The standard from which all computations are made of the distances passed through in given times by bodies whose motion is uniformly accelerated, and from which the velocity acquired is computed when the accelerating force is known, and the force is found when the velocity acquired or the rate of acceleration is known, is the velocity of a body falling to the earth. It has been established by experiment, that in this latitude near the level of the sea, a falling body in one second falls through a distance of 16.083 feet, and acquires a velocity of 32.166 feet per second; or, rather, that it would do so if it did not meet the resistance of the atmosphere. In the case of a falling body, its weight furnishes, first, the inertia, or the resistance to motion, that has to be overcome, and affords the measure of this resistance, and, second, it furnishes the measure of the attraction of the earth, or the force exerted to overcome its resistance. Here, as in all possible cases, the force and the resistance are identical with each other.

The above is, therefore, found in this way to be the rate at which the motion of any body will be accelerated when it is acted on by a constant force equal to its weight, and encounters no resistance.

It follows that a revolving body, when moving uniformly in any circle at a speed at which its deflection from a straight line of motion is such that in one second this would amount to 16.083 feet, requires the exertion of a centripetal force equal to its weight to produce such deflection. The deflection varying as the square of the time, in 0.01 of a second this deflection will be through a distance of 0.0016083 of a foot.

Now, at what speed must a body revolve, in a circle of one foot radius, in order that in 0.01 of one second of time its deflection from a tangential direction shall be 0.0016083 of a foot? This decimal is the versed sine of the arc of 3°15', or of 3.25°. This angle is so small that the departure from the law that the deflection is equal to the versed sine of the angle is too slight to appear in our computation. Therefore, the arc of 3.25° is the arc of a circle of one foot radius through which a body must revolve in 0.01 of a second of time, in order that the centripetal force, and so the centrifugal force, shall be equal to its weight. At this rate of revolution, in one second the body will revolve through 325°, which is at the rate of 54.166 revolutions per minute.

Now there remains only one question more to be answered. If at 54.166 revolutions per minute the centrifugal force of a body is equal to its weight, what will its centrifugal force be at one revolution per minute in the same circle?

To answer this question we have to employ the other extremely simple law, which I said I must explain to you. It is this: The acceleration and the force vary in a constant ratio with each other. Thus, let force 1 produce acceleration 1, then force 1 applied again will produce acceleration 1 again, or, in other words, force 2 will produce acceleration 2, and so on. This being so, and the amount of the deflection varying as the squares of the speeds in the two cases, the centrifugal force of a body making one revolution per minute in a circle of

``` 1²

one foot radius will be ---------- = 0.000341

54.166² ```

--the coefficient of centrifugal force.

There is another mode of making this computation, which is rather neater and more expeditious than the above. A body making one revolution per minute in a circle of one foot radius will in one second revolve through an arc of 6°. The versed sine of this arc of 6° is 0.0054781046 of a foot. This is, therefore, the distance through which a body revolving at this rate will be deflected in one second. If it were acted on by a force equal to its weight, it would be deflected through the distance of 16.083 feet in the same time. What is the deflecting force actually exerted upon it? Of

``` 0.0054781046

course, it is ------------.

16.083 ```

This division gives 0.000341 of its weight as such deflecting force, the same as before.

In taking the versed sine of 6°, a minute error is involved, though not one large enough to change the last figure in the above quotient. The law of uniform acceleration does not quite hold when we come to an angle so large as 6°. If closer accuracy is demanded, we can attain it, by taking the versed sine for 1°, and multiplying this by 6². This gives as a product 0.0054829728, which is a little larger than the versed sine of 6°.

I hope I have now kept my promise, and made it clear how the coefficient of centrifugal force may be found in this simple way.

We have now learned several things about centrifugal force. Let me recapitulate. We have learned:

1st. The real nature of centrifugal force. That in the dynamical sense of the term force, this is not a force at all: that it is not capable of producing motion, that the force which is really exerted on a revolving body is the centripetal force, and what we are taught to call centrifugal force is nothing but the resistance which a revolving body opposes to this force, precisely like any other resistance.

2d. The direction of the deflection, to which the centrifugal force is the resistance, which is straight to the center.

3d. The measure of this deflection; the versed sine of the angle.

4th. The reason of the laws of centrifugal force; that these laws merely express the relative amount of the deflection, and so the amount of the force required to produce the deflection, and of the resistance of the revolving body to it, in all different cases.

5th. That the deflection of a revolving body presents a case analogous to that of uniformly accelerated motion, under the action of a constant force, similar to that which is presented by falling bodies; and finally,

6th. How to find the coefficient, by which the amount of centrifugal force exerted in any case may be computed.