Determining Worm Gear Output Torque
Determining Worm Gear Output Torque
(OP)
I have been trying to figure out how to determine the output torque of a worm and pinion combination, to no avail.
The input torque, pitch of the pinion worm, the radius of the gear wheel, as well as the number of teeth on said wheel all play into the calculation. I just don't know how to put them all together to form the equation.
The motor I have in mind to turn the pinion runs at 263 RPMS and has 2527 oz/in of torque. That equates to roughly 13.16 ft/lbs. Which may be overkill; without knowing the formula to do calculations, I can't be certain.
The desired output torque is 1200 ft/lbs. Any help is greatly appreciated!
The input torque, pitch of the pinion worm, the radius of the gear wheel, as well as the number of teeth on said wheel all play into the calculation. I just don't know how to put them all together to form the equation.
The motor I have in mind to turn the pinion runs at 263 RPMS and has 2527 oz/in of torque. That equates to roughly 13.16 ft/lbs. Which may be overkill; without knowing the formula to do calculations, I can't be certain.
The desired output torque is 1200 ft/lbs. Any help is greatly appreciated!





RE: Determining Worm Gear Output Torque
http://www.technologystudent.com/gears1/worm1.htm
Ted
RE: Determining Worm Gear Output Torque
Even at 100% efficiency, you'll need a reduction ratio over 91:1. Worm gears are notoriously inefficient, maybe 60% efficient at your ratio range. If you need at least 1200 lb-ft at the output, with 13.16 ft-lb at the input, then you'll need a ratio equal 1200/(13.16x0.60) or about 152:1.
These types of numbers likely require a single start worm and large diameter gear for a single stage.
Hope that helps.
Terry
RE: Determining Worm Gear Output Torque
@tbuelna: i determined after posting (just visualizing the gearbox) that if a single stage worm were used, then the gear wheel would have to be huge. i then determined that adding a planetary setup after the worm gear would work, but would terribly detriment the speed of the movement.
therefore, i started researching linear actuators--with a few modifications to my plans i can use an actuator instead. i found a company that produces a 20" linear actuator with a 1500lb capacity and a traverse time of 50"/min, which should more than suffice for my needs. all this for just under $400. :)
thanks for your help. i am building a robotic arm, so you'll probably encounter a number of other posts by me along the way.