measure to hold one gallon of water, that is, 10 pounds or 7000 grains. A cylindrical inch of water weighs 198 28 grains, therefore a gallon measure must contain 353031 cylindrical inches, and if D be the diameter and H the height of the cylindric gallon measure, we must have

D2 H 353'031

If the diameter be given we may compute the height by formula,

And if it were desired to make such a measure with the height equal to the diameter we should have

H=D=√3 {353031}

Again, the question may arise, "what must be the diameter of a cast-iron ball that shall weigh so many pounds?"

In such a case we have the formula

W = D3 X 252 458 X specific gravity,

when the weight is taken in grains and the diameter in inches.

246. The other problems relating to specific gravity are the following:Given the quantities and specific gravities of two or more substances separately to find the specific gravity of the mixture. For this it is sufficient to multiply each quantity by its specific gravity, add the products, and divide by the total quantity. Thus, if we melt together 15 ozs. of copper, specific gravity 8.7, 5 ozs. of tin, specific gravity 74, and 2 ozs. of lead, specific gravity 114, the specific gravity of the bronze will be

supposing that no dilatation or contraction takes place. But if there is a contraction from the original bulk, say in the ratio of 1: 99, the specific gravity will be increased in the ratio 100, and will therefore be 8.74.*

*

The other practical question is: Having given the specific gravity of a mixture in unknown proportions of two substances, the specific gravities of which are known, to find the proportions in which they are mixed. This requires the solution of an algebraic equation, of which only the result is here given. Suppose, to fix our ideas, we have a lump of gold quartz, the specific gravity of which we call W, and the specific gravities of gold and quartz we call G and Q respectively. The quantity of gold in an unit of weight of the mixture will be and the quantity of quartz will be GW.

W- - Q

G Q'

Here again it is supposed that there is no dilatation or contraction.

*Technical Arithmetic and Mensuration, by Charles W. Merrifield, F.R.S. London: Longmans & Co. This work is earnestly recommended to the attention of the reader.

ᎠᎠ

ON BUNKERS, TANKS, &c.

247. Given the length, width, and height of a coal bunker, and the number of cubic feet to a ton, to find the quantity of coals it will contain.

RULE XCV.

1o. Reduce the inches to the decimal of a foot (see Rule LIII, § 159, pages 117 -118).

2°. Multiply the length, width, and height together (see Rule XLVI, page 102), the product is the contents of the bunker in cubic feet.

3°. Divide this last (2°) by the number of cubic feet to a ton; the quotient is the quantity of coal the bunker will contain, expressed in tons and decimals of a ton.

4°. Find the value of the decimals of a ton in the inferior denominations (see Rule LV, § 161, page 120).

EXAMPLES.

Ex. 1. A coal bunker is 18 feet 9 inches long, 7 feet 6 inches wide, and 8 feet 3 inches high required the quantity of coals it will contain at 38.6 cubic feet per ton (by decimals).

12)9'0

*75

12)6'0

*5

12)3'0

*25

18 ft. 9 in. 18'75 ft.; 7 ft. 6 in. = 7'5 ft.; 8 ft. 3 in. = 8.25 ft.
Contents of bunker 1875 X 75 X 8.25 = 1160*15625 cubic feet.

Ex. 2. A coal bunker is 20 feet 4 inches long, 8 feet 7 inches broad, 4 feet 3 inches in depth: required the quantity of coal it will contain, 44 cubic feet being equal to a ton (to be worked by decimals).

4 in.)4'0

*333

20 ft. 4 in.

20°333; 8 ft. 7 in. = :8.5833; 4 ft. 3 in. 4°25,

20 333 X 8.5833 (= 174°5242389 X 4'25) = 741728015325 cubic feet.

Cubic feet in 1 ton = 44.

... tons in bunker 741-72801532516-85745 tons 16 tons 17 cwt. o qrs. 16 lbs.

Ex. 3. A cross bunker is 20 feet deep, and 18 feet long, and to find its width three measurements are made, the upper one is 29 feet, the middle one 23 feet, and the lowest one 18 feet: how many tons of coal will it hold at 45 cubic feet to the ton ?

To find the mean width, add together the first and third, and 4 times the middle width, and divide by 6.

Ex. 4. A coal bunker is 6 feet 6 inches at the top and 4 feet 6 inches at the bottom, 16 feet long and 8 feet 6 inches high: what weight of coals will it contain ?

State how many cubic feet per ton you will allow.

For the mean width add width at top and at bottom together and divide by 2: thus, 6 feet 6 inches + 4 feet 6 inches 11 feet o inches,

then I feet o inches 25 feet 6 inches, or 5'5 feet mean width.
Allow 45 cubic feet to a ton.

Then 16 X 5'5 × 8·5 ÷ 45 = 16·622 tons = 16 tons 12 cwt. nearly.

Ex. 5. Two bunkers are to be partitioned off to hold 54 tons of coal at 45 cubic feet per ton: the bunker is to be 12 feet and 5 feet.

Then since Length = cubic contents divided by product of Breadth and Height (Rule LXXXIII, page 190).

Cubic contents of 1 bunker = 54 X 45 ÷ 2 = 12'15 cubic feet.
Area of section 12 X 560 square feet,

..length=116 20:25 feet, or 20 feet 3 inches, Answer.

Ex. 6. A coal bunker is 28 feet 9 inches long, by 12 feet 6 inches wide, and 8 feet high; and 2 side bunkers each 29 feet 9 inches by 9 feet 6 inches, and the same height as the cross one, how many tons of coal will each hold at 45 cubic feet to the ton? And how many baskets of coal will there be altogether if 20 baskets weigh 16 cwts. ?

Contents of cross bunker = 28.75 X 12'5 X 8 =

2875 cubic feet.
4484

248. Any two dimensions of a square or rectangular tank being given, to find the third so that the tank shall hold any number of imperial gallons.

RULE XCVI.

1o. Multiply the length by the breadth (both being expressed in feet and decimals of a foot), the product will be the square feet.

2o. Multiply the number of gallons the tank is required to contain by 16, or divide by 6 (6·25), the result in either case is the contents in cubic feet.

NOTE.-1 gallon of water 16 cubic feet, or 64 gallons I cubic foot.

3°. Divide the result obtained by No. 1, by the product found by No 2°; the quotient is the height in feet and decimals of a foot.

Ex. 1. A tank is 7 ft. long by 3 feet 4 inches broad, what height must it be to contain 900 gallons?

Ex. 2. A tank is 8.75 feet long by 44 feet broad: what height must it be to contain 1210 gallons?

In each of the following examples it is required to find the height of the tank, having given the following quantities.

249. If we have the three dimensions of a square or rectangular tank given, to find the number of gallons it would contain we may proceed thus:

RULE XCVII.

1°. Take the continued product of the length, breadth, and depth in feet and decimals of a foot.

2o. Next, multiply this product by the number of gallons to a cubic foot, viz., 61 (or 6.25), the result is the number of gallons required to fill the tank.

EXAMPLES.

Ex. 1. A tank is 13 feet 6 inches long, 5 feet 6 inches wide, and 5 feet deep: how many cubic feet will it contain, also the number of gallons required to fill it?

13 ft. 6 in.

13'5 feet.

5 ft. 6 in.
6 gallons I cubic foot.
13'5 X 5'5 X 5 = 371°25 cubic feet.
371°25 X 6=23203125 gallons.
13'5 feet.

5'5 feet.

371 25 cubic feet.

6.25

185625

Length
Width

74250 222750