 | Richard Turner - Geometry - 1787 - 264 pages
...again by ,418879 = T8T of .7854. f^ To find the Solidity of the Middle Zone of a Parabolic Spindle. To twice the Square of the Middle Diameter add the Square of the Diameter of the End ; from this Sum fubtracT: .4 (Tenths ) of the Square of the Difference of the two Diameters ; multiply... | |
 | Mathematics - 1801 - 658 pages
.... . 9 GŁ$Z I4_. . Wl:cn the ends are circular, or parallel t9 thi revolving axe. _ To dovble tiie square of the middle diameter add the square of the diameter of one end ; multiply this sum by the length of the frustum, and the product by '2618 for the content.... | |
 | Samuel Webber - Mathematics - 1808 - 520 pages
...spheroid. . CASE I. When the ends are circular, or parallel to the revoking axis. RULE.* To double the square of the middle diameter add the square of the diameter of one end ; multiply this sum by 90 •5236 12 2808 ^^^ 78 41888 iy^Url ____- -^ •c- 4^^' f -3k. 36... | |
 | Nicolas Pike - Algebra - 1808 - 470 pages
...is somewhat like a cask, two equal segments being wanting, one on each side of the axis. RULE — To twice the square of the middle diameter, add the square of the end diameter ; multiply that sum by '7S54, and that product, multiplied by one third of the length,... | |
 | Thomas Keith - 1817 - 306 pages
...89'33 12. PROBLEM XXIV. To find the Solidity of the middle Frustum of a parabolic Spindle. RULE. To twice the square of the middle diameter, add the square of the diameter of the end; and from the sum subtract four-tenths of the square of the difference between these diameters : the... | |
 | William Hawney - Geometry - 1820 - 336 pages
...thai, of either of the ends being given. RULE. To twice the «juare of the middle diameter, add tW square of the diameter of the end ; multiply the sum by the length of the zone, and the product again by 26 1 8 for the solidity. NOTE. This rule is useful in guaging. A cask... | |
 | Anthony Nesbit - Surveying - 1824 - 476 pages
...frustum of a spheroid. CASE I. When the ends are circular or parallel to the revolving axe. RULE. To twice the square of the middle diameter, add the square of the diameter of one end ; multiply the sum by the length of the frustum, and the product thence arising by .2618, for... | |
 | Thomas Curtis - Aeronautics - 1829 - 810 pages
...being given. Rule. — Case 1. When the ends are circular, or perpendicular to the fixed axis. — To twice the square of the middle diameter add the square of the diameter of either of the ends, and multiply the same by the length of the frustum, and by -2618, and the result... | |
 | John Bonnycastle - Geometry - 1829 - 256 pages
...the ends being given. CASE I. When the ends are circular, or parallel to the revolving axis. RULE* To twice the square of the middle diameter add the square of the diameter of either of the ends, and this sum * Demon. Let AO= a, DO= 6, E»=A, no= c, ro-= x, re =y, and p= 3.... | |
 | Ireland commissioners of nat. educ - 1834 - 370 pages
...18849-6. PROBLEM XXIV. To find the solidity of the middle frustrum of a parabolic spindle. RULE. To double the square of the middle diameter, add the square of the diameter of the end ; and from the sum subtract -^ of the square of the difference between these diameters ; the remainder... | |
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To twice the square of the middle diameter, add the square of the diameter of one end; multiply the sum by the length of the frustum, and the product by '2618 for the content. 
