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CB123 (Industrial) (OP)
2 Jun 05 6:47

I'm trying to calculate the input torque required in a worm gearbox to produce a known torque at the output shaft. But I'm struggling to find a calculation that will do the job.

I have a unit, for which I know the output torque and the number of teeth and starts in the worm gear box. However I don’t know the motors input torque. This unit is currently not up to the job but if I can calculate the motors input torque, then I can replace the motor with something a bit beefier.

If it's possible to go back to basics, what is the relationship between size of worm wheel / number of teeth and torque increase? Is it the same as spur gears? I.e 1/2 reduction in speed = x2 increase in torque (put simply)?

Thanks for any help

israelkk (Aerospace)
2 Jun 05 6:53
Any college machine design book such as from Shigley or Norton have the info you need.
scarecrow55 (Mechanical)
3 Jun 05 7:45
divide the number of teeth in the wheel by the number of starts on the worm. That's the ratio. If you multiply that by the input torque and efficiency (say 75%) you will get the output torque. Efficiency can vary alot on wormgears
innodave (Industrial)
25 Aug 05 21:03
I need to lift 10,000 lbs with a 40:1 gearbox does anyone know what the torque calc. is for the imput shaft. I am currently using a 11.5A geared Milwaukee drill and it's not quite there it's pulling around 6300lbs
Thanks for any imput
davidh729 (Mechanical)
30 Aug 05 8:33
innodave, It all depends on the size of the ouput pulley. Multiply the load by the distance from the axis of rotation to get the output torque. Then divide by the gear box ratio. You'll also have to take the efficiency into consideration. 40:1 sounds like a 2 stage box so it's probably between 80 and 90%.

Dave Hyman
iRobot Corp

sacem1 (Mechanical)
31 Aug 05 20:21
If the gears are parallel sounds more like 3 stages, usually the ratio per stage is not over 5:1 because of production costs.
If it is worm-gear then a single entry worm and 40 teeth gear would be enough.
To innodave; If you have a rope and drum which is 1 ft diameter (0.5 ft radius) you will need:

load x torque arm = requiered torque
10,000 lbs x 0.5 ft = 5,000 lbs-ft net absorved

Torque absorved / reducer efficiency = Torque needed

5,000 / 0.7 = 7,142 lbs-ft

Torque needed / Reducer reduction = Torque input

7,142 lbs-ft / 40 = 178 lbs-ft = 2,143 lbs-in

That in a tipical co-axial speed reducer should be around 3 HP @ 1750 rpm direct coupled motor-speed reducer.

In a typical worm gear speed reducer should be something like a 3.5" between axis distance with 2.5 HP @ 1750 rpm input.

Hope you got what you needed.


rstock (Mechanical)
3 Oct 05 17:34
Hi, there.  Just reading through the responses here, and I think it answered a question of my own.  I just wanted to check to make sure I'm not incorrectly applying the equations given.  

I have a 60" butterfly valve that is operated by an actuator through a gearbox.  I know the torque required to operate the valve (13,500 ft lb), the torque supplied by the actuator (150 ft lb) and the gearbox ratio (360:1).  I am trying to determine whether the gearbox was correctly sized.  Based on the equation given in the above post,

Torque needed/Reducer reduction = torque input

I should be able to simply input my info:

13,500/360 = torque input = 37.5 ft lb

This would mean the 150 ft lb actuator would be sufficient.

I just wanted to make sure this is a correct application of this equation.  Do you see any problems with the logic I have used?  Thanks.

sacem1 (Mechanical)
3 Oct 05 18:27

Your aplication of the formula is correct but you have to take into consideration additionally the efficiency of the gearbox which I would guess is a wormgear reducer conected to a spur gear reduction probably 40:1 wormgear and 9:1 spur gear that should give you about also a 70% efficiency rate so you get:

37.5 / 70% = 53.5 ft-lb which the 150 ft-lb actuator should handle with no problem. Usually on those setups the safety operating margin is quite high ( in this case almost 3:1) because it is designed to cope with a possibly sticking butterfly and the resulting overload of the actuator.

In most cases a 25% additional power safety margin will be enough to handle overloads.

Hope you have your answer.


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