Ex. 1. What is the solidity of a sphere whose radius is 4000 miles? Ans. 2680825173331 sol. miles? 2. What is the solidity of a sphere whose diameter is 896000 miles? Ans. 376636654912085333 sol. m.

498. PROB. 30.-To find the solidity of a spherical segment, RULE.-Multiply the half sum of the areas of the two bases by the altitude of the segment and to this product add the solidity of a sphere whose diameter is this same altitude.

NOTE-If the segment has but one base the other base is 0 and the same rule applies.

Ex. 1. What is the solidity of a spherical segment whose altitude is 86.60254 inches and the radii of whose bases are 100 and 50 inches? Ans. 2040523.848355309475191954314 sol. in.

2. What is the solidity of a segment of one base, the altitude being 13.39746 feet and the diameter of the base 100 feet?

Ans. 53870.808015040738834712352sol. ft.

499. PROB. 31.-To find the solidity of a spherical wedge or ungula,

RULE. Find the solidity of the sphere and then say as 360° is to the angle of the wedge, so is the solidity of the sphere to the solidity of the wedge.

NOTE. The angle of the wedge is the same as the angle of the lune that forms its base.

Ex. 1. What is the solidity of a wedge whose angle is 36° in a sphere whose radius is 4000 miles?

Ans. 26808251733 sol. m. 2. What is the solidity of an ungula whose angle is 45° in a sphere whose diameter is 896000 inches?

Ans. 470795818640106662sol. in.

500. PROB. 32.-To find the solid contents of a spherical pyramid or of a sector,

RULE.-Having found the area of the triangle, polygon or zone which forms the base, multiply this area by of the radius of the sphere.

Ex. 1. What is the solidity of a spherical pyramid whose base is a triangle having its angles 80°, 90° and 130°, the radius of the sphere being 4000 miles? Ans. 74467365921 sol. m.

2. What is the solidity of a pentagonal spherical pyramid in a sphere whose diameter is 5 inches, the angles of the base being 150°, 119° 30′, 75°, 145° and 170° 30'? Ans. 10.9083 sol. in.

3. What is the solidity of a spherical sector in a sphere whose radius is 12 inches, the arc of the great circle bisecting the sector, or the sectoral angle at the centre of the sphere being 120° ? Ans. 1809.556992 sol. in. 4. What is the solidity of the remainder of the hemisphere after the sector in Ex. 3 has been taken out?

501. PROB. 33. scribed in a sphere,

Ans. 1809.556992 sol. in.

To find the solid contents of a cube in

RULE. Divide the square of the diameter by 3, and the cube of the square root of this quotient will be the solidity sought.

NOTE. The diagonal of the cube is a diameter of the sphere, but the square of the diagonal of any rectangular parallelopipedon is equal to the sum of the squares of its three dimensions; i. e. in the cube, since its three dimensions are equal, the square of the diagonal is equal to three times the square of either edge; hence the rule.

Ex. 1. What is the solidity of a cube inscribed in a sphere whose diameter is 10 feet?

2. What are the solid contents of a cube inscribed in a sphere whose circumference is 18.849552 inches?

RULE.

Ans. 41.569219+sol. in.

To find the contents of a solid of any

- Immerse the solid in a vessel of known form and dimensions partly filled with water, and note the rise of the water in the vessel.

NOTE. This rule is founded on the self-evident fact that the volume or bulk of the water displaced is equal to that of the solid immersed.

Ex. 1. An irregular stone, immersed in a cylindrical vessel 10 inches in diameter, raised the water in the vessel 5 inches; what were the contents of the stone? Ans. 392.699 sol. in.

2. There is water 4 inches deep standing in a pail which is 12 inches deep, 10 inches in diameter at the bottom and 13 inches at the top — interior dimensions; what are the contents of a lobster, which, being immersed in this water, will raise it 4 inches? Ans. 415.737341 sol. in.

§ 52. GAUGING.

503. GAUGING is the art of finding the contents of casks or vessels of any form, in gallons, bushels, etc.

504. PROB.-To find the contents of kegs, barrels, etc.

It is difficult or impossible to find the exact contents of kegs, barrels, etc., in consequence of the different curvature of the staves, the difficulty in determining the interior dimensions of the cask, etc.; but, by experience it is found that all such vessels may be gauged with sufficient accuracy by the following

RULE 1.-Multiply the difference between the bung and head diameters of the cask, by numbers varying from .5 to .7, according as the staves are curved little or much, and add the product to the head diameter to obtain the MEAN DIAMETER ; then proceed as in finding the contents of a cylinder in Art. 492.

Ex. 1. What are the contents in wine and in beer gallons of a

cask whose length is 44 inches, head diameter 28 inches, and bung diameter 36 inches?

(36-28) X .75.6

28.

33.6 mean diameter.

33.62.785398=area of circle (477, Rule 2),

.. 33.62 X .785398 X 44

No. gal. wine, 1st Ans.; but if

231

both numerator and denominator of this fraction are divided by .785398, we shall have

before, both numerator and denominator be divided by .785398, 33.62 X 44

it will give

359

, very nearly. Hence,

RULE 2.-Find the mean diameter in inches as in Rule 1; then multiply the square of the mean diameter by the length of the cask in inches, and divide the product by 294 for wine and by 359 for beer gallons.

2. What are the contents in wine gallons of a cask whose length is 36 inches, and whose head and bung diameters are respectively 16 and 19 inches?

3. What are the contents in beer measure of a cask whose length is 44 inches, and whose head and bung diameters are 26 and 31 inches?

4. What are the contents in bushels of a hogshead whose length is 48 inches, and whose head and bung diameters are 32 and 40 inches?

5. What is the capacity in bushels of a cask whose length is 4 feet, and whose head and bung diameters are 3 and 34 feet?

NOTE 2.-To find the contents of vessels in the form of a cylinder, cone, frustum, sphere, etc., proceed as in the Geometrical Problems.

NOTE 3.-To find the contents of irregular vessels or cavities of any description, first fill the vessel or cavity with water, then pour its contents into a vessel of known form and dimensions, and proceed as before.

§ 53. TONNAGE OF VESSELS.

505. The tonnage of a vessel is the number of tons she will carry, and is determined by measurement.

506. The ship carpenter estimates the tonnage by one rule, and government by another.

CARPENTER'S RULE.-For a single decked vessel, multiply the length of the keel, breadth at the main beam, and depth of the hold, in feet, together; divide the product by 95, and the quotient is the number of tons.

For a double decker, take half of the breadth at the main beam for the depth of the hold, and proceed as before.

GOVERNMENT RULE.-For a single decker, take the length in feet above the deck from the fore part of the main stem to the after part of the stern post, the breadth at the widest part above the main wales on the outside, and the depth from the under side of the deck plank to the ceiling in the hold. From the length take of the breadth and the continued product of the remainder, breadth and depth, divided by 95, will give the tonnage.

For a double decker, take the length above the upper deck; for the depth take half the width and proceed as before.

Ex. 1. What is the carpenter's tonnage of a single decker whose length is 80 feet, breadth 21 feet, and depth 18 feet? Ans. 318 tons.