zontal line BF. Then let one surface be pressed or pushed down by shaking, from в to c, and the other will ascend through the equal space FG; after which let them be permitted freely to return. The surfaces will then continually vibrate in equal times between AC and EG. The velocity and times of which oscillations are therefore required.

When the surfaces are any where out of a horizontal line, as at P and Q, the parts of the fluid in QDR, on each side, below QR, will balance each other; and the weight of the Part in PR, which is equal to 2PF, gives motion to the whole. So that the weight of the part 2PF is the motive force by which the whole fluid is urged, and therefore

a

wt. of 2PF

whole wt.

is the

accelerative force. Which weights being proportional to their lengths, if I be the length of the whole fluid, or axis of the tube filled, and a = FG or BC; then is the accelerative force. Putting theref. r = GP any variable distance, v the velocity, and t the time; then PF = a— X, and

Τ

4g

2a

2x

the accelerative force; hence vv or 2gfs = 1⁄21⁄2 (ax − xx);

the fluents of which give v2 =

2ax

4g

16 (2ax

x2), and v =

√(4g × 2a =2) is the general expression for the velocity

at any term. And when x = a, it becomes v2a√ for the greatest velocity at в and F.

√(2ax-x2)

a

Again, for the time, we have i or÷=√÷ the fluents of which give t = √ x arc to versed sine and radius 1, the general expression for the time. And when r = a, it becomes t = p for the time of moving

g

from G to F, p being = 3·1416; and consequently P√ g the time of a whole vibration from G to E, or from c to A. And which therefore is the same, whatever AB is, the whole length / remaining the same.

And the time of vibration is also equal to the time of the vibration of a pendulum whose length is 1, or half the length of the axis of the fluid. So that, if the length / be 784 inches, it will oscillate in 1 second.

Scholium. This reciprocation of the water in the canal, is nearly similar to the motion of the waves of the sea. For

the

the time of vibration is the same, however short the branches are, provided the whole length be the same. So that when the height is small, in proportion to the length of the canal, the motion is similar to that of a wave, from the top to the bottom or hollow, and from the bottom to the top of the next wave; being equal to two vibrations of the canal; the whole length of a wave, from top to top, being double the length of the canal. Hence the wave will move forward by a space nearly equal to its breadth, in the time of two vibrations of a pendulum whose length is (1) half the length of the canal, or one-fourth of the breadth of a wave, or in the time of one vibration of a pendulum whose length is the whole breadth of the wave, since the times of vibration are as the square roots of their lengths. Consequently, waves whose breadth is equal to 39 inches, or 335 feet, will move over 3 feet in a second, or 195 feet in a minute, or nearly 2 miles and a quarter in an hour. And the velocity of greater or less waves will be increased or diminished in the subduplicate ratio of their breadths.

Thus, for instance, for a wave of 18 inches breadth, as 39: 39: √18: √(394 × 18) = 3√313 = 26.5377 the velocity of the wave of 18 inches breadth.

PROBLEM 50.

To determine the Time of emptying any Ditch, or Inundation, &c, by a Cut or Notch, from the Top to the Bottom of it.

Let x

AB the variable height of water at
any time;

b = AC the breadth of the cut;

d =

A

the whole or first depth of water;
the area of the surface of the water
in the ditch;

g= 16 feet.

The velocity at any point D, is as BD, that is, as the ordinate DE of a parabola BEC, whose base is AC, and altitude AB Therefore the velocities at all the points in AB, are as all the ordinates of the parabola. Consequently the quantity of water running through the cut ABGC, in any time, is to the quantity which would run through an equal aperture placed all at the bottom in the same time, as the area of the parabola ABC, to the area of the parallelogram ABGC, that is, as 2 to 3.

But gr: 2g : 2gr the velocity at Ac; therefore × 2gxx br br gr is the quantity discharged per

second

second through ABGC; and consequently

4bxgx
3A

is the ve

locity per second of the descending surface. Hence then

: * :: 1" descending.

:

-3AX 4bx √gx

= the fluxion of the time of

Now when a the surface of the water is constant, or the ditch is equally broad throughout, the correct fluent of this fluxion gives t= √d-√ for the general time of

3A

268 √dx

sinking the surface to any depth x. And when x = 0, this expression is infinite; which shows that the time of a complete exhaustion is infinite.

=

But if d 9 feet, b = 2 feet, A 21 × 1000 = 21000, and it be required to exhaust the water down to of a foot deep; then, and the above expression becomes 3 x 21000 3-4

X = 14400", or just 4 hours for that time.

