EXAMPLE.-The generating circle of a cycloid has an area of 115.45 inches; what is the area of the cycloid?

115.45x3=346.35, Ans.

Ex. 2. The area of a circle describing a cycloid is 1.625 feet; what is the area of the cycloid in inches?

Ans. 702 inches. Ex. 3. The diameter of a circle describing a cycloid is 66.5 feet; what is the area of the cyloid in inches?

Ans. 1500434.064 inches.

To ascertain the Length of a Cycloidal Curve, Fig. 48. RULE.-Multiply the diameter of the generating circle by 4, and the product will give the length of the curve.

Or, dx4 length of curve.

EXAMPLE.-The diameter of the generating circle of a cycloid, Fig. 48, is 8 inches; what is the length of the curved s c? 8x4=32=product of diameter and 4=the length required. Ex. 2. The diameter of the generating circle is 20 inches; what is the length of the cycloidal curve?

Ans. 80 inches. Centre of Gravity. At a distance from the centre, n, of the chord, d c, of the curve d s c= of the radius of the generating circle.

NOTE.-The curve of a cycloid is the line of swiftest descent; that is, a body will fall through the arc of this curve, from one point to another, in less time than through any other path.

RINGS.

Circular Rings.

Definition. The space between two concentric circles.

To ascertain the Sectional Area of a Circular Ring, Fig. 49. RULE. From the area of the greater circle, a b, subtract that of the less, c d, and the difference will be the area of the ring.

Or, a-a'=area.

EXAMPLE. The diameters of the circles forming a ring are each 10 and 15 inches; what is the area of the ring? Area of 15=176.7146

Ex. 2. The diameters of a circular ring are 10.75 and 18.25 inches; what is its area?

Ans. 171.82 inches.

Centre of Gravity. Is in its geometrical centre.

Cylindrical Rings.

Definition. A ring formed by the curvature of a cylinder.

To ascertain the Convex Surface of a Cylindrical Ring, Fig. 50.

RULE. To the thickness of the ring, a b, add the inner diameter, b c; multiply this sum by the thickness and the product by 9.8696, and it will give the surface required. Or, d+dxdx 9.8696=surface.

EXAMPLE. The thickness of a cylindrical ring, a b, is 2 inches, and the inner diameter, b c, is 18; what is the surface of it?

2+18=20=thickness of ring added to the inner diameter.

=

20 × 2 × 9.8696=394.784 the sum above obtained the thickness of the ring, and that product by 9.8696, the result required. ·

Ex. 2. The thickness of a ring of metal of 20 inches diameter (internal) is 2 inches; what is the surface of it?

Ans. 434.2624 inches.

Link.

Definition. An elongated ring.

To ascertain the Convex Surface of a Link, Figures 51. RULE.-Multiply the circumference of a section of the body, a b, of the link by the length of its axis, and the product will give the surface required.

Or, cxl-surface.

NOTE.-To ascertain the Circumference or Length of the Axes.

When the Ring is elongated. To the less diameter add its thickness, multiply the sum by 3.1416; multiply the difference of the diameters by 2, and the sum of these products will give the result required.

When the Ring is elliptical. Square the diameters of the axes of the ring, and multiply the square root of half their sum by 3.1416; the product will give the length of the body of the ring.

a

Figs. 51.

e

ан

f

EXAMPLE. The link of a chain is 1 inch in diameter of body, a b, and its inner diameters, b c and e f, are 12.5 and 2.5 inches; what is its circumference.

2.5+1×3.1416=10.9956-length of axis of ends.

12.5-2.5×2=15-length of sides of body.

Then, 10.99.56+15=25.9956-length of axis of link, which x3.1416 (cir. of 1 in.)-81.6678=result required. Centres of Gravity. Are in their geometrical centres.

Definition.

Cones.

A figure described by the revolution of a right

angled triangle about one of its legs.

For Sections of a Cone, see Conic Sections, page 228.

To ascertain the Surface of a Cone, Fig. 52.

RULE.-Multiply the perimeter, or circumference of the base, by the slant height, or side of the cone, and half the product added to the area of the base will be the surface.

EXAMPLE.-The diameter, a b, of the base of a cone is 3 feet, and the slant height, a c, 15; what is the surface of the cone?

Perimeter of 3 feet 9.4248, and

surface of side of cone.

9.4248 × 15

=70.686=

2

Area of 3 feet 7.068, and 70.686+7.068-77.754=surface required.

Ex. 2. The diameter of the base of a cone is 6.25 inches, and the slant height 18.75; what is the surface of it?

Ans. 214.757 inches. Ex. 3. The diameter of the base of a cone is 20 inches, and the slant height 14.142; what is the surface of the cone? Ans. 758.445 inches.

To ascertain the Surface of the Frustrum of a Cone, Fig. 53.

RULE.-Multiply the sum of the perimeters of the two ends by the slant height of the frustrum, and half the product added to the areas of the two ends will be the surface required. p+p'xh +a+a=surface required.

Or,

2

EXAMPLE.—The frustrum, a b c d, Fig. 53, has a slant height of 26 inches, and the circumferences of its ends are 15.75 and 22.5 inches respectively; what is its surface?

surface required.

Then, 497.25 +12.175=509.425

Ex. 2. What is the surface of the frustrum of a cone, the diameters of the ends, a c and b d, being 4 and 8 feet, and the length of the slant sides 20 feet? Ans. 439.824 feet.

Ex. 3. What is the surface of the frustrum of a cone, the diameter of the ends being 6.66 and 10 feet, and the length of the slant side 3.73 feet?

Centres of Gravity.

Ans. 210.989.

Cone or Frustrum.-At the same distance from the base as in that of the triangle or parallelogram, which is a right section of them.