There are so many gear motors to choose from eBay.  How do you know if you buy the right gear motor for the right job?  If you buy a gear motor for a good price but it doesn't do the job, it will simply become a waste of time and money. 

Some people might be curious about how a low speed, low power gear motor is able to generate a high torque.  For example, a 60W output, 50 RPM gear motor is able to generate a 11.5 N-m (8.5 ft-lb) of torque, but yet, a 60W output, 2500 RPM motor is only able to generate a 0.23 N-m (0.17 ft-lb) of torque.  The trick is speed reduction.  Torque is very important to many applications because by knowing torque, one can determine how much of load the motor is able to lift an item, to drive a wheel, to turn a conveyor and so on.  For example, if a 50 rpm motor has a torque output of 8.5 ft-lb (102 in-lb), a 4 inch diameter pulley should be able to lift 51 lb (102 in-lb / 2 in of pulley radius) of weight at 50 rpm; a 2 inch diameter pulley should be able to lift 102 lb (102 in-lb / 1 in of pulley radius) of weight at 50 rpm.  Having that being mentioned, it is NOT recommended to put weight directly on a gear motor's shaft!  An excessive radial bearing load may damage a motor's gear box quickly!  It is always safe to isolate the load by introducing a pulley or sprocket system. 

One vital formula that is needed to calculate torque of a motor or its reduced speed output is:

Power (watts) = Torque (N-m) * RPM * 0.105

and of course,

1 N-m = 0.74 ft-lbf = 8.9 in-lbf
1 hp = 0.00134 Watt

Here's an example on how to use this formula.  Joe purchased a 12v gear motor from eBay that runs 60W at 50 RPM.  He wants to know if he can lift a 200 lb of weight at 20 RPM with this motor. 

Here is how he can find out:

First, he needs a belt and pulley speed reduction to reduce the motor's 50 RPM to his desired 20 RPM.  That is, the speed reduction ratio of 50RPM  to 20 RPM or 5 to 2 ratio or 2.5 to 1 ratio.  He decides to get a 2 inch diameter pulley mounted on the motor's shaft to belt drive the 5 inch diameter pulley for the final output (of 20 RPM).  Now he has a system that outputs 60 watts at 20 RPM instead of the motor's original 50 RPM.  Also, he isolates the radial bearing load from the gear box.  With 60 watts at 20 RPM, Joe now wishes to find out the torque of his system at 20 RPM:

60 Watts = Torque (N-m) * 20 RPM * 0.105
Torque (N-m) = 60W / (20 RPM * 0.105)
Torque = 28.6 N-m = 254 in-lb

Joe knows that the smaller the pulley size he has, the more weight his system can lift.  He decides to go with a 2 inch diameter pulley and mounts it to his 20 RPM system.  That is, the output shaft that is connected to the 5 inch pulley is mounted with his 2 inch diameter pulley to lift his load.  Torque / Radius of Pulley = 254 in-lb / 1 inch = 254 lb.  Joe is pleased to know that his system is able to lift 254 lb of weight that is more than his targeted 200 lb of load at 20 RPM.

I hope this guide helps many of you save your design time and money.  Happy eBay hunting!