Ex. 2. Required the number of square inches which will cover a circular spindle, whose length is 80 and thickness 16 inches ? Anf. 2747.3166 336. Ex. 3. Required the area of a circular spindle, whose length is 12, and thicknefs y inches. Anf. 294.3621 sq. inches. PROBLEM XVII. To find the solidity of a circular spindle. Multiply the area of the revolving segment by the distance between the centres of the arch and spindle, subtract the prcduct from the cube of half the length of the spindle, then multiply the remainder by 4, and this product again by 3.1416 for the solidity. See the last figure. EXAMPLE I. Required the area of a circular spindle, whose length is 60 and diameter 4). Dd AD 1250 the area of the sector AFCB. 8.75 X 30= 262.5 the area of the triangle ACF. 937.5 the area of the rev. segm. ACB. 49375 60125 29625 29500 4320.3125 30 half the spindle. 900 3)27000 9000 one-third cube į spindle. 4 18718.7500 3.1416 1123125000 Ex. 2. Required the solidity of a circular spindle, whose length is 30, and thickness 224 inches. Anf. 7350.853125 Ex. 3. Required the solidity of a circular spindle, whose middle diameter is 36, and length 40 inches. Anf. 29919 citbic inches. PROBLEM XVIII. To find the folidity of the middle zone of a circular spindle. RULE. From the fourth part of the {quare of the length of the whole spindle, subtract the square of half the length of the middle fruitum, and multiply the remainder by the length of of the frustum : Multiply the central distance by the revolving area which generates the frustum ; then subtract this latter product from the former, and multiply the remainder by 3.1416, and twice the product will be the solidity. EXAMPLE I. Required the folidity of the frustum of a circular spindle, whose length is 40, greatest diameter 36, and least 16 inches. Draw EG parallel to mn, then EF shall be equal , mn, =20 and EF2 +FB-=EB=500 chord. EB2 500 -= 50 diameter of the generating circle. FB 10 Hence rad. BD = 25 AL-=AD2-LD-=625—495576 133.3 3 3 442.6 20 8853.3 first product. BE |