the contracted stream just below. The quantity of this rise, and of the consequent velocity below, keep both gradually increasing, as the tide continues ebbing, till at quite ow water, when the stream or natural current being the quickest, the fall under the arches is the greatest And it is the quantity of this fall which it is the object of this problem to determine. Now, the motion of free running water is the consequence of, and produced by the force of gravity, as well as that of any other falling body. Hence the height due to the velocity, that is, the height to be freely fallen by any body to acquire the observed velocity of the natural stream, in the river a little way above bridge, becomes known. From the same velocity also will be found that of the increased current in the narrowed way of the arches, by taking it in the reciprocal proportion of the breadth of the river above, to the contracted way in the arches; viz. by saying, as the latter is to the former, so is the first velocity, or slower motion, to the quicker. Next, from this last velocity, will be found the height due to 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 6 the breadth of the channel in feet; v = mean velocity of the water in feet per second ; breadth of the waterway between the obstacles. Now 25 21: :c: 21 25 256 And Fc: b::v: v, the velocity in the contracted way. 21c Also 322 : v2 :: 16:2, height fallen to gain the velocity v. 256 And 32:(v)2:: 16: ( 2 X 21c 256. 21c 250 ditto for the vel. v. 21c 256 Then () x 21c V2 v2 is the measure of the fall required. 256 Or[1]xis a rule for computing the fall. X v2 very nearly, for the fall. VOL. II. Bbb b EXAM 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 6926, c = 200, v = 31. 1.4262-c2 1217616-40000 64c2 Hence 64×40000 62=1038. ='46. Theref. 46 X 10-473ft. 4ft. 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 21 ft. per second. Here then b = 994, c = 820, v=21 = 2. 14262-0 1403011-672400 Hence 64c2 812 16 = '01722. 64 X 672400 And v2 = = 576. Theref. 01722 x 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 Blackfriar's-bridge, the fall will be much the same as that of Westminster. ADDITIONS, ADDITIONS, BY THE EDITOR, R. ADRAIN. New method of determining the Angle contained by the chords of two sides of a Spherical Triangle. See prob. v. page 77, vol. 2 THEOREM. 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. DEMONSTRATION. Let the two sides AB, AC of the spherical triangle ABC be produced till they meet in G, and let the supplements BG, CG, be bisected in D and E, also let the chords amb, Anc of the arcs AB, AC be drawn ; and the great circular arc DE will be the measure of the rectilineal angle contained by the chords AmB, anc. 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 Gc, therefore the angle at the centre GFE is equal to the angle at the circumference GAnc(theo. 49, Geom.), and therefore Anc and FE, 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 Amb and anç. A. E. D. New New method of determining the oscillations of a Variable Pendulum 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. PROBLEM. 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 vertical. SOLUTION. 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 вmon 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 r, t and T be the times of the descent of m through the vertical spaces AB, AC, and BC; g = 32 feet the measure of accelerative gravity; f = the measure of the retarding force which CHm D 'E the tension of the thread exerts on m in the direction ma, and c = the indefinitely small horizontal velocity of m at B.. fx Asr:x::f; fr fx = the vertical action of the tension on mi the true accelerative force with whichm is and theref g- fy Again, ry::f: ~ : : : : r 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, are g fx and: nd, the former of these forces tending to increase, 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 But by hypothesis, the angle mac is indefinitely small, we have 2m'g = a given quantity; our first fluxional equation therefore becomes =g-f, of which the proper fluent is x = (g-ft2; and by substituting for the value just found, our second fluxional equation Now when is less than 1, let g = √ ɲ, and in this case the correct fluent of the equation+y=0, is easily found from which equation it is manifest that as increases y also increases, so that m never returns to the vertical, and there are no vibrations. Again, when p the correct fluent of the t So that in this case also, when increases y increases, and the body m never returns to the vertical. Since in this case ɲ = 4m ==, therefore 17m' = m, 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'. m-m But |