slide. Apply this number to the lower scale on the slide, and beneath it on the rod is the mean diameter. Fix the left end of the slide to the length of the cask on the upper part of the rod; find the mean diameter on the upper line on the slide, and over it, on the upper part of the rod, is the content in gallons. NOTE. When any dimension for finding the content of a cask by the head rod is thrown off the rule, work with half that dimension, and double the answer. If the segment or mean mean diameter, and take diameter be thrown off, double the the answer. If the mean diameter and length be thrown off, double both, and take the answer. 39. To compute the content of a cask by the sliding-rule. Subtract the head from the bung diameter, and multiply the difference, when it is Add the product to the head diameter, the sum will be the mean diameter. Then set the length on B to the gauge-point on D, and against the mean diameter on D will be the content on B. 40. To compute the content of a cask by the pen. Find the mean diameter by (Art. 39), multiply its square by the proper tabular multiplier, and this product multiplied by the length will give the content. EXAMPLE 1. It is required to find the content of a cask of the first variety, whose dimensions are, head diameter 28-2, bung diameter 33·0, and length 35'6 inches. By the head rod. Set the index on the slide to 28.2 on the lower part of the rod, and over 33.0 on the same line, on the scale marked spheroid is 33; this applied to the lower scale on the slide, cuts on the lower part of the rod 31·5, which is the mean diameter. Place the left end of the slide to 35.6 on the upper part of the rod, and over 31.5 on the upper line of the slider, stands on the rod 100 gallons, the content. By the sliding-rule. First, 33.0-28.2=4·8, and 4·8×·68=3·3; hence 28.2+3·3= 31.5, mean diameter. Then set 35'6 on B to 18-7892 on D, and at 31.5 on D cuts on B 100, the content in gallons. By the pen. The mean diameter is 31.5; then 31.5×31.5×·002832×35:6=100:03 gallons, the content. EXAMPLE 2. Let the head diameter of a cask be 22.7, the bung diameter 31.3, and the length 50 inches; required its content? By the head rod. Set the index to 22.7, and against 31.3 will be 6 on the scale spd., which found on the lower scale, is against 28.7 on the rod. Then place the end of the slide to 50, on the upper part of the rod, and against 28·7 on the slide will be 116 gallons, the content. By the sliding-rule. Suppose the cask to be of the second variety; then 31.3-22.7 8.6, and 8·6 ×·64—5·5. Again, 22·7×5*5=28·2, the mean diameter. Now set 50 on B to 18-7892 on D, and beneath 28.2 is 112.6 galls., content. By the pen, The mean diameter is 28-2; then ULLAGING. 41. A cask is on ullage when the liquor it contains does not fill it, and to compute the quantity contained is ullaging. The depth of the liquor is called wet inches, the remainder of the bung diameter or length dry inches. A cask on its side is said to be lying, one on its end standing. The dimensions necessary for finding the ullage of a cask are, the bung diameter, if the cask be lying, the length, if it be standing, and the wet inches, together with the content. Should the content not be given, it must be found by the preceding rules. If it be given or ascertained, and the cask be standing, measure with a bung rod, or some other proper instrument, through a hole in one end, the length, and the inches on the rod, wet with the liquor, will be the wet inches. If the cask be lying, measure the vertical bung diameter, and the inches wet will be the wet inches, for all practical purposes. 42. To find the ullage of a lying cask by the head-rod, sliding-rule, or ullage-rule. Set the bung diameter on C to 100 on the line marked SL, and against the wet inches on C will be the segment or mean area on the line marked SL. Then set the content on B to 1 on A, and beneath the mean diameter or segment on A will be the ullage quantity on B. NOTE. When the wet inches are less than of the bung diameter, the segment or mean area is found on the upper part of the line SL, on the first form of Rule (5); and on a line marked Seg. Ly, immediately beneath the line marked SL, on the second form of Rule (5). 43. To ullage a lying cask by the pen. RULE. Divide the wet inches by the bung diameter; if the quotient be less than 500, subtract of the difference from the quotient; if the quotient exceed 500, add of the difference to the quotient; the remainder or sum multiplied by the content will give the ullage quantity. EXAMPLE. The content of a lying cask is 124 gallons, the bung diameter 360, and the wet 24.8 inches; how many gallons does it hold? By the sliding-rule. Set 360 on C to 100 on SL, and against 24.8 on C is 75 on SL, the segment; then set 124 on B to 1 on A, and below 75 on A is 93 galls. Ans. on B. By the pen. 24.8÷36='690; from '690-500—190, which ÷4='0475; and 690+0475=7375. Then '7375×124-91.45 galls. Ans. The rule by the pen is obviously only an approximation. 44. To find the ullage of a standing cask by the sliding or ullage rule. Use the side or line marked SS., instead of the side or line marked SL., and proceed as in ullaging a lying cask. (Art. 42). NOTE. When the wet inches are less than of the length, the segment or mean area is found on the upper part of the line SS, on the first form of Rule (5); and on a line marked Seg. St., immediately below the line marked SS., on the second form of Rule (5). 