 | Anthony Nesbit - 1824 - 476 sider
...the base. Find the area of a section at the eaves DC, in the same manner. To the sum of these areas, add the square root of their product; multiply this sum by the perpendicular height GF =^ HD, and ^ of the product will be the solidity of the frustum ABCD. Multiply the area of the section... | |
 | William Templeton (engineer.) - 1833 - 224 sider
...before. •» «bi2&' PROBLEM V. To find the Solid Content of the Frustum of a Pyramid. RULE. — To the sum of the areas of the two ends, add the square...multiply this sum by the perpendicular height ; and i of the product is the solid content. EXAMPLE. — Required the solid content of the frustum of a... | |
 | Samuel YOUNG (of Manchester.) - 1833 - 272 sider
...•f)l foot. PROBLEM III. To find the solidity of timber broader at one end than the other. RULE. To the areas of the two ends add the square root of their Product, which multiplied by ^ of the length will give the solidity. 9. a. . «. 6. 7. 8. 10. 11. 10 ... 12... | |
 | William Templeton - 1846 - 518 sider
...two ends add the square root of their product ; multiply this sum by the perpendicular height, and J of the product is the solid content. EXAMPLE. —...base = 144 inches, and area of the top end = 64. 144 + 64 = 208 and V 144 X 64 p = 96, then 208 + 96 X 24 _ 3 ~ cubic inches, and -f- 1728 = 1.4074 cubic... | |
 | William Templeton (engineer.) - 1846 - 236 sider
...nearly, as before. PROBLEM VI. To find the solid content of the frustum of a pyramid. RULE. — To the sum of the areas of the two ends add the square...perpendicular height, and ^ of the product is the solid content.1 . EXAMPLE. — Required the solid content of the frustum of a pyramid ABCD, whose perpendicular... | |
 | James Bates Thomson - 1847 - 432 sider
...15 ft., and base is 30 ft. square 1 638. To find the solidity of a frustum of a pyramid and cone. To the sum of the areas of the two ends, add the square root of the product of these areas ; then multiply this sum by % of the perpendicular height. (Leg. VII. 19.... | |
 | William Templeton (engineer.) - 1848 - 256 sider
...nearly, as before. PROBLEM VI. To find the solid content of the frustum of a pyramid. RULE. — To the sum of the areas of the two ends add the square...height, and -^ of the product is the solid content. height = 24 inches, the area of the base = 144 inches, and area of the top end = 64. 144 + 64 = 208... | |
 | James Bates Thomson - 1848 - 434 sider
...MEXSL'KATIOW. [SECT. XIX. 638. To find the solidity of a frustum of a pyramid and cone. To the rum of the areas of the two ends, add the square root of 'Jit product of these areas ; then multiply this sum by -J of the perpendicular heiyht. (Leg. VII.... | |
 | Thomas Rainey - 1849 - 320 sider
...necessity of the sum of the circumferences. To find the solidity of the frustum of a cone or pyramid, To the sum of the areas of the two ends, add the square root of the product of these areas; place this sum, with the vertical hight, on the right of the line, and... | |
 | James Elliot - 1851 - 152 sider
...PROBLEM III. To find the Solid Content of a Frustum of a Cone or Pyramid (Right or Oblique). RULE. To the areas of the two ends add the square root of their product. Multiply this sum by the height, and divide the product by 3. FORMULA. S = (E + e + \/ Ee) x \h. E and e being the areas of... | |
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RULE. To the sum of the areas of the two ends add the square root of their product ; multiply this sum by the perpendicular height, and \ of the product is the solid content. 
