23. What is the volume of a rectangular solid 11 ft. long, 4 ft. wide, and 4 ft. high?

24. A cask holding 256 gal. of water will hold how many bushels of wheat?

PRODUCERS' AND DEALERS' APPROXIMATE RULES. 463. To find the contents of a bin or elevator in bushels, stricken measure. RULE.-Multiply the cubic feet by .8, and add 1 bushel for each 300, or in that proportion.

To find the contents of a bin or crib in bushels, by heaped measure. RULE.-Multiply the cubic feet by .63.

REMARK.-If the crib flare, take the mean width.

To find the number of shelled bushels in a space occupied by unshelled corn. RULE.-Divide the cubic inches by 3840, or multiply the cubic feet by 45.

To find the dimensions of a bin to hold a certain number of bushels. RULE.—To the number of bushels add one-fourth of itself, and the sum will be the cubic feet required, to within one three-hundredth part.

To find the exact number of stricken bushels in a bin. RULE.-Divide the cubic inches by 2150.42.

To find the exact number of heaped bushels in a bin. RULE.-Divide the cubic inches by 2747.71.

To find the capacity of circular tanks, cisterns, etc. RULE.-The square of the diameter, multiplied by the depth in feet, will give the number of cylindrical feet. Multiply by 55 for gallons, or multiply by .1865 for barrels.

REMARK.-In tanks or casks having bilge, find the mean diameter by taking one-half of the sum of the diameters at the head and bilge.

To find the number of perches of masonry in a wall, of 24 cubic feet in a perch. RULE.-Multiply the cubic feet by .0404.

To find the number of perches of masonry in a wall, of 16 cubic feet in a perch. RULE.-Multiply the cubic feet by .0606.

REMARK.-The above is correct within part. In large contracts add of 1%. EXAMPLE.- How many perches, of 24 cu. ft. each, in a wall 150 ft. long, 50 ft. high, and 2 ft. thick?

EXPLANATION.-Short Method.-150 × 50 × 2 = 15000; 15000 .0404606; add 1000, or

.606606.606.

Extended Method.-150 X 50 X 2 = 15000; 15000 ÷ 24.75 = 606.6, same as before.

Same example, perch of 164 cu. ft.

EXPLANATION.-Short Method.-150 X 50 X 2 = 15000; 15000 X.0606 909; add 1000 = .9;

909.9909.9.

To find the number of cubic feet in a log. RULE.-Divide the average diameter in inches by 3, square the quotient, multiply by the length of the log in feet, and divide by 36.

To find the number of feet, board measure, in a log. RULE.—Multiply the cubic feet, as above obtained, by 9.

HAY MEASUREMENTS.

464. Few products are so difficult of accurate measurement as hay, owing to the pressure, or the want of it, in packing, time of settling, volume in bulk, and freedom from obstruction in packing. Plainly, the larger (higher) the stack, or mow, and the greater the foreign weight in compress, the more compact it will be.

465. The accepted measurements are of three kinds:

1st. To find the weight of hay in a load or shed loft, unpressed. RULE.-Allow 540 cubic feet for a ton.

2d. To find the weight in common hay barn, or small (low) stack. RULE.-Allow 405 cubic feet for a ton.

3d. To find the weight in mow bases in barns, compressed with grain, and in butts of large stacks of timothy hay. RULE.-Allow 324 cubic feet for a ton.

CUBE ROOT.

466. The Cube or Third Power of a number, is the product of three equal factors.

467. The Cube Root of a number is one of the three equal factors the product of which represents the cube. Thus, a cubic foot = 12 x 12 x 12, or 1728 cubic inches, the product of its length, breadth, and thickness; and since 12 is one of the three equal factors of 1728, it must be its cube root.

468. The operation of finding one of the equal factors of a cube is called extracting the cube root.

469. As shown in the explanation of extracting the square root, the first point to be settled in extracting any root is the relative number of unit orders or places in the number and its root.

=8

REMARK. From this comparison may be inferred the following:
1st. The cube of any number expressed by a single figure cannot have

less than one nor more than three places or unit orders.

2d. Each place added to a number will add three places to its cube. 3d. If a number be separated into periods of three figures each, beginning at the right hand, the number of places in the root will equal the number of periods and partial periods if there are any.

93 H 729

103 = 1000

471. To help in understanding the cube root, first form a cube and thus ascertain its component parts or elements. Take 57 as the number to be cubed.

=

472. From this result observe that 57s the cube of the tens, plus three times the square of the tens multiplied by the units, plus three times the tens, multiplied by the square of the units, plus the cube of the units; or that the cube of any number made up of tens and units = t3 + 3t2 u + 3 t u2 + u3, which for the purpose of reference we will call Formula (a). And if all orders above simple units are considered tens, Formula (a) will apply to the cube of any number.

473. To assist in understanding the operation of extracting the cube root, observe the forms and dimensions of the illustrative blocks, and the relation of each to the other in the formation of the complete cube.

EXPLANATION.-Since the block (A) is a cube, the number representing the length of its side will be its cube root.

The given number consists of two periods of three figures each, therefore its cube root will contain two places, tens and units.

