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QUEST. 59. Required the same as in the last question, when the point B is on the other side of AC, supposing AB 9, AC 12, and BC 6 furlongs; also the angle ASB 33° 45', and the angle BSC 22° 30′.

Answer.

AS 10-64, BS 15'64, cs 14.01.

QUEST. 60. It is required to determine the magnitude of a cube of gold, of the standard fineness, which shall be equal to a sum of 480 million of pounds sterling; supposing a guinea to weigh 5 dwts 94 grains. Ans. 18.691 feet.

QUEST. 61. The ditch of a fortification is 1000 feet long, 9 feet deep, 20 feet broad at bottom, and 22 at top; how much water will fill the ditch?

Ans. 1158127 gallons nearly.

QUEST. 62. If the diameter of the earth be 7930 miles, and that of the moon 2160 miles: required the ratio of their surfaces, and also of their solidities: supposing them both to be globular, as they are very nearly?

Ans. the surfaces are as 13
and the solidities as 49

to 1 nearly ; to 1 nearly.

PRACTICAL EXERCISES CONCERNING SPECIFIC

GRAVITY.

THE Specific Gravities of Bodies are their relative weights contained under the same given magnitude; as a cubic foot, or a cubic inch, &c. .

The specific gravities of several sorts of matter, are expressed by the numbers annexed to their names in the Table of Specific Gravities, at page 220; from which the numbers are to be taken, when wanted.

Note. The several sorts of wood are supposed to be dry. Also, as a cubic foot of water weighs just 1000' ounces avoirdupois, the numbers in the table express, not only the specific gravities of the several bodies, but also the weight of a cubic foot of each in avoirdupois ounces; and hence, by proportion, the weight of any other quantity, or the quantity

quantity of any other weight, may be known, as in the following problems.

PROBLEM I.

To find the Magnitude of any Body, from its Weight.

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.

EXAM. 1. Required the content of an irregular block of common stone, which weighs Icwt, or 112lb.

Ans. 12284 cubic inches. EXAM. 2. How many cubic inches of gunpowder are there in 1lb weight? Ans. 29 cubic inches nearly.

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

Ans. 38

PROBLEM II.

cubic feet.

To find the Weight of a Body from its Magnitude.

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.

EXAM. 1. Required the weight of a block of marble, whose length is 63 feet, and breadth and thickness each 12 feet; being the dimensions of one of the stones in the walls of Balbeck?

Ans. 683 ton, which is nearly equal to the burden of an East-India ship.

EXAM. 2. What is the weight of 1 pint, ale measure, of gunpowder? Ans. 19oz. nearly.

EXAM. 3. What is the weight of a block of dry oak, which measures 10 feet in length, 3 feet broad, and 2 feet deep? Ans. 43351lb.

PROBLEM

PROBLEM III.

To find the Specific Gravity of a Body.

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

As the weight lost in water,

Is to the whole weight,

Then say,

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

EXAMPLE.

A piece of stone weighed 10lb, but in water only 6lb, required its specific gravity?

Ans. 2609.

CASE 2. When the body is lighter than water, so that it will not quite sink, affix to it a piece of another body, heavier than water, so that the mass compounded of the two may sink together. Weigh the denser body and the compound mass, separately, both in water and out of it; then find how much each loses in water, by subtracting its weight in water from its weight in air; and subtract the less of these remainders from the greater. Then say,

As the last remainder,

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

PROBLEM IV.

Ans. 600.

To find the Quantities of Two Ingredients in a Given Compound. TAKE the three differences of every pair of the three specific gravities, namely, the specific gravities of the compound and each ingredient; and multiply the difference of every two specific gravities by the third. Then say, as the greatest product, is to the whole weight of the compound, so is each of the other products, to the two weights of the ingredients.

VOL. II.

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EXAMPLE.

A composition of 112lb 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?

Ans. there is 100lb of copper in the composition. and consequently 12lb of tin S

OF THE WEIGHT AND DIMENSIONS OF BALLS AND SHELLS.

THE weight and dimensions of Balls and Shells might be found from the problems last given, concerning specific gravity. But they may be found still easier by means of the experimented weight of a ball of a given size, from the known proportion of similar figures, namely, as the cubes of their diameters.

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To find the Weight of an Iron Ball, from its Diameter.

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An iron ball of 4 inches diameter weighs 91b, and the weights being as the cubes of the diameters, it will be, as 64 (which is the cube of 4) is to 9 its weight, so is the cube of the diameter of any other ball, to its weight. Or, take of the cube of the diameter, for the weight. Or, take of the cube of the diameter, and of that again, and add the two together, for the weight.

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EXAMPLES.

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EXAM. 1. The diameter of an iron shot being 6.7 inches, required its weight? Ans. 42.294lb.

EXAM. 2. What is the weight of an iron ball, whose diameter is 5.54 inches? Ans. 24lb nearly.

་ཚ PROBLEM I

To find the Weight of a Leaden Ball.

A leaden ball of 1 inch diameter weighs of a lb; therefore, as the cube of 1 is to, or as 14 is to 3, so is the cube

of the diameter of a leaden ball, to its weight. Or, take of the cube of the diameter, for the weight, nearly.

EXAMPLES.

EXAM. 1. Required the weight of a leaden ball of 6.6 inches diameter? Ans. 61.6061b.

EXAM. 2. What is the weight of a leaden ball of 5·30 inches diameter ? Ans. 32lb nearly.

PROBLEM III.

To find the Diameter of an Iron Ball.

. MULTIPLY the weight by 74, and the cube root of the product will be the diameter.

EXAMPLES.

EXAM. 1. Required the diameter of a 421b iron ball?

Ans. 6.685 inches.

EXAM. 2. What is the diameter of a 24lb iron ball?

Ans. 5.54 inches.

PROBLEM IV.

To find the Diameter of a Leaden Ball.

MULTIPLY the weight by 14, and divide the product by 3; then the cube root of the quotient will be the di

ameter.

EXAMPLES.

EXAM. 1. Required the diameter of a 64lb leaden ball?

Ans, 6 684 inches. EXAM. 2. What is the diameter of an 8lb leaden ball? Ans 3.3.3 inches.

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TAKE

To find the Weight of an Iron Shell.

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of the difference of the cubes of the external

and internal diameter, for the weight of the shell.

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That is, from the cube of the external diameter, take the cube of the internal diameter, multiply the remainder by 9, and divide the product by 64.

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EXAMPLES.

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