eighth turn from the top; but the eighth turn from the top is the ninth turn from the roll, consequently we must add 16 times the thickness of the rope to the diameter of the roll. = 32.4632 = length of the last turn that 36-1284 length of the first turn that goes off. 68.5916 4 half the number of turns to meetings. 6)274-36640 = distance from the top of the pit to meetings in feet. 45-72773 that distance in fathoms, the same as above. 45.72773 Meetings will be 5.58306 fathoms below mid-shaft. PART III. HYDROSTATICS. HYDROSTATICS treats of the pressure, weight, and equilibrium of non-elastic fluids. If a fluid be at rest in a vessel, the base of which is parallel to the horizon, equal parts of the base are equally pressed by the fluid. For, upon every part of the base there is an equal column of the fluid supported; and as all the columns are of equal weight, they must press the base equally, or equal parts of the base will sustain an equal pressure. All the parts of the fluid press equally at the same depth. The pressure of a fluid at any depth is as the depth of the fluid. For the pressure is as the weight, and the weight is as the height of a column of the fluid. If a fluid is pressed by its own weight, or by any other force, at any point it presses equally in every direction whatever. This arises from the nature of fluidity, which is, to yield to any force in any direction. If it cannot give way to any force which is applied, it will press against other parts of the fluid in the direction of that force; and the pressure in all directions is the same; for if any one was less, the fluid would move that way until the pressure was the same every way. Therefore, in a vessel containing a fluid, the pressure is the same against the bottom as against the sides, or even upwards, at the same depth. The pressure of a fluid upon the base of the containing vessel, is as the base and perpendicular altitude, whatever may be the figure of the containing vessel. The pressure of a fluid against any upright surface, as the gate of a sluice, is equal to the area of that surface multiplied by half its depth. Ex.-If the gate of a sluice be 9 feet broad and 6 feet deep, what is the pressure of water against it? The pressure against the internal surface of a cubical vessel is three times the weight of the fluid contained in it; for the pressure against one side of the vessel is equal to half the pressure on the bottom. Ex.-What pressure does the internal surface of a cubical vessel sustain, each side of the cube being 4 feet? 4000, and 4000 × 3 = 12000. 45 x 621* 1 623 lbs. is the weight of a cubic foot of water. X ON THE HYDROSTATIC PARADOX. The hydrostatic paradox may be explained upon the same principles as the mechanical powers; and an explanation conducted in this manner strips it of its paradoxical appearance. The hydrostatic paradox is expressed thus:-A quantity of fluid, however small, may be made to balance a quantity, however large. Thus, in Fig. 1 to Hydrostatics, let A B be a large vessel, and C D a small one which is connected with it. Then, if water be poured into either of them, it will stand at the same height in both; consequently there is an equilibrium between them. Now, it must be observed that there is an equilibrium between them in the same manner as in any of the mechanical powers. Take the lever, for instance: suppose 1 lb. to balance 100 lbs. ; this is exactly the same as 1 lb. balancing 100 lbs. in the hydrostatic paradox. In the former case, the length of the arms of the lever must be in proportion to each other as 1 to 100; and, in the latter, the area of the vessels must be to each other as 1 to 100. But, properly speaking, one pound does not, nor cannot balance one hundred, in any case whatever; for one pound can only balance one pound, use what means you will. Archimedes only required a fixed point to be able to sustain the whole earth; but, as Carnot very justly observed, if he had found it, it would not, in fact, have been Archimedes, but the fixed point or fulcrum, which would have sustained the earth. In the hydrostatic paradox also, one pound, instead of balancing one hundred, only balances one pound. All the rest of the weight is supported by the vessel, in the same manner as the weight is supported on the fulcrum of the lever. And, to shew that the same takes place as in the lever, fix a piston which is water-tight into either of them, suppose the greater for instance; and if the area of the greater be to the area of the lesser as 100 to 1, then, by pressing the piston down 1 inch in the large vessel, it will raise the water 100 inches in the smaller one. Now, this is a property of the lever which is well known to every one, viz. that if the arms of the lever be as 1 to 100, then the long end of the lever will move through 100 inches whilst the short one moves through 1 inch. It is upon the principle of the hydrostatic paradox that Mr. Bramah applied the hydrostatic press. Suppose the diameter of the cylinder piston to be 20 inches, and that of the pump piston which is used for forcing the water into the cylinder of an inch; then, the area of the cylinder piston is 202 × 7854 314-16 square inches, and the area of the pump piston is (1) ×·7854 TX 78540491. Then, 314-16 = = 6400; that is, the area of the cylinder piston is to the area of the pump piston as 6400 to 1. Therefore, if any force be applied to the small piston, it will produce an effect on the large one as 1 to 6400, if there be water or any incompressible fluid whatever between them. Thus, suppose, by means of a lever, we can press down the small piston with a force of 10 cwt.; then the large piston will ascend with the force of 10 x 6400 cwt. = 3200 tons; and in the same manner the power may be calculated, whatever proportion the pistons may have to each other. ON SPECIFIC GRAVITY. The specific gravity of a body is the relation of its weight, compared with the weight of some other body of the same magnitude. A body immersed in a fluid will sink if its specific gravity be greater than that of the fluid; but if it be less, the |