A TABLE

OF THE SPECIFICK GRAVITIES OF SEVERAL SOLID AND FLUID BODIES; WHERE THE SECOND COLUMN CONTAINS THEIR ABSOLUTE WEIGHT, AND THE THIRD THEIR RELATIVE WEIGHT, IN AVOIRDUPOIS OUNCES.

The use of the Table of Specifick Gravities will best appear by several Examples.

How to discover the quantity of adulteration in metals. Suppose a body be compounded of gold and silver, and it be required to find the quantity of each metal in the compound.

First, find the Specifick gravity of the compound, by weighing it in air and in water, and dividing its aerial weight by what it loses thereof in water, and the quotient will shew its specifick gravity, or how many times heavier it is than its bulk of water. Then, subtract the specifick gravity of silver (found in the Table) from that of the compound, and the specifick gravity of the compound from that of the gold: the first remainder will shew the bulk of gold, and the latter, the bulk of silver in the whole compound; and if these remainders be multiplied by the respective specifick gråvities, the products will shew the proportional weights of each met al in the body.

Suppose the specifick gravity of the compounded body be 14; that of standard silver (by the Table) is 10-535, and that of standard gold 18-888; therefore, 10-535 from 14, remains 3.465, the proportional bulk of the gold in the compound; and 14 from 18.888, remains 4-388, the proportional bulk of silver in the compound: then, 18-888, the specifick gravity of gold, multiplied by the first remainder 3.465, produces 65 447 for the proportional weight of gold; and 10 535, the specifick gravity of silver, multiplied by the last remainder, produces 51-495 for the proportional weight of silver in the whole body: So that for every 65 447 ounces or pounds of gold, there are 51.495 ounces or pounds of silver in the body.

Hence it is easy to know whether any suspected metal be genuine, or alloyed or counterfeit, by finding how much heavier it is than its bulk of water, and comparing the same with the Table; if they agree, the metal is good; if they differ, it is alloyed or counterfeited.

How to try Spiritous Liquors.

A cubick inch of good brandy, rum, or other proof spirits, weighs 234 grains; therefore if a true inch cube of any metal weighs 234 grains less in spirits than in air, it shews the spirits are proof: If it lose less of its aerial weight in spirits, they are above proof; if it lose more, they are under proof; for, the better the spirits are, the lighter they are, and the worse, the heavier.

Or, let any solid, of sufficient specifick gravity, be weighed first in air, then in water, and then in another liquid; from its weight in the air take its weight in water, and the remainder is the weight of its bulk of water. From its weight in air take its weight in the other liquid, and the remainder is the weight of the same quantity of that liquid. Divide the weight of this quantity of liquid by the weight of the same quantity of water, and the quotient will be the specifick gravity of the liquid.

All bodies expand with heat and contract with cold; but some more, and some less than others: therefore the specifick gravities of bodies are not precisely the same in summer as in winter.

The four following Problems, relating to spiritous liquors, are wrought by Alligation.

103. What proportion of rectified spirits of wine must be mixed with water, to make proof spirit, the specifick gravity of the rectified spirits being 850, that of proof spirit 925, and of water 1000 ?

9251000)75 Or equal measures.

109. What proportional weight of rectified spirits of wine and water must be mixed, to make proof spirit, the specifick gravities, as before?

1000 20

110. What is the specifick gravity of best French brandy, consisting of 5 parts, measure, of rectified spirits of wine, and 3 parts water?

850×5=4250 1000x3=3000

5+38) 7250

906 25 specifick gravity.

111. A retailer has 30 gallons of rum, whose specifick gravity is 900: How much water must he add to reduce it to standard

proof?

1000)25g. rum. g. wat. g. rum. g. wat.

925 {1000)

900/ 75 As 75 : 25 :: 30 : 10 to be added.

112. The cubick inch of common glass weighs about 1.36oz. Troy ditto of salt water 5427oz. ditto of brandy 48927oz. Suppose then, a seaman has a gallon of brandy in a bottle, which weighs 4th Troy, out of water, and to conceal it, throws it overboard into salt water: Pray, will it sink or swim, and by how much is it heavier or lighter than the same bulk of salt water?

Then, 270 7059x 5427=146-912oz.weight of salt water occupied by the bottle and brandy. And 48927 (=weight of a cubick inch of brandy) x231=113 02+oz and 113 02+54-167·02oz.

weight of the bottle and brandy. From this take the weight of the salt water, viz. 146-192oz. Ans. Supposing the bottle full, it is 20 11oz. heavier than the same bulk of salt water, and therefore will sink.

Given the weight to be raised by a balloon, to find its diameter.

RULE.

1. As the specific difference between common and inflammable air, is to one cubick foot: so is any weight to be raised, to the cubick feet contained in the balloon.

