CASE. 2, The weight of any body being given to find the folidity, and the fpecific gravity thereof,

RULE. Divide the given weight by the tabular weight, corre fponding to the name of the fame kind, and the quotient will be the folidity in cubic inches.

EXAMPLE I. How many folid feet are there in a block of marble that weighs 8 tons; and what will it come to at 5 fhillings per foot folid?

1

First, 8 tons 17920 lb.

Then,0977286) 17920,00000000 (183328,45 inches.

Now 1728)183328,45 (106,09 cubic feet (nearly) at 5s. or,251.

106,09

25

53045

E. 2.

21218

26,5225267. 10s. 54d. anfwer.

In the Spectators club of fat people, it is faid that each perfon weighed no less than 4 cwt. how many folid inches was there in one of their bodies?

Firft, 4cwt.448 lb.

Then ,041310(448.000000(10844,8 folid inches the answer.

E. 3. Suppofe that a man of war, with all its ordnance, rigging, and appointment, draws fo much water as to difplace 1300 tons of fea water, London beer measure the weight of the vessel is required? Firft, 1300 X 45200 hhds. and a hhd = 282 X 54=15228 cubic inches.

Therefore 15228 X 5200=79185600 cubic inches displaced,

Then

79185690

,037253

2375568

3959280

1583712

5542992

2375568

T. cwt. lb.

2240)2949901,156800(1316,92015 tons 1316 18 17 the weight required.

E. 4. Hiero, king of Scilly, ordered his jeweller to make him a crown, containing 63 ounces of gold; the workmen thought by fubftituting part filver therein, to have a proper perquifite, which taking air, Archimedes was appointed to examine it, who, on putting it into a veffel of water, found it raised the fluid, or that itself contained 8,2245 cubic inches of metal, and having difcovered that the cubic inch of

gold

gold more critically weighed 10,36 ounces, and that of filver but 5,85 ounces, he, by calculation, found what part of his Majefty's gold had been changed, and you are defired to repeat the process?

First, 10,36) 63,00 (6,08 108, had it been all gold.

Alfo, 5,85) 63,00 (10,76923, if all filver.

Then by Sect. 28,

Mean rate 8,2245-608108 = 2,54473

10,769032,14342

Sum 4,68815

4,68815) 2,54473 (,5428, oz. part gold
4,68815) 2,14342 (,4572, oz. part filver

oz.dwts. grs.

Then { ;5422}X63={28,8036-28 18

34,1964 34 3 22,272, G

1,728, S

Proof 63 00 00,000

CASE 3. The weight and magnitude being given, to find the specific gravity.

RULE. Divide the weight in ounces, by the folidity in cubic feet, the quotient will be the fpecific gravity.

EXAMPLE I. I have a piece of marble that contains 4 folid feet; and weighs 675 lb. what is its specific gravity?

First, 675 X 16 = 10800 ounces.

Then 10800 4 2700 the specific gravity.

E. 2. I have a piece of timber that contains 6 feet, and weighs 300 lb. what wood it it?

First, 400 X 16 4900 ounces.

Then 480006

800 the fpecific gravity.

Now, in the table of fpecific gravity, against 800 you will find dry afh or elm, the wood required.

N. B. All bodies of what nature or kind foever, being weighed in open air, and balanced by thofe whofe fpecific gravity is greateft; those bodies whofe fpecific gravity is leaft, will weigh the heaviest in

vacuo.

Thus, if a piece of lead, at the end of a nice balance, and a piece of cork at the other end, are in equilibrio in the air, and thus placed under the receiver of an air-pump, as foon as the air begins to be exhaufted, the equilibrium will begin to be deftroyed, till at last when all the air is taken away, the cork will defcend and fhew itself really heavier thạn the lead.—(And for the fame reason, a pound of feathers is heavier than a pound of lead, which may feem a paradox to fome ;) but the reason is very evident from the laws of Hydroftatics; for both bodies being

weighed

weighed in air, each would lofe the weight of an equal bulk of air, confequently the feathers will lofe a greater weight than the lead, be cause it is of greater bulk, therefore when the air is taken away from both, the weight that is reftored to the feathers, being the greateft, will cause it to preponderate or weigh down the lead in vacuo.

CASE 4. The folidity of any piece of timber being given to find how far it will fink.

