Suppose that a ([Fig. 13]) makes 1 revolution per minute, how many will c make, a having 60 teeth, b 30 teeth, and c 40 teeth? In this case we have but one driving wheel a, and one driven wheel b, the driver having 60 teeth, the driven 30, hence 60 ÷ 30 = 2, equals revolutions of b and also of c, the two latter being on the same shaft.
It will be observed then that the revolutions are in the same proportion as the numbers of the teeth or the radii of the wheels, or what is the same thing, in the same proportion as their diameters. The number of teeth, however, is usually taken as being easier obtained than the diameter of the pitch circles, and easier to calculate, because the teeth will be represented by a whole number, whereas the diameter, radius, or circumference, will generally contain fractions.
Fig. 14.
Suppose that the 4 wheels in [Fig. 14] have the respective numbers of teeth marked beside them, and that the upper one having 40 teeth makes 60 revolutions per minute, then we may obtain the revolutions of the others as follows:—
| Revolu- tions. | Teeth in first driver. | Teeth in first driven. | Teeth in second driver. | Teeth in second driven. | ||||||
| 60 | × | 40 | ÷ | 60 | × | 20 | ÷ | 120 | = | 666⁄100 |
and a remainder of the reciprocating decimals. We may now prove this by reversing the question, thus. Suppose the 120 wheel to make 666⁄100 revolutions per minute, how many will the 40 wheel make?
| Revolu- tions. | Teeth in first driver. | Teeth in first driven. | Teeth in second driver. | Teeth in second driven. | |||||||
| 6.66 | × | 120 | ÷ | 20 | × | 60 | ÷ | 40 | = | 5999⁄100 | = |
revolutions of the 40 wheel, the discrepancy of 1⁄100 being due to the 6.66 leaving a remainder and not therefore being absolutely correct.