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?