2. What is the content of a cask, whose length is 20, and bung and head diameters 16 and 12? NOTE. This is the most exact rule of any, for three dimensions only; and agrees nearly with the diagonal rod. PROBLEM VII, To find the ullage of a cask by the sliding rule, NOTE. The ullage of a cask is what it contains, when only partly filled.. And it is considered in two positions, namely, as standing on its end with the axis perpendicular to the horizon, or as lying on its side with the axis parallel to the horizon. RULE, By one of the preceding problems find the whole content of the cask. Then set the length on N to 100 on SS for a segment standing, or set the bung diameter on N to 100 on SL for a segment lying; then against the wet inches on N is a number on SS or SL, to be reserved. Next, set 100 on B to the reserved number on A; then against the whole content on B will be found the ullage on A. EXAMPLES. 1. Required the ullage answering to 10 wet inches of a standing cask, the whole content of which is 92 gallons, and length 40 inches. Set Set 40 on N to 100 on SS; then against 10 on N is 23 on SS, the reserved number. Then set 100 on B to 23 on A, and against 92 on B is 21°2 on A, the ullage required. 2. What is the ullage of a standing cask, whose whole length is 20 inches, and content 11 gallons; the wet inches being 5? Ans. 2'65 gal. 3. The content of a cask being 92 gallons, and the bung diameter 32, required the ullage of the segment lying, when the wet inches are 8. Ans. 164 gal. PROBLEM VIII. To ullage a standing cask by the pen. RULE. Add together the square of the diameter at the surface of the liquor, the square of the diameter of the nearest end, and the square of double the diameter taken in the middle between the other two; then multiply the sum by the length between the surface and nearest end, and the product by 0004 for ale gallons, or by *0005 for wine gallons, in the less part of the cask, whether empty or filled. 1 EXAMPLE. EXAMPLE. The three diameters being 24, 27 and 29 inches, required the ullage for 10 wet inches. Divide the wet inches by the bung diameter; find the quotient in the column of versed sines, in the table of circular segments, taking out its corresponding segment. Then multiply this segment by the whole content of the cask, and the product by 14 for the ullage required, nearly. EXAMPLE. EXAMPLE. Supposing the bung diameter 32, and content 92 ale gallons; to find the ullage for & wet inches. 32)8(25, whose tab. seg. is 153546 92 307092 1381914 14°126232 is 3531558 17657790 answer. NOTE. The capacity of any vessel, having its cavity in the form of any solid, contained in MENSURATION OF SOLIDS, may be determined by means of the rule, given. for finding the content of such solid. END OF GAUGING. |