53. HYPERBOLOID. An hyperboloid is a solid generated by the revolution of an hyperbola about its axis, which remains fixed. See ¶ 30. The hyperboloid is also called the hyperbolic conoid. A frustum of an hyperboloid is a portion of it contained between two planes perpendicular to the axis. To find the solidity of an hyperboloid :— To the square of the radius of the base add the square of the diameter of a section midway between the base and vertex ; multiply this sum by the altitude, and the product by .5236. To find the solidity of a frustum of an hyperboloid : Add together the squares of the radii of the two ends, and the square of the middle diameter between them; multiply the sum by the altitude, and this product by .5236. EXAMPLES. 1. Find the solidity of an hyperboloid, the altitude of which is 25, the radius of the base 26, and the middle diameter 34. Ans. 23980.9. 2. What is the solidity of an hyperboloid, the radius of the base being 6, the diameter of the middle section 10, and the altitude 20? Ans. 1424.19. 3. What is the solidity of the frustum of an hyperboloid, the diameters of the ends being 6 and 10, the middle diameter 81, and the height 12? Ans. 667.59. 4. Find the solidity of a frustum of an hyperboloid, the diameters of the ends being 3 and 5, the middle diameter 4.25, and its height 8. Ans. 111.26. 5. What is the solidity of an hyperbolic frustum, the end diameters being 12 and 20, the middle 17, and height 18? Ans. 4005.46. 54. CYLINDRIC RINGS. A cylindric ring is a solid generated by the revolution of a sphere through a circular orbit; or, it is a cylinder bent into the form of a circle. The interior diameter of the ring is a line passing through its centre, and limited by the interior surface; and the external diameter is a line passing through its centre, and terminating at its exterior surface. Half the sum of the two diameters is the diameter of the axis of the ring; and half the difference of the diameters is the diameter or thickness of the ring. AB is the exterior diameter, cd the interior diameter, mon the axis, and Ac the thickness of the ring. The diameter of the axis of the ring multiplied by 3.1416 gives its length; and the square of its thickness multiplied by .7854 gives the area of a cross-section of the ring, which being multiplied by the length of its axis, gives the solidity; and said length, multiplied by the circumference of a cross-section of the ring, will give its surface, or superficial area. Or, To find the solidity of a cylindric ring: Multiply the square of the thickness by the diameter of the axis, and that product by 2.467412. To find the surface of a cylindric ring: Multiply the diameter of its axis by its thickness, and that product by 9.8696. EXAMPLES. 1. Find the solidity of a cylindric ring, its diameters being 16 and 24. Ans. 789.57. 2. Find the solidity of a cylindric ring, its diameters being 8 and 14 inches. Ans. 244.274 inches. 3. The interior diameter of a cylindric ring is 26, and its thickness 8 inches; what is its solidity? Ans. 5369 inches. 4. What is the surface of a cylindric ring, whose thickness is 1 inch, and its inner diameter 9 inches? Ans, 98.696 square inches. 5. Find the surface of a cylindric ring, whose diameters are 36 and 52. Ans. 3474.1. 6. What is the surface of a cylindric ring, whose thickness is 6 inches, and inner diameter 24 inches? Ans. 1776.528. 55. SPINDLES. A spindle is a solid generated by the revolution of a segment of an ellipse, circle, parabola, or hyperbola, about the chord or diameter, which cuts off the segment; and said chord forms the axis of the spindle. The central distance of a circular spindle is the distance between the centre of the circle and the centre of the spindle.— Thus, vs is the central distance. The length of the spindle and half its middle diameter being given, (since these are the chord and versed sine of the generating circular segment,) the radius of the circle may be found, and the area of the segment, by the rules given in paragraphs 24 and 25. THE CIRCULAR SPINDLE. n To find the solidity of the circular spindle : Multiply the area of the generating segment (cbe) by the central distance (vs), and subtract the product from one-twelfth of the cube of the length of the spindle, and multiply the remainder by 6.2832. Or, To find the solidity of the circular, the elliptic, or the hyperbolic spindle :— To the square of the middle diameter (bo) add the square of double the diameter at one-fourth the length, (viz. nr;) multiply the sum by the length, and the product by .1309, and the result will be the solidity, nearly.-(nr is called the quarter diameter.) The parabolic spindle is just eight-fifteenths of its least circumscribing cylinder. Therefore, To find the solidity of the parabolic spindle : Multiply the square of the middle diameter by the length of the spindle, and the product by .41888. EXAMPLES. 1. The length of a circular spindle is 8, and its middle diameter 6; what is its solidity? Ans. 138.516. 2. Find the solidity of an elliptic spindle, its length being 20, the middle diameter 6, and its quarter diameter 4.75. Ans. 330.5225. 3. What is the solidity of an hyberbolic spindle, its length being 10, its middle diameter 8, and its quarter diameter 4.4 ? Ans. 185.14496. 4. The length of a parabolic spindle is 30, and its middle diameter 17; what is its solidity? Ans. 3631.68. 5. Find the solidity of a parabolic spindle, whose length is 18, and middle diameter 6 feet. Ans. 271.434. 6. What is the solidity of a parabolic spindle, whose length is 50, and middle diameter 10 inches? Ans. 2094.42 inches. 7. The middle diameter of a parabolic spindle is 16, and its solidity 4289.33; what is its length? Ans. 40. An elliptical spindle is nearly four-sevenths of its least circumscribing cylinder, and its solidity may be found accordingly. 56. MIDDLE FRUSTUMS OF SPINDLES. n α The figure represents the middle frustum of an elliptic spindle. The diameter, ab, at half the distance from the centre of the frustum to its end, is called the quarter diameter. |