For example, let the quadrant AB in the following figure represent the path of the crank, and the line AO that of the piston. Let OF be the position reached by the crank. AOF is the angle formed by the crank with line of centers, and supposed to be 60°. FE normal to AO is the sine of this angle, and AE the versed sine. The latter is the distance traveled by the piston from the point A, and is .5, the length of the crank being 1.
Secondly, we ascertain how far the piston must advance for every degree or minute or second of the revolution of the crank in its quadrant by merely subtracting from its versed sine that of the preceding one. Thus the versed sine of 60° being .5, and that of 59° being .4849619251, the difference .0150380749 is the motion of the piston, or its mean velocity while the crank is traversing the 60th degree of its revolution.
Thirdly, we want to know the rate at which the motion of the piston is accelerated during any interval.
This acceleration is found by subtracting from the motion during each interval that during the preceding one. For example, the motion of the piston during the 60th degree being, as already seen, .0150380749, and that during the 59th degree being .0148811893, the difference between them, .0001568856, is the acceleration or amount of motion added during the 60th degree.
By this simple process we find the acceleration of the piston during the first degree of the revolution of the crank to be .0003046096, and that during the 90th degree to be .0000053161. But this latter is the amount by which the acceleration was reduced during the preceding degree. Therefore at the end of this degree the acceleration has ceased entirely.
Now, by erecting on the center line AC, at the end of each degree, ordinates which are extensions of the sine of the angle, and the lengths of which represent the acceleration during that degree we find that these all terminate on the diagonal line CO. Thus, when the crank has reached the 60th degree, and the piston has advanced half the distance to the mid-stroke or to E, Fig. 32, the acceleration during the 60th degree has been .0001523049, or one half of that during the first degree.
But how do we know the amount of the accelerating force exerted by the crank at the beginning of the stroke? This question is answered as follows:
We find that for the first three degrees the accelerating force is, for the purpose of our computations, constant, the diminution not appearing until we have passed the sixth place of decimals.
Let us now suppose the crank 1 foot in length to make 1 revolution per minute, so moving through 6° of arc in 1 second. At this uniform rate of acceleration the piston would be moved in 1 second the versed sine of 1° .0001523048 × 6² = .0054829728 of a foot.
A falling body uniformly accelerated by a force equal to its own weight moves in 1 second 16.083 feet. Therefore this uniform stress on the crank is .0054829728 16.083 = .000341, which is the well-established coefficient of centrifugal force—the centrifugal force of one pound making one revolution per minute in a circle of one foot radius.