of a = a,

But the area

a2 = w, and the area

Now put the required thickness AG or EH = ', and the altitude AE or GH = a as before. And because in such cases the slope of the wall is usually made equal to of its altitude, that is FH = AE or a, the lever FM will be and the lever FN FH + EH = a + x. of GHF GH X HF = a × √α = AH = AE × AG=ax=w; these two drawn into the respective levers FM, FN, give the two momenta, aw Στα χ aa3, and (a + 4x) × ax = za2x + {ax*; theref. the sum of the two, ({ax2+}u2x+75a3)n must be =75a3m, or dividing by lan, x2 + }ax + za2 = ža2 × ing to both sides to complete the square, the equation becomes x2+zax +1⁄2za2±ža2. — + a2, the root of which

Z

m

m

m

=

-; now add

m

is x + ‡a = a√(13+), and hence x=a√(23+)− za.

9n

9n

Now, for a brick wall, m = n nearly, and then the breadth x = a√ (23 + 1) — zα = 1}α/34-ja·189a, or almost

for the thickness AG at top, in stone walls.

In the same manner we may proceed when the slope is supposed to be any other part of the altitude, instead of as used above. Or a general solution might be given, by assuming the thickness = part of the altitude.

1

REMARK.

Thus then we have given all the calculations that may be necessary in determining the thickness of a wall, proper to support the rampart or body of earth, in any work. If it should be objected, that our determination gives only such a thickness of wall, as makes it an exact mechanical balance to the pressure or push of the earth, instead of giving the former a decided preponderance over the latter, as a security against any failure or accidents. To this we answer, that what has been done is sufficient to insure stability, for the following reasons and circumstances. First, it is usual to build several counterforts of masonry, behind and against thè wall, at certain distances or intervals from one another; which contribute very much to strengthen the wall, and to resist the pressure of the rampart. 2dly. We have omitted to include the effect of the parapet raised above the wall; which must add somewhat, by its weight, to the force or resistance of the

wall.

wall. It is true we could have brought these two auxiliaries to exact calculation, as easily as we have done for the wall itself: but we have thought it as well to leave these two appendages, thrown in as indeterminate additions, above the exact balance of the wall as before determined, to give it an assured stability. Besides these advantages in the wall itself certain contrivances are also usually employed to diminish the pressure of the earth against it: such as land-ties and branches, laid in the earth, to diminish its force and push against the wall. For all these reasons then, we think the practice of making the wall of the thickness as assigned by our theory, may be safely depended on, and profitably adopted; as the additional circumstances, just mentioned, will sufficiently insure stability; and its expense will be less than is incurred by any former theory.

PROBLEM V.

To determine the Quantity of Pressure sustained by a Dam or Sluice, made to pen up a Body of Water.

By art. 313 Hydrostatics, vol. 2, 6th edit. the pressure of a fluid against any upright surface, as the gate of a sluice or canal, is equal to half the weight of a column of the fluid, whose base is equal to the surface pressed, and its altitude the same as that of the surface. Or, by art. 314 of the same, the pressure is equal to the weight of a column of the fluid, whose base is equal to the surface pressed, and its altitude equal to the depth of the centre of gravity below the top or surface of the water; which comes to the same thing as the former article, when the surface pressed is a rectangle, because its centre of gravity is at half the depth.

Ex. 1. Suppose the dam or sluice be a rectangle, whose length, or breadth of the canal, is 20 feet, and the depth of water 6 feet. Here 20 x 6 = 120 feet, is the area of the surface pressed; and the depth of the centre of gravity being 3 feet, viz, at the middle of the rectangle; therefore 120 x 3 360 cubic feet is the content of the column of water. But each cubic foot of water weighs 1000 ounces, or 624 pounds; therefore 360 x 1000 = 360000 ounces, or 22500 pounds, or 10 tons and 100 lb, is the weight of the column of water, or the quantity of pressure on the gate or dam.

Ex. 2. Suppose the breadth of a canal at the top, or surface of the water, to be 24 feet, but at the bottom only 16 feet, the depth of water being 6 feet, as in the last example: required the pressure on a gate which, standing across the canal, dams the water up?

Here

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 6 20 x 6 = 120 the area of the sluice, the same as before in the 1st 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 k of the trapezoid AD, produce the two

-

sides AC, BD, till they meet in G; also draw CKE 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 = 6 = EG gives F the centre of gravity of the triangle ABG, and F1 = 4 = 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 ▲ ABC A CDG: A CDG :: IF FK, or by theor. 88 Geom. GE2 - GF2: GF2:: IF : FK, that is 182 122 to 122 or 32-22 to 22 or 5: 4 :: IF = 4: 10 = 35 = FK; and hence EK = 6 24 = 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.

';

I 4

Ex. 3. 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 AC2.

1

Ac'. Therefore the resistance or the strength of the gate must be as the reciprocal of this Ac2.

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, pc, 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 AD2 × BD.

If now there be put AB = a, and BD = x, then AD2 AB - BD2 = q2-x; conseq. ADX BD = (a2x2)× x=a2x-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, ax-3x2x=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.

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 the 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, will be BF, to the radius CF; therefore the force of the water against the rudder will be as BF2, by art. 3 pa. 366 vol. 2. But the force BF resolves into the two BG, GF, of which the latter 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, therefore CF CB BF2: the force upon the rudder to

turn the ship about.

BC. BF2

CF

Now put a = CF, x = BC; then BF2 = α2 x2, and the a2x-13 x(a3—x2) — a3r—13; and, to have this a maxi

force

BC. BF2

CF

a

=

-;

=

mum, its flux. must be made to vanish, that is, a'x-3x2x=0; and hence ra√ 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 (v — v)v= Vu 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 vv-v2 a maximum, or the greatest, its fluxion must vanish, that is vi-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.

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, (vv) the effect. Now the flux. of this product is (v — v)'v - (v — v) × 2vv=0';

hence