3.49 seconds, the time of descending in air. And as 16: √195: 32: 8√ 195 # 111·71 feet, the velocity at the end of that time, or with which the ball enters the water.

Again, by prob. 22 of vol. 2, art. 2, the space s = × hyp.

1

log. of, or rather x hyp. log. of

26

being decreasing and e2 greater than a) =

m

26

26

(the velocity

xcom. log. of

-a where N 11325 the density of lead, n = 1000 that

3n

= ?

at entering the water, and v the velocity at any time afterwards, also d the diameter of the ball 4 inches, and m = 2-302585 the hyp. log. of 10.

Hence then N11325, n = 1000, N— n = 10325, d = 256d(N-n) 256. 10325

therefore s = 602 = x log, of=5mx log.

This theorem will give s when v is given, and by reverting it will give in terms of s in the following manner.

Dividing by 5m, gives log. ofns, by putting

5m

5m

; therefore, the natural number is 10" ±

and v = √(a + 1), which, by substituting the numbers above mentioned for the letters, gives v = 17-134 for the last velocity, when the space s➡ 603, or when the ball arrives at the bottom of the water.

But now to find the time of passing through the water, putting t any time in motion, and s and v the corresponding space and velocity, the general theorem for variable forces gives i = -. But the above general value of s being ×

hyp. log. or 5 × hyp. log., therefore its fluxion

which when = 17.134, or s = 603, gives 2·6542 seconds,

for the time of descent through the water.

PROBLEM

PROBLEM 12.

Required to determine what must be the diameter of a water-wheel, so as to receive the greatest effect from a stream of water of 12 feet fall?·

B

In the case of an undershot wheel, put the height of the water AB = 12 feet, and the radius BC or CD of the wheel, the water falling perpendicularly on the extremity of the radius CD at D. Then AC or AD = cx, and the velocity due to this height, or with which the water strikes the wheel at D, will be as (ax), and the effect on the wheel being as the velocity and as the length of the lever CD, will be denoted by x(a−x) or (ax2-r3), which therefore must be a maximum, or its square ar2 3 a maximum. In fluxions, 2axx F 3x2 = 0; and hence a za 8 feet, the radius. But if the water be considered as conducted so as to strike on the bottom of the wheel, as in the annexed figure, it will then strike the wheel with its greatest velocity, and there can be no limit to the size of the wheel, since the greater the radius or lever BC, the greater will be the effect.

In the case of an overshot wheel, a-2x will be the fall of water, √(a−2,x) as the velocity, and (a - 2.x) or

(ax-2.x3) the effect, then ax2 - 2x3 is a maximum, and 2axx-6x2 ÷ = 0; hence a 4 feet is the radius of the wheel.

B

But all these calculations are to be considered as independent of the resistance of the wheel, and of the weight of the water in the buckets of it.

PROBLEM 13.

What angle must a projectile make with the plane of the horizon, discharged with a given velocity v, so as to describe in its flight a parabola including the greatest area possible?

By the set of theorems in art. 92 pa. 156 vol. 2, for any proposed angle, there can be assigned expressions for the horizontal range and the greatest height the projectile rises to, that is the base and axis of the parabolic trajectory. Thus, putting s and c for the sine and cosine of the angle of eleva

tion;

=

tion; then, by the first line of those theorems, the velocity being v, the horizontal range R is sc2; and, by the 4th or last line of theorems, the greatest height H is =

522. But, by the parabola, of the product of the base or range and the height is the area, which is now required to be the greatest possible. Therefore R x HSCU2X22 must be a maximum, or, rejecting the constant factors, s3c a maximum. But the cosine c, of the angle whose sine is s, is (1-2); therefore s3c s3/(1 − s2) = √√/ (5o — s) is the maximum, or its square Só 58 a maximum. In fluxions 655's 857803 4s2; hence 4s2 = 3, or s2 = 4, and s=√3 = 8660254, the sine of 60°, which is the angle of elevation to produce a parabolic trajectory of the greatest

area.

PROBLEM 14.

Suppose a cannon were discharged at the point ; it is required to determine how high in the air the point c must be raised above the horizontal line AB, so that a person at c letting fall a leaden bullet at the moment of the cannon's explosion, it may arrive at в at the same instant as he hears the report of the cannon, but not till th of a second after the sound arrives at B: supposing the velocity of sound to be 1140 feet per second, and that the bullet falls freely without any resistance from the air?

be

Let r denote the time in which the sound passes to c; then will the time in passing to B, and a the time also the bullet is falling through CB. Then, by uniform motion, 1140x=AC, and 1140x-114 AB, also by descents

of gravity, 12: x2 :: 16 : 16x2 = BC. Then, by right-angled triangles, AC2-BC2=AB2, that is 11402x2 - 16°1=11402x2224 × 1140x + 1142, hence 224 × 1140x - 162x2 = 1142, or 1015 3x-x+50°77, the root of which equa. is x=10.03 seconds, or nearly 10 seconds; conseq. BC=16x2=1610 feet nearly, the height required.

