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Square 2,25x,7854-1,76715 area of the base, * P

X 20 length
. Ans. 35,34300 solid content.
Or, 18 inches.

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

'Ave. - £0 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 fruin end to end, will coutain when hewn square.

RULE. Múltiply twice the square of its semi-diameter in nches 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 ?

11x11x2x20;144=33,6+ feet, the solidity when hewn square. Art. 11. To find how niany 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 snuare, by the last article-Then say, As the thickness of the board including the saw call : is to the sulid feet : : so is 12 (inches) to the number of feet of boards.

EXAMPLE. llow many feet of square edged boards, 14 inch thick, including the saw call, can be sawn from a lug 20 feet Į long and 24 inches diameter ?

12x12x2x20+144-40 feet, solid content,

As 11 : 40 : : 18 : 384 feet, the

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, thru pecks. Ans. ART. 13. The dimensions of the walls of a brick build.

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

RULE. From the whole circumference of the wall measured round on the outside, subtract four times its thickness, then multiply the remainder by the height, and that pro. duct 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 fout. 44+40+44+40=168 feet, whole length of wall.

- 4 four times the thickness.

164 remains. Multiply by 20 height.

$280 solid feet in the whole wall. Multiply by 21,6 bricks in a solid foot. . Product, 70848 bricks. Ans.

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

RULE. Multiply the length of the keel by the breadth of the. beam, and that procluct by the depth of the hold, and di. vide the last product Ly 05, and that quotient by the tonnaye.

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 kee! by the breadth of the beam, and that product by half the breadth of the beam, and divide by 95.

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

84x28x14+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 21 ewt. of what weight must an anchor be for a 12 inch cable ?

As 6x6x6: 2 cwt. :: 12.X 12x12: 18cwt. 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 83 cwi. ?

cuct. As 18 : 12x12x12 : : 2,25 : 21672166 Ans. ART. 16. Having the dimensions of two similar built 1 ships of a different capacity, with the burthen of one

of them, to find the burthen of the other.

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

EXAMPLE If a ship of 300 ton's burthen be 75 feet long in the keel, I deinand'the burthen of another ship, whose keel is 100 feet luny ? As 75x75x75 : 300 : : 100x100x100 : 711 2 0 24+



OR CROSS MULTIPLICATION, ..', Is a rule made use of by workmen and artificers in cast. ing up the contents of their work.

RULE. 1. Under the multiplicand write the corresponding de. noininations of the multiplier. '

2. Multiply each terın into the multiplicand, beginning at the lowest, by the higl.est denoinination in the multiplier, and write the result of each under its respective term; observing to carry an unit for every 12, from each lower denuinination 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. : :


F. I. :: F. I. Multiply as

4 6 97 4 7 3 9 5 8 . 97. 29 O 27 9 9 256

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llow many square feet in a buaru 16 feet 9 inches long, and 2 feet 3 inches wideBy luodecimals.

By Decimals. I
F. 1.
16 9

-.169x16,75 feet.
:'2:3 2,25 i

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