PROBLEM 30.

To find the solidity of a spherical segment having two bases.

RULE. To the sum of the squares of the radii of the two bases add one-third of the square of the distance between them; then multiply this sum by the breadth, and the product by 1-5708, and the result will be the solidity.

1. Required the solid content of the odds zone A D F E, the diameter of whose greater base A E is equal to 20 inches, and the less diameter D F 15 inches, and the distance between the two bases 10 inches.

Sphere.

2. What is the solidity of the middle zone of a sphere, the diameter of whose greater base is 24 inches, the less diameter 20 inches, and the distance between the bases 4 inches? Ans. 1566-6112 solid inches.

PROBLEM 31.-THE SPHEROID.

The form of the earth is an oblate spheroid, the axis about which it revolves being about 34 miles shorter than the diameter perpendicular to it.

To find the solidity of an ellipsoid.

RULE.-Multiply the fixed axis by the square of the revolving axis, and the product by the decimal 5236; the result will be the required solidity.

1. In the prolate spheroid A B C D, the transure or fixed axis A C, is 3 feet, and the conjugate or revolving axis D B,. is 2 feet; what is the solidity? Thus, 22 4 x 3 = 12 x 5236 = 6.2832 feet, the solidity. Ans.

=

2. What is the solidity of a prolate math B spheroid whose fixed axis is 100, and re

volving axis 6 feet? Ans. 1884-96.

Prolate Spheroid.

C

3. What is the solidity of an oblate spheroid whose axes are

20 and 10, (202 x 10 x 5236.)

Ans. 2094.4.

PROBLEM 32.

To find the solidity of a paraboloid.

RULE.-Multiply the area of the base by half the altitude, and

the product will be the solidity.

1. What is the solidity of a paraboloid, A CBA, whose height C m is 7 feet, and the diameter A B, its circular base, 4 feet? Thus, 42 x 7854 x (7 ÷ 2) .7854 × 2.5 = cubic feet. Ans.

12.5664 x 3.5 =

= 16 x

43.9824

2. Required the solidity of a paraboloid whose height is 50 inches, and the diameter. of its base 40 inches. Ans. 10,9 ft.

3. Required the solidity of a paraboloid whose height is 50 inches, and the diameter at its base, 100 inches.

Ans. 113.6284 feet.

1. The Tetraedron, or equilateral pyramid, is a solid bounded by four equal triangles.

2. The Hexaedron, or cube, is a solid bounded by six equal squares.

3. The Octaedron is a solid bounded by eight equal triangles.

4. The Dodecaedron is a solid bounded by twelve equal pentagons.

5. The Icosaedron is a solid bounded by twenty equal triangles.

The above five regular bodies may be represented by making the figures of pasteboard, and cutting the lines half through, so that the parts may be folded and sewed together.

The following table shows the surface and solidity of each of the regular solids, when the linear edge is unity.

To find the surface of a regular solid when the length of the linear edge is given.

RULE.-Multiply the square of the linear edge by the tabular number in the column of surfaces, and the product will be the surface required.

1. The linear edge of a tetraedron is 3; what is its surface? The tabular area is 1.73205; then 32 9, and 1-73205 × 9

= 15.58845. Ans.

2. The linear edge of an octaedron is 5; what is its surface? The tabular area is 3.46410 (X 52 = 25)=86.6025 =su

face.

3. The linear edge of an icosaedron is 6; what is its surface? The tabular area is 8:66025 × (62 = 36) = 311·769

PROBLEM 34.

= surface.

To find the solidity of a regular solid when the length of the linear edge is known.

RULE.-Multiply the cube of the linear edge by the tabular number in the column of solidities, and the product will be the solidity required.

1. What is the solidity of a regular tetraedron whose side is 6? The tabular number is 0.11785 × (63 = 216) = = 25.4556. Answer.

2. What is the solidity of a regular octaedron whose linear edge is 8?

The tabular number is 0-47140 × (82 = 512) = 241-3568. Answer.

3. What is the solidity of a regular dodecaedron whose linear edge is 3?

The tabular number is 7·66312 × (33 = 27) = 206·90424 solidity.

=

4. Required the solidity of an octaedron whose side is 10 feet. Ans. 471.4 feet. 5. Required the solidity of a dodecaedron whose side is 4 feet. Ans. 490-4396 feet.

6. Required the solidity of an icosaedron whose side is 3 feet. Ans. 58-9056 feet. 7. What is the solidity of a dodecaedron whose side is 9 feet? Ans. 5586-4144 feet.

Method of finding the decimal numbers used in this volume.

=

418879; 07958÷ 2 = 2821; 1.3:1416 ·031832; 2/78548862; 2057071; 433013 area of an equi= lateral triangle.

NOTE.-In the above table, the exact products or remainders cannot always be obtained, but by approximation the result will be sufficiently correct for all purposes of practical utility. In some cases the circumference is given, 3.141592: this will make