Sizing up an electric motor - Page 1
Forum section: electronics / projects

author: TimNJ
author_rank: Super Contributor
timestamp: September 17, 2013, 08:28:19 pm
content: For my first year engineering project, I need a mechanism that can push a 2kg load around 3feet, and do it as quickly as possible.I'm looking for a motor that would be able to do this, and I'm not quite sure what to make of the numbers.Probably one of the more popular motors I have seen is this Mabuchi model (http://www.jameco.com/Jameco/Products/ProdDS/2144673.pdf).It has a "full load torque" of 380gm*cm. And it has a "stall torque" of 1360gm*cm.I'm not entirely sure what these terms refer to. I'm guessing stall torque is the torque the motor produces from a standstill. But what is full load torque defined as? The force exerted at the full load speed?If what I have said is true, how can you translate these torque figures into 2D motion? As in calculating the acceleration of the load, final velocity, etc. I know this is probably more of a physics question than anything, but any ideas?Thanks.

author: Kremmen
author_rank: Super Contributor
timestamp: September 17, 2013, 08:55:05 pm
content: OK, firstly a specification such as "as quickly as possible" does not really make sense. You'd be surprised what modern science and technology makes possible...So you need to come up with specific numbers.Stall torque is the torque of a non-rotating (locked) shaft when full voltage is applied to the motor. You should not try this in practice except for the smallest motors. Full load torque means the torque produced at nominal speed and nominal power.Torque translates to force via the moment arm. Torque in SI units is measured in newtonmeters so you get the linear force in newtons by dividing the torque with the length of the moment arm (in practice the effective radius of the driven wheel). From the force you can then find the rest by applying simple kinematics, Newton's laws and all that.

author: bilko
author_rank: Frequent Contributor
timestamp: September 17, 2013, 09:19:11 pm
content: Quote from: TimNJ on September 17, 2013, 08:28:19 pmTake for examplehttp://en.wikipedia.org/wiki/RailgunIs this what you are looking for ?

author: TimNJ
author_rank: Super Contributor
timestamp: September 17, 2013, 10:19:14 pm
content: Well I wrote as quickly as possible because I am in the preliminary stages and I'm not sure what i can do with the resources at hand. The project cannot exceed $50 in parts. I'll use 3 seconds as a guess. I'm going to do my calculations as a write this, so the 3 seconds might be gone by the end of this post.I decided to draw what I think is the speed-curent/torque graph. If it makes any sense to me, it looks as if this motor can only be used safely between around 4k-5kRPM. If it's loaded too much where it's going slower (than ~4kRPM) it will have more torque but also an unmanageable current use. Am I correct in saying this?In my case, since the motor would only be rotating on the order of a 100s of RPMs (MAX), then would this motor still be useful? Provided the entire sequence only lasts a few seconds, I cannot imagine the motor burning out within that timeframe. Operating at stall current is obviously not good, but since it's not operating at close to stall current for a prolonged time, is it okay?I would get roughly a torque of 0.133Nm at the pivot point. Dividing by a moment of perhaps 0.05m for a wheel, would give me a linear force of 2.66N. This is assuming that the motor is operating at around stall current the entire time, since it will probably never reach over 100RPM which is tiny in comparison to its max and nominal speeds. Using F=ma, that would mean the 2kg object would experience an acceleration of 1.33m/s which means that it could push the 2kg object over 3ft in under 1 second? That seems rather quick actually. Is this possible? It's only a $3 motor.If so I guess my next question would be, is 2kg enough mass to keep the wheels from spinning out. Perhaps the voltage would have to be ramped up so that there's no slippage between the wheels and the ground.The other issue is the enormous current demand between 10-12A. Perhaps a large bank of low ESR capacitors charged to 12V will be needed to provide the momentary burst. I think a 10-15A AC/DC power supply would be very expensive. If possible, it would be nice to use a low current power supply to charge the capacitors rather slowly. I have not done any calculations or research on that so I am not sure if that is feasible or a good approach at all.

