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the weight of a cubic foot of eacb, in Avoirdupois ounces ; and hence, by proportion, the weight of any other quantity, or the quantity of any other weight, may be readily known.

PROBLEM I.

To find the magnitude of a body, from its weight being

given.

RULE.

As the tabular specific gravity of the body, is to its weight in Avoirdupois ounces,

So is one cubic foot, or 1728 cubic inches, to its content in feet, or inches, respectively.

EXAMPLES.

1. Required the content of an irregular block of common stone, which weighs 1 cwt. or 112 lbs.

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c. in.

c. in.

Hence 2520 : 1792 :: 1728 : 12284. Ans.

2. How many cubic inches of gunpowder are there in one pound weight?

Aps. 30 nearly. 3. How many cubic feet are there in a ton weight of dry oak?

Ans. 38136

PROBLEM II.

To find the weight of a body from its magnitude being

given.

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RULE.

As one cubic foot, or 1728 cubic inches, is to the content of the body,

So is its tabular specific gravity, to the weight of the body.

EXAMPLES.

1. Required the weight of a block of marble, whose length is 63 feet, and its breadth and thickness each 12

; these being the dimensions of one of the stones in the walls of Balbec.

feet;

Here 122 x 63=9072 c. ft.=content of the body.

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Hence 1 : 9072 :: 2700 : 244944 oz.=638 16 tons, which is equal to the burthen of an East India ship.

2. What is the weight of a pint of gunpowder, ale measure ?

Ans. 19 oz. nearly. 3. What is the weight of a block of dry oak, which measures 10 feet in length, 3 feet in breadth, and 23 feet deep?

Ans. 43351. lbs.

PROBLEM III.

To find the specific gravity of a body.

RULE.

Case 1. When the body is heavier than water, weigh it both in water and out of water, and the difference will be the weight lost in the water.

Then, as the weight lost in the water, is to the whole weight,

So is the specific gravity of water, to the specific gravity of the body.

EXAMPLE.

A piece of stone weighed in air 10 pounds, but in water only 64 pounds. Required its specific gravity.

10
63

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Case 2. When the body is lighter than water, so that it will not quite sink, affix to it another body heavier than water, so that the mass compounded of the two may sink together.

Weigh the heavier body and the compound mass separately both in water and out of it, and find how much each loses in water, by subtracting its weight in water from its weight in air.

Then as the difference in these remainders is to the weight of the light body in air,

So is the specific gravity of water to the specific gravity of the body.

EXAMPLE.

Suppose a piece of elm weighs in air 15 pounds, and that a piece of copper which weighs 18 pounds in air, and 16 pounds in water, is affixed to it, and that the compound weighs 6 pounds in water; required the specific gravity of the elm

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To find the quantities of two ingredients in a given

compound

RULE. Take the difference of every pair of the three specific gravities, viz., of the compound and each ingredient, and multiply the difference of every two by the third.

Then as the greatest product is to the whole weight of the compound, so is each of the other products to the weight of the two ingredients.

EXAMPLE.

A composition of 112 pounds being made of tin and copper, whose specific gravity is found to be 8784; required the quantity of each ingredient: the specific gravity of tin being 7320, and of copper 9000. 9000 9000

8784
7320
8784

7320

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14757120)1475712000(100 Ans. 1001b. of copper ) in the composition.

and 12 lb. of tin

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