Since the given number is a product of its root taken three times as a factor, the first figure, or highest order of the root, must be obtained from the first left hand period, or highest order of the power; therefore, find first the greatest cube in 135; since 185 comes between 125 (the cube of 5) and 216 (the cube of 6) the tens of the root must be 5 plus a certain remainder; therefore, write 5 in the root as its tcns figure.

Subtracting the cube of the root figure thus found (5 tens, or 50)3

=

125000, by taking 125 from the left hand period, 185, and so obviate the necessity of writing the ciphers; to this remainder bring down the next, or right hand period, 193, thus obtaining as the entire remainder 60193.

Referring to Formula (a), observe that, having subtracted from the given number the cube of its tens (t3), the remainder, 60193, must be equal to 3 t2 u + 3 tu2 + u3.

If a cube (B), 50 inches in length on each side, is formed, its contents will equal 125000

7

50

Λ

cubic inches, and it will be shown that the remaining 60193 cubic inches are to be so added to cube (B) that it will retain its cubical form. In order to do this, equal additions must be made to three adjacent sides, and these three sides, being each 50 inches in length and 50 inches in width, the addition to each of them in surface, or area, is 50 50, or 502, and on the three sides, 3 (502), or 3 t2, as in the squares (C). It will also be observed that three oblong blocks (D) will be required to fill out the vacancies in the edges, and also the small cube (E), to fill out the corner. Since each of the oblong blocks has a length of 5 tens, or 50, inches, the three will have a length of 3 X 50 inches, or 3 t. Observe, now, the surface to be added to cube (B), in order to include in its contents the 60193 remaining cubic inches, has been nearly, but not exactly obtained; and since cubic contents divided by surface measurements must give units of length, the thickness of the three squares (C), and of the three oblong pieces (D), will be determined by dividing C0193 by the surface of the three squares, plus the surface of the three oblong blocks, or by 3 t2+3 t; this division may give a quotient too large owing to the omission in the divisor of the small square in the corner; hence such surface measure taken as a divisor, may with propriety, be called a trial divisor. So using it, 7 is obtained as the second, or unit figure of the root.

Assuming this 7 to be the thickness of the three square blocks (C), and both the hight and thickness of the three oblong blocks (D), gives for the solid contents of the three square blocks (C), 52500, and for the solid contents of the three oblong blocks (D), 7350, or 3 t u+3 t u2: 59850; and by reference to the Formula (a), observe that the only term or element required to complete the cube of (tu) is the cube of the units (u 3).

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Now, by reference to the illustrative blocks, observe that by placing the small cube (E) in its place in the corner, the cube is complete. And since (E) has been found to contain 7 X 7 X 7, or 343 cubic inches, add this to the sum of 3 t u 3 tu2 u3, and obtain 3 t2 u +3 tu+u3 60193; and if to this t3, or 1250C0 is added, the result is t3+3 tu+3 t u2+u3 185193, Formula (a); then subtracting 60193 from the remainder, 60193, nothing remains.

=

=

This proves that the cube root of 185193 is 57. By the operation is also proved the correctness of Formula (a): The cube of any number equals the cube of its tens, plus three times the square of its tens mul tiplied by its units, plus three times its tens multiplied by the square of its units, plus the cube

of its units.

Rule.-I. Beginning at the right, separate the given number into periods of three figures each.

II. Take for the first root figure the cube root of the greatest perfect cube in the left hand period; subtract its cube from this left hand period, and to the remainder bring down the next period.

III. Divide this remainder, using as a trial divisor three times the square of the root figure already found, so obtaining the second or units figure of the root; next, subtract from the remainder three times the square of the tens multiplied by the units, plus three times the tens multiplied by the square of the units, plus the cube of the units.

REMARKS.-1. In examples of more periods than two, proceed as above, and after two root figures are found, treat both as tens for finding the third root figure. For finding subsequent root figures, treat all those found as so many tens.

2. In case the remainder, at any time after bringing down the next period, be less than the trial divisor, place a cipher in the root and proceed as before.

3. Should the cube root of a mixed decimal be required, form periods from the decimal point right and left. If the decimal be pure, point off from the decimal point to the right, and if need be annex decimal ciphers to make periods full.

4. To obtain approximate roots of imperfect cubes, to any desired degree of exactness, annex and use decimal periods.

5. The cube root of a common fraction is the cube root of its numerator divided by the cube root of its denominator.

6. The cube root of any common fraction may be found to any desired degree of exactness, either by extracting the root of its terms separately (adding decimal periods if need be) or by first reducing the common fraction to a decimal and then extracting the root.

7. The 4th root can be obtained by extracting the square root of the square root.

8. The 6th root is obtained by taking the cube root of the square root, or the square root of the cube root.

5.

6.

Find the cube root of 2146, to three decimal places. Find the cube root of 119204, to two decimal places. 7. Find the cube root of 46982, to one decimal place.

12.

Find the cube root of, to one decimal place.

00

Find the cube root of 1988, to two decimal places.

13. Find the cube root of 25.416237, to two decimal places. 14. Find the cube root of 3496.25, to three decimal places.

15.

Find the cube root of .4106, to three decimal places.

16. Find the decimal equivalent of the cube root of 11, to two decimal places, by reducing the fraction to a decimal of six places and extracting the root of the decimal.