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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. 8181 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 figment of a sphere, whose height is 1 foot 9 inches, the diameter being 5 feet.

Anf. 27.489 19. 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.

By Rule I,

16 AE
16

8)250

32=EF
8=CE

40=CF
3

120 treple prod, of diameter.
64 sq. of the frust. height,

480 720

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3485.08.6 folid inches as before. Ex. 2. Required the solidity of a segment, whose base dia meter is 100, and its height 13.5 inches.

Ans: 54302.75235 cubic inches. Ex. 3. How many folid miles are in either frigid zone, the height being 329 miles, and diameter of its base 3168 miles ?

Anf. 1315766512 folid miles.

PROBLEM XV.

To find the folidity of the middle zone of a sphere.

RULE I.

When the ends are unequal, add into one sum the squares of the radii of both ends, and the square of the zone's height;

multiply

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