Here the gate is in form of a trapezoid, having the two parallel sides AB, CD, viz, AB 24, and CD 16, and depth 6 feet. Now, by mensuration, problem 3 vol. 2, (AB + CD) x 620 x 6120 the area of the sluice, the same as before in the Ist example: but the centre of gravity cannot be so low down as before, because the figure is wider above and narrower below, the whole depth being the same. Now, to determine the centre of gravity of the trapezoid AD, produce the two AIE sides AC, BD, till they meet in G; also draw GKE and CH perp. to AB: then AH: CH:: AE: GE, that is, 4:6:: 12: 18 GE; and EF being 6, theref. FG 12. Now, by Statics art. 229 vol. 2, EF = 6EG gives F the centre of gravity of the triangle ABG, and FI4 FG gives i the centre of gravity of the triangle CDG. Then assuming K to denote the centre of AD, it will be, by art. 212 vol. 2, as the trap. AD: A CDG: IF: FK, or A ABC A CDG: A CDG :: JF FK, or by theor. 88 Geom. GE - GF': GF2:: IF: FK, that is 182 122 to 122 or 32-22 to 22 or 5: 4 :: IF = 4: I 4 3 FK; and hence EK = 63324 = is the distance of the centre K below the surface of the water. This drawn into 120 the area of the dam-gate, gives 336 cubic feet of water the pressure, 336000 ounces = 21000 pounds 9 tons 80 lb, the quantity of pressure against the gate, as required, being a 15th part less than in the first case. ' Ex. 8. Find the quantity of pressure against a dam or sluice, across a canal, which is 20 feet wide at top, 14 at bottom, and 8 feet depth of water? PROBLEM VI. To determine the Strongest Angle of Position of a Pair of Gates for the Lock on a Canal or River. Let AC, BC be the two gates, meeting in the angle c, projecting out against the pressure of the water, AB being the breadth of the canal or river. Now the pressure of the water on a gate AC, is as the quantity, or as the E D extent or length of it, Ac. And the mechanical effect of that pressure, is as the length of lever to the middle of Ac, or as AC itself. On both these accounts then the pressure is as 4C AC. Therefore the resistance or the strength of the gate must be as the reciprocal of this Ac... Now produce AC to meet BD, perp. to it, in D; and draw CE to bisect AB perpendicularly in E; then, by similar triangles, as AC: AE:: AB: AD; where, AE and AB being given lengths, AD is reciprocally as AC, or AD2 reciprocally as Ac2; that is, AD2 is as the resistance of the gate AC. But the resistance of AC is increased by the pressure of the other gate in the direction BC. Now the force in BC is resolved into the two BD, DC; the latter of which, DC, being parallel to AC, has no effect upon it; but the former, BD, acts perpendicularly on it. Therefore the whole effective strength or resistance of the gate is as the product AD X BD. If now there be put AB = a, and BD = x, then AD2 — AB2 -BD2 = α2x2; conseq. ADX BD = (a2x2) x x =α2x − x3 for the resistance of either gate. And, if we would have this to be the greatest, or the resistance a maximum, its fluxion must vanish, or be equal to nothing: that is, a2x-3x2±0; hence a2 = 3x2, and x = a√√ ÷ = }a√3 = •57735a, the natural sine of 35° 16': that is, the strongest position for the lock gates, is when they make the angle A or B 35° 16′, or the complemental angle ACE or BCE 54° 44', or the whole salient angle ACB = 109° 28'. Scholium. the Allied to this problem, are several other cases in mechanics: such as, the action of the water on the rudder of a ship, in sailing, to turn the ship about, to alter her course; and f action of the wind on a ship's sails, to impel her forward; also the action of water on the wheels of water-mills, and of the air on the sails of wind-mills, to cause them to turn round. Thus, for instance, let ABC be the rudder of a ship ABDE, sailing in the direction BD, the rudder placed in the oblique position BC, and consequently striking the water in the A direction CF, parallel to BD. Draw BF perp. to BC, and BG perp. to CF. Then the sine of the angle of incidence, of the direction of the stroke of the rudder against the water, willbe BF, to the radius CF; therefore the force of the water against the rudder will be as BF, by art. 3. pa. 366 vol. 2. But the force BF resolves into the two BG, GF, of which the an datter is parallel to the ship's motion, and therefore has no effect effect to change it; but the former BG, being perp. to the ship's motion, is the only part of the force to turn the ship about and change her course. But BF BG :: CF : CB, therethe force upon the rudder to fore CF CB; BF2: turn the ship about. Now BC. BY2 73 CF = CF, x = BC; then BF2=a2 CF and a; and, to have this a maxi a mum, its flux. must be made to vanish, that is, ax-3x2x=0; and hence ra1 = BC= the natural sine of 35° 16′ = angle F; therefore the complemental angle c 54° 44' as before, for the obliquity of the rudder, when it is most efficacious. The case will be also the same with respect to the wind acting on the sails of a wind-mill, or of a ship, viz, that the sails must be set so as to make an angle of 54° 44′ with the direction of the wind; at least at the beginning of the motion, or nearly so when the velocity of the sail is but small in comparison with that of the wind; but when the former is