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Art. 1.2. The length, breaci 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.

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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 time 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 inultiplied by the number of bricks contained in a solid foot, gives the answer.

EXAMPU.

HIow many bricks & inches long 4 inches wide, and 2} inchies thick, will it take to build a house 44 lect long, 40 feet wide, and cö feet high, and the walls to be one foot thick ?

8X4x2,5=So sa inches in a brick, then 1728---80 21,6 bricks in a solid foot. 44+40+-4-1-1-40=168 feet, whole length of wall.

-- four times the chickness.

164 remains. Multiply by 20 height.

5280 solid seet in the who.e wall. Multiply by 21,6 bricks in a solid foot.

Product, 70848 bricks. Ans.

EX'.

ART. 14. 'To find tic tonnage of a sliip.

NULE. Multipis the .ength of the keel by the breadth of the beam, and that pri) uct by the depth of the hold, and divide the last product by is, and the quotient is the tunnage.

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

70x94x10---95-218,0-t-ions. Ans.

ISLA 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 ?

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.

EXAMPLE.

EXAMPLES.

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1. If a cable 6 inches about, l'equire an anchor of 2% cwt. of what weight mustan anchor be for a 12 inch cable?

As 6x6x6:2fcut. : : 12x19x19:18cut. Ans.

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2. If a 12 inch cable require an anchor of 18 cwt. wirat must the circumference of a cable be, for an anchor of 21 cwt. ? c?ct.

cul. As 18 : 12x12x12 : : 2,25 : 216 3216=6 Ans. ART. 16. Having the dimensions of two similar built

ships of a different capacity, with the burthen of one oc than, co find the burtien of the other,

211.

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

EXAMPLE.

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

T.cut.grs.li. As 75x75x15 : 300 :: 100X100X100 :711 2 0 eiro

OR

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DUODECIMALS,

CROSS MULTIPLICATION, Is a rule made use of by workmen and artificers in cas ing up the contents of their werk.

RULE. 1. Unier the multiplicand write the corresponding de noninations of the muitiplier.

2. Alutsiply cach term into the multiplicand, beginn at the lowest, by the highest denomination in the mar plier, and write the resul? enchanter its respect terin ; verving to cara y ment for every 12, from ea. lower denomination to its nexé superior.

S. !!! the same manner multiply all the multiplice by the inches, or second denomination, in the multipli ani sei the result of eacii terin one place renoveil to: right hand of those in the inakliplican...

4. Do the same with tlie seconds in t'ie multiplier, ting the result of each terin two places to the right häiri of those in the multiplicand, &c.

EXAMPLES.
F. I.

F. 1. Multiply 1 S By

F. I.
16
5 8

F. I. 97 9

S

29

25 6

0

Son

91 16

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11

TEET, INCHES AND SECONDS.

H. I. Multiply 9 8 6 By 7 9S

[tiplier. 67 11 6 111 ==prod. by the feet in the mul7 3 4 6 "=ditto by the inches.

2 5 1 6=ditto by the seconds.

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How many square feet in a board 16 feet 9 inches long, and 2 feet S inclies wide. By Duodecimals.

By Thecimais.
F. 1.

. 1.
16 9

16 G=10,75 feet. 2 S

2 S 29,95

33 6
4.2 3

8375 SS50 3350

Ans. S7 & 3

Ans. 37,6875 37 8 9

a

TO MEASURE LOADS OF WOOD.

RULE, Miultiply 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 5 inches high?

7 ft. 6 in.=7,5 (171d 4 ft. 2 in.=4,167 and 2 ft. 3 in= 2,25; then, 7,5X4,167=51,2525X2,25=70,318125 solid feet, Aris.

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 cords 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 Decimcls. F.in.

F. 3 9

8.75 2 6

2,5

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9 4 6 9,575

F. in. ins. 4 8 S 4,6875=4 84, or half a cord and

81 inches over. the foregoing method is concise and easy to those who are well requainted with Duodecimals, but the following Table will give tho cortent of any load of wood, by inspection only, sufficiently end For common practice ; whicb will be found very convenient

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