2. Divide the cubick feet by 5236, and the cube root of the quotient will be the diameter required, to balance it with common air; but, to raise it, the diameter must be somewhat greater, of the weight somewhat less.

113. I would construct a spherical balloon, of sufficient capacity to ascend with 4 persons, weighing, one with another, 160, and the balloon and a bag of sand weighing 60: Required the diame. ter of the balloon?

By the Table of Specifick Gravities, page 388, I find a cubick foot of common air weighs 1.25 ounces Avoirdupoie, and a cubick foot of inflammable air 12 of an ounce Avoirdupois; therefore,

1·25-121.13oz. difference. And 160x4+60=700=11200.

oz. cub. foot. oz. cub. feet. As 1.13: 1 :: 11200: 9911-5044. And

9911-5044

3

Given the diameter of a balloon, to find what weight it is capable of

raising.
RULE.

1. Multiply the cube of the diameter by 5236, and the product will be the content in cubic feet.

2. As one cubick foot is to the specifick difference between common and inflammable air; so is the content of the balloon to the weight it will raise.

114. The diameter of a balloon is 26 65 feet: What weight is it capable of raising?

cub. foot.

26.65×26-65×26.65X 5236=9911-4+ cubick feet. And

As 1 : 1-13: 99114+: 11199-882=700

nearly.

If the magnitude of any body be multiplied by its specifick gravity, the product will be its absolute weight.

115. What weight of lead will cover a house, the area of whose roof is 6000 feet, and the thickness of the lead of a foot? 6000×50 cub. feet, and its specifick gravity 11325×50= tons. cwt. qrs. lbs. Oz.

566250 ounces=15 15 3 26 10 Ans.

To find the magnitude of any thing when the weight is known.

Divide the weight by the specifick gravity in the Table, and the quotient will be the magnitude sought.

116. What is the magnitude of several fragments of clear glass, whose weight is 13 ounces?

13-2600 005 of a cubick foot, and 005×1728=8.640 cubick inches, Ans.

Having the magnitude and weight of any body given, to find its spe cifick gravity.

Divide the weight by the magnitude, and the quotient will be the specifick gravity.

117. Suppose a piece of marble contains 8 cubick feet, and weighs 13531 or 21656 ounces: What is the specifick gravity? 21656-82707 the specifick gravity required, as by the Table. "To find the quantity of pressure against the sluice or bank, which pens

water.

Multiply the area of the sluice, under water, by the depth of the centre of gravity, (which is equal to half the depth of the water) iu feet, and that product again by C23 (the number of pounds Avoir

dupois in a cubick foot of fresh water) or by 64.4 (the Avoirdupois weight of a cubick foot of salt water) and the product will be the number of pounds required.

118. Suppose the length of a sluice or floom be 30 feet, the width at bottom 4 feet, and the depth of the water 4 feet; what is the pressure against the side of the sluice?

30×4 120 feet the area of the bottom, and 120×2 (the depth of the centre of gravity) gives 240 cubick feet, and 210×62·5= 15000=6T. 13cwt. 3qrs. 20 Ans.

The perpendicular pressure of fluids on the bottoms of vessels is estimated by the area of the bottom multiplied by the altitude of the fluid.

119. Suppose a vessel 3 feet wide, 5 feet long, and 4 feet high, what is the pressure on the bottom, it being filled with water to the brim?

3X5 15 square feet, the area of the bottom, and 15×4=60 cubick feet, and 60×62·5=3750=33 cwt. 1 qr. 26.

THE USE OF THE BAROMETER.

The Barometer is so formed, that a column of quicksilver is supported within it to such a height as to counterbalance the weight of a column of air, of an equal diameter, extending from the barometer to the top of the atmosphere.

120. At the surface of the earth, the height of this column of quicksilver is, at an average, almost 30 inches; when the barometer is at that height; what is the pressure of atmosphere on a square foot, and on the surface of a man's body, estimated at 14 square feet?

As the cubick foot of quicksilver is 13600 ounces, Avoirdupois, and as the height in the barometer, is 2.5 feet, therefore 13600×2.5 34000 ounces, 2125 pounds on a square foot; and 2125×14= 29750 pounds on a man's body.

121. If the mercury in a barometer, at the bottom of a tower, be observed to stand at 30 inches, and, on being carried to the top of it, be observed at 29.9 inches: What is the height of the tower? Divide 13600, the specifick gravity of quicksilver, by 1.25, the specifick gravity of air, and the quotient will be the height of the tower, in tenths of an inch.

The number of feet, in height, of the atmosphere, corresponding with of an inch on the barometer is variable, depending on the. temperature and density of the atmosphere.

The variation, depending on the temperature, is shewn in the following Table, calculated for every 5 degrees, from 32 to 80, Fahrenheit's Thermometer, from whence it may be easily calculated, for the intermediate degrees by allowing of a foot for each degree.