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? 50×40×60-2150,425=55,84+ or 55 bushels, three

pecks. Ans.

ART. 13. The dimensions of the walls of a brick building 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 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.

EXAMPLE.

How many bricks 8 inches long, 4 inches wide, and 2 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 ?

8×4×2,5=80 solid inches in a brick, then 1728÷80 21,6 bricks in a solid foot.

44404-44-40-108 feet, whole length of wall.

-4 four times the thickness.

164 remains.

Multiply by 20 height.

3280 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 .ength of the keel by the breadth of the beam, and that product by the depth of the hold, and divide the last product by 95, and the quotient is the tannage.

EXAMPLE.

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

72x24x194-95218,2+ions. Ans. RULE HI.

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 ?

84×28×14-95-350,29 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.

EXAMPLES.

1. If a cable 6 inches about, require an anchor of 21 cwt. of what weight must an anchor be for a 12inch cable? As 6×6×6: 24cwt.:: 12×12x12: 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 24 cwt. P

crct.

As 18

12x12x12: : 2,25 : 216216 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,

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 10 feet long? T.cwt.qrs.li. As 75×75X75: 300 :: 100x100x100:711 2 0 21

ls

DUODECIMALS,

OR

CROSS MULTIPLICATION,

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, beginnit at the lowest, by the highest denomination in the mal plier, and write the result of each under its respect term; observing to carry an unit for every 12, from eas lower denomination to its next superior.

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

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

How many square feet in a board 16 feet 9 inches

long, and 2 feet 3 inches wide ?

By Duodecimals.

By Decimals.

16 9=16,75 feet.

2 3

2,25

Ans. 37,687537. 8. 9

TO MEASURE LOADS OF WOOD.

RULE.

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

EXAMPLE.

How many solid feet are contained in a load of wood, 7 feet 6 inches long, 4 feet 2 inches wide, and 2 feet 3 inches high?

7 ft. 6 in. 7,5 and 4 ft. 2 in.=4,167 and 2 ft. 3 in= 2,25; then, 7,5x4,167 $1,2525×2,25=70,318125 solid feet, Ans.

But loads of wood are commonly estimated by the foots allowing the load to be 8 feet long, 4 feet wide, and then 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 increased 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

RULE.

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

EXAMPLE.

Required the content of a load of wood which is 3 feet. 9 inches wide and 2 feet 6 inches high.

By Duodecimals. By Decimals.

The foregoing method is concise and easy to those who are well -quainted with Duodecimals, but the following Table will give the content of any load or wood, by inspection only, sufficiently eract for common practice; which will be found very convenient.