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1. What are the contents of the frustum of a square pyramid, the sides of whose bases are 18 and 25 feet, and the altitude 15 feet?

Ans. 6995 sq. ft. 2. How many cubic feet in a log 45 feet in length, the radius of one end being 21 feet and of the other 71 ft.?

Ans. 3828.825 cu. ft.

THE SPHERE. 1051. A Sphere is a volume bounded by a curved surface, every point of which is equally distant from a point within called the centre.

1052. The Diameter of a sphere is a line passing through its centre and ending in the surface. The radius is half the diameter.

Rule.- To find the surface of a sphere, multiply the circumference by the diameter; or square the radius and multiply by 4 times 3.1416.

1. Required the surface of a sphere whose diameter is 36 inches.

Ans. 4071.5136 cu. in. 2. The circumference of the earth is nearly 25000 miles; what is its surface ?

Ans. 198937500 sq. miles. 1053. Rule.- To find the contents of a sphere, multiply the cube of the diameter by 1 of 3.1416, or by .5236.

1. Required the contents of a sphere whose diameter is 84 inches.

Ans. 310339.8144 cu. in. 2. What is the weight of a cannon ball 9 inches in diameter, the metal weighing 6953 oz. per cubic foot ?

Ans. 1535.8742 oz. 1054. Rule.- To find the edge of a cube which may be cut from a given sphere, square the diameter, divide by 3, and extract the square root of the quotient.

1. Required the edge of a cube that can be cut out of a sphere whose radius is 12 inches.

Ans. 13.856 in. 2. Required the contents of a cube inscribed in a sphere having a circumference of 15.7085 inches.

Ans. 24.05+cu. in.

THE SPHEROID 1055. A Spheroid is a volume formed by the revolution of an ellipse about one of its axes.

1056. A revolution about the longer axis forms a prolate spheroid ; about the shorter axis, an oblate spheroid.

Rule.- To find the contents of a spheroid, multiply the square of the revolving axis by the fixed axis, and that product by t of 3.1416, or by .5236.

1. What are the contents of a balloon in the shape of a prolate spheroid, the longer axis being 15 feet and the shorter 10 feet?

Ans. 785.4 cu. ft. 2. The earth is an oblate spheroid, the longer axis being about 7925 miles and the shorter 7898 miles ; what are its contents ?

Ans. 259,725,929,424.5 cu. miles.

IRREGULAR BODIES. 1057. Rule.-To find the contents of an irregular body, immerse the body in a vessel of known dimensions, containing water ; note the rise in the water, and calculate accordingly.

1. A stone was thrown into an empty cylindrical vessel, which was then filled with water; when the stone was taken out, the water fell 4.75 in.; what was the volume of the stone, the diameter of the vessel being 9 in.? Ans. 302.18+cu. in.

2. A lump of iron ore being put into a vessel 1 cubic foot in capacity, it was found that it took 21 gallons to fill the vessel; required the volume of the ore. Ans. 12081 cu. in.

GAUGING. 1058. Gauging is the process of ascertaining the capacity of casks and other vessels.

1059. Barrels and casks differ from cylinders in bulging out in the middle. It is necessary, therefore, first to ascertain the approximate mean diameter of the cask or barrel, and the capacity can then be obtained like that of a cylinder. Rule 1.— To find the mean diameter of a barrel or cask, add to the head diameter {, or, if the staves are not much curved, 3, of the difference between the head and bung diameters.

Rule II.- To find the capacity in gallons, multiply the square of the mean diameter by the length (both expressed in inches), and this product by .0034.

NOTE.—The contents of a cylinder are found (Art. 1044) by multiplying together the length, the square of the diameter, and .7854. To reduce to gallons, we divide this product by 231 (Art. 505), or, which is the same thing, multiply the length and the square of the mean diameter, by (.7851:231) or .0031.

1. What is the capacity in gallons of a cask whose head diameter is 30 inches, bung diameter 38 inches, and length 42 inches?

Ans. 178.2778 gal. 2. How many gallons in a barrel of cider, with staves slightly curved, the head diameter being 2 ft., the bung diameter 2 ft. 3 in., and the length 2 ft. 10 in.?

Ans. 76.947 gal.

LUMBERMEN'S PRACTICAL RULE. 1060. In lumbering it is convenient to be able to determine the amount of square-edged inch-boards that can be sawed from a round log. The most convenient method of doing this is by the following rule, known as Doyle's Rule:

Rule.- From the diameter in inches subtract 4; the square of the remainder will be the number of square feet of inch boards yielded by a log 16 feet in length.