author: NiHaoMike
author_rank: Super Contributor
timestamp: September 18, 2013, 03:30:07 am
content: No need for caps at that level. Some common NiMH batteries can supply very high peak currents, and RC lithiums even more.You can easily buy a cordless drill or two from Harbor Freight for well under your budget.

author: TimNJ
author_rank: Super Contributor
timestamp: September 18, 2013, 04:44:42 am
content: Thank you for the your input.Since part of the criteria is how heavy the total system is, I'm trying to minimize the weight as well. A 10-12 cell array of Nimh will be pretty heavy while I could get away with only 4 small li-ion cells. I think I remember Nimh having better current handling, but I'm hoping to get away with Li-ion.The RC lithiums look ideal, but their chargers are pretty expensive. Not to say that I wouldn't pay good money for a charger any other time, but here we are tight on cash. A complete system might cost in the area of $40, leaving $10 for the rest of the system which includes 2 motors, some sort of control circuitry, and I didn't mention before but a bridge that spans a 3-foot gap.This might sound ridiculous but is there a way I could get something that can provide 10-15A peak current @12V for around 5 seconds...all for under $20?I priced up some super caps. Around $30-35 just for the caps alone.I saw these, but I am wary of their abilities. Do you think they could handle some short term abuse?http://www.amazon.com/TangsFire%C2%AE-Li-ion-Rechargeable-Battery-charger/dp/B00BS9DBRG/ref=pd_sim_e_7

author: peter.mitchell
author_rank: Super Contributor
timestamp: September 18, 2013, 04:55:24 am
content: Buy a cheap cheap cordless drill kit, the brushed motors have incredibly high low RPM torque, meaning they can be geared quite heavily. You also get the battery and charger too. Somethign likehttp://www.harborfreight.com/18-volt-cordless-12-drilldriver-with-keyless-chuck-68850.html

author: TerraHertz
author_rank: Super Contributor
timestamp: September 18, 2013, 05:06:59 am
content: Do the requirements say anything about electric motors? Or safety?If not, then what's wrong with explosives?Or... not quite so dramatic, stored energy of other kinds. Small high pressure cylinder of gas (supermarket CO2 bulbs for eg), a piston (balloon in a cardboard cylinder, and some kind of bursting diaphragm valve. Just give the load a really sharp powerful kick, and let it coast (fly) the rest of the way by itself.Does 'as fast as possible' mean from when you turn your device on, or when the load starts moving?Because if the latter, you can do the clockwork thing - wind up a big spring, then let the spring release it's energy to the load.

author: TimNJ
author_rank: Super Contributor
timestamp: September 18, 2013, 05:35:58 am
content: Sorry I did not mention the specifics.The machine cannot be human powered while it is performing it's job. It doesn't HAVE to be electrically powered. I just took that as the most obvious route.It has to be self triggering, as in, once you put the object onto machine, it automatically starts and carries/pushes/transports the object across the distance and places it into a 1sq ft area. The object is a calculus book by the way. When the book reaches the other side, it must be sitting physically on the table surface. It cannot be on a cart or attached to anything else. There can be "stoppers" at the end to help with positioning, as in, the book can run into it and come to a stop. Initially the book can be at most 1" off the surface of the table, and at the end, it must lie directly on the table.----For something like the CO2 canister, I am assuming you are thinking about those cars that you attach a balloon to and they push air out the back? Interesting. The book is around 5lb (2kg). On wheels it is much easier, could a small CO2 canister do that? With what I am thinking now (with electric motors), the cart would be pushing the book directly on surface like a snow plow (in this case, it will be on a PTFE coated track to lower friction). For the air canister approach, you'd definitely be on wheels. So then comes the problem of dismounting the book from the cart.Maybe there are other mechanical means of moving the book. I don't know much about springs, but that seems interesting. Perhaps a cable under tension attached on the other side (at the endpoint) that when triggered pulls the book across. How it would be triggered, I have no idea. And how the actually mechanism works, I am not sure. I was thinking something like that in a lawnmower pullcord assembly.Thanks.

