# torque requirement

I wish to determine the required torque of a 55rpm gear drive motor assembly to drive a fishing reel with a 4.6:1 turn ratio that will yield 27"of line retrieval per turn or 1 crank revolution. Lift weight minimum 1 lb.

I wish to direct couple a 12vdc 55rpm right angle gear drive rated @ 8kg-cm to a fishing reel crank shaft that has a 1:3.4 ratio turning a 4 cm spool. Line is retrieval is 12" per crank revolution. How much can it lift before stall. I'm looking for a continuous lift of at least 1 lb.

• What's the inertia of the crank/reel mechanism? How much friction is there in the system (especially where the line leaves the last eyelet)? You said "minimum" lift weight 1lb - what's the maximum weight? Or are you looking for just enough torque to reel in a hook with no fish? Feb 14 '16 at 1:22
• Some other bit of info needed is the relation of torque on the input shaft to the force applied to the line. Once you have that multiply with the max lift weight you want and add some 20% margin for friction losses. Feb 14 '16 at 2:05
• @mercman If you explain with diagram then it would be more better. and tell a maximum weight that u want not the minimum. Feb 14 '16 at 4:43

DC motors typically have a torque to speed curve that's linear:

The rated torque is usually kept at the very low end of the stall torque, simply to prevent overheating, and breaking:

That said, $8 kg_f cm$ isn't especially high of a torque. If the 1:3.4 ratio was reversed, so it took more turns of the motor to turn the crank, this could lift 3 pounds and maintain the 55 rpm. However, it's more likely with more current you could overload the motor and guarantee even more - though that point is based upon the stall torque of the motor, which is not specified here.