+ d. sin. (C+A)

2 sin. a

c sin. y + b`sin. B

x {a sin. (C + E) — e sin. (B+D)

c. sin. (B + E)+ b. sin. (D + A)} which gives the actual position of the point a.

From a, thus determined, draw ab, making with AB the a, and meeting BC in b, which is the first point of reflection. In like manner the points c, d, e, may be found, and therefore the perpetual path required

The same process will apply to find the perpetual path on a table of any odd number (2n + 1) of sides whatever.

In the case of an even number (2n) of sides, it will conduct to

which shews that when the problem is possible (in this case) the polygon must be inscriptible in a circle. We leave the further discussion to the student.

322. Let A, Ar, Ar2 .... A-1 be the balls in geometric progression, a1, a2 a-1 the distances from one to the other in the same order, and v be the velocity given to A. Then since (Wood, p. 117,) the several velocities communicated by A to Ar, Ar to Ar2, &c., are

the intervals a,, a, ... a-1 will be described in the corresponding

whole time required to put A-1 in motion, will therefore be

t = o/1 × { 1 + (1 + r) + (1 + r)2 + ... (1 + r)*3 }

THE GRAVITATION OF BODIES.

: 323.

Since DB 2AB, the time along BD

time down AB; consequently, the whole time 2 time down AB = time down (4. AB).

Hence, if AB be produced to B' until AB′ = 4AB, and upon AB' be described a semicircle intersecting BD in D', and AD' be joined, AD' will be the plane required.

For the time down AD'

time down AB'. (Wood, p. 153).

324.

Let a be the length of the plane, and its inclination to the horizon, and let it be required to find the time down any portions of it (m, n,) measured from the higher and lower ends respectively.

The space (s) fallen through from rest in the time (t) being g sin..(Wood or Whewell)

S=

2

325. Let it be required to find the chord of a circle, whose diameter is (a), through which the body falling may acquire a ve

locity= 1. that acquired through the diameter.

n

Let x be the chord required.

Then since the times down the chords c as their lengths, we have

the chord required.

n

326.

From the given point P draw

of such a length that the time through it may

horizon, a line

time down the plane, and upon this line describe a semicircle cutting the given plane in the point P'; then P, P' being joined, PP' will be the line required.

For the time down the chord PP' time down the diameter time down the given plane.

Let x be the plane required. The moving force =

327. 2B - B = B,

328. At the end of the time (t) the heights to which the bodies A, a will have risen, are

tV g

S = ¿V = 2. t, s = tv \—

and the distance (D) of the centre of gravity from A is

329.

s=

Let x be part required, then by the expression

where s is the distance, descended from rest by the force of gravity (g) in the time t, along a plane inclined to the horizon by the 40, we have

330.

Let AB (Fig. 87) be the vertical diameter of the O, and let Mp drawn parallel to AB meet the tangent in M, and the curve in P, p; also let the chords PQ, pq, drawn 1 AB intersect it N, n; then the time down MP + time along PQ with the velocity acquired time down Mp + time along pq.

For, putting AB = 2a, and PM = x we easily get

x= t2, pM = 2a — x =