A spheroid is a solid described by the revolution of an ellipse about either of its axes.

about the shorter axis CD, the solid described is called an oblate spheroid.

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

PROBLEM XI.

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.

ABX CDX.5236-90x4900 X.5236-230907.6 cubic feet, which is the solidity.

Mensuration of Cylindrical Rings.

2. What is the solidity of a prolate spheriod, whose fixed

axisis 100, and revolving axis 6 feet?

3. What is the solidity of an oblate

axis is 60, and revolving axis 100?

4. What is the solidity of a prolate are 40 and 50?

Ans. 1884.96.

spheroid, whose fixed

Ans. 314160.

spheroid, whose axes

Ans. 41888.

5. What is the solidity of an oblate spheroid, whose axes are 20 and 10?

Ans. 2094.4.

6. What is the solidity of a prolate spheroid, whose axes are 55 and 33 ? Ans. 31361.022.

7. What is the solidity of an oblate spheroid, whose axes are 85 and 75?

Ans.

A cylindrical ring is formed by bending a cylinder until the two ends meet each other. Thus, if a cylinder be bent round until the axis A takes the position mon, a solid will

be formed, which is called a cylindrical ring.

RINGS.

B

The line AB is called the outer, and cd the inner diameter.

PROBLEM XII.

To find the convex surface of a cylindrical ring.

RULE.

I. To the thickness of the ring add the inner diameter.

II. Multiply this sum by the thickness, and the product by 9.8696, the result will be the area.

2. The thickness of a cylindrical ring is 4 inches, and the inner diameter 18 inches: what is the convex surface?

Ans. 868.52 sq. in.

3. The thickness of a cylindrical ring is 2 inches, and the inner diameter 18 inches: what is the convex surface?

PROBLEM XIII.

Ans. 394.784 sq. in.

To find the solidity of a cylindrical ring.

RULE.

I. To the thickness of a ring add the inner diameter

II. Multiply this sum by the square of half the thickness, and the product by 9.8696, the result will be the required solidity.

EXAMPLES.

1. What is the solidity of an anchor ring, whose inner diameter is 8 inches, and thickness in metal 3 inches?

2

8+3=11: then, 11×(3) × 9.8696=244.2726, which expresses the solidity in cubic inches.

2. The inner diameter of a cylindrical ring is 18 inches, and the thickness 4 inches: what is the solidity of the ring? Ans. 868,5248 cubic inches.

Mensuration of Cylindrical Rings.

3. Required the solidity of a cylindrical ring whose thickness is 2 inches, and inner diameter 12 inches?

Ans. 138.1744 cubic in.

4. What is the solidity of a cylindrical ring, whose thickness is 4 inches, and inner diameter 16 inches?

Ans. 789.568 cubic in.

5. What is the solidity of a cylindrical ring, whose thickness is 8 inches, and inner diameter 20 inches?

Ans.

6. What is the solidity of a cylindrical ring whose thickness is 5 inches, and inner diameter 18 inches?

Ans.

REMARK. In the following table, in the nine right hand columns of each page, where the first or leading figures change from 9's to O's, points or dots are introduced instead of the O's, to catch the eye, and to indicate that from thence the two figures of the Log. arithm to be taken from the second column, stand in the next line below.

1.838849

1.973128

1.845098

1.977724

1.851258

1.982271

1.857333 97

1.986772

1.863323

98

I.991226

1.869232

1.995635

1.875061 100

2.000000