L E. Required the content of a spheroidal cask, whose bung and head diameters are 32 and 24, and length 40 inches. By * For, by prob. 12 sect. 5 part 3, the content in inches is (2 82 + H2) x { L 11, which being divided by 282 and 231, becomes 2 B2 + H2 XL or (2 B2 + H2) X '000928371 in the one case, 1077.157 2 B? + H2 and X L or (2B? + H2) x .00113333 in the other; B being the bung and h the head diameter, and I the length of the calk. And in working by the sliding rule, it need only be remarked that 32•82 and 29°7 are the roots of 1077•157 and 8E2-355, 882355 {"0001}={118-95 wine By the Pen. Here (2 x 32? + 2+) X 40 X s 97.44 ale gallons, the content "Oon required. By the Sliding Rule. Having set 40 on c to 32.82 on d, against 32 and 24 on D, stand 38 and 21-3, as near as can be judged, on c; then 2 X 38 + 2103 76 + 21'3 = 97.3 ale gallons. And having set 40 on c to 2967 on D, against 32 and 24 on D, stand 46.5 and 26•1 on c ; then 2 X 46:5 + 26'1 = 93 + 26:1 = 119-1 wine gallons. To find the Content of a Cask of the Second or Parabolic Spindle form. *To the square of the head diameter, add double that of the bung diameter, and from the sum take or 40 of the square of the difference of the said diameters; then multiply the remainder by the length, and the product multiplied or divided by the fame numbers as in the rule to the last problem, will give the content. BY THE SLIDING RULE. As in the last problem, set the length on c to 32.82 or 29'7 on D, and on d find both the bung and head diameters, and also their difference, taking out the three 'numbers opposite to them on c; then if to twice the first be added the second, and its of the * For, by prob. 18 feat. 6 part 3, the content in inches is 8?.+ 4 BH + 3 H ? 10 F:+4 BH + 5H220+ 2 H2, 3 5 = 1.9 X (2 B2 + H2 X (B 11)?), and įn will give the same numbers as in the lait problem. |