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EXERCISES.

1. If the side of a square is 49 inches, what is its content in imperial gallons? Ans. 8.66. 2. What is the content of a regular octagon, whose side is 150 inches in imperial gallons? Ans. 391-317. 3. Find the content of a circular tun in imperial gallons, its diameter being 72 inches. Ans. 14-684.

152. PROBLEM II.—To gauge areas one inch deep, when the superficial content can be expressed by the product of two factors.

'Find the superficial content, and divide or multiply it by the proper divisor or multiplier, for the required de

nomination.'

By the Sliding Rule.

I. Set the corresponding divisor on A to one of the factors on B, then opposite to the other factor on A will be the content on B."

Or,

For let a, b = the two dimensions, then c =

mc = ab, and m: a=b:c (54.)

ab

m

II. 'Find a mean proportional between the two factors, and use it as the side of a square by the rule in the preceding problem' (article 57.)

EXAMPLE. Find the area of a rectangular cistern in imperial bushels, its length and breadth being 72 and 42 inches.

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Or, set 2218 on A to 42 on B, then opposite to 72 on A is 1.36 on B.

Or, by article 57, set 42 on C to 42 on D, then opposite to 72 on C is the mean proportional 55-2 on D. Then set 47-1 on D to 1 on C, and opposite to 55.2 on D is 1·36, the content on C.

EXERCISES.

1. Find the content in pounds of hard soap of an oblong vessel, its length being 201 and its breadth 60 inches. Ans. 444-36. 2. Find the area of a triangular vessel in imperial gallons, its base being 100 inches, and the perpendicular on it 80 inches. Ans. 14.426.

3. Required the content of a parallelogram in pounds of hard soap, the length being 84 inches, and the perpendiAns. 99.04.

cular breadth 32 inches.

4. What is the area in imperial bushels of a trapezoid, the parallel sides being 60 and 145, and the perpendicular breadth 80 inches. Ans. 3.7. 5. Find the area of a quadrilateral in pounds of dry starch, one of its diagonals being 80, and the perpendiculars on it from the opposite angles being 24.6 and 14.4.

Ans. 38.97.

6. What is the area in imperial gallons of an oval figure, whose transverse diameter is 85 inches, and five equidistant ordinates, whose common distance is 15 inches, being in order 40.6, 44.3, 50·4, 50·1, 42-7, and 38-2, and two segments at each end, whose bases are the extreme ordinates, and heights 5 inches, and nearly of a parabolic form?

Ans. 13.22.

153. PROBLEM III.-To find the area of an ellipse when its two axes are given.

'Find the area by the rule in article 386, Part I., and divide or multiply it by the proper divisor or factor, and the result will be the required area; or find the product of the axes, and multiply or divide it by the circular factor or divisor, and the result is the area.'

By the Sliding Rule.

I. Set the circular divisor for the proposed measure on A to one of the axes on B, then opposite to the other axis on A will be the content on B.

II. 'Find a mean proportional between the axes by art. 57, and use it as the diameter of an equal circle, as in problem I.'

EXERCISES.

1. Find the area of an ellipse in imperial gallons, its axes being 99 and 75. Ans. 21.03. 2. Find the content of an ellipse in imperial gallons, its axes being 108 and 75.

Ans. 22.94.

154. PROBLEM IV.-To gauge solids, whose bases are regular figures.

I. Find the cubic content, then multiply or divide it by the proper multiplier or divisor corresponding to the required measure or weight."

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II. ' Find the square of the given side or diameter, and divide or multiply it by the proper tabular divisor or multiplier for the given figure of the base" (151.)

By the Sliding Rule.

Set the corresponding gauge point on D to the depth on C, then opposite to the given side or diameter on D is the content on C.!

This rule is evident from that in article 151, which refers to the content at one inch of depth.

When the vessel is a pyramid or cone, apply the same rules, observing that, instead of the whole height, only onethird of it is to be used. Or, in the case of conical vessels, use the whole height, and the multipliers, divisors, and gauge points in the table for conical vessels.

EXAMPLE. Find the content of an octagonal prism in imperial bushels, its side being 60 inches and depth 75. By 334, Part I., v = 4·8284 × 602 × 75 = 1303668, v 1303668 2218.19

and

c=

m

=587.67 bushels.

Or, c = n,s2h 002177 × 602 × 75587·8 bushels.

Or set the gauge point 21.4 on D to the depth 75 on C, then opposite to the side 60 on D is the content 588 on C.

EXERCISES.

1. Find the content in imperial gallons and bushels of a vessel with a square bottom, each side being 30 inches, and its depth 40. Ans. 129-83 and 16.229.

2. What is the content in imperial bushels of a cylindric vessel, whose diameter is 48 inches and depth 64 inches? Ans. 52.2 bushels.

3. Find the content in imperial bushels of a regular pentagonal prismatic vessel, a side of its base being 54 inches, and its depth 80 inches. Ans. 180.94. 4. Find the content in imperial gallons of a conical vessel, the diameter of its base being 27 inches, and its height 60 inches.

Ans. 41.3.

5. What is the content in imperial bushels of a pyramidal vessel, whose base is a regular hexagon, the length of its side being 40 inches, and the height of the vessel 72 inches? Ans. 44.976.

6. Find the content of a conical vessel in imperial gallons, the diameter of its base being 60 inches, and its height 60 inches. Ans. 203.94.

7. What is the content in imperial bushels of a pyramidal vessel, whose base is a regular octagon, its side being 105 inches, and its depth 120 inches. Ans. 959.94.

155. PROBLEM V.-To find the content of a spherical vessel.

Find the volume of the sphere, and multiply or divide it by the proper multiplier or divisor for the required measure or weight; or,

Divide or multiply the cube of the diameter by the corresponding tabular divisor or multiplier" (145.)

By the Sliding Rule.

'Set the spherical guage point on D to the diameter of the sphere on C, then opposite to the diameter on D is the content on C.

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that is, and

g

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or =

d = g2'

92: d2=d: c, (using 9 for 93)

2 Lg 2LdLd~ Lc (50 and 51.)

EXERCISES.

1. Find the content of a spherical vessel, whose diameter is 34 inches in imperial gallons.

Ans. 74.2. 2. Find the content in imperial bushels of a spherical vessel, whose diameter is 68 inches. Ans. 74.2.

156. PROBLEM VI.-To find the content of a spheroid. 'Find its volume, and multiply or divide it by the proper multiplier or divisor for the required measure or weight; or, Multiply the square of the equatorial diameter by the polar diameter, and divide or multiply the product by the divisor or multiplier for spherical vessels.'

For, if b is the equatorial diameter, and a the polar diameter of a spheroid, and a that of a sphere; v the volume of the spheroid, and v' that of the sphere, then (405, Part I.) v=5236 ab2, and v′ = •5236 a3 ; v: v′ = b2: a2;

and hence,

from which the rules are evident.

By the Sliding Rule.

'Set the spherical gauge point on D to the polar diameter on C, then opposite to the equatorial diameter on D is the content on C."

EXERCISES.

1. Find the content in imperial gallons of a prolate spheroid, its polar diameter being 72 inches, and its equatorial 50. Ans. 339-91.

2. What is the content in imperial bushels of a prolate spheroid, whose diameters are 70 and 90? Ans. 104.1.

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