1. A wheel 96 inches in diameter, having 42 revolutions per minute, is to drive a shaft 75 revolutions per minute ; what should be the diameter of the pinion?

96 X 42

----- = 53.76 inches. 75

2. If a pinion is to make 20 revolutions per minute, required the diameter of another to make 58 revolutions in the same time.

58 /20 = 2.9 = the ratio of their diameters. Hence, if one to make 20 revolutions is given a diameter of 30 inches, the other will be 30 / 2.9 = 10.345 inches..

3. Required the diameter of a pinion to make 12½ revolutions in the same time as one of 32 inches diameter making 26.

32X26

---- = 66.56 inches. 12.5

4. A shaft, having 22 revolutions per minute, is to drive another shaft at the rate of 15, the distance between the two shafts upon the line of centres is 45 inches; what should be the diameter of the wheels ?

Then, 1st. 22 + 15 : 22 :: 45 : 26.75 = inches in the radius of the pinion.

2d. 22 + 15 : 15 :: 45 : 18.24 = inches in the radius of the spur.

5. A driving shaft, having 16 revolutions per minute, is to drive a shaft 81 revolutions per minute, the motion to be communicated by two geared wheels and two pulleys, with an intermediate shaft; the driving wheel is to contain 54 teeth, and the driving pulley upon the driven shaft is to be 25 inches in diameter; required the number of teeth in the driven wheel, and the diameter of the driven pulley.

Let the driven wheel have a velocity of √16 x 81 =36, a mean proportional between the extreme velocities 16 and 81.

Then, 1st. 86 : 16 :: 54 : 24 = teeth in the driven wheel.

2d. 81 : 36 :: 25 : 11.11 = inches diameter of the driven pulley.

6. If, as in the preceding case, the whole number of revolutions of the driving shaft, the number of teeth in its wheel, and the diameters of the pulleys are given, what are the revolutions of the shafts ?

Then, 1st. 18 : 16 :: 54 : 48 = revolutions of the intermediate shaft. 2d. 15 : 48 :: 25 : 80 = revolutions of the driven shaft.