2. A piece of scantling measures 19 feet 4 inches long, 10 inches wide and 7 inches thick-How many feet does it con

tain, board measure ?

3. What is the content in board measuring each 21 ft. 9 in. long, 21 inches thick?

Ans. 112 ft. 4 in. measure of 7 planks, inches wide and 2 Ans. 666 ft. 1 in.

4. How many feet are there in a rafter, which measures 16 feet long, 4 inches wide at one end, and 31⁄2 at the other, and 3 inches thick? Ans. 16 ft.

5. Required the number of feet in 4 pieces of scantling, one measuring 9 ft. 5 in. long, and 7 inches by 5; another 10 feet 6 inches long, 6 by 4; a third 11 feet 2 inches long, 6 by 5, and a fourth 10 feet 3 inches long, 6 by 4.

CASE III.

Ans. 102 ft. 2 in. 9"

To find the superficial contents of a round piece of timber, when hewn or sawn square.

RULE.

Multiply half the square of the diameter in inches by the length in feet: divide the product by 12, and the quotient will be the superficial content.

Examples.

1. Required the superficial content of a round piece of timber, whose length is 25 feet, and its mean diameter 20 inches.

20×20=400,

and 400÷2=200= half square of the diameter,

Then 200×25=5000, and 5000÷12=416 ft. 4in. Ans. 2. How many feet of square edged boards, one inch thick, including the saw gap, can be made from a log 16 feet long and 20 inches in diameter? Ans. 266 ft. 8 in.

SOLID BODIES.

Solid bodies are such as consist of length, breadth and thick ness, as stone, timber, globes, &c.

Note. In the common use of geometry, the term solid does not only apply to absolute density, as is understood regarding

the timber in a beam, but a chest or box, though empty, is considered a solid. If the object contains length, breadth and thickness, it is sufficient to constitute a solid.

OF A CUBE.

A cube is a square solid, comprehended under six geometrical squares, being in the form of a die.

To find the solidity.

RULE.

Multiply the side of the cube by itself, and that product again by the side; the last product will be the solidity.

Example.

1. A cellar is to be dug whose length, breadth and depth are each 11 feet 6 inches-How many solid feet does it contain, and what will it cost digging at 12 cents per solid yard ?

11 ft. 6 in. =

and 11.5×11.5×11.5=1520.875 feet. Answer.

11.5 feet,

OF A PARALLELOPIPEDON.

A parallelopipedon is a solid having six rectangular sides, every opposite pair of which are equal and parallel.

To find the solidity.

RULE.

Multiply the length by the breadth, and that product by

the depth.

Examples.

1. Required the number of cords in a pile of wood 100 ft. long, 8 feet high, and 4 feet wide.

2. I demand the number of cords in a pile of wood 60 feet long, 6 feet high and 4 feet wide.

Ans. 11

cords. 3. In a stack of bark measuring 24 feet 6 inches long, 19 feet 6 inches wide, and 12 feet high, how many cords? Ans. 44 cords, 101 feet. 4. Required the solid content of a bale measuring 6 feet 6 inches long, 5 feet 6 inches wide, and 4 feet deep.

and 143÷40 3 tons, 13 feet. Ans.

5. I demand the solid content of a load of bark measuring 13 feet 6 inches long, 3 feet 4 inches wide, and 2 feet 101 inches deep. Ans. 1 cord, 1.475 feet.

OF A CYLINDER.

A cylinder is a round solid, having its base circular, equal and parallel, in form of a roller used for rolling land.

To find the solidity.

RULE.

Multiply the area of the base by the length, and the product is the solid content.

Examples.

1. If a piece of timber be 6 feet in circumference, and 25 feet long, how many feet of timber are contained in it, supposing it to be perfectly cylindrical?

6×6×.07958

=

2.86488,

71.622 feet. Answer.

and 2.86488×25

=

2. I have a rolling stone, 14 inches in diameter, and 6 feet long, required the solidity.

OF A CONE.

Ans. 6.4141 feet.

A cone is a solid, having a circular base, and growing smaller and smaller, till it ends in a point, and may be nearly represented by a sugar loaf.

To find the solidity.

RULE.

Multiply the area of the base by a third part of the perpendicular height, and the product will be the solid content.

Note. To find the perpendicular height, add the square of the slant height to the square of half the diameter of the base, the square root of this sum is the perpendicular height.

Examples.

1. What is the solidity of a cone, whose slant height is 10 feet, and diameter 16?

100 square of the slant height.
64 square of half the diameter.

2. A heap of grain on a barn floor, (in a conical form,) measures 10 feet in diameter, and the perpendicular height is 3 feet-I wish to know how many bushels it contains.

Ans. 63 bu. 3 qts.+ 3. How many bushels of oats in a heap that measures 20 feet in diameter, and the perpendicular height 6 feet?

Ans. 504 bu. 3 pecks, 4 qts.+

OF A GLOBE OR SPHERE.

A globe or sphere is a round, solid body, having every part of its surface equidistant from a certain point within it, called its centre.

To find the solidity.

RULE.

Multiply the cube of the diameter by .5236, and the product will be the solidity required.

Examples.

1. The diameter of the earth is about 7964 miles, required its solidity in cubic miles.

and 505119057344.5236 = 264480338425.3184 Ans. 2. The circumference of a globe is 31.416-What is the solidity?

Ans. 523.6.

Ship's tonnage by carpenters' measure.

RULE.

Multiply the length, breadth at the main beam, and depth of the hold together, and divide the product by 95, the quotient will be the required tonnage for a single decked vessel : for double decked vessels, take half the breadth of the main beam for the depth of the hold, and work as for a single decked vessel.

Examples.

1. Required the tonnage of a single decked vessel, measuring as follows, viz. length 72 feet, breadth 24 feet, and depth of the hold 10 feet.

72×24×10

=

17280.

and 17280-95-181.85 tons.

Answer.

2. I demand the tonnage of a double decked vessel, whose length is 100 feet and breadth 32 feet.

Ans. 539 tons, nearly.

By government measure.

RULE.

Multiply the length, less three-fifths of the breadth, by the breadth, and this product again by the depth of the hold; divide by 95, and the quotient will be the tonnage required. If the vessel be double decked, take half the breadth for the depth of the hold, and work as for a single decked vessel.

Examples.

1. Required the government tonnage of a single decked vessel, whose length is 80 feet, breadth 25 feet, and depth 10 feet.

of 25 feet 15 feet, and 80-15-65 feet. Then 65×25×10=16250,

and 16250÷95-171 tons. Answer.

2. I demand the government tonnage of a double decked vessel, whose length is 90 feet, and breadth 30 feet.

Ans. 341 tons.

Note 1. For ships of war, divide the continual product of the length, breadth and depth in feet by 100, and the quotient will be the tonnage required.