## Input/output torque calculations

## Input/output torque calculations

(OP)

Hi,

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

BigMuma

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

BigMuma

## RE: Input/output torque calculations

## RE: Input/output torque calculations

## RE: Input/output torque calculations

Thanks for any imput

## RE: Input/output torque calculations

Dave Hyman

iRobot Corp

www.irobot.com

## RE: Input/output torque calculations

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.

SACEM1

## RE: Input/output torque calculations

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.

rstock

## RE: Input/output torque calculations

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.

SACEM1