M. B. BY THE SLIDING RULE. Length. Breadth. Area. As 2150.42 on A: 128.6 on B: 85.4 on A; 5.11 'n B. Depth. Content. As 1 on A: 5.11 on B :: 52,8 on A: 269.65 ›n B. Length. Or, Depth. Breadth. Content. As 128.6 on B: 52.8 on MD :: 85.4 on A : 269.65 'n B. EXAM. 3. Here 5.107 x 43.7 =223.1759 bushels, the conent required. PROBLEM II. To gauge a couch of malt contained in a rectangular frame. EXAM. 2. BY THE PEN. To find the area and content. Here 148 x 125 2150.42 185002150.42 8.602 bushels, the area; and 8.602 x 26.8 230.5336 bushels, the content required. M. B. BY THE SLIDING RULE. Length. Breadth. Area. As 2150.42 on A: 148 on B:: 125 on A ; 8.6 on B. Unity. And, Area. Depth. Content! As 1 on A 8.6 on B:: 26.8 on A : 230.53 on B. Length. Or, Depth. Breadth. Contex As 148 on B: 26.8 on MD :: 125 on A: 230.59 on B. EXAM. 3. Here 8.602 × 18.7 = 160.8574 bushels, the con tent required. PROBLEM III. To gauge a couch of malt, not in a frame, but las upon the floor, in a square or rectangular form. with one or more of its sides slanting. EXAM. 2. BY THE PEN. To find the area and content. Here 138 x 1132150.42 = 15594 ÷ 21504 7.251 bushels, the area of the couch; and 7.251 x 28.6207.3786 bushels, the content required. BY THE SLIDING RULE. Length. Depth. Breadth. Content As 138 on B: 28.6 on MD:: 113 on A: 207.5′′ on B. PROBLEM IV. To gauge a couch of malt either in the form of a cone, or a conical frustum. when the frustum is reduced to a cylinder; then, 104 x 104 x 28.2 10816 x 28.2 and 305011.22738 malt bushels, 305011.2; 111.399, the content in Length. BY THE SLIDING RULE. Depth. As 435 on B: 5.2 on MD:: 218 on A: 229.31 on B. EXAM. 3. BY THE PEN. Here 485 × 224 × 3.4 = 108640 × 3.4 = 369376; and 3693762150.42 = 171.769, the content malt bushels. BY THE SLIding rule. Breadth. Depth. Length. Conter As 224 on B: 3.4 on MD:: 485 on A: 17179 on B.. PROBLEM VI. To find the content of a rectangular floor of malt another method. EXAM. 2. BY THE RULE. Inches 435 length. 109 half the breadth. 3915 435 47.415 product. 5.2 depth. 94830 237075 246.5580 product. .930 factor. 739674 2219022 229.29894 content. Here half the length of the floor is 50; then if we multiply 250 by 50, and cut off three figures, as decimals, we obtain 12.5 for the false area. This being multiplied by 6, gives 75 bushels, for the false content. Lastly, if we multiply 75 by 7, and cut off two figures for a decimal, we obtain 5.25 bushels for the deduction; hence the true content of the floor is 69.75 bushels. Y 2 |