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EXAMPLES.

Note.—When the circumference and diameter both are given, you may multiply half the circumference by half the diameter, and that product multiplied by the length, will give the solid contents.

1. What is the solid contents of a round stick of timber of equal bigness from end to end, whose diameter is 21 inches, (=lft. 9in.) and length 20 feet ? Thus, 1 foot 9 inches=1,75ft.

X1,75

square of diameter =3,0625X,7854=2,4052+ area of the base

20

Answer, solid contents =48,1040+ feet.
Or, 21 in.X21in.X,7854=346,3614in. area of the base.

20=length in feet.

144)6927,2280(48,106 nearly. PROB. V.-To measure Pyramids and Cones, or the frus

trum of a Pyramid or Coñe. Solids which decrcase gradually from the base until they end in a point are called pyramids or cones. If the base be square, it is called a square pyramid. If the base be a triangle, it is called a triangular pyramid.—But if the base be round, it is called a cone. The top is called the vertex; and a perpendicular line from the vertex to the base is called the perpendicular height:

Art. 1. The solid contents of any pyramid cr cone may be found by multiplying the area of the base by of its perpendicular height.

The frustrum of a pyramid or cone is what remains after the top is cut off, by a plane parallel to the base.

Art. 2. To find the solid contents of the frustrum of a square py. ramid, or tapering stick of square timber.

RULE. Multiply the side of the greater base or end, by the side of the less. er base or end, and to the product add } of the square of the differ. ence of the sides of the bases, or ends; then multiply this sum by the perpendicular height or length.

EXAMPLE. There is a tapering, square stick of timber, the side of the greater base or end of which measures 18 inches, the side of the lesser end 12 inches, and its length 30 feet; what is its solid contents ? Thus, greater end 18 inches.

18 less do.X12 inches. -12

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228

3)36=square of the diff. 30 length in feet.

12=} of square of do. 144)6840(47,5 feet Answer.

ART. 3. To find the solid contents of the frustrum of a cone, or ta. pering stick of round timber.

RULE. Multiply the diameters of the bases or ends together, and to the pro. duct add one third of the square of the difference of the diameters ; then mùltiplying this sum by ,7854 gives the mean area between the two bases or ends, which multiplied by the length gives the solid contents.*

EXAMPLE If the diameter of one end of a tapering stick of round timber be 21 inches, and the diameter of the other end 12 inches, and the length 30 feet, what is the solid contents ? Thus, diameter of greater end 21

21 less end x 12

12 diff. of diameters 9 252

X9 } sq. of the diff. of the diameters+27

square of diff. 279

of do. 27 Then, 279X,7854x30_144=45,65+ solid feet, Ans. PROB. VI.—To find how many solid feet a round stick of timber, of equal bigness from end to end, will contain when

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hewn square.

RULE. Multiply twice the square of the semi-diameter in inches by the length in feet, and divide the product by 144, and the quotient will be the answer.

EXAMPLE.

If the diameter of a round stick of timber be 21 inches, and its length 20ft., how many solid feet will it contain when hewn square ? Diameter 21in2=10,5in, semi-diameter.

Then 10,5x10,5x2X20-144=30,625 feet, Anș. PROB. VII.—To find the solid contents of a globe or sphere.

RULE. Multiplying the circumference and diameter together gives the area, which multiplied by one-sixth of the diameter gives the solid contents.

* This is a correct way of measuring round logs, or timber, which taper gradually from end to end. But the old English method of measuring round timber was, to girt the stick in the middle, and call one-fourth of this girt the side of a square equal to the circumference. This one-fourth part squared, and the square multiplied by the length, they called the solid contents, which was an erroneous method ; for, the girt in the middle is not the mean between the ends, nor one-fourth of the girt equal to the side of a square of equal area with circumference; and from this old erronepus practice of measuring timber was introduced the custom of calling 40 feet of round timber and 50 feet of hewn timber a ton, for 40 feet of round timber, measured by this method, will actually make about 50 feet of hewn timber.-We suppose, that when timber is accurately measured, 40 feet of every kind should make a ton.

EXAMPLE.

What is the solid contents of a globe or ball whose diameter is 18 inches ?

Thus, the diameter being 18 inches, the circumference is found to be 56,57+ inches. (See Rule, page 235.) Then the circumference, 56,57 inches, multiplied by the diameter, 18 inches, =1018,+ which multiplied by one-sixth of the diameter, (=3 inches) gives 3054 solid inches, =Ans. 1 solid foot 1326in.

PROB. VIII.—To find how many bushels, or gallons, will be contained in a vessel of given dimensions, whether it be cubic, cylindric, or globular.)

RULE. If it be a cubic vessel, find its contents or capacity in inches, by Prob I. or II. p. 236. If it be cylindric, find its contents by Prob. IV. If it be globular, by Prob. VII. Then, as 2150,4 cubic inches make a bushel, therefore, dividing by 2150,4 will give the bushels; and di. vide by 231, the number of inches in a wine gallon, gives the num. ber of wine gallons ; or, divide by 282, and you will have the beer gallons.

EXAMPLES 1. How many bushels, and how many wine gallons, will a cistern hold that is 3 feet long, 2 feet wide, and 2 feet deep?

Thus, 3 feet =36 inches, and 2 feet =24 inches.
Then 36x24x24=20736 cubic inches.
And 20736-2150,4=9,64 bushels, Answer.

And 20736:231=Ans. 89 wine gallons, 177in. rem. How many wine gallons are contained in a cylindrical vessel whose diameter is 18 inches, and depth 12 inches ?

Thus, 18X18X,7854x12, (by Prob. 4, page 237,)=3053,6352 cubic inches, and 3053,6352-231=13,2+ gallons, Ans.

