Spherical Sector. Definition. A figure generated by the revolution of a sector of a circle about a straight line drawn through the vertex of the sector as an axis. NOTE.-The arc of the sector generates the surface of a zone termed the base of the sector of a sphere, and the radii generates the surfaces of two cones, having a vertex in common with the sector at the centre of the sphere. To ascertain the Contents of a Spherical Sector, Fig. 109. RULE.-Multiply the surface of the zone, which is the base of the sector, by one third of the radius of the sphere, and the product will give the contents required. Or, a ×=S, a representing the area of the base. Fig. 109. b g EXAMPLE.-What are the contents of a spherical sector, Fig. 109, generated by the sector c a h, the height of the zone a b c d being, a o, 12 inches, and the radius, g h, of the sphere 15 inches? 12 × 94.248=1130.976=height of zone X circumference of sphere=surface of zone (see p. 90). 1130.976×30=5654.88=product of surface of zone and of diameter (=of radius)=result required. Ex. 2. The diameter of a sphere is 10 inches, and the height of the zone or base of a sector is 5 inches; what are its contents? Ans. 261.5 inches. NOTE. The surface of a spherical sector the sum of the areas of the zone and the two cones. SPINDLES. Definition. Figures generated by the revolution of a plane area bounded by a curve, when the curve is revolved about a chord perpendicular to its axis, or about its double ordinate, and they are designated by the name of the arc from which they are generated, as Circular, Elliptic, Parabolic, etc. Circular Spindle. To ascertain the Contents of a Circular Spindle, Fig. 110. RULE.-Multiply the central distance, o e, by half the area of the revolving segment, a c ef. Subtract the product from one third of the cube of half the length, fe; then multiply the remainder by 12.5664, and the product will give the contents required. (1÷2)3 ×12.5664=S, I representing the length of the spindle, and a the area of the revolving segment. EXAMPLE. What are the contents of a circular spindle, when the central distance, o e, Fig. 110, is 7.071067 inches, the length, fe, 14.14213, and the radius, o c, 10 inches? NOTE. The area of the revolving segment, fe, being the side of the square that can be inscribed in a circle of 20, is 202 × .7854-14.142132 +4=28.54. 7.071067 × 14.27=100.9041=central distance half area of revolving segment. 100.9041 7.071673 =16.947=remainder of above product and of cube of half the length. 16.947 × 12.5664=212.9628=result required. Ex. 2. The central distance of a circular spindle is 3 inches, the area of the revolving segment is 11.5 inches, and the length of the spindle 8 inches; what are the contents of it? Ans. 51.3086 inches. Ex. 3. The length of a circular spindle is 24 inches, and its diameter 18; what are the contents of it? Ans. 3739.585 cubic feet. The chord of the arc, 24, and the versed sine (18), 9, being given, the diameter of the circle is found by rules p. 75, 76. Area of revolving segments, per table, p. 136, 137, 159.094. Frustrum or Zone of a Circular Spindle.* To ascertain the Contents of a Frustrum or Zone of a Circular Spindle, Fig. 111. RULE. From the square of half the length, ƒ g, of the whole spindle, take of the square of half the length, i c, of the frustrum, and multiply the remainder by the said half length of the frustrum; multiply the central distance, o c, by the revolving area which generates the frustrum; subtract this product from the former, and the remainder, multiplied by 6.2832, will give the contents required. NOTE.-The revolving area of the frustrum can be obtained by dividing its plane into a segment of a circle and a parallelogram. *The middle frustrum of a circular spindle is one of the various forms of casks. 2 Or, l÷2 3 2 -(cxa) x 6.2832=S, 1 and repre senting the lengths of the spindle and of the frustrum, a area of the revolving section of frustrum, and c the central distance. EXAMPLE. The length of the middle frustrum of a circular spindle, i c, is 6 inches, the length of the spindle, fg, is 8 inches, the central distance, o e, is 3 inches, and the area of the revolving or generating segment is 10 inches; what are the contents of the frustrum? (8-2)2 (6÷2)2=13×3=39=product of half the length of the frus 3 trum and the remainder of the square of half the length of the frustrum subtracted from the square of half the length of the spindle. 39-3×10=9=product of the central distance and the area of the segment subtracted from preceding product. 9×6.2832=56.5488=last product × 6.2832=result required. Ex. 2. The length of a circular spindle is 1.333 feet, the length of the middle frustrum of it is 1 foot, the central distance is 6 inches, and the area of the revolving segment is 40 inches; what are the contents of the frustrum? Ans. 452.3904 cubic inches. Ex. 3. The length of the middle frustrum of a circular spindle is 12 inches, the length of the spindle being 24, the central distance is 3.5, and the area of the revolving segment is 96 inches; what are the contents of the frustrum? Ans. 2865.1392 cubic inches. Centre of Gravity. Is in its geometrical centre. Segment of a Circular Spindle. To ascertain the Contents of a Segment of a Circular Spindle, Fig. 112. RULE.-Subtract the length of the segment, i c, from the half length, i e, of the spindle; double the remainder, and ascertain the contents of a middle frustrum of this length. Subtract the result from the contents of the whole spindle, and half the remainder will give the contents of the segment required.* Or, C-c÷2 S, C' and c representing the contents of the spindle and middle frustrum. EXAMPLE. The length of a circular spindle, i a, Fig. 112, is 14.14213, the central distance, o e, is 7.07107, the radius of the arc, o a, is 10, and the length of the segment, i c, is 3.53553 inches; what are its contents? 14.14213 2 −3.53553×2=7.07107-double the remainder, of the length of the segment subtracted from half the length of the spindle-length of the middle frustrum. NOTE.-The area of the revolving or generating segment of the whole spindle is 28.54 inches, and that of the middle frustrum is 19.25. The contents of the whole spindle is.... 66 Hence...... 66 middle frustrum is.. 25.0323=the contents required. 212.9628 cubic in. 162.8982 66 66 50.0646-2= *This rule is applicable to the segment of any spindle or any conoid, the volume of the figure and frustrum being first obtained. |