Gear Motors

Hello
I know this is not a plc question but many of your probably have the knowledge to answer this. I have a Dayton gear motor rated for 43inch lbs spinning clockwise. When it is spinning slowly, which is what I need it to do, I can make it spin clockwise faster by hand. I need it to have more resistance in the clockwise rotation. Do I get a motor with more torque?? How do I fiqure the torque in that direction?? The motors are rated for torque in the oppisite direction the motor is turning.

What are the torque requirements for the application? What kind of gear motor?

I need the motor to retain its set RPM while a torque of 87in. lbs. is put on it clockwise, the motor will be turning clockwise as well. It will be a 90vdc parallel shaft gearmotor.

Sounds like your 43in/lb gearmotor is undersized/geared for the application. A higher gear ratio may help if your application can handle the effect that would have on top speed.

My top speed is very low. Maybe 10rpm. How would I figure out which one is enough torque/ratio?

Sounds like you should use a gear reducer that's non-backdriveable, like a worm gear drive.



-Eric

torque

in/lbs * 12 = ft/lbs
torque*rpm/5252 = hp
If the speed needs to be real close you may need some feed back from the motor.

Eric Nelson said:

Sounds like you should use a gear reducer that's non-backdriveable, like a worm gear drive.



-Eric

Click to expand...

Eric seemed to me the issue was more about being FORWARD driven...ie the loads torque over drives the motor.

I would say you need to determine the actual torque needed for the application. Since it is a Dayton gearmotor it was probably obtained at Grainger. They offer a variety of 90vdc gearmotors with different speeds and torque capabilities.

Your quoted specs would be something like this (I assumed the 10 rpm wasnt max speed rating of the motor):
http://www.grainger.com/Grainger/productdetail.jsp?xi=xi&ItemId=1611777458&ccitem=

There are numerous methods to determine torque required, if you get a chance take a look at this thread, specifically post 80.
http://www.plctalk.net/qanda/showthread.php?t=11015&highlight=Learning

Since you know what speed you want then try to obtain a gearmotor that is rated to run at 10rpm with the required torque.

rsdoran said:

Eric seemed to me the issue was more about being FORWARD driven...ie the loads torque over drives the motor.

Click to expand...

...

Although, a worm drive might work in this case as well...



-Eric

This is really a braking application. I would determine the braking torque needed(something was said about 78 in-lbs but its not clear to me whether this is the net torque or not) and then purchase a DC motor capable of that torque in reverse as well as forward. Then a reversing-regen drive if such a thing exists at 90VDC.

As mentioned, this system would work a lot better if there was speed feedback on the motor.

The problem with high ratio worm gears is that they tend to lock up when back driven. Whenever I've tried to do any kind of motor braking thru one of them, all I got was violent jerking which seemed ready to tear the gearbox apart. I'd stay away from worm gears.

I reread the original post and a few points do not make sense.

The motors are rated for torque in the oppisite direction the motor is turning.

Click to expand...

These 90vdc motors are usually reversible and rated torque is avaialable in either direction. The shaft should not turn faster in either direction unless the torque rating is exceeded.

When it is spinning slowly, which is what I need it to do, I can make it spin clockwise faster by hand.

Click to expand...

As I mentioned these things are usually hard to turn by hand BUT depending on what is attached to the output shaft you could easily exceed 43 in lb, you could easily exceed 43 foot pounds. The 90°units are usually worm gear and harder to turn by external forces then the parallel shaft models.

These are small subfractional HP motors and the applications rarely need anything like regen or feedback. They are used in applications where you may need to slow run/jog for setup and run speed may vary slightly. In the old days LABEL applicators were nothing more than one of these motors and a DC controller, you matched the speed to match the line speed which could vary from product to product. I have used them on small feeders where different feed rates were needed.

Whatever the application is, if the original poster can determine the overhung load by using weight or the torque using a guage, torque wrench etc then he will be closer to figuring out what HP motor he needs. POSTER can you describe and/or show a picture of the application?

It may be the case that the motor is fine for the application but what ever is attached to the shaft gives him the leverage to exceed the 43 in lb rating.

Definitely undersized

Unregistered said:

Hello
I know this is not a plc question but many of your probably have the knowledge to answer this. I have a Dayton gear motor rated for 43inch lbs spinning clockwise. When it is spinning slowly, which is what I need it to do, I can make it spin clockwise faster by hand. I need it to have more resistance in the clockwise rotation. Do I get a motor with more torque?? How do I fiqure the torque in that direction?? The motors are rated for torque in the oppisite direction the motor is turning.

Click to expand...

I need the motor to retain its set RPM while a torque of 87in. lbs. is put on it clockwise, the motor will be turning clockwise as well. It will be a 90vdc parallel shaft gearmotor.

Now that I put the numbers together I see the problem clearly. You have an overhauling load (imagine a crane) of 87 in-lb with a 43 in-lb driver. A descriptin of the process would help. You dont say what final drive RPM is. What you need is more of a brake than a motor. Fourtunately this is a small unit of 87 in-lb or almost 7.5 ft-lb

The overhauling load will make the motor run faster.

You can use a work gear to control this
BUT you have to be careful about this because if the pitch is wrong on the worm they CAN be driven backward contrary to common knowledge. So maybe a worm gear drive WITH a properly pitched worm rated at 87 in - lb (or more) will control this OK.

How critical is the speed? Do you really want to spend the time and money to overcome this problem?

Dan Bentler

Like Eric mentioned before, a "none-backdrivable" worm gear will help to prevent the overhauling condition, however you may end up sacrificing speed. If speed is important, then you need a bigger gearmotor.

In general if you have a very low speed and will never need to exceed that speed you should size the gear reduction to meet the maximum gearbox output speed. For instance if you are using a motor with a max speed of 1800 rpm and will run the output shaft at 10 rpm you could theoretically use a 180:1 gear reduction, of course in practice you need some safety zone for speed so a reduction of 150:1 or so will give you a lot of torque and help control the inertia on the output shaft. In general I believe the inertia at the output shaft is reduced by the square of the gear reduction. If you need very tight speed control it gets expensive because you need feed back to the DC drive so you know how fast the load side is really moving.

Something that bears mentioning here is that the torque seen by a motor shaft isn't dictated by the size of the motor. It is dictated by the size and direction of the load!

A 1 hp motor at 1800 rpm will produce UP TO 35 in lb of torque at the motor shaft. If the load only produces half of that torque resisting the motor direction then the motor only puts out 17.5 in lb of torque.

If the load resists movement by twice that torque, then the motor will stall out and no movement occurs.

If the "load" is in the same direction as the motor movement then the motor doesn't have to produce any torque for the load to move. This is often called an overrunning load.

If you have a gear reduction the torque the motor can provide to the load is multiplied by the gear ratio. The torque seen by the motor is divided by the gear ratio. And of course, you have to overcome friction in the gear box going both ways. This friction is subtracted from torque in both directions.

Spur and helical gears have high efficiencies, and so it is possible for the load to drive the motor backwards if it produces enough torque in the opposite direction of the motor rotation. It is possible for the load to drive the motor faster than it's rated speed if the load produces enough torque in the same direction as the motor rotation.

Because of the angle between the worm and worm gear and because of the sliding between the worm and the gear, backdriving a high ratio worm gear reduction is difficult or impossible if the ratio exceeds 30:1. However, if this is a hoist or other life threatening situation, don't count on the worm gear to stop rotation. Vibration can cause the worm and gear to loose contact and allow the load to move backwards.

Look at the mechanics of the drive train and figure out what is going on.