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EXAMPLE.

A composition of 1121b 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

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.

PROBLEMI.

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

An iron ball of 4 inches diameter weighs 9lb, 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.

EXAMPLES.

EXAM. 1. The diameter of an iron shot being 6o7 inches, required its weight?

Ans. 42.2941b. Exam. 2. What is the weight of an iron ball, whose diameter is 5.54 inches ?

Ans. 241b nearly.

PROBLEM II. To find the Weight of a Leaden Ball. A leaden ball of 1 inch diameter weighs of a lb; therefore as the cube of i is to 14, or as 14 is to 3, so is the cube

of

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. 61.606lb. 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 7, and the cube root of the product will be the diameter.

EXAMPLES.

a

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.

a

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 diameter

EXAMPLES.

a

Exam. 1. Required the diameter of a 641b leaden ball?

Ans. 6.684 inches. Exam. 2. What is the diameter of an 8lb leaden ball ?

Ans. 3•343 inches.

PROBLEM V.

To find the Weight of an Iron Shell. TAKE 27 of the difference of the cubes of the external and internal diameter, for the weight of the shell.

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.

EXAMPLÉS.

EXAMPLES.

Exam. 1. The outside diameter of an iron shell being 12.8, and the inside diameter 9.1 inches; required its weight?

Ans. 188.941 lb. EXAM. 2. What is the weight of an iron shell, whose external and internal diameters are 9.8 and 7 inches ?

Ans. 8441b.

PROBLEM VI.

To find how much Powder will fill a Shell. Divide the cube of the internal diameter, in inches, by 37.3, for the lbs of powder.

EXAMPLES.

Exam. 1. How much powder will fill the shell whose internal diameter is 9.1 inches ?

Ans. 135lb nearly. Exam. 2. How much powder will fill a shell whose internal diameter is 7 inches?

Ans. 6lb.

PROBLEM VIL.

To find how much Powder will fill a Rectangular Box. Find the content of the box in inches, by multiplying the length, breadth, and depth all together. Then divide by 30 for the pounds of powder.

EXAMPLES.

EXAM. 1. Required the quantity of powder that will fill à box, the length being 15 inches, the breadth 12, and the depth 10 inches ?

Ans. 60lb. Exam. 2. How much powder will fill a cubical box whose side is 12 inches?

Ans. 573b.

PROBLEM VIII.

To find how much Powder will fill a Cylinder. MULTIPLY the square of the diameter by the length, then divide by 88.2 for the pounds of powder.

EXAMPLES

Exam. 1. How much powder will the cylinder hold, whose diameter is 10 inches, and length 20 inches ?

Ans. 52 lb nearly:

Exam. 2

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ExAM. 2. How much powder can be contained in the cylinder whose diameter is 4 inches, and length 12 inches?

Ans. 5 & lb.

PROBLEM IX.

To find the Size of a Shell to contain a given Weight of Powder.

MULTIPLY the pounds of powder by 57-3, and the cube root of the product will be the diameter in inches.

EXAMPLES.

Exam. 1. What is the diameter of shell that will hold 13 lb of powder ?

Ans. 9.1 inches. Exam. 2. What is the diameter of a shell to contain 6lb of powder ?

Ans. 7 inches.

a

PROBLEM X.

a

To find the Size of a Cubical

Box, to contain a given Weight of
Powder.

MULTIPLY the weight in pounds by 30, and the cube root of the product will be the side of the box in inches.

EXAMPLES.

a

EXAM. 1. Required the side of a cubical box, to hold 50lb of gunpowder ?

Ans. 11.44 inches. EXAM. 2. Required the side of a cubical box, to hold 400lb of gunpowder ?

Ans. 22.89 inches.

PROBLEM XI.

To find what Length of a Cylinder will be filled by a given

Weight of Gunpowder. MULTIPLY the weight in pounds by 38.2, and divide the product by the square of the diameter in inches, for the length.

EXAMPLES.

ExAM. 1. What length of a 36-pounder gun, of 6 inches diameter, will be filled with 12lb of gunpowder ?

Ans. 10°314 inches. Exam. 2. What length of a cylinder, of 8 inches diameter, may be filled with 20lb of powder ? Ans. 111 inches.

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OF THE PILING OF BALLS AND SHELLS. Iron Balls and Shells are commonly piled by horizontal courses, either in a pyramidical or in a wedge-like form; the base being either an equilateral triangle, or a square, or a rectangle. In the triangle and square, the pile finishes in a single ball; but in the rectangle, it finishes in a single row of balls, like an edge.

In triangular and square piles, the number of horizontal rows, or courses, is always equal to the number of balls in one side of the bottom row. And in rectangular piles, the number of rows is equal to the number of balls in the breadth of the bottom row. -Also, the number in the top row, or edge, is one more than the difference between the length and breadth of the bottom row.

PROBLEM I.

To find the Number of Balls in a Triangular Pile. MULTIPLY continually together the number of balls in one side of the bottom row, and that number increased by 1, also the same number increased by 2; then Ź of the last product will be the answer.

nin+1.n + 2 That is,

is the number or sum, where

6 n is the number in the bottom row.

EXAMPLES.

Exam. 1. Required the number of balls in a triangular pile, each side of the base containing 30 balls ? Ans. 4960.

Exam. 2. How many balls are in the triangular pile, each side of the base containing 20 ?

Ans. 1540.

PROBLEM II.

To find the Number of Balls in a Square Pile. MULTIPLY continually together the number in one side of the bottom course, that number increased by 1, and double the same number increased by 1; then of the last product will be the answer.

nin+1.2n + 1 That is,

is the number. 6

VOL. II.

T

EXAMPLES

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