Newbie Question on DC Motors

JohnBot115

Hi,
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I’m new to the forums and Robotics generally. I’ve been reading and experimenting with some basic stuff. Primarily ‘Building Robotic Drive Trains’ by Clark and a few basic electronic experiment type books.
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Question for you:
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I understand that if I have a gearhead DC motor rated at, let’s say, 250 ounce-inches (15 pounds), that I can lift 250 ounces on a string attached to a gear / pulley one inch in diameter. And that at two inches, that torque is cut in half, at 3 inches a third and so on.
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1)····· What if the gear is attached to an axle 1 foot long. How do I calculate whether or not the axle will turn?
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For example, let’s say I have a catapult device design to launch a tennis ball. There is an 2-fot arm attached to an axle, and at the end of the arm, a small cup holds a tennis ball. The axle rotates the arm forward. When it stops, the tennis ball launches forward.
Assuming the power for the launch comes from another source, let’s say, nylon or surgical tubing that pulls the arm forward. And also assuming the purpose of the motor is to rotate the axle and arm back to the ‘ready’ position (i.e. work against the force of the nylon tubing)… how would I calculate the force, or ounce-inches needed.
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So far, I’m guess I will get a force gauge, and measure how much power it takes to pull back the arm against the tubing. If that is, let’s say, 10 pounds of force, then I need a motor capable of generating 10 or more pounds of torque. But I’m also guess that the length and weight of the axle and arm will complicate things. I’m just not sure how to proceed to calculate these variables, or even look for the formulas.
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Any help would be much appreciated.

Thanks.

John
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CannibalRobotics

Sounds like a unit problem. In foot-pounds your motor has a torque of 1.3 Ft-Lbs. (Let's call it Tm)
A hooks law spring has an axial force of F=-kx where x is the displacement and k is the 'spring constant' with units of (force/distance). The spring constant comes with the spring or can be calculated by hanging a known weight on the spring and measuring the displacement. The negative sign is there to show that a spring wants to go the 'other-way' from a 0 based displacement. Now, if you want to cock a spring loaded device with a level arm, like a throwing device, all of the math works at the pivot of the throwing arm. The torque from the spring is (-kx)*rs where rs = distance from the spring attachment to the pivot. Knowing that torque, you can decide where to attach the line from your motor. The force from the motor is Tm * rm where rm is the distance from the pivot to the string attach point on the lever.

Hopefully this helps a little, I'll put together a drawing and upload a little later.
Jim-

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SRLM

A note: the length of the axle has no effect on the (ideal) solution. The axle could be one inch or 1000 inches, and nothing would be different in the calculations. This is because it is considered to be a rigid object, and so it's moment of inertia simply depends on the mass and distribution of that mass with respect to the rotational axis.

As for your calculations, I would look at what rotational kinematics has to offer. The wikipedia article on torque would be a good place to start (en.wikipedia.org/wiki/Torque), as would almost any physics text book (introductory). But unless you have to do the calculations as part of a presentation or paper or something, I would just guestimate and experiment. Non conservative and unpredictable forces like friction and air resistance would probably significantly throw off your results anyway.

Also, in situations like this you normally gear down the motor before you attach thinks to the output shaft, since most motors are made to run at several thousand RPM and their output power decreases non-linearly with decreasing rotational speed (IIRC). You may also want to consider some way to store the energy (such as rubber bands) and just use the motor to input energy slowly into the system, and the bands to release it quickly. Nevermind, I see that's your plan already...

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erco

After building lots of mechanisms and studying motor torque/rpm plots, I can shed some light on this. To avoid a disappointing undergear/undertorque/underpower situation, only expect HALF of the torque you calculate at any particular RPM. Mechanical losses in gears and pivots add up more quickly than you think. My half-power rule is pretty darn close most of the time. Assuming you started out trying to operate at peak power (half no-load rpm times half the stall torque), this will yield good results. Worst case, you're overgeared with torque to burn, no danger of stalling the motor and burning brushes, and you will operate with higher efficiency, which is typically near 75 or 80 percent of no-load speed.

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CannibalRobotics

OK, Here's the drawing. This is an idealized system and as the angles change, so to do the forces. Erco's comment about efficiency is also a very important one. Let us know how it goes!

Jim-


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JohnBot115

·Hi all,
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Thanks so much for the replies. This is helpful.
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Jim, that is a great summary. It’s a little over my head, though, so I better hit the books and study some basic physics (any articles/websites that you can recommend for motor specific calculations would be great). In the meantime, maybe I’ll just invest $30 bucks in the motor and see what I can come up with. One clarifying question: you said the motor has a torque of 1.3 foot-pounds. Interesting. I had thought that if ·the motor specs said 250 ounce-inches that all I would do is say 250 ounces = 15.6 pounds. And that a one inch pulley diameter, that motor can lift 15.6 pounds. But maybe that’s not correct? Am I converting wrong?
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Also, my next idea is to get a book on basic robotics. I’m considering either the Evil Genius Robotics book or Robot Builders Bonanza. Does anyone have an opinion on these, or know of a better alternative? I do need something basic, but would like it to ramp up eventually so I can actually tinker around and build something
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Anyway, thanks so much for the responses. It’s exactly what I needed- advice from people who are already building robots!
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Thanks again.
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John

CannibalRobotics

Your not wrong, watch·the units. Your motor does have a torque of 15.6 pound-inches.·This is correct but pound-inches is·not a widely used unit. For foot-pounds you need to divide by 12 again to turn the inches into feet. Dimensional analysis goes like this:

··················· · 1lb··············· 1 ft
250 oz in· *

---

·· *·

---

= 1.3 ft lb
··················· 16 oz············ 12 in

Cheers,

Jim-

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SRLM

Don't forget your units: 15.6 is not just a value (like coefficient of friction), it is 15.6 inch-pounds. Multiply by 1 foot/12 inches and you get 1.3 foot-pounds.

