20+2÷2=11×3.1416=34.5576=sum of diameters÷2 and x 3.1416. 34.5576 × 10=345.576, and 345.5762=119422.77=square of the product of the circumference and number of revolutions. √119422.77+102=345.72=the square root of the sum of the above product and the square of the height of the spiral the result required. Ex. 2. The greater and less diameters of a conical spiral are 1.5 and 8.75 feet, its height 6 feet, and the number of its revolutions is 5; what is the length of it? Ans. 80.725 inches. Ex. 3. The greater and less diameters of a conical spiral are 3 and 9 feet, its height 12.5 feet, and the number of its revolutions 10; what is the length of it? Ans. 188.91. Centres of Gravity. Plane Spiral.-It is in its geometrical centre. Conical Spiral.-It is at a distance from the base of the line joining the vertex and centre of gravity of the base. NOTE.* This rule is applicable to winding engines where it is required to ascertain the length of a rope, its thickness, the number of revolutions, diameter of drum, etc., etc. Illustration. The diameter over the roll of a flat rope upon the drum of a winding engine shaft is 134.5 inches, the diameter of the drum is 94.5 inches, and the number of revolutions 20; what is the length of the rope and what is its thickness? Ans. Length of the rope, 7194.247 inches. = Area of 134.5 in. 14208.049 66 66 94.5 = 7013.802 7194.247 Then, 7194.247-20=359.712=area of rope÷revolutions= area of each thickness. 134.5-94.52+94.5=114.5 and 114.5 x 3.1416-359.712 =circumference of the mean diameter of each thickness. Hence, 359.712÷359.712=1 inch the width or thickness of the rope. * For Rules to ascertain the elements of Winding Engines, see Haswell's Engineers' and Mechanics' Pocket-book, p. 263-4. SPINDLES. Definition. Figures generated by the revolution of a plane area, 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 or curve from which they are generated, as Circular, Elliptic, Parabolic, etc., etc. Circular Spindle. To ascertain the Convex Surface of a Circular Spindle, Fig. 60. RULE.-Multiply the length, fc, by the radius, o c, of the revolving arc; multiply this arc, fa c, by the central distance, o e, or distance between the centre of the spindle and centre of the revolving arc; subtract this product from the former, double the remainder, multiply it by 3.1416, and the product will be the surface required. 2 C Or, lxr—(a× √r2— (2) *)×2 p; a representing the length of the arc, c the chord, and p 3.1416. Fig. 60. a EXAMPLE. -What is the surface of a circular spindle, Fig. 60, the length of it, ƒ e, being 14.142 inches, the radius of its arc, o c, 10, and the central distance, o e, 7.071? 14.142 × 10=141.42-length × radius. Length of arc, by rules, p. 76, 78=15.708. 15.708 × 7.071=111.0713-length of arc x central distance. 141.42-111.0713-30.3487-difference of products. 30.3487 × 2 60.6974 x 3.1416 190.687 = the remainder doubled x3.1416, which is the result required. Ex. 2. The length of a circular spindle is 28.284 feet, the radius of its arc 20, and the distance between the centre of the spindle and the centre of the revolving arc is 14.142; what is the surface of it? Ans. 762.7484 feet. Centre of Gravity. Is in its geometrical centre. To ascertain the Convex Surface of a Zone of a Circular Spindle, Fig. 61. RULE.-Multiply the length, i c, by the radius, o a, of the revolving arc; multiply the arc, d a b, by the central distance o e; subtract this product from the former, double the remainder, multiply it by 3.1416, and the product will be the surface required. 2 Or, 1×r—(a × √r2 — (2) 3) × 2 p, 1 representing the length of the zone. EXAMPLE.-What is the convex surface of the zone of a circular spindle, Fig. 61, the length of it being 7.653 inches, the radius of its arc 10, the central distance 7.071, and the length of its side or arc, d b, 7.854 inches? 7.653 × 10=76.53=length radius. 7.854×7.071=55.5356=length of arc × central distance. 76.53-55.5356=20.9944=difference of products. 20.9944 × 2=41.9888 x 3.1416=131.912 the remainder doubled 3.1416, which is the result required. Ex. 2. The zone of a circular spindle is 23 inches in length, F the radius of its arc 30, its central distance 21.2, and the length of its side 23.56; what is its convex surface? Ans. 1197.2255 inches. Centre of Gravity. Is in its geometrical centre. To ascertain the Convex Surface of a Segment of a Circular Spindle, Fig. 62. RULE.-Multiply the length, i c, by the radius of the revolving arc, o a; multiply the arc by the central distance, o e; subtract this product from the former, double the remainder, multiply it by 3.1416, and the product will be the surface required. 2 Or, lxr−(a× √r2 — (2) 3) × 2 p, 1 representing the length of EXAMPLE. What is the convex surface of a segment of a circular spindle, Fig. 62, the length of it being 3.2495 inches, the radius of its arc 10, the central distance 7.071, and the length of its side, i d, 3.927 inches? 3.2495 × 10 32.495-length radius. 3.927×7.071=27.7678=length of arc× central distance. 32.495-27.7678=4.7272=difference of products. 4.7272×2=9.4544×3.1416-29.702, which is the result re quired. Ex. 2. The segment of a circular spindle is 14.142 feet in length, the radius of its arc is 20, and the distance between the plane of the segment, i e, and the centre of the revolving arc, o e, Fig. 62, is 14.142, and the length of its side, i d, is 15.708; what is its convex surface? Ans. 381.3745 feet. For Surface of a Circular Spindle, Zone, or Segment. General Formula. S=2(lr-ac) p, l representing length of spindle, segment, or zone, a the length of its revolving arc, r the radius of the generating circle, and c the central distance. Illustration. The length of a circular spindle is 14.142 inches, the length of its revolving arc is 15.708, the radius of its generating circle is 10, and the distance of its centre from the centre of the circle from which it is generated is 7.071; what is its surface? 2x (14.142 × 10-15.708 x 7.071) x 3.1416-190.6869= result required. Centre of Gravity. See Appendix, page 283. NOTE.-The surface of the frustrum of a spindle is obtained by the division of the surface of a zone. Cycloidal Spindle. To ascertain the Convex Surface of a Cycloidal Spindle, Fig. 63. RULE.-Multiply the area of the generating circle by 64, and divide it by 3; the quotient will give the surface required. ax 64 3 Or, =surface. Fig. 63. of a EXAMPLE. The area of the generating circle, a b c, cycloidal spindle, d e, is 32 inches; what is the surface of the spindle? |