 | Charles Haslett - Engineering - 1855 - 482 pages
...the base is circular or parallel to the revolving 'axis. RULE. From triple the fixed axis take double the height of the segment; multiply the difference by the square of the height, and by •5236 ; then say, as the square of the fixed axis is to the square of the revolving axis, so is... | |
 | Charles Haslett - 1855 - 544 pages
...the base is circular or parallel to the revolving axis. RULE. From triple the fixed axis take double the height of the segment; multiply the difference by the square of the height, and by •5'23ti ; then say, as the square of the fixed axis is to the square of the revolving axis, so... | |
 | Charles W. Hackley - Engineering - 1856 - 530 pages
...the base is circular or parallel to the revolving axis. RULE. From triple the fixed axis take double the height of the segment ; multiply the difference by the square of the height, and by •5236 ; then say, as the square of the fixed axis is to the square of the revolving axis, so is... | |
 | Charles Haynes Haswell - Measurement - 1858 - 350 pages
...and the product, multiplied by .5236, will give the contents required. Or, (Sr2 + A2) x Ax -5236 =S. 2. From three times the diameter of the sphere subtract twice the height of the segment ; multiply this remainder by the square of the height, and the product, multiplied by .5236, will give the contents... | |
 | Anthony Nesbit - Measurement - 1859 - 494 pages
...198944286 square miles. PROBLEM XH. To find the solidity of a segment of a sphere. RULE I. KULE II. From three times the diameter of the sphere subtract twice the height of the segment ; multiply lhe remainder by the square of the height, and that product by .52361 for the solidity. Note 1 . The... | |
 | Alfred Newsom Niblett - 1861 - 204 pages
...height; and this sum multiplied by the height, and the product again by '5230, will give the solidity. 2. From three times the diameter of the sphere subtract twice the height of the segment; multiply the remainder by the square of the height, and that product by -52361 for the solidity. NOTK 1, The surface... | |
 | Oliver Byrne - Engineering - 1863 - 600 pages
...and this sum multiplied by the height, and the product again by -5236, will give the solidity. Or, From three times the diameter of the sphere subtract twice the height of the segment, multiply by the square of the height, and that product by -5236 ; the last product will be the solidity. The... | |
 | Whiting Griswold - Railroad engineering - 1866 - 144 pages
...or parallel to the revolvingaxis or least diameter. RULE 10. From triple the fixed axis take double the height of the segment ; multiply the difference by the square of the height, and by '5236. Then say, as the square of the fixed axis is to the square of the revolving axis, so is the... | |
 | Alfred Hiley - 1871 - 184 pages
...when the diameter of the sphere and the height of the segment are gwen. From three times the diameter subtract twice the height of the segment ; multiply the difference by the square of the height of the segment, and the product by ^. (3) To find the curved or convex surface. (4) To find the whole... | |
 | James E. Ryan - Arithmetic - 1877 - 212 pages
...Sphere EQUALS TJie cube of the diameter x .5236. XL. a. THE VOLUME OF A Segment of a Sphere. RULE. — From three times the diameter of the sphere subtract twice the height of the segment ; multiply this remainder by the square of the height, and the product by .5236. 451. XL1. To COMPUTE THE VOLUME... | |
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From three times the diameter of the sphere subtract twice the height of the segment; multiply this remainder by the square of the height and the product by 0.5236. 
