View Full Version : Screw speed reduction torque gain?

AVRnj

12-03-2013, 08:13 PM

I have an application where I am using a linear actuator that I am building, basically a DC motor with a lead screw coupled to the shaft, a linear guide system and my own encoder system for position control.

Before I buy my motors, I am trying to do some calculations.

I am assuming that like gears, a screw will convert speed to torque, minus the friction loss, but I cannot find any math on this, or I cannot even verify this accurate.

Can anyone first validate that I will gain torque, and if so, has anyone seen any calculations on this?

Th232

12-03-2013, 08:30 PM

If it's a linear actuator run by a motor (input being rotational movement), shouldn't the output (linear movement) be force and not torque?

As for calculations, power = force x linear velocity = torque x rotational velocity. Power in = power out + inefficiencies.

AVRnj

12-04-2013, 07:39 AM

Thanks for the reply! I suppose I don't know enough physics to understand the difference between force and torque, and when either applies.

When I think of torque I think of it in terms of how much weight can be moved a certain distance away, kg/cm etc.

For the linear actuator, I am still trying to think of it in that same manner, in terms of how much weight can the linear actuator push a certain distance away.

Th232

12-04-2013, 07:56 AM

The units of torque are kg.cm (or to use SI notation, N.m), not kg/cm. This is a common mistake in a lot of servo specifications for some reason.

Torque is similar to force in the same way that rotational velocity is similar to linear velocity, torque and rotational velocity are both dependent on the distance a point is from an axis of rotation. E.g. if you have two spools winding up thread at the same RPM, if one is of a larger diameter it will wind up more thread.

Thus your question of how much weight a linear actuator can push from a certain distance away is rather much an inconsistency in your comparisons since there is no axis of rotation. It either can or cannot push the weight. Think of it like towing something with a car, it doesn't matter if the object is 1 metre from the car or 10 metres, if it moves in the first case then it will move in the second.

As for how much weight you can push, regardless of distance, more information is needed. In particular, losses due to friction and the direction of travel are important. Pushing 1 kg across a surface requires much less power than lifting 1 kg, and if I'm at an ice rink I can easily push several hundred kg, but I definitely can't do it over concrete.

NB: This may not make sense, it's 1 am here and I really should be sleeping.

AVRnj

12-04-2013, 03:16 PM

This kind of makes sense, but again, my Physics is really lacking.

Thanks for trying to help me, I appreciate it.

jwatte

12-04-2013, 10:19 PM

The math works something like this (without considering loss):

1) The motor generates X amount of torque.

2) The gearbox to the worm drive gives you mechanical advantage and rotational slowdown; still in torque.

3) Once the worm drive rotates, there is a torque acting at the worm drive contact point radius from the center of the worm screw.

4) If the angle is exactly such that the thread moves as much as the rotational radius when the screw turns, the amount of linear force you get equals the torque at that radius of the worm screw, divided by that radius.

5) When the screw is angled less (which is common,) you get proportionally more force, and less speed.

If I remember the math right, the actual torque:force multiplier is (2*pi*radius)/(pitch*radius) so 2*pi/pitch.

Thus, if you have a 2 N.m motor with a nominal speed of 3000 rpm, a 100:1 gearbox gives 2 N.m, and then a 20 mm pitch worm screw gives you a 2 N.m/0.020m == 100 N force. The speed would be 3000 rpm / 100 gear-ratio / 60 spm * 20 mm/r == 50*20mm/100 s == 10 mm/s.

Or something like that. Probably should check a mechanics textbook instead to be sure :-)

AVRnj

12-06-2013, 05:38 AM

Guys, Thanks for both of your help! This gives me something to work with, much appreciated!

jwatte

12-06-2013, 11:16 AM

2 N.m/0.020m == 100 N force

... which did not include the 2*pi factor, so that really should be 628 N. If my math is right ;-)

tician

12-06-2013, 11:27 AM

"power screw formula" is a search phrase that produces several decent results.

Or simply jump to a wikipedia (http://en.wikipedia.org/wiki/Leadscrew#Mechanics).

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