In managing agricultural machinery, it is sometimes useful to know the amount of torque being transmitted by a component of a power train. For example, it would be useful to know the amount of torque and the horsepower available at shafts A and B in Figure 6.7.
Using the previous discussion of the relationship between torque and speed and the torque equation the maximum torque available at each shaft can be determined:
To solve for ToB the values for the three variables must be known. The two rpm's were calculated in the previous section. The remaining variable is the torque supplied by the motor. Horsepower will be discussed in greater detail in a later chapter, but rotary horsepower can be determined by the horsepower equation used in the previous section.
Because we know the horsepower and the rpm being produced by the engine, we can rearrange this equation to solve for the engine torque:
PHp x 5252 50 hp x 5252
Toeng = -Hp-=---= 119.36... or 120 lb-ft eng N 2,200 rpm
Once the torque produced by the motor is known, the torque on shaft B can be calculated.
N1 2,200 rpm
The torque at shaft A can also be calculated:
Nmotor 2,200 rpm Toa = Tomotor x = 120 lb-ft x --— = 44 lb-ft
A motor Na 6,000 rpm
Note that because we already knew the speed of shaft A, we used the speed ratio to solve for the torque. If we did not know the speed of shaft A, we would solve for it first.
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