The Mechanic's, Machinist's, and Engineer's Practical Book of Reference

PROBLEM III.

To find the surface and solid content of a cone or pyramid.

RULE 1. Multiply the circumference of the base by the slant height, and half the product will be the slant surface; to which add the area of the base, and the product will be the whole surface. RULE 2. Multiply the area of the base by the perpendicular height, and of the product will be the solid content.

EXAMPLE 1. Required the convex surface of a cone whose base A B = 20 inches, and slant height BD = 29.5.

3.1416 x 20 x 29.5

=926 772 square inches,

and divided by 144 6435 superficial feet. EXAMPLE 2. Required the solidity of the cone as above, the perpendicular CD being 28 inches. 202 x 7854 × 28

1-697 cubic feet.

2932.16 cubic inches, and divided by 1728 =

PROBLEM IV.

To find the surface of the frustum of a cone or pyramid.

RULE. Multiply the sum of the perimeters of the two ends by the slant height, and half the product will be the slant surface; to which add the areas of the two ends, and the product will be the whole surface.

EXAMPLE. Required the convex surface of

the frustum of a cone ABCD, whose base
AB 20 inches, the slant height BC= 19,
and top end CD:
= 11.

3-1416 x 20 + 31416 x 11 x 19

= 925·2012 square inches, and divided by 144 B 6.425 feet nearly.

PROBLEM V.

To find the solid content of the frustum of a cone.

RULE. To the product of the diameters of the two ends add the sum of their squares; multiply this sum by the perpendicular height and by 2618; the product is the solid content.

EXAMPLE 1. Required the solid content of the frustum in Problem IV., whose perpendicular E F = 18 inches.

20 × 11 220, and 220 + 202 + 11a × 18 x 2618=3491.8884 cubic inches, and divided by 172820208 cubic feet nearly.

EXAMPLE 2. Required the content, in imperial gallons, of the inverted frustum of a cone ABCD, whose inner dimensions are 34 feet deep, 18 inches diameter at bottom, and 22 inches diameter at top.

Or, 13238-7024 × 0.003606544775 gallons nearly, as before.

PROBLEM VI.

To find the solid content of the frustum of a pyramid. RULE To the sum of the areas of the two ends add the square root of their product; multiply this sum by the perpendicular height, and of the product is the solid content.

EXAMPLE Required the solid content of the frustum of a pyramid ABCD, whose perpendicular height: 24 inches, the area of the base 144 inches, and area of the top end = 64.

144 64 208, and 144 × 6496; then 20896 × 24

2432 cubic inches, and 1728

1.4074 cubic feet nearly.

PROBLEM VII.

To find the solidity of a wedge.

RULE. To the length of the edge add twice the length of the base; multiply that sum by the height, and by the breadth of the base, and one-sixth of the product will be the solidity.

EXAMPLE. Required the content in cubic inches of the wedge ABCDE, whose base ABC 12 inches long and 4 inches broad, the length of the edge DE 10 inches, and perpendicular height r E = 20 inches.

To find the convex surface and solid content of a sphere or globe.

RULE 1. Multiply the square of the diameter by 3.1416; the product will be the convex superficies.

RULE 2. Multiply the cube of the diameter by 5236, and the product is the solid content.

EXAMPLE 1. Required the convex surface of a sphere, whose diameter AB = 25 inches.

25.52 x 3·1416 2042-8254 square inches, ÷ 144 14-1862 square or superficial feet. EXAMPLE 2. Required the solid content of a sphere whose diameter A B : = 25 inches.

25.53 × ·5236 = 8682-00795 cubic inches; ÷ 1728 5·0243 cubic feet.

PROBLEM IX.

To find the convex surface and solid content of the segment of a

sphere.

RULE 1. Multiply the height of the segment by the whole circumference of the sphere, and the product is the curved surface. RULE 2. Add the square of the height to three times the. square of the radius of the base; multiply that sum by the height, and by 5236, and the product is the solid content.

EXAMPLE 1. The diameter A B of the sphere B ABCD: 20 inches; what is the convex surface of that segment of it whose height ED 8 inches?

3.1416 x 20 × 8 = 502·656 square inches; ÷ 144 = 3·49 superficial feet.

EXAMPLE 2. The base FG of the segment FDG = 18 inches, and perpendicular ED = 8; what is the solid content?

8264, and 92 × 3

cubic inches, ÷ 1728

243; then 243+64 × 8 × 5236 = 1285.9616 = 7441 cubic feet.

EXAMPLE 3. Suppose ABCD to be a sugar-pan, and that the diameter of the mouth A B is 4 feet, the depth DC being 25 inches, how many imperial gallons will it contain?

252 625, and 242 x 3 = 1728; then B

111.084 gallons.

PROBLEM X.

To find the solidity of a spheroid.

RULE. Multiply the square of the revolving axis by the fixed axis, and by 5236, and the product will be the solidity.

EXAMPLE 1. Required the solid content of the prolate spheroid ABCD, whose fixed axis AC is 50, and revolving axis BD 30.

EXAMPLE 2. What is the solid content of an oblate spheroid, the fixed axis being 30, and revolving axis 50?

50 x 30 x 523639270, the solid content.

PROBLEM XI.

To find the solidity of the segment of a spheroid when the base is circular or parallel to the revolving axis.

RULE. From triple the fixed axis take double the height of the segment; multiply the difference by the square of the height, and by 5236; then say, as the square of the fixed axis is to the square of the revolving axis, so is the former product to the solidity.

EXAMPLE 1. Required the solid content of the segment ABC, whose height Br is 10; the revolving axis E F being 40, and fixed axis B D 25.

10 x 2 55, and 55 × 102 × 2879.8. Then, as 252 : 402 ::

2879-87372-3 nearly.

EXAMPLE 2. What is the solid content of the segment of a spheroid whose height 20 inches, the revolving axis being 25, and fixed

axis 50?

230384; then,

50 × 3 20 × 2= 110, and 110 × 202 × 5236 = as 502 252 :: 230384: 5759 6 inches, the solid.content.

PROBLEM XII.

To find the convex surface and solid content of a cylindric ring. RULE 1. Multiply the thickness of the ring added to the inner diameter by the thickness and by 9.8698, and the product will be the convex surface.

RULE 2. To the thickness of the ring add the inner diameter; multiply that sum by the square of the thickness and by 2·4674, and the product will be the solid content.