author: TimNJ
author_rank: Super Contributor
timestamp: September 18, 2013, 05:43:47 am
content: Actually I really like the sound of a lawnmower recoil type thing. Unsure of triggering, but I'll sleep on it thanks.

author: TerraHertz
author_rank: Super Contributor
timestamp: September 18, 2013, 07:36:12 am
content: No, I was thinking of the balloon being in a cardboard cylinder, with a cardboard plunger that 'threw' the load. The balloon just being used to avoid the need for airtight piston action.When they say 'stops', can that be a pile of bricks?Also, does the book need to be intact afterwards?Even with the more complete requirements, I still think explosives would be the simplest way. You know there's very few problems that can't be solved with the application of appropriate high explosives.For extra points, you could launch it in ballistic trajectory, with a three axis accelerometer and some postage stamp processor, and have it integrate the area under its own curve. If you get to provide the book, there's also all kinds of fun ways to cut blocks out of the insides of the book and put things in it. Like that processor. Or hobby rocket motors.I hope you can tell I'm not really trying to help?

author: peter.mitchell
author_rank: Super Contributor
timestamp: September 18, 2013, 08:24:47 am
content: a rubber belt or something for the book to sit on controlled by a motor would be pretty doable. Just put a flap on the belt so the belt does not slide under the book, should be able to launch the book no worries.

author: Kremmen
author_rank: Super Contributor
timestamp: September 18, 2013, 08:37:59 am
content: Quote from: TimNJ on September 18, 2013, 05:35:58 amOh just make a pendulum with a massive hammer that smacks the book along. Depending on the size of the room you perform the experiment in (limiting the arm length), you can get quite some speed from the impact. Spring assist possible of course...

author: amyk
author_rank: Super Contributor
timestamp: September 18, 2013, 10:13:45 am
content: Quote from: TerraHertz on September 18, 2013, 05:06:59 amGood to see I'm not the only one who thought "pyrotechnics" when I read the initial requirement "I need a mechanism that can push a 2kg load around 3feet, and do it as quickly as possible."

author: TimNJ
author_rank: Super Contributor
timestamp: September 18, 2013, 12:37:17 pm
content: Quote from: peter.mitchell on September 18, 2013, 08:24:47 amThat was actually one of our proposed ideas. Didn't think of the flap but that's intelligent. Do you reckon the acceleration calculations would be the same?

author: mzzj
author_rank: Super Contributor
timestamp: September 18, 2013, 06:12:59 pm
content: Quote from: TimNJ on September 17, 2013, 10:19:14 pmInstead of the 0,05m wheel use only tiny wheel so that you will reach full RPM at the 1 meter (3 feet distance). Or use gearing with your bigger wheel.0,038 Nm of torque with a 5 mm radius "wheel" gives you 7,6 newtons and 3,8m/s acceleration with a top speed of 1,9m/s and max motor speed of ~3600 RPM.

author: Smokey
author_rank: Super Contributor
timestamp: September 18, 2013, 09:45:53 pm
content: http://www.amazon.com/Comprehensible-Guide-Servo-Motor-Sizing/dp/0976511614More than you will ever want to know about servo motor sizing.

author: TimNJ
author_rank: Super Contributor
timestamp: September 19, 2013, 02:40:40 am
content: Quote from: mzzj on September 18, 2013, 06:12:59 pmThat's a very valid point. Why on earth does a smaller wheel provide more pushing force? That doesn't make sense to me.What is the downfall of smaller wheels? Less traction? Make them wider?I wish I could figure out how to make them stay on the ground without spinning out.Thanks.

