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By Rule I.

3.1416

8

18.8496 the circumference.

6 the diameter.

Ans. 113.0976 sq. inches.

By Rule II. -7854

36 sq. diam.

47124 23562

28.2744 area of a great circle.

4
Anf. 113.0976 as above.

Ex. 2. Required the surface of a sphere, whose diameter is 5 feet 6 inches.

Anf. 95.0334/9. feet. Ex. 3.

What is the surface of a ball, whose diameter is inch?

Anf. 3.1416 inches. Ex. 4. How many inches will cover a globe of 12 inches diameter ?

Anf. 452.3904. Ex. 5. Required the surface of a globe of 18 inches diameter.

Ans: 7.0686 sq. feet. Ex. 6. Required the superficies of the terraqueous globe, it: diameter being 7958 miles. And if only one fourth part of its surface be dry land, and two acres sufficient to produce food for one person, how many persons can live on the earth at one time?

198956786.5824 sq. miles. Ans. 349739196.6456 dry land.

15916542927 persons. Note. A square mile contains 640 acreş.

PROBLEM

PROBLEM XII.

To find the folidity of a sphere.

RULE I.

Multiply the cube of the diameter by --5236, and the product will be the folidity.

Rule 2. A globe may be considered as composed of an infinite number of cones, whose bases are in the surface of the sphere, and common vertex in the centre ; therefore the folidity of the globe may be found thus :-Multiply its surface by

the diameter, and the product will give the solidity.

Rule 3. Find the folidity of a cylinder, of equal diameter and altitude with the globe, and the result will give the folis dity of the globe.

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EXAMPLE I.

Required the folidity of a globe, whose diameter is 50 incha

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By RULE II.

50
50

2500
3.1416

15708000 62832

7854.0000 surface.

50

6)392700

65450 Ans

By RULE III.

50 50

2500 .7854

3927000 15708

1963.5000

50

98175.0000 solid cylin,

2

3)196350

65450 Anf. as above.

ter.

Ex. 2. Required the solidity of a sphere of 10 inches diame

Anf. 523.6 Ex. 3. Required the content of a sphere, whose diameter is

Anf. 81815 cubic feet. Сс

25 feet.

Ex. 4. What is the solidity of a sphere, whose diameter is 3 feet 1 inch ?

Ans. 15.3483 cubic feet. Ex. 5. Required the folidity of a globe, its diameter being 8 feet 4 inches.

Anf. 303.0092 Ex. 6. How many solid miles are in the terraqueous globe, its diameter being 7958 miles ? Anf. 263883017937.1232.

PROBLEM XIII.

To find the furface of any zone, or fegment of a sphere.

RULE.

Multiply the circumference of a great circle of the sphere by the segment's height, and the product will be the superficies.

EXAMPLE I.

Required the superficies of a zone, whose height is'3 inches, the diameter of the sphere being 12 inches.

3.1410

12

37.6992 circumference.

3 the zone's height. 113.0976 Anf. in square inches.

Ex. 2. Required the surface of a fegment of a sphere, whose height is 1 foot 9 inches, the diameter being 5 fcet.

Anf. 27-489 lg. feet. Ex. 3. How many square inches will cover a segment, whose height is i inch, the diameter of the sphere being 3 inches ?

Anf. 9.4248 fq. inches.

PROBLEM

PROBLEM XIV.

To find the folidity of a spherical segment.

RULE I.

From the treple product of the diameter of the sphere, multiplied by the square of the segment's height, subtract twice the cube of the height, and the remainder, multiplied by .5236, will give the solidity.

Rule 2. To thrice the square of the radius of the fegment's base, add the square of its height; then multiply the sum by its height, and the product again by ·5236, the last product, is the solidity.

EXAMPLE

Required the solidity of a spherical segment, whose height is & inches, and the radius of its base 16 inches.

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