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then www::ss, which proportion will give either of those specific gravities, the one from the other.
s, the specific gravity of the body;
s, the specific gravity of the fluid.
So that the specific gravities of bodies, are as their weights in the air directly, and their loss in the same fluid inversely.
325. Corol. 6. And hence, for two bodies connected together, or mixed together into one compound, of different specific gravities, we have the following equations, denoting their weights and specific gravities, as below, viz.
H = weight of the heavier body in air,
5th, b +
dividing the absolute weight of the
body by its loss in water, and multiplying by the specific gravity of water.
But if the body L be lighter than water; then / will be negative, and we must divide by L + / instead of L 1, and to find / we must have recourse to the compound mass c; and because, from the 4th and 5th equations, L
the absolute weight of the light body, by the difference be tween the losses in water, of the compound and heavier body, and multiply by the specific gravity of water. Or thus,
as found from the last equation.
Also, if it were required to find the quantities of two ingredients mixed in a compound, the 4th and 6th equations would give their values as follows, viz.
the quantities of the two ingredients H and L, in the compound c. And so for any other demand.
To find the Specific Gravity of a Body.
326. CASE I.-When the body is heavier than water: weigh it both in water and out of water, and take the difference, which will be the weight lost in water. Then, by corol. 6, prop. 64, s = where B is the weight of the body out of water, b its weight in water, s its specific gravity, and w the specific gravity of water. That is,
As the weight lost in water,
Is to the whole or absolute weight,
EXAMPLE. If a piece of stone weigh 10 lb, but in water only 62 lb, required its specific gravity, that of water being 1000? Ans. 3077.
327. CASE II. When the body is lighter than water, so that it will not sink: annex to it a piece of another body, heavier than water, so that the mass compounded of the two may sink together. Weigh the denser body and the compound mass, separately, both in water, and out of it; then find how much each loses in water, by subtracting its weight in water from its weight in air; and subtract the less of these remainders from the greater. Then say, by proportion,
As the last remainder,
Is to the weight of the light body in air,
That is, the specific gravity is s=
(c — c)
6, prop. 64. EXAMPLE. Suppose a piece of elm weighs 15 lb in air; and that a piece of copper, which weighs 181b in air and 16 lb in water, is affixed to it, and that the compound weighs 6 lb in water; required the specific gravity of the elm ?
Ans. 600. 328. CASE
328. CASE III. For a fluid of any sort.-Take a piece of a body of known specific gravity; weigh it both in and out of the fluid, finding the loss of weight by taking the difference of the two; then say,
As the whole or absolute weight,
So is the specific gravity of the solid,
EXAMPLE. A piece of cast iron weighed 35 ounces in a fluid, and 40 ounces out of it; of what specific gravity is that fluid?
329. To find the Quantities of Two Ingredients in a Given
TAKE the three differences of every pair of the three specific gravities, namely, the specific gravities of the compound and each ingredient; and multiply each specific gravity by the difference of the other two. Then say, by proportion,
As the greatest product,
Is to the whole weight of the compound,
To the weights of the two ingredients.
That is, H =
.c≈ the one, and L = the other, by cor. 6, prop. 64.
(s - f) s
EXAMPLE. A composition of 112 lb being made of tin and copper, whose specific gravity is found to be 8784; required the quantity of each ingredient, the specific gravity of tin being 7320, and that of copper 9000?
Answer, there is 100 lb of copper, in the composition, and consequently 12 lb of tin,
330. The specific gravities of several sorts of matter, as found from experiments, are expressed by the numbers annexed to their names in the following Table:
331. Note. The several sorts of wood are supposed to be dry. Also, as a cubic foot of water weighs just 1000 ounces avoirdupois, the numbers in this table express, not only the specific gravities of the several bodies, but also the weight of a cubic foot of each, in avoirdupois ounces; and therefore, by proportion, the weight of any other quantity, or the quantity of any other weight, may be known, as in the next two propositions.
332. To find the Magnitude of any Body, from its Weight.
As the tabular specific gravity of the body,
Is to its weight in avoirdupois ounces,
So is one cubic foot, or 1728 cubic inches,
Example 1. Required the content of an irregular block of
common stone, which weighs I cwt, or 112 lb?
Ans. 12281 cubic inches.
Example 2. How many cubic inches of gunpowder are
there in 1 lb. weight?
Ans. 29 cubic inches nearly.
Example 3. How many cubic feet are there in a ton weight Ans. 38 cubic feet.
of dry oak?
333. To find the Weight of a Body from its Magnitude.
As one cubic foot, or 1728 cubic inches,
Is to the content of the body,
So is the tabular specific gravity,
To the weight of the body.
Example 1. Required the weight of a block of marble, whose length is 63 feet, and breadth and thickness each 12 feet; being the dimensions of one of the stones in the walls of Balbeck?
Ans. 683 ton, which is nearly equal to the burden of an East-India ship.
Example 2. What is the weight of 1 pint, ale measure, of gunpowder? Ans. 19 oz. nearly. Example 3. What is the weight of a block of dry oak, which measures 10 feet in length, 3 feet broad, and 24 feet deep or thick? Ans. 4335
334. HYDRAULICS is the science which treats of the motion of fluids, and the forces with which they act upon bodies.
335. If a Fluid Run through a Canal or River, or Pipe of various Widths, always filling it; the Velocity of the Fluid in different Parts of it AB, CD, will be reciprocally as the Transverse Sections in those Parts.
THAT is, veloc. at A : veloc. at CCDAB; where AB and CD denote, not the diameters at A and B, but the areas or sections there.
For, as the channel is always equally full, the quantity of water running through AB is equal to the quantity running through CD, in the same time; that is, the column through