secured by means of the formula used in Example 6, in cases where the arc approaches near to a semicircle.

Remark. When the arc is greater than a semicircle, the remaining part of the circumference must be found by Rule II., with the help of the formula used in Example 6, if necessary. This remark does not apply to cases where the degrees of the arc are given, which are solved by Rule I.

PROBLEM VIII.

To find the diameter of a circular zone, its two parallel chords A B, CD, and its breadth E F, being given.

C2

2

d = √ { P2 + 2 (C° + c ) + ( ~ ̄ ̄ °)"},

1. The parallel sides of a circular zone are 6 and 8 feet, and its breadth 7 feet; required the diameter of the circle.

By the first Formula the diameter

Here, the diameter d is first found, as above; then by the second and third Formulæ,

3. The parallel chords of a zone are the same as in the Example 1 and its breadth 1 foot; required the diameter.

Ans 10 fect.

NOTE. In this example the two chords are both on the same side of the centre of the circle.

4. The two parallel chords of a circular zone are 16 and 12 feet, and the diameter of the circle 20 feet; required the breadth of the zone. Ans. 14 feet.

NOTE. 1. The breadth of the zone, in this example, is found by squaring and transposing the first formula, whence there results a quadractic equation, from which the value of b is found.

NOTE 2. When the chord B D = A C, and the height G H have been found, the lengths of the equal arcs A C, B D are found by the Prob. VII.

PROBLEM IX.

In an ellipse are given any three of the four following parts to find the fourth, viz. the transverse axis TR, the conjugate axis CO, the abscissa HQ, and the ordinate PQ.

1. The transverse axis is 30, the conjugate 20, and the abscissa 3 feet.

By the second formula,

PQ = y

32
1152
=

2. The transverse TR

=

70 feet, the conjugate CO and the ordinate P Q = 20; required the abscissa H Q.

Ans. By the first formula, HQ = 21 feet,

50,

3. The transverse is 180 inches, the ordinate 16, and the abscissa 54; required the conjugate.

=

Ans. By the fourth formula, the conjugate 40 inches. 4. If the conjugate be 50 feet, the ordinate 20, and the abscissa 21; what is the transverse?

Ans. By the third formula, the transverse

=

70 feet.

=

5. The transverse TR 100 yards, and the conjugate CO 60; required the distance of the foci Ff from the centre H. Ans. By the last formula, H F = Hƒ= 40 yards.

6. The ratio of the major and minor axes of the earth's orbit is as 1 to n, the former being about 190,000,000 miles 2 a, How much is the earth nearer to the sun in winter than in summer?

Ans. The distance here required is twice the focial distance from the centre of the earth's elliptical orbit, which, by the last formula is found to be 2 a√T — n2.

7. Required the distance of the foci of an elliptical section, passing through the poles of the earth, the earth's axes being 7926 and 7899 miles.

Ans. 654 miles, or 327 miles each from the earth's centre.

PROBLEM X.

The axes of an ellipse are given to find its circumference.

RULE I.-Multiply half the sum of the two axes by 3.1416, and the product will give an approximate length of the circumference, which will be found near enough for most practical purposes.

RULE II. To half the sum of the two axes add the square root of half the sum of their squares, and multiply half the sum by 3.1416 for the circumference very nearly.

FORMULA (see last figure).

Let 2a and 26 represent the axes, as in the last problem, 3.1416; then,

and = Circumf.

(a + b), or = 1⁄2 π (a + b + √ 2 (a2 + b2)).

EXAMPLES.

1. The axes of an ellipse are 15 and 10 feet; required the circumference by Rule I.

Ans. 39 feet 3 inches.

2. The axes being the same as in the last example; required the circumference by Rule II.

Ans. 39 feet 7 inches nearly.

3. Find the meridional circumference of the earth, the axes being as given in the last example of Prob. IX.

Ans. 24,858 miles nearly.

PROBLEM XI.

In a parabola I V H, the focus of which is F, any two of the three following parts, viz., the parameter PQ, the abscissa V G, and the ordinate G H being given, to find the third part.

1. The parameter P Q of a parabola is 50, and its ordinate 60 feet, required the abscissa V G. Ans. By the first formula a

G H

=

=

602

GH =

72.

50

Ans. 1.6.

2. The parameter of a parabola is 10, and its ordinate 4; required the abscissa.

3. The abscissa of a parabola is 4, and its corresponding ordinate 10; required the parameter.

PROBLEM XII.

Ans. 25.

To find the length of the arc of a parabola, its ordinate and abscissa being given. (See last figure.)

Let x

FORMULE.

and y represent the same parts, as in the last Problem ;

then

The arc V H

=

2+ y2 nearly

EXAMPLES.

1. Required the half arc V Q of a parabola, V F being = 3 feet, and F Q 6.

2. The abscissa is 2, and the ordinate 6; required the length of the half arc of the parabola.

Ans. 6.4291.

NOTE. 1. The parabola is the path of projectiles in vacuo; it is also used in the astronomical theory of comets.

NOTE 2. The student who wishes for further information concerning this curve, as well as concerning the ellipse and hyperbola, may consult the various works on conic sections.

PART III.

"ON OF SUPERFICIES OR SURFACES. Try surface is estimated by the number of squares Vard to its thickness; the side of those 10, nebet, one vard, &c. Hence the area is *Space Dienes, or square feet, or square yards, &c. F SQUARE MEASURE.