RULE. Divide the fpecific gravity of the timber by the fpecific gravity of the water: multiply this fum by the depth of the timber, and it will give the inches under water.

EXAMPLE 1. How many inches will a cubic foot of elm fink in common water?

First, 1,000),8000 (,800

12

9,600 inches the answer.

E. 2. How many inches will a cubit foot of deal fink in common water?

First, 1,000),6570 (,657

12

7,884 inches under water.

CASE 5. The folidity of any timber being given, to find how much it will carry.

RULE. Subtract the fpecific gravity of the timber from that of the water, the remainder is the number of ounces that one folid foot will carry.

EXAMPLE I.

How much weight is juft neceffary to fink a cubic

foot of deal in common water?

E. 2.

1,000
,657

16),343 (21,437 lb. the answer.

How much will a raft, made of 12 pieces of yellow deal, carry in fea water, if each piece be a foot square, and 20 feet long?

Firft, 1033-657=376, and 12 X 20 = 240

Then 240 X 376,90240

.*. 9024÷16≈ 5640 lb. the answer.

To make a deceitful balance, or pair of fcales, whofe beam will hang in equilibrio without the fcales, or with the empty scales ; and yet fhall also be in equilibrio when unequal weights are placed

in the scales; fo as to cheat in any proportion intended, in making the balance at firft. See Plate 1. Fig. 32.

To the beam A B = 23 inches long, whofe arm CB, of 11 inches in length, keeps in equilibrio about the point C, the arm AC; of 12 inches in length, by being made fo much thicker, or having fo much more matter, as may make amends for its being fhorter; hang the feales D, E, in fuch a manner, that D, which weighs one part in twelve lefs than E, fhall hang at the longeft end of the beam, and they will keep each other in equilibrio; then placing 12 pounds weight at G, in the fcale E, it will keep in equilibrio no more than 1 pounds of F, the commodity to be fold, if placed in the fcale D; becaufe then, F will be to G in a reciprocal proportion of BC to A°C.

Now, though fuch a balance may be fo nicely made as to deceive the eye, the cheat is immediately difcovered by changing the weights, and the commodity F, from one fcale to another; for then, the owner of the balance, muft either confefs the fraud, or add to the commodity he fells, &c. not only what was wanting, but also as much as he intended to cheat him of; and a fraction of the added weight, proportionable to the inequality of the arms of the balance, that is, in this cafe, the buyer, instead of the 11 pounds offered him for 12, his due, will have, by changing the fcales, 13 pounds. For whereas, in the firft pofition of the balance F=11×12 A C was equal to G=12X11=BC when G or 12 pounds is placed in the fcale D, then 12X 12 will be equal to no less than CB=11X13 G. Or,

As the arm CB, 11 inches long, is to the arm CA, 12 inches long, fo is F, or the weight 12, placed in the fcale D: to G 13, or the weight of the commodity keeping the weight in equilibrio.

And therefore, as this analogy gives a reciprocal proportion between the weights and their velocities, the momenta will be equal; which, with contrary directions, destroy one another,

N. B. In all these cafes, we fuppofe the weight to hang freely from those ends of the balance to which they are faftened.

À TABLE,

A TABLE by which the quantity and weight of water in a cylindrical pipe of any given diameter of bore, and perpendicular height may be found; and confeqently, the power may be known that will be fufficient to raise the water to the top of the pipe, in any pump, or any or any other hydraulic machine..

For tens of feet high, remove the decimal points one place forward; for hundreds of feet, two places, for thousands, three places, and so on. Then multiply the fums by the fquare of the diameter of the given bore, and the products will be the answer.

EXAMPLE. The quantity and weight of water in a cylindric pipe, 85 feet high, and 10 inches diameter.

Which number 80110,6 of cubic inches being divided by 231, the cubic inches in a wine gallon, gives 342,6 for the number of gallons: and the respective weights 42255,489 and 46360,3, being divided, the former by 12, and the latter by 16, give 3521,29 for the number of troy pounds, and 2897,5 for the number of avoirdupoife pounds, that the water in the pipe weighs. So much power would be required to balance or fupport the water in the pipe, and as much more to work the engine, as the friction thereof amounts to.

In all pumps, the preffure of the column of water, or its weight felt by the working power, when raised to any given height above the furface of the well, is in proportion to the height of the column, confidered throughout, as if it were equal in diameter to that part the bore, in which the pifton or bucket works.