Also, 831104-882.36=941.910, the content in wine gallons; and 831104÷8214=101.181, the content in malt bushels. By Rule II. Inches. 72 bottom diameter. 2880 product. Inches. 72 bottom diameter. 112 sum. 112 sum. 224 112 112 12544 square of their sum. 86 perpendicular depth. 57984 77312 831104 the same as before. Hence we obtain the contents by division, as in the last Rule. By the Sliding Rule. By Problem X., Part II., find a mean diameter between the two given diameters; then find areas from these three diameters, by Problem XIII., Part IV. Multiply the sum of these areas by of the depth; and the product will be the content. Note. The content in malt bushels, may be found by a similar process. 2. The top diameter of a guile-tun is 60 inches, the bottom diameter 40 inches, and the perpendicular depth 74 inches; what is the content in ale gallons? Ans. 522.118 ale gallons. 3. What is the content, in malt bushels, of a vessel whose top diameter is 40 inches, bottom diameter 36 inches, and depth 52 inches? Ans. 27.449 malt bushels. PROBLEM IX. To find the content of a vessel in the form RULE, By the Pen. To the sum of the areas of the two ends, add four times the area of a section parallel to, and equally distant from both ends; multiply this sum by the perpendicular depth, and of the product will be the content in cubic inches. Divide this content by 282 for ale gallons, 231 for wine gallons, and 2150.42 for malt bushels. Note 1. The length of the middle section is equal to half the sum of the lengths of the two ends; and its breadth is equal to half the sum of their breadths. 2. If the ends be elliptical, the transverse diameter of the middle section will be equal to half the sum of the transverse diameters of the two ends; and the conjugate diameter equal to half the sum of the conjugate diameters of the two ends. 3. If one end be an ellipse and the other a circle, add the transverse diameter of the elliptical end to the diameter of the circular end; and take half the sum for the transverse diameter of the middle section. The conjugate diameter of the middle section must be found in a similar manner. 4. If the square of the diameter of a circle be multiplied by .7854, the product will be the area in square inches; and if the rectangle of the two diameters of an ellipse be multiplied by .7854, the product will be the area in square inches. (See Problems XIII. and XVI., Part IV.) SCHOLIUM. The Rule given in this Problem for a prismoid, is very clearly demonstrated in Simpson's Fluxions, page 178, first edition; and also in a similar manner at page 302, of Holliday's Fluxions. There is likewise a very elegant demonstration given in Proposition III., Section I., Part IV., of Dr. Hutton's Mensuration, third edition, in which it is shewn to be true for all frustums whatever, and for all solids whose parallel sections are similar figures; and Mr. Fletcher, in the second edition of his Universal Measurer, Part III., page 54, and Mr. Moss, in his Gauging, page 175, third edition, say, that it is nearly true for any other solid, whatever may be its form. EXAMPLES. 1. Required the content, in ale and wine gallons, and malt bushels, of the prismoidal vessel ABCDEFG; the length AB of its bottom being 40, and its breadth AG 21 inches; the length CF of its top 52, and its breadth FE 25 inches, and the perpendicular depth An 60 inches? Here 40×21=840, the area of the bottom; and 52x 25=1300, the area of the top of the vessel. 2 92 =46, the length of the middle section; =23, the breadth of the middle section; then 46 × 23 × 4=1058 × 4=4232, four times the area of the middle section; whence (840+1300+4232)× 60 =6372 X 6 10=63720, the content in cubic inches; and 63720÷282= 225.957, the content in ale gallons; 63720÷231=275.844, the content in wine gallons; and 63720-2150.42=29.631, the content in malt bushels. By the Sliding Rule. By Problem II., Part IV., find the areas of the two ends; and also the area of the middle section. Add the areas of the two ends, and four times the area of the middle section, together; then say, as six on A, is to the sum of these areas on B; so is the perpendicular depth on A, to the content on B, :: 21 15.00 four times the area of ditto. S25. S 4.61 area of the top. 22.59 sum. As 6: 22.59 :: 60: 225.90 content. |