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Also 322 v 16:v, height fallen to gain the velocity v

:

256

256

21c

256

And 322: (5)2:: 16: (1), ditto for the vel. 21.

256

21c

Then (2) ×

256

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vv

is the measure of the fall required.

Or [1-1]X is a rule for computing the fall.

21c

64

1.4262-c2

Or rather

64c2

Xv very nearly, for the fall.

EXAM. 1. For London-bridge.

By the observations made by Mr. Labelye in 1746, The breadth of the Thames at London-bridge is 926 feet; The sum of the waterways at the time of low-water is 236 ft. ; Mean velocity of the stream just above bridge is 3 ft. per. sec. But under almost all the arches are driven into the bed great numbers of what are called dripshot piles, to prevent the bed from being washed away by the fall. These dripshot piles still further contract the waterways, at least of their measured breadth, or near 39 feet in the whole; so that the waterway will be reduced to 197 feet, or in round numbers suppose 200 feet.

Then b = 926, c = 200, v = 3 19

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Theref. 46 × 10 = 4·683 ft. 4 ft. 8 in. the fall required. By the most exact observations made about the year 1736, the measure of the fall was 4 feet 9 inches.

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Though the breadth of the river at Westminster-bridge is 1220 feet; yet at the time of the greatest fall, there is water through only the 13 large arches, which amount to but 820 feet; to which adding the breadth of the 12 intermediate piers, equal to 174 feet, gives 994 for the breadth of the river at that time; and the velocity of the water a little above the bridge, from many experiments, is not more than 21 ft. per second.

Here then b = 994, c = 820, v = 21 = 1,

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Theref. 01722 × 5·0872 ft. = 1 in. the fall required; which is about half an inch more than the greatest fall observed by Mr. Labelye.

And, for Blackfriars-bridge, the fall will be much the same as that of Westminster, or rather less.

See farther on this subject, Gregory's Mathematics for Practical Men, p. 308, &c.

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APPENDIX.

A dissertation on the times of exhausting vessels of a fluid, through holes or apertures in their bottoms, parallel to the horizon.

THEOREM I.

=

Supposing ABCD to be any vessel containing a fluid; and putting m = 323 feet 386 inches, n the area of the aperture in the bottom vc, x = EF the altitude of the surface of the fluid above the bottom, and z = the area of the des. cending surface GH; then, the time of exhausting the fluid CDGH will be equal to the fluent of

zi

n√mx

Demonstration. By Sir I. Newton's Principia, lib. ii. prop. 36, the velocity of the issuing fluid at E, is equal to that which is acquired by a body in falling through Er

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or ; but the velocities of falling bodies are as the roots of the spaces fallen, and m is the velocity (per second) acquired by falling through the space m; hence m:x::m:

mx the velocity of the issuing fluid at E; but the velocities at the orifice and of the descending surface GH will be inversely as their areas (for since the quantity or solidity is the same, the velocities, or altitudes of the equal column will be inversely as the sections); therefore zn:: /mx: n✓mx = the velocity (per second) of the descending surface

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Gн, or the space through which GH would uniformly descend in one second with the velocity it hath at the altitude r. Now in descending the space i, the velocity may be considered as uniform; and uniform descents are as their times, wherefore zi the time of descending n ✓mr

n✓mx

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::: 1 second :

space, or the fluxion of the time of exhausting.

Scholium.

Q. E. D.

In the above investigation, the resistance made by the air to the issuing fluid, is neglected, the exhaustion being supposed to be made in a vacuum; which will occasion a small difference (though scarce perceptible) between the calculations and experiments made in the air. But this consideration must by no means be neglected, when the density of the medium, into which the fluid issues, bears any considerable proportion to that of the fluid.

It will make no difference in the account, in whatever part of the base the aperture is placed, the altitude of the surface above it being the only consideration; nor is it material what the figure of it is, whether circular, triangular, square, &c. regular, or irregular, the area of it alone being the only necessary consideration.

As we have above found the fluxion of the time, so by taking the fluent of it we obtain the time itself, substituting first the value of z instead of it, as found in terms of x from the equation of the figure of the vessel, as in the following problems, except when it is a prism, for then z is constant, and does not affect the fluent.

PROBLEM I.

To find the time of emptying a prism.

When the vessel is a cylinder or any other prism, then the section GH or z is constant, and considering it as such, 2z I

the fluent of

is ✔, which is the time of exhaust

zi n√mx n

m

ing a prism whose base is z, and altitude x.

Cor. 1. If the apertures of two prisms therefore be equal, the times of emptying them, will be as their bases drawn into the square roots of their altitudes. If their bases bè also equal, the times will be as the square roots of their heights. But when their altitudes are equal, the times will be as their bases.

Cor. 2. Putting a for the altitude EI of the prisms ABCD, and b for the area of the base CD; then the time of emptying

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the part ABHG at E or D is X

n √m
26

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blem, the time of emptying ABCD is

above.

Cor. 3. By the problem, the time of exhausting ABHG at

G or F is

26 a

n

m

; comparing this therefore with the last

corollary, it appears that the times of emptying the same part ABHG, at the depth D and G, will be respectively as √a — √x to (a-x).

Cor. 4. By writing, in the general theorem (

for 2, a for X, and a- for i, we obtain

n/ma

zi

b

n√mx

> b for the

time in which AB would descend to GH with the first velocity, or the time in which a quantity equal to ABHg would run out supposing the vessel to be kept always full by a supply from without; comparing this therefore with corol. lary 2, it will appear that the time of emptying the part ABHG at D, when there is no supply from without, is to the time in which an equal quantity would run out when the vessel is kept always full, as a- √x to or as 2 x (a-ax) to a-x. And when x is nothing, or when ABHG becomes the whole prism, the proportion is that of 2a to a or 2 to 1.

PROBLEM II.

a-x
2

To determine the time of exhausting any pyramidal vessel.

Putting, as before, a = the altitude E1, b = the base or area of the bottom DC, t = the area of the top AB; and n = area of the aperture, m = 321 feet; also any other altitude EF, and z the area of the section GH.

Then, by the nature of the figure, whether the greater or less end is uppermost, we shall have as IE: EF:: the dif ference between the lines AB, CD, to the difference between the lines GH, CD; that is a :x:: √b - √t: √b —√2 = Xx; hence z = √b. Xx, and z=b

a

a

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