of the centre of percussion, since it would receive, at the same distance from the axis, the same increments of velocity per second that the centre of percussion does, it would manifestly move exactly as that point does, and perform its oscillations in the same time that the body does. Therefore, &c.

112. The centres of suspension and oscillation are reciprocal.

Let O represent the centre of oscillation of a body when suspended from the axis A; also let G be its centre of gravity. Let AO=L, AG=G, OGG,; also let the radius of gyration about A be represented by K2, and that about G by k2. Therefore (equation 59), K2=G2+k2;

Now let the body be suspended from O instead of A; when thus suspended it will have, as before, a centre of oscillation. Let the distance of this centre of oscillation from O be L1,

Since then the centre of oscillation in this second case is at the distance L from O, it is in A; what was before the centre of suspension has now therefore become the centre of oscillation. Thus when the centre of oscillation is converted into the centre of suspension, the centre of suspension is thereby converted into the centre of oscillation. This is what is

meant, when it is said that the centres of oscillation and sus

pension are reciprocal.

PROJECTILES.

113. To determine the path of a body projected obliquely

in vacuo.

Suppose the whole time, T seconds, of the flight of the body to any given point P of its path, to be divided into equal exceedingly small intervals, represented by AT, and conceive the whole effect of gravity upon the projectile during

each one of these intervals to be collected into a single impulse at the termination of that interval, so that there may be communicated to it at once, by that single impulse, all the additional velocity which is in reality communicated to it by gravity at the different periods of the small time ▲T.

Let AB be the space which the projectile would describe, with its velocity of projection alone, in the first interval of time; then will it be projected from B at the commencement of the second interval of time in the direction ABT with a velocity which would alone carry it through the distance BK=AB in that interval of time; whilst at the same time it receives from the impulse of gravity a velocity such as would alone carry it vertically through a space in that interval of time which may be represented by BF. By reason of these two impulses communicated together, the body will therefore describe in the second interval of time the diagonal BC of the parallelogram of which BK and BF are adjacent sides. the commencement of the third interval it will therefore have arrived at C, and will be projected from thence in the direction BCX, with a velocity which would alone carry it through CX=BC in the third interval; whilst at the same time it receives an impulse from gravity communicating to it a velocity which would alone carry it through a distance represented

I

by CGBF in that interval of time. These two impulses together communicate therefore to it a velocity which carries it through CD in the third interval, and thus it is made to describe all the sides of the polygon ABCD... P in succession. Draw the vertical PT, and produce AB, BC, CD, &c. to meet it in T, N, O. . ., and produce GC, HD, &c. to meet BT in K, L, &c.

Now, since BC is equal to CX, and CK is parallel to XL, therefore KL is equal to BK or to AB.

Again, since CD is equal to DZ, and DL is parallel to ZM, therefore LM is equal to KL or to AB; and so of the

rest.

If therefore there be n intervals of time equal to AT, so that there are n sides AB, BC, CD, &c. of the polygon, and n divisions AB, BK, &c. of the line AT, then AT, =nAB and BT=(n−1)AB,

... TN=(n−1) KC=(n−1) BF.

Similarly CN=(n-2)CX, therefore NO=(n-2) DX= (n−2)BF; and so of the remaining parts of TP.

Now these parts of TP are (n-1) in number, therefore TP=(n−1)BF+(n−2)BF+(n−3) BF+... {(n-1) terms}; or TP= {(n−1)+(n−2) + ... }BF.

Therefore, summing the series to (n-1) terms,

Now g represents the additional velocity which gravity would communicate to the projectile in each second, if it acted upon it alone. gAT is therefore the velocity which it would communicate to it in each interval of AT seconds. gAT is therefore the velocity communicated to the body by each of the impulses which it has been supposed to receive from gravity.

Now BF is the space through which it would be carried in the time AT by this velocity,

Now this is true, however small may be the intervals of time AT, and therefore if they be infinitely small, that is, if the impulses of gravity be supposed to follow one another at infinitely small intervals, or if gravity be supposed to act, as it really does, continuously.

But if the intervals of time AT be infinitely small, then the number n of these intervals which make up the whole finite time T, must be infinitely great. Also when ʼn is in

In the actual case, therefore, of a projectile continually deflected by gravity, the vertical distance TP between the tangent to its path at the point of projection, and its position P after the flight has continued T seconds, is represented by the formula

Moreover AT=nAB, and AB is the space which the body would describe uniformly with the velocity of projection in the time AT, so that nAB is the space which it would describe in the time n. AT or T with that velocity. If therefore V equal the velocity of projection, then

AT=V. T.... (90);

so that the position of the body after the time T is the same as though it had moved through that time with the velocity of its projection alone, describing AT, and had then fallen through the same time by the force of gravity alone, describing TP (see Art. 101.).

M

114. Let AM=x, MP=y, angle of projection TAM=a, velocity of projection =V.

=AT=V . T, .'. T=

.. x sec. α=.

x sec. a

x tan. a-y=MT-MP=TP=1gT2. . . . . (91). Substituting the value of T from the preceding equation,

Let H be the height through which a body must fall freely by gravity to acquire the velocity V, or the height due to that velocity; then V2=2gH (Art. 47.), therefore 4H

2V2 g

therefore, by substitution,

y=x tan. a

sec.2a
4H

x2

(92).

115. To find the time of the flight of a projectile.

It has been shown (equation 91), that if T represent the time in seconds of the flight to a point whose co-ordinates are z and y, then

If the projectile descend again to the horizontal plane from which it was projected, and T be the whole time of its flight,