the third 5; then the pressure upon the first and second taken together is by addition 4: upon the first, second, and third, it must be 9; and upon the first, second, third, and fourth, it will be 16; but 4, 9, 16, are the squares of 2, 3, 4. Emma. And the pressure upon the whole side a b c d must be 36 times greater than that upon the small part a 1 b 7.

Charles. And if there are three vessels, for instance, whose depths are as 1, 2, and 3, the pressure against the side of the second will be four times greater than that against the first; and the pressure against the side of the third will be nine times greater than that against the first.

Father. You are right: the beautiful simplicity of the rule, and its being the same by which the accelerating velocity of falling bodies is governed, will make it impossible that you should hereafter forget it.

The use that I shall hereafter call you to make of the rule, induces me to put a question to Emma.

In two canals, one five feet deep, and the other 15, what difference of pressure will there be against the sides of these canals?

Emma. The pressure against the one will be as the square of 5, or 25; that against the other will be as the square of 15, or 225; now the latter number divided by the former gives 9 as a quotient, which shows that the pressure against

VOL. II.-E

the sides of the deep canal is 9 times greater than that against the sides of the shallow one. Can this principle be proved by an experiment?

Father. By a very simple one: Plate II. Fig. 16. is a vessel of the same size as the last, the bottom and the side bare wood mortised together; the front and opposite side are glass carefully inserted in the wooden parts, and made water-tight. A thin board c hangs by two hinges xy, and is held close to the glass panes by means of the pulley and weight w. The board is covered with cloth and made watertight.

Now observe the exact weight which is overcome when the water is poured in and rises to the line 1; then hang on four times that weight, and you will see that water may be poured into the vessel till it rise to the line 2, when the side c will give way and let part of it out.

Emma. But why does only a part run away? Father. Because when a small quantity of the water has escaped, the weight w is greater than the pressure of the water against c, therefore the door c will be drawn close to the glass panes, and confine the rest within the vessels.

You may now hang on a weight nine times greater than the first, and then the vessel will contain water till it rise up to the mark 3, when the side will give way by the pressure, and part of the water escape.

Charles. You have explained the manner of estimating the pressure of fluids against the sides of a vessel; by what rule are we to find the pressure upon the bottom?

་

Father. In such vessels as those which we have just described: that is, where the sides. are perpendicular to the bottom, and the bottom parallel to the horizon, the pressure will be equal to the weight of the fluids.

Emma. If then the vessel y z hold a gallon of water, which weighs about eight pounds, and if the bottom were made moveable like the side, would a weight of eight pounds keep the water in the vessel ?

Father. It would: for then there would be an equilibrium between the pressure of the water and the weight. And the pressure upon any one side is equal to half the pressure upon the bottom: that is, provided the bottom and sides are equal to one another.

out.

Charles. Pray, Sir, explain how this is made

Father. The pressure upon the bottom is, as we have shown, equal to the weight of the fluid. But we have also shown that the pressure on the side grows less and less continually, till at the surface it is nothing. Since then the pressure upon the bottom is truly represented by the area of the base multiplied into the altitude of the vessel; the pressure upon the side will be

represented by the base multiplied into half the altitude.

:

Emma. Is the pressure upon the four sides equal to twice the pressure upon the bottom? Father. It is consequently the pressure of any fluid upon the bottom and four sides of a cubical vessel is equal to three times the weight of the fluid.

Can you, Charles, tell me the difference between the weight and the pressure of a conical vessel of water standing on its base?

Charles. The weight of a conical vessel of any fluid is found by multiplying the area of the base by one third part of its perpendicular height; but the pressure is found by multiplying the base by the whole perpendicular height; therefore the pressure upon the base will be equal to three times the weight.

NOTE. The rule for finding the solidity of a cone or a pyramid is this, "Multiply the area of the base by one third of the height, and the product will be the solidity."

See Bonnycastle's Introduction to Mensuration, &c. p. 132. Second Edition.

CONVERSATION VIII.

Of the Motion of Fluids.

Father. We will now consider the pressure of fluids with regard to the motion of them through spouting-pipes, which is subject to the same law.

If the pipes at 1 and 4 (Fig. 15.) be equal in size and length, the discharge of water by the pipe at 4 will be double that at 1. Because the velocity with which water spouts out at a hole in the side or bottom of a vessel, is as the square-root of the distance of the hole below the surface of the water.

Emma. What do you mean by the squareroot?

Father. The square-root of any number is that which being multiplied into itself produces the said number. Thus the square-root of 1 is 1; but of 4 it is 2; of 9 it is 3; and of 16 it is 4, and so on.

Charles. Then if you had a tall vessel of water with a cock inserted within a foot of the top, and you wished to draw the liquor off three times faster than it could be done with that, what would you do?

Father. I might take another cock of the