Very basic question regarding reduction drives.

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SUMMARY

The discussion centers on the mechanics of speed-reduction drives, specifically the relationship between torque and speed under load conditions. When a prime mover rated at 1.4 mN torque at 2700 rpm is used with a reduction ratio of 27:1, the rear wheel velocity is calculated to be 100 rpm with a torque of 37.8 mN. The key conclusion is that if the load resistance equals the torque input, the rear wheel maintains a constant speed of 100 rpm; otherwise, the shaft rpm will increase until a failure occurs or the load and torque input balance out.

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motivehunter
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Hi!

In a speed-reduction drive, because the power transmitted is almost constant, as the speed decreases the torque increases. Is this reduced rear wheel/shaft speed the same when the loaded condition is considered?

Let's assume that the prime mover is rated to produce 1.4 mN torque at 2700 rpm by the manufacturer. Using a reduction ratio of 27:1, our rear wheel velocity would be 100 rpm and torque would be 37.8 mN. My question is, when the prime mover is loaded, does it still make the rear wheel rotate at 100 rpm or at a much lower speed?

I'm sorry for posting such a simple question but its one of those things about which i can't seem to think too clearly!
Thanks!
 
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If the resistance is equivalent to the torque input, then the rpm will be constant. If it is lower, then the shaft rpm will accelerate until something will break (or somehow the load and torque input equalize).
 

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