PROBLEM 15.

Required the quantity, in cubic feet, of light earth, necessary to form a bank on the side of a canal, which will just support a pressure of water 5 feet deep, and 300 feet long. And what will the carriage of the earth cost, at the rate of 1 shilling per ton?

This question may be considered as relating either to water sustained by a solid wall, or by a bank of loose earth. In the former case, let ABC denote the wall, sustaining the pressure of the water behind it. Put the whole altitude AB a, the base BC or thickness at bottom b, any variable depth AD = x, and the thickness there DE=y. Now the effect which any number of particles of the fluid pressing at D have to break the wall at B, or to overturn it there, is as the number of particles AD or r, and as the lever BD — x; therefore the fluxion of the effect of all the forces is (a — x)xx axx Xx3 the fluent of which is arr3, which, when ra, is a3 for the whole effect to break or overturn the wall at B; and the effects of the pressure to break at B and D will be as AB3 and AD3. But the strength of the wall at D, to resist the fracture there, like the lateral strength of timber, is as the square of the thickness, DE. Hence the curve line AEC, bounding the back of the wall, so as to be everywhere equally strong, is of such a nature, that 3 is always proportional to y, or y as a and is therefore what is called the semicubical parabola. Now, to find the area ABC, or content of the wall bounded by this convex curve, the general fluxion of all are as y* becomes, the fluent of which is 3x + =xy, that is of the rectangle AB X BC; and is therefore less than the triangle ABC, of the same base and height, in the proportion of to, or of 4 to 5.

But in the case of a bank of made earth, it would not stand with that concave form of outside, if it were necessary, but would dispose itself in a straight line Ac, forming a triangular bank ABC. And even if this were not the case naturally, it would be proper to make it such by art; because now

E

neither is the bank to be broken as with the effect of the lever, or overturned about the pivot or point c, nor does it resist the fracture by the effect of a lever, as before; but, on the contrary, every point is attempted to be pushed horizontally outwards, by the horizontal pressure of the water, and it is resisted by the weight or resistance of the earth at any part, DE. Here then, by hydrostatics, the pressure of the water against any point D, is as the depth AD; and, in the triangle of earth ADE, the resisting quantity in DE is as DE,

which

which is also proportional to an by similar triangles. So that, at every point D in the depth, the pressure of the water and the resistance of the soil, by means of this triangular form, increase in the same proportion, and the water and the earth will everywhere mutually balance each other, if at any one point, as B, the thickness BC of earth be taken such as to balance the pressure of the water at B, and then the straight line AC be drawn, to determine the outer shape of the earth. All the earth that is afterwards placed against the side AC, for a convenient breadth at top for a walking path, &c, will also give the whole a sufficient security.

But now to adapt these principles to the numeral calculation proposed in the question, the pressure of water against the point B being denoted by the side AB = 5 feet, and the weight of water being to earth as 1900 to 1984, therefore as 1984: 1000 :: 5 : 2·52 = BC, the thickness of earth which will just balance the pressure of the water there; hence the area of the triangle ABC = AB × BC = 2 × 2·52 =63; this mult. by the length 300, gives 1890 cubic feet for the quantity of earth in the bank; and this multiplied by 1984 ounces, the weight of 1 cubic foot, gives, for the weight of it, 3749760 ounces = 234360lbs = 104.625 tons; the expense of which, at 1 shilling the ton, is 51. 4 s. 7 d.

PROBLEM 16.

A person standing at the distance of 20 feet from the bottom of a wall, which is supposed perfectly smooth and hard, desires to know in what direction he must throw an elastic ball against it, with a velocity of 80 feet per second, so that, after reflection from the wall, it may fall at the greatest distance possible from the bottom, on the horizontal plane, which is 21 feet below the hand discharging the ball?

In the annexed figure let DR

be the wall against which the
ball is thrown, from the point
A, in such a direction, that it B
shall describe the parabolic

curve AE before striking the

wall, and afterwards be so reflected as to describe the curve EP. Now if Es be the tangent at the point E, to the curve AE described before the reflection, and EF the tangent at the same point to the curve which the ball will describe after.reflection, then will the angle REF beCES; and if the curve E be produced, so as to have GF for its tangent, it will.meet Ac produced in B, making BC AC, and the curve AE will be

similar