PROBLEM XIV.

To determine the Content of a Segment of a Circle an Inch in Depth.

RULE.

By the Pen.

If the segment be small in comparison of its circle, and if great accuracy be not required, multiply the base by two-thirds of the altitude, and divide the product by the number of cubic inches in the proposed integer; the quotient will be the content nearly.

But if great accuracy be required, find the area of the segment by the Rule and Table given in PRob. xiv. SECT. I. OF MENSURATION, and when found, divide it by the number of cubic inches in the proposed integer,

for the content.

By the Sliding Rule.

Set the gage-point for the proposed integer, on A, to two-thirds of the altitude on B, and opposite to the base on A, will be the content nearly, on B, provided the segment be less than a semicircle.

EXAMPLE.

Let ABC be the segment of a circle, whereof the base AC is 120 inches, and the altitude DB, 40 inches; the content in wine gallons, ale gallons, and malt bushels is required, for an inch in depth.

B

SOLUTION.

Less accurately.

80 two-thirds the base
40 altitude

W.G. divisor 231) 3200 (13.85 wine gallons.

A. G. divisor 282) 3200 (11·34 ale gallons.

M. B. divisor 2150) 3200 ( 1.49 malt bushels.

G. P. 282 to 80 against 40 are 11.34 A. G.

2150

More correctly.

1.49 M. B.

60 half the chord or base

60 half the base

Altitude 40) 3600 product

Quotient 90 altitude of the opposite segment
Add 40 altitude of the given segment

Sum 130 diameter of the circle.

Again,

Diameter Altitude Quotient

130) 40 (333 tabular versed sine.

Now entering the Table of Circular Segments with 333, we find the corresponding area to be ·228858. Wherefore it will be,

•228858 tabular area

130 × 130 = 16900 square of the diam.

205972200 1373148 228858

W.G. divisor 231) 3867-700200 (16·74 wine gallons.

A. G. divisor 282) 3867-700200 (13.71 ale gallons.

M. B. divisor 2150)3867-700200 (1.79 malt bushels.

By the first of these methods (which at best is only an approximation) the result is considerably less than the true content of the segment; the given altitude, however, in this instance was not diminutive in comparison of the diameter of the circle.

PROBLEM XV.

To determine the Content of an Annular Space an Inch in Depth.

RULE.

By the Pen.

Multiply the sum of the outer and inner diameters,

by their difference, and divide the product by the circular divisor for the proposed integer.

By the Sliding Rule.

Set the circular gage-point, for the proposed integer, on D, to the sum of the diameters, on C, and opposite to the difference of the diameters on D, will be the content on C.

Or,

Set the circular divisor on A, to the difference of the diameters on B, and against the sum of the diameters on A, will be found the content on B.

EXAMPLE.

Let DEFG be a ring or annular space, whereof the outer diameter DF is 40 inches, and the inner diameter, 30 inches; the content, at an inch deep, is required in wine gallons, ale gallons, and malt bushels.

Wherefore it will be,

W.G. cir. divisor 294.12) 700 (2.38 wine gallons.

Cir. G. P. 18.95 to 70 against 10 are 1.95 A. G.

[52-32

0.25 M. B.

Or,

Cir. Divis. 359 to 10 against 70 are 1.95 A. G. 0.25 M. B.

2738

PROBLEM XVI.

To determine the Content, for an Inch in Depth, of the Segment of an Ellipsis, the Base of the Segment being parallel either to the transverse or to the conjugate Diameter, both of which must be given.

CASE 1.

When the segment's base is parallel to the transverse axis.

RULE.

Find the content of a circular segment of the same altitude, for an inch in thickness, the diameter of the