AC2. 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, 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 AD2 X BD.

X

If now there be put AB = a, and BD = x, then AD = AB - BD2 = a2-x; conseq. ADX BD = (a2-x2) x 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, a2x-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 OF 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.

A

B

CG

D

E

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 direction cr, parallel to BD. Draw BF perp. to BC, and BG perp. to cr. 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

1

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:: BF:

turn the ship about.

BC. BF2

CF

the force upon the rudder to

Now put a = CF, x = BC; then BF2 = a2 x2, and the

mum, its flux. must be made to vanish, that is, a2x - 3x2x=0; and hence x = a = 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 v 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-20ὺ=0, hence v = v; 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, (v-v) = the effect. Now the flux. of this product is (v - v) -- (ν - υ) x 200 =0; hence v = = 20, or v = 3v, and v = v, or the velocity of the wheel equal only to of the velocity of the water.

PROBLEM VII.

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 the 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 a purose, I must enter my decided protest; as it 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 some degree dried, and also by opening the joints of the voussoirs at one end. Instead of the circular arch therefore, we shall in 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.

For this purpose, put a = DK the thickness of the arch at the top, x = any absciss DP of the required arch ADCM, U = KR the corresponding absciss of the given exterior line KI, and y = PC = RI their equal ordinates. Then by the principles of arches, in my tracts on that subject, it is found that CI or w = a + x

supposing a constant

quantity, and where a is some certain quantity to be determined hereafter. But KR oru is = ty, if t 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=

: but at D the value of w is = a, and w = 0, the curve

at D being parallel to KI; therefore the correct fluent is wve

τω

a=

2102
32

Hence then j2

=

Ωτος

the correct fluent of which gives y = a x hyp. log. of

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 w = cr becomes = AL OF MN = the given quantity e suppose, and y = AQ or a = b suppose, in which position the last equation becomes b = a x hyp. log. ; and hence it is found that the value of the constant quantity h.l.ct(c) which being substituted for it, in

Vo, is

b

c+(c2-a2)

the above general value of y, that value becomes

a

+(w2-a2)

log. of

from which equation the value of the ordinate PC may always be found, to every given value of the vertical cI.

But if, on the other hand, PC be given, to find cr, which will be the more convenient way, it may be found in the log. of a, and c = x log. of

following manner: Put A = e+(e2-a2)

+3

a

b

; then the above equation gives cy + A = log. of w+ (w2 a2); again, put in = the number whose log. is cy + A; then n = w + √(w2 - a2); and hence w =

Now, for an example in numbers, in a real case of this nature, let the foregoing figure represent a transverse vertical section of a magazine arch balanced in all its parts, in which the span or width AM is 20 feet, the pitch or height na is 10 feet, thickness at the crown DK = 7 feet, and the angle of the ridge LKN 112° 37', or the half of it LKD = 56° 18+, the complement of which, or the elevation KIR, is 33° 41', the tangent of which is = 3, which will therefore be the value of t in the foregoing investigation. The values of the other letters will be as follows, viz, DK=a=7; AQ=b=10; DQ=h=10; AL=C=10= ; A = log. of 7=8450980;

CI W

* log. of + √(a)

a2 + n2
2n

=

a 2n

a

3

Val. of y Val. of w

or CP.

log. of 2.56207 = 0408591; cy + A = 0408591y + •8450980 = log. of n. From the general equation then, viz, ++n, by assuming y successively equal to 1, 2, 3, 4, &c, thence finding the corresponding values of cy + a or 0408591y+8450980, and to these, as common logs. taking out the corresponding natural numbers, which will be the values of n; then the above theorem will give the several values of wor ci, as they are here arranged in the annexed table, from which the figure of the curve is to be constructed, by thus finding so many points in it. Otherwise. Instead of making n the number of the log. cy + A, if we put m = the natural number of the log.

cy only; then m =

w =

m2 + 1

το +(2-12)

a

or cr.

or by squaring, &c, a2m2-2ame + w2 = w2 - a2, and hence × a; to which the numbers being applied, the very same conclusions result as in the foregoing calculation and table.

2m

PROBLEM VIII.

To construct Powder Magazines with a Parabolical Arch.

It has been shown, in my tract on the Principles of Arches of Bridges, that a parabolic arch is an arch of equilibration, when its extrados, or form of its exterior covering, is the very same parabola as the lower or inside curve. Hence then a parabolic arch, both for the inside and outer form, will be