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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.


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 holds 50 x 40 x 60+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 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.



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 divide the last product Ly 95, and that quotient by the ton


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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.

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

84 x 28 x14+95–350,29 tons. Ans. Art. 15. From the proof of any cable, to find the

strength of another.

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 ? 1 As 6x6x6 : 21cwt. : : 12x12x12 : 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 B} cw.? cut.


in. As 18 : 12x12x12 : : 2,25 : 216/216=6 Ans. dArt. 16. Having the dimensions of two similar built

ships of a different capacity, with the burthen of onc 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 500 tons burthen be 75 feet long in the keel, I demand the burthen of another ship, whose keel is 100 feet long? As 75X75X75 : 300 : : 100X100X100 : 711 2 0 24 +



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 denoininations of the multiplier.

2. Multiply each terin into the multiplicand, beginni at the lowest, by the high.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 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 resalt of each term two places to the right hand of those in the multiplicand, &c.

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Product, 36 10. 7.

-48 16

69 10 %


F. 1. Multiply 9 8 6 By 7 9 3

[tiplier. 67 11 6 =prod. by the feet in the mud 7 3. 4. 6 "Eilitto by the inclicy.

2 5 1 6=dittu by the seconds.

75 5 3 7 6 Ans.

F. I.
Multiply 194 lu

F. I.
$5: 6.7
899. 10

7 8.9

Product, 55 2 9 3 9

48% 11

2 8 10

llow many square feet in a board 16 feet 9 inches long, and 2 leet 3 inches wide : By Duodecimals.

By Decimals.
F. I.

K. 1.
16 9

16916,75 feet. 33

3 2,25

33 60

8375 3350 SS50

* F. L $7.6875 8


RULE. Multiply the length by the breadth, and the produet by the depth or height, which will give the content in solid feet; of which 64 make half a cord, and 128 a cord.


How many solid fret arr contained in a load of woodi 7 feet 6 inches long, 4 fret 2 inches wide, and 2 feet s inches high?

7 ft. 6 in. 7,5 and 4 ft. 2 in. *4,167 and 2 ft. S in= 2,25; then, 7,54,167 -31,2525x2,25=70,318125 solid feet, Ans.

But loads of wood are commonly estimated by the foot, allowing the lead to be 8 feet long, 4 feet wide, and theu 2 feet high will make half a cord, which is called 4 feet of wood; but if the breadth of the load be less than 4 feet, its height must be iacreased so as to make half a cord, which is still called 4 feet of wood.

By measuring the breadth and heighth of the load, the content may be found by the following

RUIE. Multiply the breadth by the height, and half the product will be the content in feet and inches.


Required the content of a load of wood which is s feet 9 uches wide aad 8 feet 6 inches high. By Duodecimals. By Idecimals.

F. in.

7 6
1 10 6


9 4 6


F. in.

Ans. 4 8

4,6875-4 8t, or half a cord and

84 inches orer. The furngoing zoethod is concin and ousy wo those who are well acquainted with Iruodecimals, but the following Table will give the coutent of say loud of wood, by inspection only, suficiendy aust of common praotico ; which will be found vory convenien

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