NOTE.—This is quite a close approximation to a scientific rule; and though it favors the buyer in small logs and the seller in large ones, yet, since logs are often crooked, no rule averages a more correct result.

1. How many square feet of lumber can be cut from a log 44 in. in diameter and 24 ft. long? SOLUTION.—No. of square feet=40x40xį=2400.

2. How many square feet of square-edged lumber in a log 12 in. in diameter and 18 ft. long? Ans. 72 sq. ft.

3. What is the yield of a log 36 in. in diameter and 20 ft. long?

Ans. 1280 sq. ft. NOTE.Doyle's Rule is the basis of the tables in Scribner's Lumber and Log Book, which is a recognized standard among lumbermen.

SECTION XV.

ARITHMETICAL ANALYSIS.

1061. We present a few problems and solutions under the head of Arithmetical Analysis.

NotE.—For an analysis of many of the old problems which present such excellent combinations of conditions as to be regarded as classic, see the author's Normal Written Arithmetic.

CASE I.

1. If an article had cost 20% less, the gain would have been 30% more; what was the gain per cent. ?

SOLUTION.—The second cost is 100% - 20%, or 80% of the first cost. If on 100% the amount is a certain rate, on 1% the rate will be 100 times as great, and on 80% it will be go of 100, or į times as great; hence

- 4, or = 30%, the difference in the rate, and 1=120%, the rate at first cost; hence the gain per cent. Was 20.

2. If an article had cost me 10% less, the gain would have been 12% more; what was the gain per cent. ?

Ans. 8%. 3. If the cost had been 4% less, the gain would have been 43% more; what was the gain per cent. ?

Ans. 12%.

CASE II. 1. If an article had cost 20% more, the gain would have been 25% less; what was the gain per cent. ?

SOLUTION.—The second cost is 120% of the first cost, and therefore on it the amount will be as great a rate per cent. as on the first cost; hence -, or =25%, the difference in the rates, hence=150%, the rate at first cost, and the gain is 50%.

2. If the cost of certain goods had been 25% more, the gain would have been 30% less; what was the gain

Ans. 50%. 3. If an invoice of calicoes had cost 15% more, the gain would have been 12% less; what was the gain per cent. ?

Ans. 8% loss. 4. If I had paid 10% more for my fall stock, the profit would have been 10% less; what was the gain per cent. ?

Ans. 10%

per cent. ?

CASE III.

1. A merchant sold cloth at 20% gain, but had it cost $49 more, he would have lost 15% by selling at the same price; what did the goods cost?

SOLUTION.-The cloth was sold for 120%, or of the cost, but had it cost $49 more, it would have been sold for 85%, or 17 of the cost; hence of the first cost equals 17 of the second cost, and 14 of the first cost equals zo of the second cost; but the difference between the first cost and the second cost is $49; hence 14-17, or 14 of the first cost equals $49, and the cost was $119.

2. A quantity of goods were sold at 25% gain, but if they had cost $40 less, the gain at the same selling price would have been 35%; what was the cost of the goods ?

Ans. $540. 3. A farmer lost 10% on his wheat crop, but if it had cost him $50 more he would have lost 20%; what was the cost of the crop ?

Ans. $400. 4. A commission merchant sold flour for his principal at a loss of 10%, but if the flour had cost $1 a barrel less, he would have gained 5%; what was the cost of the flour per barrel?

Ans. $7.

CASE IV. 1. A father willed $43,500 to his two sons, A and B, aged 12 and 15 years respectively, to be divided in such a manner that the two parts, on interest at 6%, would amount to equal sums when they became of age; what were the parts ?

SOLUTION.—A's money was on interest 9 years, and B's 6 years. For 6 years at 6%, it of the principal equals the amount; hence it of B's share equals his amount ; and in the same way we see that 77 of A's share equals his amount. Now, since the amounts are equal, 37 of A's share equals jf of B's, from which we find B's share=of A's; hence

of A's + 73 of A's, or 14 of A's=$43,500; os of A's=$300,

- $20,400, and 75 of A's, or B’s=$23,100. 2. A gentleman divided $84,700 among his three sons, aged 11, 14, and 17 years respectively, so that the different shares, being on interest at 5%, should amount to equal sums when they became of age; what were the shares ?

Ans. $25,200; $28,000; $31,500.

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