H = weight of the heavier body in air, h = weight of the same in water, S its specific gravity; L = weight of the lighter body in air,s its specific gravity; 7 weight of the same in water, C = weight of the compound in air, c = weight of the same in water, 1st, (Hh) S = Hw, 4th, 5th, H h + l 6th, S specific gravity of water. Then its specific gravity But if the body L be lighter than water; then I will be negative, and we must divide by L+ l instead of L—l, and to find / we must have recourse to the compound mass C; and because, from the 4th and 5th equations, L -1 = C CHh, therefore s = that is, divide the absolute weight of the light body, by the difference between the losses in water, of the compound and heavier body, and multiply by the specific gra vity of water. Or thus, s = SSL as found from the last equation. CS. 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. PROP. VI. To find the specific gravity of a body. 8. CASE L.-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 Bw water. Then, by corol. 6, prop. 4. 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, So is the specific gravity of water, 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. 9. 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 con pound 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, As the last remainder, Is to the weight of the light body in air, So is the specific gravity in water, To the specific gravity of the body. Example. Suppose a piece of elm weighs 15 lb. in air; and that a piece of copper, which weighs 18 lb. 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. 10. 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, Is to the loss of weight, So is the specific gravity of the solid, To the specific gravity of the fluid. That is, the specific gravity w = B-b s, by cor. 6, prop. 4. Example. A piece of cast iron weighed 34 ounces in a fluid, and 40 ounces out of it; of what specific gravity is that fluid? Ans. 1000. PROP. VII. 11. To find the quantities of two ingredients in a given compound. 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, As the greatest product, Is to the whole weight of the compound, That is, the one H = by cor. 6. prop. 4. C; and the other L= (S-) 8 C, (S-8) 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, 12. SCHOLIUM.-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 : 13. 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. 550 Charcoal 240 13 PROP. VIII. 14. 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, To its content in feet, or inches, respectively. Example 1. Required the content of an irregular block of common stone, which weighs 1 cwt. or 112 lb.? Ans. 1228 cubic inches. Example 2. How many cubic inches of gunpowder are there in 1 lb. weight? Ans. 30 cubic inches nearly. Example 3. How many cubic feet are there in a ton weight of dry oak ? Ans. 381 cubic feet. 15. 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 its 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 the breadth and thickness each 12 feet; being the dimensions of one of the stones in the walls of Balbec? Ans. 683 ton, which is nearly equal to the burthen of an East 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 2 feet deep? Ans. 433518 lb. OF HYDRAULICS. 16. HYDRAULICS is the science which treats of the motion of fluids, and the forces with which they act upon bodies. PROP. X. 17. 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. 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 AB is equal to the column through CD, in the same time; or AB × length of its column = CD x length of its column; therefore AB: CD :: length of column through CD: length of column through AB. But the uniform velocity of the water, is as the space run over, or length of the columns; therefore AB: CD: velocity through CD: velocity through AB. Corol. Hence, by observing the velocity at any place AB, the quantity of water discharged in a second, or any other time, will be found, namely, by multiplying the section AB by the velocity there. But if the channel be not a close pipe or tunnel, kept always full, but an open canal or river; then the velocity in all parts of the section will not be the same, because the velocity towards the bottom and sides will be diminished by the friction against the bed or channel; and therefore a medium among the three ought to be taken. So, If the velocity at the top be That at the bottom And that at the sides Dividing their sum by 3 gives 60 50 3) 210 sum; 70 the mean velocity, which is to be multiplied by the section, to give the quantity discharged in a minute. PROP. X1. 18. The velocity with which a fluid runs out by a hole in the bottom or side of a vessel, kept always full, is equal to that which is generated by gravity through the height of the water above the hole; that is, the velocity of a heavy body acquired by falling freely through the height AB. DIVIDE the altitude AB into a great number of very small parts, each being 1, their number a, or a = the altitude AB. B Now, by prop. III. the pressure of the fluid against the hole B, by which the motion is generated, is equal to the weight of the column of fluid above it, that is the column whose height is AB or a, and base the area of the hole B. Therefore the pressure on the hole, or small part of the fluid 1, is to its weight, or the natural force of gravity, as a, to 1. But since the velocities generated in the same body in any time, are as those forces; and because gravity generates the velocity 2 in descending through the small space 1, therefore 1:a :: 2 : 2a, the velocity generated by the pressure of the column of fluid in the same time. But 2a, is also, (formerly shown,) the velocity generated by gravity in descending through a or AB. That is, the velocity of the issuing water, is equal to that which is acquired by a body in falling through the height AB. Corol. 1. The velocity, and quantity run out, at different depths, are as the square roots of the depths. For the velocity acquired in falling through AB, is as AB. Corol. 2. The water spouts out with the same velocity, whether it be downwards or upwards, or sideways; because the pressure of fluids is the same in all directions, at the same depth. And therefore, if the adjutage be turned upwards, the jet will ascend to the height of the surface of the water in the vessel. And this is confirmed by experience, by which it is found that jets really ascend nearly to the height of the reservoir, abating a small quantity only, for the friction against the sides, and some resistance from the oblique motion of the water in the hole. Corol. 3. The quantity run out in any time, is equal to a column or prism, whose base is the area of the hole, and its length the space described in |