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Specific Gravity (Foot in diameter and feet high)

foot in diameter and 10 feet high, would contain 10 cubic feet, 10 X 1000 X 7.207 = 72.070 oz. = 4500 lbs.

Specific Gravity of Solids. 1. By the Pitcher.- a pitcher, or similar vessel, brim full, put in the body, it will displace its own bulk of water; catch this water as it overflows and weigh it. Divide the weight of the body by that of the water displaced, the quotient will be its specific gravity. A very neat instrument for performing this process accurately has been contrived by Messrs. Eckfeldt and Dubois, of the United States Mint. 2. By the Hydrostatic Balance. - Weigh the body, fasten it, preferably by a horse-hair, immerse it in water, and note the loss of weight. The weight in air divided by the loss of weight in water the s. g. 3. When the Body is Lighter than Water. - Attach to it some heavy body of known weight in air and water. Weigh the two together, first in air and then in water; note the loss. The loss of weight of the heavy body in water being known the weight of the light body in air, will give the specific gravity. Thus, a bit of wood weighed in air 200 grains, attached to a piece of copper the two weighed in air 2247 grains, and in water 1620 grains, suffering a loss of 627 grains, the copper alone loses in water 230 grains, 627-230 = 397, the loss of the wood; 200397.504, s. g. of the wood. the difference between these losses divided into When the Solid is Soluble in Water.

Take its s. g. in regard to some liquid which does not dissolve it, multiply this by the s. g. of the liquid.

Thus, a piece of sugar weighed in air 400 grs., it lost in oil of turpentine 217.5. 400+ 217.5 = 1.84. The S. g. of turpentine is .87; 1.84 X .87 = 1.6., s. g. of the sugar.

When the Body is in Powder.

Introduce it into a counterpoise bottle, of which the capacity is known. Fill the bottle with pure water at 60°. It will hold as much less as is equal to the bulk of the powder, and the weight of the powder in air divided by this difference will give the s. g.

Thus, the bottle holds 1000 grs. of water; 100 grs. of emery are introduced, and the bottle filled up with water. If no water were displaced the two should weigh 1100 grs., they really weigh 1070; the difference, 30 grs. = the weight of water displaced; 100+30= 3.333, s. g of the emery.

When the Solid is Compound, As a nugget of gold and quartz. Take the s. g. of the nugget, that of gold and quartz being known, then apply the following formulæ : s. g. gold s. g. nugget - s. g. quartz s. g. gold X s. g. quartz s. g. nugget of nugget = weight of gold in nugget. s. g. gold - s. g. nugget s. g. quartz s. g. gold - s. g. quartz s. g. nugget of nugget = weight of quartz in do.

X X weight This method will do approximately, but not accurately for alloys of metals generally.