PROB. IX.—The dimensions of the walls of a brick build. ing given, to find how many bricks are sufficient to build it.

RULE. From the whole extent of the wall, measured round on the outside, subtract four times its thickness, and multiply the remainder by the height, and that product by the thickness of the wall, gives the solid contents of the whole, which multiplied by the number of bricks in a solid foot, gives the answer.

Note. To find the number of bricks in a solid foot, multiply the length in inches by the breadth in inches, and that product by the thickness; then divide 1728 by this product.

EXAMPLES.

How many bricks, 8 inches long, 4 inches wide, and 24 inches thick, will it take to build the walls of a house 40 feet long, 30 feet wide, and 20 feet high, the walls to be one foot thick ?

8X4X21-80 solid inches in a brick ; then 1728+80=21,6 bricks in a solid foot.

Then 40+40+30-+30=140 four times the thickness

136

height X20 solid feet in the whole wall 2720 number of bricks in a solid foot X21,6 Answer

58752 bricks. PROB. X.—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 divide the product by 95, the quotient will be the tonnage. NotE.—If the vessel be doubledecked, half the breadth of the mainbeam is accounted the depth of the hold.

EXAMPLES.

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What is the tonnage of a vessel 65 feet in length by the keel, the breadth of her beam 22 feet, and the depth of the hold 10ft. 6in,= 10,5ft. ?

Thus, 65X22X10,5-95=Ans. 158 tons, 5 rem. PROB. XI.

To find the number of gallons, &c., that are contained in a vessel in the form of the frustrum of a cone, or a tub, whose top and bottom diameters are unequal,

RULE. Find the cubic contents of the given vessel in inches, (by the rule, Prob. V. Art. 3, for finding the contents of the frustrum of a cone,) which divided by 231 will give the wine gallons, &c.

Ex. How many wine gallons are contained in a tub, whose bottom diameter is 27 inches, top diameter 36 inches, and depth 50 inches ? Thus, 36—27=9 diff. x 9=81, sq. of diff. ;3=27, ; of

And top diam. 36 x bottom diam. 27=972,+ 27, 1 sq. of diff.=999. Then 999 x,7854 x 50=39230,73 cubic inches, which — by 231 gives 169,83 gallons, Ans.

PROB. XII, To guage a cask, or to find how many gallons it will hold.

To guage a cask, you must measure the head diameter and the bung diameter, (taking the measure within the cask,) and the length of the cask, making allowance for the thickness of the heads; then take the difference between the head and bung diameter, and when the staves are about an ordinary curve, add about 6 tenths of the difference to the head diameter, which will reduce the cask to a cylinder; that is, it will give the mean diameter.* Then multiply

the square of the mean diameter in inches by ,7854, and that product by the length in inches, (Prob. 4.) gives the cubic

sq. of diff.

* If the diameters of the heads are unequal, take their mean diameter, and if the staves are very much curved, take ,66 instead of six tenths; but if they be nearly straight, take 55, &c.

quotient (231

contents in inches, which divided by 231, gives the wine gallons, and by 282 gives he beer gallons.

But as the square of the mean diameter is always to be multiplied by ,7854 and divided by 231 for wine gallons, we may contract the operation, and multiply by their ,7854

thus, for wine gals. multiply by 34, pointing off 4 fig=,0034;) ures for the decimals; and for ale or beer gals., (2835,0028)

by 28, pointing off four figures for decimals. Hence the following

RULE. Multiply the square of the mean diameter by the lengtn, and mul. tiply this product by ,0034 for wine, and by ,0028 for beer gallons.

Ex. There is a cask whose head diameter is 25 inches, bung diam. eter 30 inches, and whose length is 38 inches. How many wine galļons, and how many beer gallons does it contain ?

Thus, the diff. between the head and bung diameter is 5 inches; this multiplied by 6 tenths gives 3in. to be added to the head diameter.

25+3=28in. mean diam. X 28in.=784,sq. of mean diam. x 38in. length,=29792. Then 29792 x,0034=101,2928 wine gals. ; and 29792,0028=83,4176 beer gals.

PROB. XIII.--Of Mechanical Powers. Art. 1. Of the Lever. It is a fundamental principle in mechanics, that the power, and weight which will be raised, are to each other in versely as the spaces which they pass over. Hence to find what weight may be raised by a given power.

As the distance between the body to be raised, or balanced, and the fulcrum, or prop, is to the distance between the prop and the point where the power is applied, so is the power, to the weight which it will balance.

EXAMPLE. 1. There is a lever 10ft. long, and the fulcrum or prop, on which it turns, is 2ft. from one end ; how many pounds weight at the end 2ft. from the prop, will be balanced by a power of 56lbs. at the end 8ft. from the prop ?

As 2ft. : 8ft. :: 56lbs. Ans. 224lbs. 2. What weight inust be applied to the above lever eight feet from the prop, to balance one thousand pounds two feet from the prop ?

As 8ft. : 2ft. :: 1000lbs. Ans. 250lbs. 3. If 1000lbs. be placed 2 feet from the prop, on the above lever, how far must 250lbs. be placed from the prop on the other side, to balance the 1000lbs. ?

As 250lbs. : 1000lbs :: 2ft. Ans. 8ft. Art. 2. Of the Wheel and Axle.-The spaces passed over are as their diameters or circumferences.

There is a windlass, the wheel of which is 60 inches in diameter, and the axis around which the rope coils is 6 inches in diameter ; how many pounds on the axle will be balanced by 160lbs. at the wheel ?

As hin. : 60in, :: 160lbs. Ans. 1600lbs. 2. How many pounds at the wheel of this windlass will balance 1200lbs. on the axle ? As 60in. : 6in :: 1200lbs. Ans. 120lbs.

ART. 3. Of the Screw.-The power is to the weight which will be

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