PROPERTIES OF THE HYPERBOLA.

1. The square root of the sum of the squares of the semi-axes, is equal to the distance of the focus from the middle of the transverse axis.

2. If half the transverse axis be subtracted from the distance between the focus and the middle of the said axis, the remainder will be the distance of the focus from the vertex.

3. The difference of two lines drawn from the two foci to any point in the curve, is equal to the transverse axis.

4. The parameter, latus rectum, or double ordinate at the focus, is equal to the square of the conjugate divided by the

transverse.

Note. Those who desire to obtain a complete knowledge of the properties of Conic Sections, are referred to the works of Emerson, Hutton, and Simson en that subject.

PROBLEM XI.

To find the solidity of a spheroid.

RULE.

Multiply the square of the revolving axis by the fixed axis, and the product thence arising by .5236; and the last product will be the solidity.

Note 1. The multiplier .5236 is of 3.1416.

2. A semi-spheroid is equal to

same base and altitude.

of a cylinder, or to double a cone of the

EXAMPLES.

1. What is the solidity of the prolate spheroid ABCD, whose transverse or fixed axis AC is 50, and the conjugate or revolving axis BD 30?

Here 30 x 50 x .5236 = 900 x 50 x .5236 = 45000 X.5236 = 23562, the solidity required.

2. What is the content of an oblate spheroid, whose axes are 50 and 30? Ans. 39270. 3. Required the solidity of a prolate spheroid, whose axes are 45 and 95. Ans. 100727.55.

PROBLEM XII.

To find the solidity of the segment of a spheroid.

CASE I.

When the base is circular, or parallel to the revolving axis.

RULE.

From triple the fixed axe, take double the height of the segment; multiply the difference by the square of the height, and the product thence arising by .5236; then say, as the square of the fixed axe is to the square of the revolving axe, so is the last product to the solidity required.

EXAMPLES.

1. Required the solidity of the segment ABC of a prolate spheroid, the height Bm being five, the fixed axe 2, BD 50, and the revolving axe EF 30.

2. The axes of a prolate spheroid are 100 and 60; what is the solidity of that segment whose height is 10, and its base parallel to the revolving axe?

Ans. 5277.888. 3. The axes of an oblate spheroid are 50 and 30; what is the solidity of that segment whose height is 6, and its base parallel to the revolving axe?

CASE II.

Ans. 4084.08.

When the base is elliptical, or perpendicular to the revolving axe.

RULE.

From triple the revolving axe, take double the height of the segment; multiply the difference by the square of the height, and the product thence arising by .5236; then say, as the revolving axe is to the fixed axe, so is the last product to the solidity required.

EXAMPLES.

1. What is the solidity of the segment ABC of a prolate spheroid, the height Bm being 6, the fixed axe EF 50, and the revolving axe BD 30?

B

Here (30 x 3

--

× 36 × .5236

=

D

90-12

6 × 2) × 6o ×.5236 78 × 36 × .5236 = 2808 × .5236 = 1470.2688; then, as 30; 50: 1470.2688: 2450.418, the solidity required.

2. The axes of an oblate spheroid are 50 and 30; what is the solidity of that segment whose height is. 6, and its base perpendicular to the revolving axe? Ans. 1560.74688.

PROBLEM XIII.

To find the content of the middle frustum of a spheroid.

CASE I.

When the ends are circular or parallel to the revolving axe.

RULE.

To twice the square of the middle diameter, add the square of the diameter of one end; multiply the sum by the length of the frustum, and the product thence arising by .2618, for the solidity.

Note 1. A cask in the form of the middle frustum of a prolate spheroid is, by Gaugers, called a cask of the first variety; hence this Rule is useful in cask gauging.

2. An ale gallon contains 282, a wine gallon 231, and a Winchester bushel 2150.42 cubic inches.

EXAMPLES.

1. What is the solidity of the middle frustum EGHF of a prolate spheroid; the middle diameter CD being 30, the diameter of each end EF or GH 18, and the length AB 40?

Here (30 x 2 + 182) × 40 × .2618 = (900 × 2 + 324) × 40 × .2618 (1800 + 324) × 40 × .2618 = 2124 x 40 x .2618 = :84960 × .2618 = 22242.528, the solidity required.

2. What is the solidity of the middle frustum of an oblate spheroid; the diameter of each end being 40, the middle diameter 50, and the length of the frustum Ans. 31101.84.

18?

3. The length of a spheroidal cask is 30, the bung diameter 24, and the head diameter 18 inches; what is its content in ale and wine gallons?

Ans. 41.1081 ale, and 50.184 wine gallons.

CASE II.

When the ends are elliptical, or perpendicular to the revolving axe.

RULE.

To twice the product of the transverse and conjugate diameters of the middle section, add the product of the transverse and conjugate of one end; multiply the sum by the height of the frustum, and the product thence arising by .2618, for the solidity.

EXAMPLES.

1. In the middle frustum EFGH, of an oblate spheroid, the diameters of the middle section AB are 50 and 30; those of the end EF or GH 40 and 24; and its height mn 18; what is the solidity?

Here (50 x 30 x 2 + 40 x 24) x 18 x .2618 = (3000+960) x 18 x .2618= 3960 x 18 x .2618 = 71280 x .261818661.104, the solidity required.

2. What is the solidity of the middle frustum of a prolate spheroid; the diameters of the middle section being 100 and 60, those of each end 80 and 48, and its height 36? Ans. 149288.832.