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it as before, that is, the height to be freely fallen through by gravity, to produce it. Then the difference of these two heights, thus freely fallen by gravity, to produce the two velocities, is the required quantity of the waterfall in the arches; allowing however, in the calculation for the contraction, in the narrowed passage, at the rate as observed by Sir I. Newton, in prop. 36 of the 2d book of the Principia, or by other authors, being nearly in the ratio of 25 to 21. Such then are the elements and principles on which the solution of the problem is easily made out as follows.

Let b the breadth of the channel in feet;

v = mean velocity of the water in feet per second; c = breadth of the waterway between the obstacles.


Now 25:21::c: c, the waterway contracted as above.




-c:b::v: 21c", the velocity in the contracted way.

Also 322 v2 :: 16:2, height fallen to gain the velocity v.





256 21c

And 322 : (210) 3 v)2 :: 16: 16: ()×402, ditto for the vel. vel.v.

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is the measure of the fall required.

Or [(21) -14 is a rule for computing the fall.

Or rather


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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 31 feet per second. 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 ≈ 31:


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Theref. 46 X 103 4.73 ft. 83 in. the fall required. By the most exact observations made about the year 1736, the measure of the fall was 4 feet 9 inches.

EXAM. 2. For Westminster-bridge.

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 24 feet per second.

Here then b = 994, c = 820, v = 24 1.4262-c2 1403011-672400


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And v2

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Theref. 01722 × 51


·0872 ft. 1 in. the fall requir

ed; which is about half an inch more than the greatest fall

observed by Mr. Labelye.

And, for Blackfriar's-bridge, the fall will be much the same as that of Westminster.


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New Method of determining the Angle contained by the chords of two sides of a Spherical Triangle.


See prob. v. page 77, vol. 2.

If any two sides of a Spherical Triangle be produced till the continuation of each side be half the supplement of that side, the arc of a great Circle joining the extremities of the sides thus produced will be the measure of the angle contained by the chords of those two sides.

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Let the diameter AG be the common section of the planes of abg, acg, and F the centre of the sphere, from which draw the straight lines FD, FE.

Since, by hypothesis, GE is the half of ac, therefore the angle at the centre GFE is equal to the angle at the circumference Ganc (theo. 49. Geom.) and therefore anc and F¤, being in the same plane, are parallel: in like manner, it is shown that FD and Amв are parallel, and therefore the rectilineal angles BAC and DFE, are equal, and consequently, since DE is the measure of the angle DFE, it is also the measure of the angle contained by the chords Amв and anc.

Q. E. D.

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New method of determining the Oscillations of a Variable

The principles adopted by Dr. Hutton in the solution of his 45th problem, page 537, vol. 2, are, in my opinion, erroneous. He supposes the number of vibrations made in a given particle of time to depend on the length of the pendulum only. without considering the accelerative tension of the thread; so that by his formula we have a finite number of vibrations performed in a finite time by the descending weight, even when the ascending weight is infinitely small or nothing. Besides, the stating by which he finds the fluxion of the number of vibrations, is referred to no geometrical or mechanical principle, and appears to be nothing but a mere hypothesis. The following is a specimen of the method by which such problems may be solved according to acknowledged principles.


If two unequal weights m and m' connected by a thread passing freely over a pulley, are suspended vertically, and exposed to the action of common gravity, it is required to investigate the number of vibrations made in a given time by the greater weight m, supposing it to descend from the point of suspension, and to make indefinitely small removals from the verti




Let the summit A of a vertical ABCDE be the
point from which m descends, в any point in AE
taken as the beginning of the plane curve BmDN
described by m, which is connected with m' by
the thread am. Let me be at right angles to ae,
and put ac=x, cm=y, am=r; also let , t and r
be the times of the descent of m through the ver-
tical spaces AB, AC, and вc; g=321 feet, = the
measure of accelerative gravity; f the mea-
sure of the retarding force which the tension of
the thread exerts on m in the direction ma, and c =
definitely small horizontal velocity of m at B.

Asr :x ::


the in

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and theref. g
m is urged in a vertical direction.

Again, r : y : : ƒ :


the horizontal action on m produced


by the tension of the thread am. Thus the whole accelerative forces by which m is urged in directions parallel to x and y, fy, the former of these forces tending to in


are g

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crease x, and the latter to diminish y; and therefore by the general and well known theorem of variable motions (See Mec. Cel. B. 1, Chap. 2), we have the two equations

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У fy

But by hypothesis, the angle mac is indefinitely small, we have

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therefore=1, and ƒ= = a given quantity; our first

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of which the proper fluent is x1(g-ƒ) t2: and by substituting for x the value just found, our second fluxional equation becomes

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Now when p is less than 1, let q=√✓1-p, and in this case the

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from which equation it is manifest that as t increases y also increases, so that m never returns to the vertical, and there are no vibrations. Again, when p4, the correct fluent of the same fluxional equation is

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So that in this case also, when t increases y increases, and the body m never returns to the vertical. Since in this case p 1, therefore 17mm, and therefore by this case



and the preceding, there are no vibrations performed by the descending weight m when it is equal to or greater than 17 times the ascending weight m'.

But when pis greater than, put n = √p — 1, and in case the correct equation of the fluents is

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