4 x 417 And if it be required to depress it 8 feet, or till 1 foot depth of water remain in the ditch, the time of sinking the water to that point will be 43′ 38′′.

Again, if the ditch be the same depth and length as before, but 20 feet broad at bottom, and 22 at top; then the descending surface will be a variable quantity, and, by prob. × 20000; hence in this case the

16 vol. 2, it will be

flux. of the time, or

90+x 90

correct fluent of which is t =

90-d

00-4) for

the time of sinking the water to any depth x. Now when x = 0, this expression for the complete exhaustion becomes infinite.

x = 1 foot, the time t is 42′ 56′′. And when foot, the time is 3h 50′ 28′′.

PROBLEM 51.

To determine the Time of filling the Ditches of a Fortifi cation 6 Feet deep with Water, through the Sluice of a Trunk of 3 Feet Square, the Bottom of which is level with the Bottom of the Ditch, and the Height of the supplying Water is 9 Feet above the Bottom of the Ditch.

Let ACDB represent the area of the vertical sluice, being a square of 9 square feet, and AB level with the bottom of the ditch. And suppose the ditch filled to any height AE, the surface being then at EF.

VOL. III.

BB

Put

Then g√x :: 2g : 2√√/gx the velocity with which the water presses through the part AEFB; and theref. 2/gx x AEFB = 2b√g.x(ax) is the quantity per second running through AEFB Also, the quantity running per second through ECDF is 2gr × ECDF = ybgx(ba + x) neatly. For the real quantity is, by proceeding as in the last prob. the difference between two parab. segs. the alt. of the one being x, its base b, and the alt. of the other a − b; and the medium of that dif. between its greatest state at AB, where it is AD, and its least state at CD, where it is 0, is nearly ED. Consequently the sum of the two, or žb√gx (a+116x) is the quantity per second running in by the whole sluice ACDB. Hence then b√gx ×

A

v is

the rate or velocity per second with which the water rises in

=

-6A

: = -

the ditches; and so v ; the fluxion of the time of filling to any height AE, putting c = a + lib.

Now when the ditches are of equal width throughout, a is a constant quantity, and in that case the correct Яuent of this fluxion is t

6A

bgc

√ √ thege

NC + Nx2

neral expression for the time of filling to any height AE, or ax, not exceeding the height AC of the sluice.

Again for filling to any height GH above the sluice, denoting as before a AG the height of the head above GH, 2 gr will be the velocity of the water through the whole sluice AD: and therefore 2b2gr the quantity per second,

and

vthe rise per second of the water in the ditches;

consequently v : – * : : 1′′ : i = − − general fluxion of the time; the correct fluent of which,

A

being 0 when x = a − b = d, is t = √(√d - √x) the time of filling from CD to GH.

Then the sum of the two times, namely, that of filling from AB to CD, and that of filling from CD to GH, is

6

++/log. ( +] for the √d

whole time required. And, using the numbers in the prob., A √6-√3 6

= 0·03577277A, the time in terms of A the area of the length and breadth, or horizontal section of the ditches. And if we suppose that area to be 200000 square feet, the time required will be 7154", or 1h 59 14".

And if the sides of the ditch slope a little, so as to be a little narrower at the bottom than at top, the process will be nearly the same, substituting for A its variable value, as in the preceding problem. And the time of filling will be very nearly the same as that above determined.

PROBLEM 52.

But if the Water, from which the Ditches are to be filled, be the Tide, which at Low Water is below the Bottom of the Trunk, and rises to 9 Feet above the Bottom of it by a regu lar Rise of One Foot in Half an Hour; it is required to ascertain the Time of Filling it to 6 Feet high, as before in the last Problem,

Let ACDB represent the sluice; and when the tide has risen to any height GH, below CD the top of the sluice, without the ditches, let FF be the mean height of the water within. And put b 3 = AB = AC;

Then

g= 16;

A = horizontal section of the ditches;
x = AG;

ZAE.

C

G

E

A

g:EG :: 25:2√/g(x-z) the velocity of the water through AEFB; and √g: √EG::g: g(x-x) the mean vel. through EGHF; theref. 2bxg(x-2) is the quantity per sec. through AEFB; and b(x-z)g(x-z) is the same through EGHF;

conseq. 3bgx (2x + z) √√ (x −z) is the whole through AGHB per second. This quantity divided by the surface A,

gives × (2x + z) √(x-2)=v the velocity per second

3A