pretty considerable in respect of the latter, then the angle ought to be proportionally greater, to have the best effect, as shown in Maclaurin's Fluxions, pa. 734, &c. A consideration somewhat related to the same also, is the greatest effect produced on a mill-wheel, by a stream of water striking upon its sails or float-boards. The proper way in this case seems to be, to consider the whole of the water as acting on the wheel, but striking it only with the relative velocity, or the velocity with which the water overtakes and strikes upon the wheel in motion, or the difference between the velocities of the wheel and the stream. This then is the power or force of the water; which multiplied by the velocity of the wheel, the product of the two, viz, of the relative velocity and the absolute velocity of the wheel, that is (vv)v= v2, will be the effect of the wheel; where v denotes the given velocity of the water, and the required velocity of the wheel. Now, to make the effect vo- a maximum, γυ of the or the greatest, its fluxion must vanish, that is vu-2vv=0, hence vV; or the velocity of the wheel will be equal to half the velocity of the stream, when the effect is the greatest; and this agrees best with experiments by mi 9jou A former way of resolving this problem was, to consider the water as striking the wheel with a force as the square of the relative velocity, and this multiplied by the velocity of the wheel, to give the effect; that is, (v) the effect. Now the flux of this, product is (y—v)u— (v—v)x2vv=0; hence ་ hence v qo #26, orey 130, and ov, or the velocity of the wheel equal only to of the velocity of the water.de 19d sgablo biE PROBLEM VII. H ༢༠ ། ཙྭ༦ དྷཱི'༩ To determine the Form and Dimensions of Gunpowder Magazines. In the practice of engineering, with respect to the erection of powder magazines, the exterior shape is usually made like the roof of a house, having two sloping sides, forming two inclined planes, to throw off the rain, and meeting in an angle or ridge at e top; while the interior represents a vault, more or less extended, as the occasion may require and the shape, or transverse section, in the form of some arch, both for strength and commodious room, for placing the powder barrels. It has been usual to make this interior curve a semicircle. But, against this shape, for such a purpose, I must enter my decided protest; as it is an is an arch the farthest of any from being in equilibrium in itself, and the weakest of any, by being unavoidably much thinner in one part than in others. Besides, it is constantly found, that after the centering of semicircular arches is struck, and removed, they settle at the crown, and rise up at the flanks, even with a straight horizontal form at top, and still much more so in powder magazines with a sloping roof; which effects are exactly what might be expected from a contemplation of the true theory of arches. Now this shrinking of the arches must be attended with other additional bad effects, by breaking the texture of the cement, after it has been in so some de gree dried, and also by opening the joints of the voussoirs at one end. Instead of the circular arch therefore, we this place give an investigation, founded on the true principles of equilibrium, of the only just form of the interior, which is properly adapted to the usual sloped roof. 30 For this purpose, put a = DK the thickness of the arch at the top, x = any absciss DP of the required arch ADCM, KR the corresponding L absciss of the given exterior line KI, and y — PC — RT their equal ordinates. Then by the principles of abov arches, in my tracts on that subject, at is found that cartoon that ยก P to 1 do pyr—xy Tot L name lași săi quaine 25 bow da u = QX ➡orax supposing a constant quantity, and where a is some certain quantity to be determined hereafter.) But KR or is ty,ift be put to denote the the tangent of the given angle of elevation KIR, to radius 1; and then the equation is w = a + x-ty Now, the fluxion of the equation w = a + x - ty, is w = » – tỷ, and the 2d fluxion is ; there fore the foregoing general equation L at D being parallel to KI; therefore the correct fluent is 1 a2 = = 24. Hence then j2 = 2002 w2 or j = 2012 ༧(u° — @༧)→ the correct fluent of which gives ya x hyp. log. of w+ √(w2-a2) a Now, to determine the value of a, we are to consider that when the vertical line cr is in the position AL or MN, then wcI becomes AL or MN the given quantity c suppose, and y = AQ or aм suppose, in which position the b c+ √(c2 ~ a2) last equation becomes ba x hyp. log. and hence it is found that the value of the constant quantity ✔a, is h. l.c+ √(C2=q2); which being substituted for it, in the above general value of y, that value becomes ს a ; from which equation the value of the ordinate PC may always be found, to every given value of the vertical cI. 1 But if, on the other hand, rc be given, to find cr, which will be the more convenient way, it may be found in the following manner: Put A = log. of a, and c = x log. of (a); ; then the above equation gives cy + a = log. a);. again, put n the number whose log. is cyA; then w√(w2 a2); and hence w Now, for an example in numbers, in a real case of this nature, |
