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Ex. 4.-Supposing the cubic inch of common glass weighs 1-4921 ounces troy, the same of sea-water ·59542, and of brandy 5368; then a seaman having a gallon of this liquor in a glass bottle, which weighs 3-84 lbs. out of water, and, to conceal it from the officers of the customs, throws it overboard. It is proposed to determine, if it will sink, how much force will just buoy it up?

3.84 lbs. × 12 = 46-08 ounces, and

46:08 1.4921


cubic inches in the bottle.


If we suppose a gallon of brandy to contain 231 solid inches, then 231 +30-8826 = 261.8826 cubic inches in both.

261-8826 x 59542 155-9311 ounces nearly, the weight of the salt water occupied by both bottle and brandy.

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Ex. 5.-A given cone is immersed in water, with its vertex downwards; what part of the axis will be immersed if the specific gravity of the fluid be to that of the cone as 8 to 5?

Now, by the principles of hydrostatics, the magnitude of the whole cone is to the magnitude of the part immersed, as the specific gravity of the fluid is to that of the body, or as 8 to 5. But the whole cone and part immersed are similar; their contents are as 8 to 5. Hence,

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of the axis immersed, the whole being 1.

Ex. 6.—Hiero, king of Sicily, ordered his jeweller to make him a crown, containing 63 ounces of gold. The workmen thought that substituting part silver was only a


proper perquisite; which taking air, Archimedes was appointed to examine it; who, on putting it into a vessel of water, found it raised the fluid 8-2245 cubic inches: and having discovered that the inch of gold more critically weighed 10.36 ounces, and that of silver but 5·85 ounces, he found by calculation what part of the king's gold had been changed. And you are desired to repeat the process, 10-36 1:63: 6.081 cubic inches.

5.85 1 63: 10-77 cubic inches.


By Alligation:

(6·081 ) 2-5455 cubic inches gold. 8-2245 {10-77) 2-1435 cubic inches silver.

2.5455+2.1435 = 4.689.

Then, 4-689: 63 :: 2·5455 : 34-2 ounces gold.
4-689 63: 2-1435: 28-8 ounces silver.

Ex. 7.-What must be the thickness of a right-angled cone of copper, the inner diameter of which is 20 inches, so that it may just float with its edge level with the surface of the fluid; the specific gravities of the copper and fluid being as 9 to 1, and the interior and exterior surfaces having a common base?

Let r inner radius, t


thickness, and p = 3·1416. Then, since the cone is right-angled, the radius of the base is equal to the altitude; therefore,

p (r + t)2 × (r + t) p [(v + t)3 — r3]

an inch.


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1,000 Air at the Earth's Sur

** Since a cubic foot of water, at the temperature 40° Fahrenheit, weighs 1000 oz. avoirdupois, or 62 lbs. the numbers in the preceding Tables exhibit very nearly the respective weights of a cubic foot of the several substances tabulated. Dr. Gregory's Edition of Dr. Hutton's Course.


Hydraulics or Hydrodynamics treats of the motion of fluids, and the forces with which they act upon bodies against which they strike, or which move in them.

The velocity with which a fluid issues from a very small orifice in the bottom or side of a vessel that is kept constantly full, is equal to that which a heavy body would acquire by falling from the level of the surface of the fluid to the level of the orifice.

Therefore, if h = height of the fluid above the orifice, g the velocity acquired by a falling body in one second, and the velocity with which the water issues,

v = √2 gh.

The quantity of water that issues in one second through a given orifice is equal to a column of water having the area of the orifice for its base, and the velocity with which the fluid issues for its altitude.

That is, if A= the area of the orifice, and Q = the quantity of fluid running out in one second,

Q = A √2 g h.

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Experiments do not exactly agree with this theory as to the quantity of water run out; for the vein of water that issues through the small orifice suffers a contraction, by which its section has been found to be diminished in the ratio of nearly 5 to 7. Therefore, instead of Q = A √2gh, we must take Q A √2 g h. But is nearly



; therefore, we may take Q

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which a heavy body acquires in falling through ; con


sequently, the velocity of the water at the orifice is found equal to that which a heavy body would acquire in falling freely through half the altitude.

The experiments of Bossut show that the actual discharge through a hole made in the side or bottom of a vessel, is to the theoretical as 1 to 62, or nearly as 8 to 5. Consequently, the theoretical discharge must be diminished in this ratio to have the true discharge. Also, if a pipe: from 1 to 2 inches long be inserted in the aperture, the contraction of the vein is prevented, and the actual discharge is to the theoretical as 4 to 5.

The quantities discharged are as the square root of the depth multiplied into the areas of the orifices.

The following is extracted from Mr. Banks's Treatise on Mills:

When the water is discharged through perpendicular sections, the velocity at the bottom of the orifice is rather greater than that at the top of the orifice; but if the depth and breadth of the orifice be but small compared with its depth below the surface of the dam, we may take the depth of the centre of the hole for the mean depth, without any sensible error.

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