CASE VIII.

To find the distance that any object may be seen at sea, elevated at any height above the level of the water.

RULE. Add to the earth's semi-diameter in feet, the height of the object; square the sum, next square the number of miles in the earth's semi-diameter, take the difference of those squares, then by (Euclid 47, Lib. 1,) the square root of the difference of those squares will be the distance required in feet.

1. There is a point of the Andes in South America which is 4 miles above the level of the sea, to what distance could a person see from the top of such an elevated point, provided the atmosphere was perfectly clear, and not assisted by refraction.

Illustration.-If we put the earth's semi-diameter at 4000 miles, then 4004 x 4004 - 4000 x 4000 = 178.93 miles, the distance required, which is about twice as far as a person could see elevated from a point one mile above the level of the sea.

2. Suppose a ladder 40 feet long be so planted as to reach a window 33 feet from the ground on one side of the street, and without moving it at the foot, will reach a window on the other side 21 feet high, what is the breadth of the street? Ans. 56, 64+ feet.

Cubes. 1 8 27 64 125 216 343 512 729

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RULE 1.-Point every third figure beginning at the units place, then find the nearest cube to the first point, and subtract it therefrom; put the root in the quotient, bring down the figure in the next point to the remainder for a dividend.

2. Square the quotient and multiply it by 3, for a divisor; find how often the divisor is contained in the dividend, rejecting units and tens, and place the number of times in the quotient.

3. Square the last figure placed in the quotient, and place the result to the right hand of the divisor, for a defective divisor.

4. Then multiply the last figure placed in the quotient by the other figures, and that product by 30; add the last product to the (defective divisor,) placing units under units and tens under tens, for a complete divisor. Illustration.-Extract the cube root of

3. What is the cube root of 32461759? 4. What is the cube root of 84604519? 5. What is the cube root of 259694072?

RULE.

tion.

CASE III.

Ans. 319.

66 439.

To extract the Cube Root of a fraction.

66 638.

Reduce the fraction to the lowest denominaThen extract the Cube Root of the numerator for a new numerator, and also of the denominator for a new denominator.

To extract the Cube Root of a mixed number.

RULE. Reduce the fractional part to its lowest terms, and then the mixed number to an improper fraction, extract the roots of the numerator and denominator for a new numerator and denominator, but if the mixed number given be a surd, reduce the fractional part to a decimal, annex it to the whole number, and extract the root therefrom.

1. If a cubical piece of timber be 47 inches long, 47 inches broad and 47 inches deep, how many cubical inches does it contain? Ans. 103823.

2. There is a cellar dug, 12 feet in every way, length, breadth and depth, how many solid feet of earth are taken out of it? Ans. 1728.

3. The solid content of a cube is 389017 feet, what is the superficial content of one of its sides? Ans. 5329.

Well known principles assumed.

Circles are to one another as the squares of their diameters. Spheres are to each other as the cubes of their diameters. Cubes and all similar solid bodies as the cubes of their diameters or homologous sides. Whatever constitutes length, breadth and thickness or depth is a solid.

EXAMPLES.

1. If a ball 3 inches in diameter weigh 4 lbs., what will be the weight of a ball that is 6 inches in diameter? Ans. 32 lbs.

2. The solid content of a cellar, which is alike in length, breadth and depth, is 100 cubic yards, required the length of its side. Ans. 13.95 feet +

CASE V.

The side of a cube being given to find the side of a cube which shall be double, treble, &c. in quantity to the given cube.

RULE.-Cube the given number and multiply it by 2, 3, &c. the cube root of the product, is the side sought.

1. There is a cubical vessel whose side is 12 inches, it is required to find the side of another vessel that is to contain 3 times as much? Ans. 17.306 inches. 2. If a ship of 400 tons burden be 80 feet long in the keel; what is the burden of another ship the keel of which is 100 feet long? Ans. 781 tons, 5 cwt.

3. The dimensions of a ship are, viz: keel 125 feet long, beam 25 feet, depth of hold 15, what dimensions should a ship of similar forma have, to carry 3 times the burden? Ans. length of keel 180,28 feet, breadth of beam 36.05, depth of hold 21.63.

4. Find the dimensions of a similar ship that shall contain, or carry just half the burden of that whose dimensions are given. Ans. length of keel 99.21 feet, beam 19.81, hold 11.09.

5. Suppose a cannon ball of 4 inches diameter, weighs 18 lb.; what is the diameter of another that weighs 42 lbs? Ans. 5.30 inches.

6. Suppose a mortar shell of 8 inches diameter, weighs 50 lb; what is the diameter of a shell that weighs 100 lbs? Ans. 10.08 inches.

CASE VI.

To find mean proportionals between two given numbers. RULE.-Multiply the square of the lesser extreme by the greater, the cube root of the product will be the lesser mean. Again, multiply the square of the greater extreme by the lesser extreme, the cube root of the product will be the greater mean.

Example. Required to find 2 mean proportionals between 4 and 256. Illustration, 4 x 4 x 256 =4096

4096 16. Again, 256 x 256 x 4 = 262144, the 3/262144 = 64, hence 16 and 64 are the mean proportionals required.

SINGLE POSITION.

This rule is called Position because by using supposed numbers according to the conditions of the question, the answer is obtained.

RULE.-As the sum of the errors is to the given sum, so is the supposed number to the true one required. PROOF. Add the several parts of the result together, and if it agrees with the given sum it is right.

1. A person, after spending and of his money had 60 dollars left; what had he at first? Ans. $144.

Suppose he had 24