45. To find the ullage of a standing cask by the pen. RULE. Divide the wet inches by the length of the cask; if the quotient be under 500, subtract of the difference from the quotient; if above 500, add of the difference to the quotient. The remainder, or sum multiplied by the content of the cask, will give the ullage quantity. EXAMPLE. Find the ullage quantity in a standing cask, whose content is 120 gallons, length 430, and wet 176 inches. By the sliding-rule. Set 43.0 on C to 100 on SS, and beneath 17·6 on C is 400 on SS; then place 120 on B to 1 on A, and beneath 40-0 on A will be on B 48 gallons ullage. By the pen. Now 17.6÷43'0=409, and 500-409-091. Then -091 ÷10=0091, and ·409—·009=400, and 400×120=48 gallons, the ullage quantity. MISCELLANEOUS QUESTIONS. 1. Required the divisor, multiplier, and gauge point for a cone of flint glass. Ans. Divisor 33, multiplier 030303, gauge point, 5·7445. 2. What are the divisor, multiplier, and gauge point for a pentagon in bushels? Ans. Divisor 1289-2889, multiplier 0007756, gauge point 35.9067. 3. What is the circular gauge-point for gallons, when the middle area of a regular frustum is taken? Ans. 46 024. 4. Multiply 74 by 027, by the sliding rule? Ans. 1·998. 5. A malt couch is 7.44 bushels in area, and mean depth 18-3 inches; required the content in bushels, and in net malt? Ans. 110-96 net. J 136.15 bush. 6. What is the quantity of malt on a floor whose length is 766, breadth 164, and depth 03.1 inches. Ans. 175.5 bush. 7. A soap-frame 45 long and 15 inches broad, has in it hot silicated soap to the depth of 51-4 inches,-how many pounds does it contain? Ans. 1443.2 lb. 8. Given a plate glass pot whose depth is 21 inches, and cross diameters at 3.5 from the mouth, are 24·2 and 23′9; at 10.5 are 25.3 and 25·5; and at 17.5 are 27.4 and 28·1; to determine the pounds weight of metal it will hold? Ans. 967-44 lb. NOTE Plate glass pots are tabled for dry inches. 9. A floor of malt is found to be 250 bushels, and its corresponding couch was 158.8 bushels; from which would the duty charge arise, and upon how many bushels net? Ans. Couch; 129-42 bush. net. 10. The length of a cask of the first variety is 43'0, bung 36.5, and head 28.2 inches; how many gallons is its content? Ans. by the rule, 141. by the pen, 140-7 galls. 11. Given a cask whose content is 108 gallons, bung diameter 33.5, and wet 24.1 inches, to find the ullage quantity? Ans. By the rule, 84.5 By the pen, 83-6 galls. Ans. 4. 12. Divide 228 by 57 on the sliding rule? 13. There is a mash tun whose top diameter is 70, bottom diameter 504, and depth 40.0 inches; required the content, and tabulation at each inch, in qrs., bushels, and gallons? Ans. Content, 6 qrs. 3 bush. 6 galls. 14. What is the area in gallons, of an ellipse whose diameters are 72 and 50 inches? Ans. 10-1952 galls. 15. Required the content in bushels, of a frustum of a cone, of which the top diameter is 24, the bottom diameter 36, and the depth 53 inches? Ans. 17.11 bush. 16. The length of a cooler is 212, the breadth 148, and the depths at different places, 38, 3·6, 4·4, 4·4, 3·9, 4·0, 3′5, 4·2, 4.3, and 3.9 inches; how many gallons does it contain? Ans. 452.63 galls. 17. A copper with a rising crown has to be inched and tabulated, for dry inches, in barrels, firkins, and gallons, the dimensions of which are-depths 48, from the centre of the crown, and 54 from the rising of the crown to the top of the vessel, the cross diameters at 6 from the top 75.0 and 74.6, at 17, 74.8 and 74-4, at 26, 69.6 and 69′0, at 33, 65.5 and 65.5, at 39, 61.6 and 61.2, and at 45, 57.1 and 57.2 inches, and the quantity to cover the crown 22 gallons: what is the content of the copper? Ans. 18 bar. 2 firk. 5·04 galls. 18. Of a standing cask the length is 460, wet 19.8 inches, and the content 207 gallons; how much does it contain? Ans. 874 galls. 19. The depth of a flint glass pot is 18.5 and its mean diameter 29.9 inches; find the area and content. Ans. Area 81.2736 lb., content 1503-5616 lb. 20. Suppose the dimensions of a guile tun, in the form of the frustum of a square pyramid, to be depth 30 inches; the length of a side 5 inches from the bottom, 32:5, at 15 inches, 38.5, and at 25 inches, 42.5 inches; it is required to find the areas in gallons of each frustum. Ans. 1. 3.8094 gallons. 2. 5.3458 do. 3. 6.5143 do. 21. Required the content of a cask whose head diameter is 17.4, bung diameter 19.6, and length 23.7 inches? Ans. 24 nearly. 22. How many bushels of malt are there in a floor, of which the length is 863, breadth 285, and depth 1.7 inches? Ans. 188-5 bush. 23. What quantity of hard soap, hot, is contained in a cylinder, whose diameter is 36.5 and depth 71·4 inches? Ans. 2667-81 lb. 24. Let the altitude of the globular part of a still be 8 inches, and the cross-diameters at the top and bottom be 27·2 and 26.8, and 55.2, and 54.8; also the cross diameters of the body of the still at 4.5 inches down be 59.8 and 60·2; at 13.5 inches 63.8 and 64.4; at 22.5 inches 64.0 and 64·6, and at 32.5 inches 62.0 and 62.4 inches; and let the quantity required to cover the crown be 35 gallons; required the content in gallons? Ans. 500-968 galls. |
