Definition.

Pyramids.

A figure, the base of which has three or more

sides, and the sides of which are plane triangles.

To ascertain the Surface of a Pyramid, Fig. 54.

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

EXAMPLE. The side of a quadrangular pyramid, a b, Fig. 54, is 12 inches, and its slant height, a c, 40; what is its surface? 12×4=48=perimeter of base.

Then

12 × 12+960=1104 surface required.

Ex. 2. The sides of a hexagonal pyramid are 12.5 inches, and its slant height 62.5; what is its surface?

Ans. 2749.703 inches. Ex. 3. The sides, a b, b c, of an oblong quadrangular pyramid, Fig. 55, are 15 and 17.5 inches, and its slant height, a d, 36; what is its surface? Ans. 1432.5 inches.

Ex. 4. The sides of an octagonal pyramid are 4 feet 2 inches, and its slant height 6 feet 9 inches; what is its surface? Ans. 196.326 square feet.

Ex. 5. What is the surface of a pentagonal pyramid, its slant height being 12 feet, and each side of its base 2 feet? Ans. 66.882.

To ascertain the Surface of the Frustrum of a Pyramid, Fig. 56.

RULE.-Multiply the sum of the perimeters of the two ends by the slant height or side, and half the product added to the area of the ends will be the surface.

EXAMPLE.-The sides, a b, c d, Fig. 56, of a quadrangular pyramid are 10 and 9 inches, and its slant height, a c, 20; what is its surface?

10×4=40

Then

9x4=36

76=sum of perimeters.

1520

76×20=1520, and =760=area of sides.

2

10×10=100, and 9×9=81.

100+81+760=941, the surface required.

Ex. 2. The ends of a frustrum of a quadrangular pyramid are 15 and 9 inches, and its slant height 40; what is its surface? Ans. 2226 inches. Ex. 3. The ends of a frustrum of a triangular pyramid are 20 and 10 inches, and its slant height 50; what is its surface? Ans. 2466.5 inches.

Ex. 4. The sides of a frustrum of a hexagonal pyramid are 15 and 25 inches, and its slant height 20; what is its surAns. 32.002 square feet.

face?

Centres of Gravity. Pyramid or Frustrum.—At the same distance from the base as in that of the triangle or parallelogram, which is a right section of them.

Helix (Screw).

Definition. A line generated by the progressive rotation of a point around an axis and equidistant from its centre.

To ascertain the Length of a Helix, Fig. 57.

RULE. To the square of the circumference described by the generating point, add the square of the distance advanced in one revolution, and the square root of their sum multiplied by the number of revolutions of the generating point will give the length of the line required.

Or, √(c2+h2)×n=length of line, n representing the number of revolutions.

Fig. 57.

EXAMPLE.-What is the length of a helical line running

3.5 times around a cylinder of 22 inches in circumference and advancing 16 inches in each revolution?

222+162=740=sum of squares of circumference and of the distance advanced.

√/740×3.5=95.21=square root of above sum × number of revolutions-length of line required.

Ex. 2. What is the length of the helical line described by a point in a screw in one revolution at a radius from its axis

of 11.3 inches, the progression of the line or pitch of the screw being 17 inches? Ans. 73.

Ex. 3. What is the length of the helical line described by a point on the periphery of a screw of 10 feet in diameter, having a pitch of 20 feet?

Ans. 37.242.

Centre of Gravity. Is in its geometrical centre.

Spirals.

Definition. Lines generated by the progressive rotation of a point around a fixed axis.

A Plane Spiral is when the point rotates around a central point.

A Conical Spiral is when the point rotates around an axis

or a cone.

To ascertain the Length of a Plane Spiral Line, Fig. 58.

RULE.-Add together the greater and less diameters,* divide their sum by two, multiply the quotient by 3.1416, again by the number of revolutions, and the product will give the length of the line required.

Or, when the circumferences are given, take their mean length, multiply it by the number of revolutions, and the product will give the length required.

=

x3.1416 x n length of line, n representing the

number of revolutions.

Fig. 58.

bd

* When the spiral is other than a line, measure the diameters of it

from the middle of the material composing it.

EXAMPLE.-The less and greater diameters of a plane spiral spring, as a b, c d, Fig. 58, are 2 and 20 inches, and the number of revolutions 10; what is the length of it?

2+20
2

=11=sum of diameters÷2.

11×3.1416=34.5576=above quotient × 3.1416.

34.5576 × 10=345.576=above product number of revolutions the length of line required.

Ex. 2. The greater and less diameters of a plane spiral are 4 and 30 inches, and the number of revolutions 5; what is the length of it? Ans. 267.036 inches.

To ascertain the Length of a Conical Spiral, Fig. 59. RULE. Add together the greater and less diameters;* divide their sum by two, and multiply its quotient by 3.1416. To the square of the product of this circumference and the number of revolutions of the spiral, add the square of the height of its axis, and the square root of the sum will be the length required.

EXAMPLE.-The greater and less diameters of a conical spiral, Fig. 59, are 20 and 2 inches, its height, c d, 10, and the number of revolutions 10; what is the length of it?

* See Note to Rule, page 117.