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Worm gear torque ratios

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METG2009

Mechanical
Oct 23, 2008
1
HI,

If you were to use a hydraulic motor with an output of around 1500 ft.lbs of torque. What type of a worm gear set up would you have to use to put out 20 000 ft.lbs, if that is even possible.

Any information would be a great help
 
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That's a ratio of about 13:1, which is no big deal. You could probably even do it with an ordinary, non-worm reducer. You'll have to use a little bit higher ratio than 13 to figure for losses due to efficiency and maybe add an additional safety factor, depending on what type of load you have.

Remember that an increase in torque means a corresponding decrease in rpm.

Also remember that the motor rating is just the *maximum* torque possible on the motor end. The actual torque depends on the load. Keep this in mind when selecting a gearbox in case you ever encounter a locked-rotor condition. You don't want the output torque to become so high that you destroy something. For example, if you simply guessed and picked out a 50:1 worm drive to have plenty of torque, a locked-up mechanism would then see about 50 x 1500 = 75,000 ft-lbs of torque (neglecting efficiency). This might be more than some parts of your gear train can handle.

Don
Kansas City

 
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