As for books, I haven't found any books that I really like. I've read Robot builders bonanza, and it was okay but it just wasn't right. It was certainly informative, but it didn't seem to help much in the way of learning how to build stuff. IMHO the best (and so far only) way to do so is just to experiment. This has the downside of being more expensive than buying a book, but oh the skills are so much more valuable. IMHO.

Anyway, welcome to the world of robotics. Prepare for long days away from the light of the sun and glossy looks from former friends...

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JohnBot115

Ah...ok. I think I see what you're saying.

So, the standard units (SI) are important. And it's saying the same thing, just in a different way. So:

The motor can lift 15.6 pounds on a lever 1 inch long or the motor can lift 1.3 pounds of a lever 1 foot long. But the SI is important, so we'll use the Foot/Pounds.

This is probably why 15.3 / 12·= 1.3. I.E... the lever can lift 15.6 pounds at one inch, 7.8 pounds at two inches... and carry that forward... it can life 1.3 pounds at 12 inches.

SRLM

Just a note, but feet and inches are not SI. SI is things like meters and kilograms, so the SI unit for torque is newton-meters.

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CannibalRobotics

Really, at the end of the day, it does not matter what you use as long as your consistant.
HOWEVER, when you start to talk to others about the project, it's nice to use standardized and well defined units. It's easy to loose sight of unit diferences in a conversation.
See "Mars Observer" for demonstration of this in action.

Jim-


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Scope

CannibalRobotics said...
OK, Here's the drawing.

I have a lamp that looks just like that.

JohnBot115

One other quick question:
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I’m using nylon tubing to power the forward motion of the arm, and the motor to retract the arm to ready position. Obviously, the motor will turn it’s shaft, and therefore rotate the axle to retract the arm. However, when it releases, that will spin the shaft in the opposite direction.
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I thought it would be best to use a one-way bearing on the axle (or on the sprocket if I use a chain between the motor and axle). So that the bearing will be fixed in one direction (i.e. it won’t turn, so when the shaft rotates, it pulls the chain), but free-wheeling in the other (so that when the arm moves in the opposite direction, it can move full speed and not rotate the shaft).
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My question is: do I need the one-way bearing or can the shaft rotate freely in the other direction. From my understanding, it’s not a good idea to manually rotate the shaft. Thanks! ·

SRLM

For your application it probably doesn't matter if you rotate the motor shaft. It might wear a little bit more, but unless you are doing tens of thousands of launches you won't notice a thing. If you have a gear box it might slow down the launch, which could be fixed by simply running the motor in the opposite direction.

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CannibalRobotics

Back EMF could be an issue. At the moment you start turning the shaft, the motor becomes a generator and current starts moving in your circuit. If you can disconnect the throwing rotation from the cocking rotation it would be good.
For parts, go find an old laser printer. Most of them have solenoid powered shaft clutches that do exactly what you are working on.
Good luck!
Jim-

PS: Scope, does your lamp throw stuff?

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GWJax

Yes electrical-mechanical clutch with a gear head would work perfectly in this devise. You can find these in copiers such in the feed drive section, Registration section and the fusing output section. I repair copiers all day and have been for around 26 years. Laser printers also have them but really would not have the strength to hold while the winding is taking place. Also make sure you get at lease a 24V clutch for this type of torque unless you step down the gears to stay under 5 f/p/si then you can use either a 12 volt or even a 5 volt clutch if you can find them.

Jax

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Al2

Hi:

No one here has seen the error?

"I understand that if I have a gearhead DC motor rated at, let’s say, 250 ounce-inches (15 pounds), that I can lift 250 ounces on a string attached to a gear / pulley one inch in diameter."

It would be able to lift 500 ounces ! You have to look at the pulley as a·lever that is 1/2" long.


Al at http://diyrobots.webng.com
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JohnBot115

Thanks so much, this is great advice. But if you don’t mind, could someone explain what a electro-mechanical clutch solenoid is, in a little detail? I’m familiar with solenoids. I have a couple of 12 volt, mini push solenoids. But I haven’t heard of a clutch type solenoid. I’m guessing it connects and disconnects some sort of grip (clutch?) to the shaft when it needs to rotate? Sounds perfect. But I’m not familiar with it at all…? I did just order a clutch-type bearing seen at this link. Perhaps it’s similar?
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http://www.allelectronics.com/make-a-store/item/CH-10/ONE-WAY-ROLLER-BEARING-AND-SPROCKET/1.html
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Also, I think I discovered a fatal flaw in my design:
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If the torque is 250 oz-in, that means by definition -it can lift 250 ounces on a ·one-inch lever off the motor shaft. But in this case, the throwing arm coming off the shaft / axle works like the lever. That throwing arm needs to be around 18 inches long. The rubber band generating the forward force will attach to the arm a little more than half way up- let’s say at the 12” mark. That means that the resistance the motor needs to overcome is at a point 12” up the lever. The motor can lift 250 ounces at one inch. But at 12 “ it can only lift 1/12 of that, correct? ·So in reality, 1/12 of 250 ounces is 20 ounces… just over a pound. That may not be nearly enough torque.
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Does that sound right? If so, my fix is to instead attach a small rope to the back of the throwing arm. Have that run over a small pulley and down to the motor. That way, the motors torque is pulling directly on the arm, at the point of contact with the band, and it doesn’t lose a lot of power. See the simple diagram I attached.
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Thanks again for all the help.
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