Many machines use power trains that are more complex than what has been discussed up to this point. Figure 6.7 represents a power train designed to allow an engine or motor to supply the power for two different components. These components are attached to output shafts A and B. This figure illustrates the application of direction and speed of rotation for more complex power trains.
Problem: What are the speed and the direction of rotation of shaft A and shaft B in Figure 6.7?
Solution: In this power train the output shaft of a 50-horsepower motor powers gear 1. Gear 2 is powered by gear 1, and because gear 3 is attached to the same shaft, it is also powered by gear 1. Gear 4 is powered by gear 3, and gear 5 by gear 4. Shaft A will turn at the same speed and direction of rotation as gear 5, and shaft B will turn at the same speed and direction of rotation as gears 2 and 3.
More than one approach can be used to determine the speed of shaft A and shaft B. The first one that we will show uses the gear equation. Because there is more than one pair of gears, the equation must be used more than once. The speed of gear 2 is:
Because gear 2 is fixed to shaft B, shaft B and gear 3 turn at 4,400 rpm.
To determine the speed of shaft A, the gear equation is used two more times, first to find the speed of gear 4:
16 x 4,400 N4 =---= 8,800 rpm and then the speed of gear 5:
Therefore, shaft A turns at 5,900 rpm.
Remembering the discussion of speed ratios, we can develop an alternative approach for solving this problem. If the speed of shaft B is determined by the speed of the driver gear times the gear ratio of the two gears, then the speed of shaft B can be determined by:
T1 24
and the speed of shaft A can be determined by:
When the speed ratio approach is used, it is important to place the number of teeth of the driver in the numerator and the number of teeth of the driven in the denominator.
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