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

x 20 length.

Or, 18 inches.

18 inches.

$24X,7854=254,4696 inches, area of the base.

20 length in feet.

144)5089,3920(35,343 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 in inches by the length in feet, then divide the product by 144, and the quotient will be the answer.

EXAMPLE. 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 ?

11X11 X2X20---144=33,6+ feet, the solidity when hewn square. 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 thickness of the board including the saw calf : is to the solid feet :: so is 1.2 (inches) to the number of feet of boards.

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

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

As 14 : 40 : : 12 : 384 feet, the Ans.

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Art. 14. To find the tornage of

RULE.
Multiply the length of the keel by the
beam, and that product by the depth of the
vide the last product by 5, and that quoti

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

RULE. ultiply the length by the breadth, and that product ne depth, divide the last product bý 2150,425 the inches in a statute bushel, and the quotient will be 118 WOR.

EXAMPLE. ere is a square box, the length of its bottom is 50 s, breadth of ditto 40 inches, and its depth is 60 s; how many bushels of corn will it hold? X40X60-2150,425 = 55,84+ or 55 bushels, three . Ans. 13. The dimensions of the walls of a brick buildbeing given, to find how many bricks are necesa y to build it.

RULE. om the whole circumference of the wall measured

on the outside, subtract four times its thickness, multiply the remainder by the height, and that proby the thickness of the wall, gives the solid content

whole wall; which multiplied by the number of - contained in a solid foot, gives the answer.

EXAMPLE. v many bricks 8 inches long, 4 inches wide, and hes thick, will it take to build a house 44 feet long, t wide, and 20 feet high, and the walls to be one

Multiply the length of the keel by the
beam, and that product by half the breadt
and divide by 95.

* EXAMPLE.
Aship 84 feet by the keel, 28 feet byt
is the tonnage ?

84x28x143-95=350.
Art. 15. From the proof of any cab!

strength of another

RULE,
The strength of cables, and consequent
of their anchors, are as the cube of their
Therefore; As the cube of the periphery

Is to the weight of its anchor;
So is the cube of the periphery of
To the weight of its anchor.

EXAMPLES
1. If a cable 6 inches about, require
cut. of what weight must an anchor be fun
As 6x6x6:2 cut. ; : 12x12x12 : 1

2. If a 12 inch cable require an ancho
must the circumference of a cable be,

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4X2,5=80 solid inches in a brick, then 1728780
5 bricks in a solid foot.
-40-444-440=163 feet, whole length of wall.

-4 four times the thickness

- cut.

As 18 : 12x12x12 :: 2,25 : 216.
Art. 16. Having the dimensions of
slips of a different capacity, with t
of them. In find the barthen of the

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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 di. vide the last product by 95, and that quotient by the tonnage.

EXAMPLE 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.

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

84x28x147-95=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.

EXAMPLES
1. If a cable 6 inches about, require an anchor of 24
cwt. of what weight inust an anchor be for a 12 inch cable?

As 6x6x6:21cut. : : 12x12x12 : 18c?et. Ans.

2. If a 12 inch cable require an anchor of 18 cwt. what must the circumference of a cable be, for an anchor of 24.cw ? cwt.

cwt. As 18 : 12x12x12 : : 2,25 : 216 3/216=6 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.

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

EXAMPLE. If a ship of 500 tons burthen be 75 feet long in the keel, I demand the burthen of another ship, whose keel is 100 feet long?

T.cwt.grs.lb. As 75x75x75 : 300 :: 100x100x100: 711 2 0 24 +

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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 de: nominations 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 lower denomination to its next superior. :

3. In the same manner multiply all the multiplicand 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.

EXAMPLES. F. I. F. I. ů. Ì. F. I. Multiply even 3 :

5

$ 6. 9 By 4 Top 3 9 5 8 9 . 29 0

27 9 9 25 6 91 10 1 4 2 9

Product, 53 29

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FEET, INCHES AND SECONDS.

F. 1. 1 Multiply 9.86 By7 9 3.

.tiphier. 67 11 6 m =prod. by the feet in the mule 7 3 4 6 =ditto by the inches.

2 5 1 6=ditto by the seconds. 75 5 376 Ans."

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How many square feet in a board 16 feet 9 inches long, and 2 feet 3 inches wide ? ;. By Duvdecimals.

By Decimals. A
F. 1.

: F. 1.
16 9

16 9=16,75 feet. 2 3

23= 2,25

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