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2. A piece of timber being 16 inches broua, in inches thick, and 20 fect long, to find the content?

Breadth 16 inches.
Depth 11

Prod. 176x20=3520 then, S520-144=24,4 feet,

the Answer. 3. A piece of timber 15 inches broad, 8 inches thick, and 25 feet long; how many solid feet doth it contain ?

Ans. 20,8+feet. Art. 8. When the breadth and thickness of a piece of

timber are given in inches, to find how much in length will make a solid foot.

RULE. Divide 1728 by the product of the breadth and depth, and the quotient will be the length making a solid foot.

EXAMPLES.

1. If a piece of timber be 11 inches broad and 8 inches deep, how many inches in length will make a solid foot?

11x8=88)1728(19,6 inches, Ans. 2. If a piece of timber be 18 inches broad and 14 inches deep, how many inches in length will make a solid foot ? 18x14=252 divisor, then 252)1728(6,8 inches, Ans

Art. 9. To measure a Cylinder. Definition.-A Cylinder is a round body whose bases are circles, like a round column or stick of timber, of equal bigness from end to end.

RULE. Multiply the square of the diameter of the end by ,7854 which gives the area of the base ; then multiply the area of the base by the length, and the product wil be the solid content.

EXAMPLE. What is the solid content of a round stick of timber of equal bigness from end to end, whose diameter is 18 inches and length 20 feet?

18 in.=1,5 ft.

X1,5

Square 2,25X,7854=1,76715 area of the base.

x 20 length.

Ans. 35,54300 solid content.

Or, 18 inches.

18 inches.

EXAMPLE.

324 X,7854=254,4696 inches, area of the base.

20 length in feet.

144)5089,3920(85,345 solid feet. Ans. ART. 10.. To find how many solid feet a round stick of

timber, equally thick from end to end, will contain when hewn square.

RULE. Multiply twice the square of its semi-diameter innches by the length in feet, then divide the product by .144, and the quotient will be the answer.

If the diameter of a round stick of timber be 22 inches and its length 20 feet, how many solid feet will it contain when hewn square ?

11x11x2x20=144=33,6+ feet, the solidity when ART. 11. To find how many feet of square edged boards

of a given thickness, can be sawn from a log of a given diameter.

RULE. Find the solid content of the log, when made square, by the last article-Then say, As the thiekness of the board including the saw calf : is to the solid feet : : so is 12 (inches) to the number of feet of boards,

How many feet of square edged boards, 14 inch thick, including the saw calf, can be sawn from a log 20 feet long and 24 inches diameter ?

12x12x2x20---144=40 feet, solid content.

As 1+ : 40 : : 12 : 384 feet, the Ans,

hewn square.

EXAMPLE,

ART. 12. The length, breadth and depth of any square box being given, to find how many bushels it will contain..

RULE. Multiply the length by the breadth, and that product -by the depth, divide the last product by 2150,425 the solid inches in a statute bushel, and the quotient will be the answer.

EXAMPLE

There is a square box, the length of its bottom is 50 inches, breadth of ditto 40 inches, and its depth is 60 inches; how many bushels of corn will it hold?

50X40X60-2150,425=55,84+ or 55 bushels, three pecks. Ans. Art. 13. The dimensions of the walls of a brick build

ing being given, to find how many bricks are necessary to build it.

RULE. From the whole circumference of the wall measured round on the outside, subtract four times its thickness, then multiply the reinainder by the height, and that product by the thickness of the wall, gives the solid content of the whole wall; which multiplied by the number of bricks contained in a solid foot, gives the answer.

EXAMPLE.

How many

bricks 8 inches long, 4 inches wide, and 24 inches thick, will it take to build a house 44 feet long, 40 feet wide, and 20 feet high, and the walls to be one foot thick ?

8x4X2,5=80 solid inches in a brick, then 1728-80 21,6 bricks in a solid foot. 44+40+44+40=168 feet, whole length of wall.

-4 four times the thickness.

164 remains. Multiply by 20 height.

S280 solid feet in the whole wall.
Multiply by 21,6 bricks in a solid foot.

Product, 70848 bricks. Ans.

EXAMPLE.

ART. 14. To find the tonnage of a ship.

RULE. Multiply the length of the keel by the breadth of the beam, and that product by the depth of the hold, and divide the last product Ly 95, and that quotient by the tonnage.

Suppose a ship 72 feet by the keel, and 24 feet by the beam, and 12 feet deep; what is the tonnage ?

72x24x12+-95=218,2+tons. Ans.

RULE II. Multiply the length of the keel by the breadth of the beam, and that product by half the breadth of the beam, and divide by 95.

A ship 84 feet by the keel, 28 feet by the beam; what is the tonnage ?

84 x 28 x 14+95=350,29 tons. Ans. Art. 15. From the proof of any cable, to find the

strength of another.

RULE.
The strength of cables, and consequently the weights
of their anchors, are as the cube of their peripheries.
Therefore; As the cube of the periphery of any cable,

Is to the weight of its anchor;
So is the cube of the periphery of any other cable,
To the weight of its anchor.

EXAMPLE.

EXAMPLES.

1. If a cable 6 inches about, require an anchor of 2 cwt. of what weight must an anchor be for a 12 inch cable?

As 6x6x6 : 24cwt. : : 12x12x12 : 18cwt. Ans.

2. If a 12 inch cable require an anchor of 18 cit. what inust the circumference of a cable be, for an anchor of 21 cwt, ? cwt.

cwt. As 18 : 12X12X12 : : 2,25 : 216791636 Ans. Art: 16. Having the dimensions of two similar built

ships of a different capacity, with the burthen of one of them, to find the burthen of the other,

in.

RULE. The burthens of similar built ships are to each other as the cubes of their like dimensions.

EXAMPLE.

If a ship of 300 tons burthen be 75 feet long in the keel, I demand the burthen of another ship, whose keel is 100 feet long?

T.cwt.qrs.lb. As 75X75X75 : 300 :: 100X100X100 : 711 2 0 244

DUODECIMALS,

CROSS MULTIPLICATION, Is

a rule made use of by workmen and artificers in casting up the contents of their work.

RULE. 1. Under the multiplicand write the corresponding denominations of the multiplier.

2. Multiply each term into the multiplicand, beginning at the lowest, by the highest denomination in the multiplier, and write the result of each under its respective term; observing to carry an unit for every 12, from each Jower denomination to its next superior.

3. In the same manner multiply all the niultiplicand by the inches, or second denomination, in the multiplier, and set the result of each term one place removed to the right hand of those in the multiplicand.

4. Do the same with the seconds in the multiplier, setting the result of each term two places to the right hand of those in the multiplicand, &c.

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