[decoyplankton]

I would like to calculate the amount of torque required by a motor to move an object linearly. Imagine the motor is driving a rack and pinion gearbox. The object is on wheels and will be attached to the rack to receive the linear force from the motor. I already know that it takes about 35 N force to move the object.Is the torque required from the motor the same i.e 35Nm ?

Thanks

Richard

[ebaines]

The torque that the motor must provide depends on the gearing that you will be using between the drive shaft of the motor and the mechanism that must receive the 35N of force. For example, if the final gear is 2 cm in radius then the motor and gear train must deliver 35 N times 2 cm = 70 N-cm of torque. If the motor is connected directly to this 2 cm radius gear, then that's your answer. Or you could insert a gear train that amplifies the motor's torque, such that a lower torque motor is needed but at the expense of the final gear turning at a slower rate than the motor. So - first figure out how you're going to connect the output of the motor to your mechanism.

[decoyplankton]

Thanks, that makes sense. So I simply think of the radius of the gear on the pinion as a lever and multiply by the length of the lever ?

What if my motor is turning a threaded shaft (like a leadscrew) and the load is moving along the threaded shaft. Obviously the speed that the load moved along the thread would be rpm * pitch, but would the torque required be 35N multiplied by the shaft radius ? Also if I used a gear train as you suggested, would the torque required reduce proportionally with the ratio of the gearbox or is there some other relationship ? I'm not too worried about speed at the moment as long as I understand the maths, I can jiggle with the figures later.

Thanks again,

Richard

[ebaines]

So I simply think of the radius of the gear on the pinion as a lever and multiply by the length of the lever ?

Yes.

What if my motor is turning a threaded shaft (like a leadscrew) and the load is moving along the threaded shaft. Obviously the speed that the load moved along the thread would be rpm * pitch, but would the torque required be 35N multiplied by the shaft radius ?

No - the motor's torque would be effectively multiplied by the ratio of how far the motor turns to how far the load moves.

Also if I used a gear train as you suggested, would the torque required reduce proportionally with the ratio of the gearbox or is there some other relationship ?

You're on the right track. The fundamental principal is that due to conservation of work the further the motor turns compared to the load being moved the less torque the motor needs to produce. Same concept as to how a lever arm works. So if you use a gear set that has a ratio of a:b (meaning 'a' rotations of the input shaft yield 'b' rotations of the output shaft) then the torque of the motor is multiplied by a/b. This a:b ratio can be calculated by counting teeth on the gear: it's equal to the number of teeth on the output gear divided by the number on the input gear. Or the ratio of the gear diameters.

[decoyplankton]

No - the motor's torque would be effectively multiplied by the ratio of how far the motor turns to how far the load moves.

So if the leadscrew moved the load 2cm per revolution, I would calculate the ratio by finding the circumference of the shaft in cm and dividing it by 2cm (distance moved by load). This would give me a ratio which I need to multiply by 35N (the original force needed to move the load). That would give me a figure of torque required at the shaft.

Having arrived at a figure for torque required to turn the leadscrew, I can then multiply this by the ratio of my gearbox which I am using to connect the motor to the shaft. Then I should have a final figure for torque required from the motor for the whole system.

Is that correct ?

Forgive me but I've long since forgotten my mechanical science lectures from about 20 years ago!

[ebaines]

So if the leadscrew moved the load 2cm per revolution, I would calculate the ratio by finding the circumference of the shaft in cm and dividing it by 2cm (distance moved by load). This would give me a ratio which I need to multiply by 35N (the original force needed to move the load). That would give me a figure of torque required at the shaft.

Close! You would divide (not multiply) by the ratio you described. Remember - if you decrease the distance the object travels per revolution of the motor shaft it makes it easier for the motor and hence less torque is needed. So in your example you take 35N and divide by the ratio of screw circumference to pitch per revolution.

Having arrived at a figure for torque required to turn the leadscrew, I can then multiply this by the ratio of my gearbox which I am using to connect the motor to the shaft. Then I should have a final figure for torque required from the motor for the whole system.

Is that correct ?

Yes! The only slight complication is that no gear set is perfectly efficient, so plan on losses of about 10% of so. In other words build in a safety factor.

Forgive me but I've long since forgotten my mechanical science lectures from about 20 years ago !!

No problem. However - the last mechanics course I took was over 30 years ago! The good news is that things like gears and levers haven't changed much in the mean time.

[decoyplankton]

OK thanks very much. I think I understand that now.

It's been a long time since I have had to figure something like this out. Also good advice about the 10% safety factor. At least now I will have an idea of what size motor to be specifying for my project.

Thanks for your help

Richard

Spalding. UK