Need help with DC motor selection

[tombstone]

I am a newbie to these forums, and was hoping you folks could help me with a question I have. How do you determine the torque required to move an object of a certain weight?

I am trying to build a simple DC motor powered sliding drawer which will hold a typical sized computer tower. I guess-timate the total weight to be about 5 kilograms max. The drawer will be on wheeled tracks to minimize friction.

Is there a formula or something that I can use to figure out the grams/cm minimum torque rating for the DC motor I would need to accomplish this project?

Thanks in advance!

 

[stevez]

While you might apply some theoretical approaches you will have to make an awful lot of estimates or assumptions. If this is a one time effort I'd advocate an approach that might be overkill - as long as overkill didn't generate an unworkable solution.

You probably won't know much about the friction. You probably can't be sure that the rails won't sag creating a low spot. Are the rails level and can they be maintained that way - you probably don't know or can't control that precisely. A good design will have to anticipate this - and will result in a larger motor.

The speed at which the drawer is to open could be an issue - the faster you want it to move - the more power required. Go too fast and you'll need to worry about braking, acceration, deceleration if you don't want to slam into stops and ruin the tower. Too slow and a lot of patience is required.

The force to get something moving might be thought of as breakaway - it's more than the force necessary to keep something moving once it is moving. A motor starting at zero rpm has a predictable amount of torque available in excess of the torque it needs to get itself going - that's got to be figured in to things too.

More in the next post - this will get too long otherwise.

 

[stevez]

To get things going let's imagine the drawer is going to move on a horizontal, level plane and it will be pulled by a rope that is exactly horizontal. Theoretically if there is no friction then the force is more of an F=ma thing - you need to apply enough force to get it moving at the rate you want. Force equals mass times acceleration. The thing we know for sure is that the force is not zero.

Now, if you provided enough force and you pulled up on the rope you'd lift the drawer. Assuming no friction you know how much force you need - a little more than 5 kg. We know that's not the answer but what we now know is that it's more than zero and less than 5 kg of force.

What you might do is think about the worst possible slope that the rails might be at and calculate the amount of force required to pull the drawer up this slope in a frictionless state. A slope of 1 in 10 is about 6 degrees. If it were my project I'd start here. At this slope you'd need 0.5 kg of pull in a direction parallel to the slope (6 deg) to hold the drawer still - a little more to get it moving. If you wanted to plan on 15 degrees you'd have to plan on 5 kg x 0.25 or about 1.25 kg of pull. We could just decide to work with 1 kg.

More in the next post - I am doing this at work between calls.

 

[stevez]

I need to switch to IP. I live and work in an IP environment. In engineering school (1969) they told us that everything would soon be metric. If they are correct then all I know is "soon" is more than 37 yrs.

1 kg is about 2.2 lbs. Let's call it 2 lbs of force is what we need.

Lets imagine that we'll pull on a cable that winds itself on a drum that is 2" in diameter ( 1 inch radius). To generate a 2 lbs force on the cable we would need a torque of 2 in-lbs - 2 lbs with a lever arm of 1 inch.

It seems unlikely that we'd connect the 2" drum directly to a motor. We'd probably reduce the motor speed which increases the available torque. Remember that we have to think about the running torque of the motor once we're moving - and the starting torque to get things started.

We have to decide on the speed the drawer will move. Let's start with 1 inch per second - it's in the ballpark, so to speak. The drum is 2" in diameter and will take up 6.28 inches of rope or cable on each revolution. The drum will have to rotate at 1/6.28 rev per second - that's just under 10 rpm.

If the nominal load speed of the motor is to be 2000 rpm and the max speed of the drawer is 1 inch per second - you'll need a reducer of 200:1 to get to 10 rpm. The torque at the motor shaft will be 1/200th the torque at the drum

 

[stevez]

I have to get back to work. The approach shown isn't too much different from a ball screw or some other arrangement. There is a more direct way to calculate horsepower or the metric equivalent but it won't get you the information you need to see if your motor will actually start.

Does this give you a start?

 

[stevez]

Sorry for the fragmented stuff.

One other approach - find a drawer that is similar and load it with about 5 kg. Take a level and create a deliberate slope in a way that increases the force - an amount that might represent what you'd consider to be a typical case or maybe worst case. With a fisherman's scale - pull on it - making note of the breakaway force and the force while in motion. This may not exactly represent what you have but it will help give you a place to start.

Then - describe the mechanism that you'll use to connect between the motor and drawer. If it's a screw, we'll need to know the pitch.

 

[tombstone]

Fantastic explanation. That is exactly what I needed to get me going. As I was reading through it, I started getting flashbacks of the physics courses I took in university.

Basically, given that I will likely purchase a Gear Head DC motor (built-in gear reduction), I would need a model with a minimum of 1.25 Kg/cm (assuming a 1 cm radius gear/drum), and appoximately 24 rpm rotation speed ( 2 cm diameter gear/drum would pull 6.28 cm of cable per revolution, so for 1 inch per second, it would need to turn 2.54/6.28 revolutions per second). Sorry to drag this back into metric, but the Gear Head motors I'm looking at have their torque ratings in grams/cm.

For my purposes (and assuming I got it right - and noting the fact that I always try to err on the high side of my estimates), I located a Gear Head DC motor, with 2000 grams/cm of torque, 132:1 gear ratio, and a speed of 42 rpm (which would allow me to achieve close to 2 in/s - best case senario).

The drawer will be built into my desk, so it should be fairly close to level (allowing for the 6 degree incline should compensate for any slope in the floor of this ancient house I'm in). Also, the control circuit I designed for this uses a latching gate switch to supply power to the motor (bi-directional for open/close using a single switch) which resets via 2 roller switches placed at the extremities of the allowable drawer positions. The roller switch positions are adjustable in the assembly to allow coast-to-stop adjustments so the drawer doesn't halt abruptly when opening/closing. The only thing I haven't figured out yet is how to detect a stall condition, for cases where the assembly has jammed, or there is an obsticle in the way too heavy to push. I assume it would have something to do with monitoring the current through the motor, and resetting if it reaches the max rating of 340 mA.

Thank you very much!