PROBLEM IX. To find the solidity of the frustum of a paṛabolic conoid, the greater diameter AB, the lesser CD, and the perpendicular height EF being given. To the square of the diameter of the greater end AB, add the square of the diameter of the lesser end CD; multiply the sum by 3927, and the product by the height EF, will give the solidity required. EXAMPLE. What is the solidity of a parabolic frustum, the diameter of the greater end being 60 feet, the lesser end 48 feet, and the distance of the ends 18 feet? 6024825904 the sum of the squares of the ends. Then 59043927 x 18 = 41733 0144 the solidity required. OF AN ANNULUS, OR CYLINDRIC RING. I. DEFINITIONS. If a circle is carried round a right line as an axis, and in the same plane with the circle, either touching the axis, or at a given distance from it, it will generate a solid, called an annulus, or cylindric ring. II. The diameter of the generating circle is called the thickness of the ring. III. Twice the distance of the generating circle, from the axis of rotation, is cailed the inner diameter. PROBLEM I. To find the solidity of an annulus, or ring, whose thickness and inner diameter are known. METHOD I. To the thickness of the annulus, add the inner diameter; multiply the sum by the square of the thickness, and the product by 2.4674, will give the solidity sought. METHOD METHOD II. Multiply the circumference round the middle of the annulus, or that circle generated by the centre of the generating circle, by the area of the generating circle, and the product will give the solidity. Note. This last method will give the solidity of any part of an annulus, or ring, comprehended between any two planes passing through the fixed axis. EXAMPLE.. What is the solidity of an annulus, whose inner diameter is 8 inches, and the thickness of the annulus 3 inches? Then 83 X 32 X 24674 = 244 2726 the solidity. PROBLEM II. To find the solidity of a hollow cylinder, the exterior and interior diameters being given, and the perpendicular height. Multiply the sum of the diameters CD and EF by their difference, and the product by 7854; then by the altitude GH of the cylinder, and you will have the solidity required. EXAMPLE. Required the solidity of a hollow cylinder, the exterior diameter CD being 14 feet, the interior diameter EF 12 feet, and the perpendicular height AC 10 feet. To find the solidity of a frustum of a hollow cone, of an equal thickness; the exterior and interior diameters at each end being given, and the perpendicular altitude. Multiply together the sum of the two interior diameters, and the difference of the diameters at either end, by 7854, and by the perpendicular altitude; then by the said difference, and the continued product will give the solidity. required. EXAMPLE. Required the solidity of a frustum of a hollow cone, the bottom diameters, AB and GH, being respectively 35 and 32 feet; the top diameters, CD and EF, respectively 29 and 26 feet; and the perpendicular altitude IK, 25 feet. Then 35 32 3 the difference of the diameter. And 32 + 26 + 3 × 7854 × 3 × 253593·205 the solidity. PROBLEM PROBLEM IV. To find the solidity of a hollow segment of a globe, the bottom diameters AB and CD being given, and the perpendicular altitudes DF and EG of the exterior and interior segments. Find the solidity of each segment by Prob. III. p. 199, subtract the lesser segment from the greater, and the difference will give the solidity of the hollow segment. PROBLEM V. To find the solidity of a hollow frustum of a zone; given the bottom diameters AB and CD, the top diameters EF and GH, and the perpendicular altitude IK. Add into one sum, twice a rectangle under the difference of the diameters at the bottom, by the less diameter; twice a rectangle of the difference of the top diameters by the less, and the squares of the difference of both ends; then multiply the sum by 7854, and the product by the perpendicular altitude will give the solidity. EXAMPLE. Required the solidity of a hollow zone, the diameters AB and CD, at the bottom, being respectively 26 and 24 feet; and the top diameters, EF and GH, 23 and 20 feet respectively; and the perpendicular height IK, 4 feet. 2 2× 26-24x24+2×23-20×20+26-24 +23-20 X7854×4=719-4264 the solidity. |