author: Kremmen
author_rank: Super Contributor
timestamp: September 19, 2013, 06:18:19 am
content: Quote from: TimNJ on September 17, 2013, 10:19:14 pmandQuote from: TimNJ on September 19, 2013, 02:40:40 amThere are several factors that define what kind of performance you will get out of an electric motor drivetrain. For a specific application like this, some pointers:- For a brief sprint ( few seconds) you won't burn the motor easily. Its thermal mass will be enough to absorb gross overcurrents momentarily provided you allow it to cool down before repeating. Current itself breaks nothing, it is the heat dissipation associated with losses that does. Overvoltage also does not break anything provided you don't exceed the withstand voltage of the windings and commutator parts. That will happen earliest at several times the nominal voltage, so doubling or tripling that is quite doable.- The torque vs speed curve of a typical DC motor at full voltage is a more or less linearly descending straight line. At w = 0 (zero speed) the torque is the stall torque. At no-load the motor is turning a max revs, producing zero torque. You can assume that the stall torque increases quite linearly as you increase the supply voltage, until the magnetic circuit saturates and can sustain no more flux.- The motor speed is (almost) linearly dependent on the supply voltage. Doubling the voltage will double the speed etc etc. Thus overvolting the motor say 3x will increase the stall torque to nearly 3 x and idle speed to nearly 3x as well. You would have to experiment to see how much overrating is practical, and go with that,- The output shaft power vs speed curve is an inverted parabola, zero at stall and idle speeds, and maximum point halfway, i.e. at half speed.--> You need to dimension the drivetrain transmission ratio such that when your vehicle crosses the finishing line, it has just accelerated to half the max motor speed. At this instant the motor reaches its maximum power, minimizing the travel time to target.To calculate the optimum solution involves some basic calculus, but seeing how your payload is a book of calculus, this should not be out of reach. I won't begin with that because a) this forum is helpless with math presentation and b) i'm too laz^H^H^H busy to go through that. Besides, its your project anyway. But maybe a few pointers: Calculate the problem in rotational coordinate system. You want to minimize the time to accelerate a load, i.e. a moment of inertia to a specific speed, applying a torque that is a linear function of instantaneous angular velocity. The moment of inertia to accelerate is a combination of the inertia of the moving parts of the motor, plus the inertia of the drivetrain and payload, reduced to the motor shaft. A mass in linear motion coordinate system can be reduced into rotational reference frame directly from the definition of moment of inertia, i.e. multiplying with the square of the offset radius of the center of mass from the axis of rotation. In this case it is the radius of the driven wheel. This also answers your question of a smaller wheel: the shorter the radius, the smaller the reduced moment of inertia and the faster the acceleration.Your final comment about traction is a very valid one. There is a point where increasing torque at the driven wheel will break it loose and you don't want that. Optimal solution is critically dependent on maintaining static friction between the driven wheel and the surface. Classical mechanics has the trivial formula for friction force, involving the coefficient of friction and normal force between sliding surfaces. This will get you a long way but there are other factors at play, as we know from F1 racing. Those cars will easily achieve cornering forces far beyond what is predicted by this classical formula. Making your driven wheel sticky may be a good ideaSo this then becomes an optimization problem in linear algebra: minimize the acceleration time such that your motor shaft speed is half maximum at the instant of crossing the finish line, and that the linear force transmitted from the drive wheel to the track surface does not exceed the limit of static friction.

author: TimNJ
author_rank: Super Contributor
timestamp: September 26, 2013, 05:09:15 am
content: Kremmen,Thank you for your response. I know it's been a very long time since I started this thread but it's been a hectic week.Thank you for your clarification on calculating the optimal running conditions and the tips on how to get the most out of the motor.As things would have it, our group changed it's mind on the approach we would like to take. Instead of pushing from behind with some sort of car, we think it would make more sense to have a winch that pulls the cart from the adjacent side until it reaches its destination.I was looking at different DC motors today. I'm actually confused by the torque specification. If torque is dependent on lever arm, how can the manufacturer define a torque? Is the "lever arm" just the 1/4" shaft in this case?

author: IanB
author_rank: Super Contributor
timestamp: September 26, 2013, 05:31:49 am
content: Quote from: TimNJ on September 26, 2013, 05:09:15 amBut torque is not dependent on lever arm. Torque is expressed as force x distance, where the distance is measured from a fulcrum point (a center of rotation), and the force is tangential to the center of rotation (at right angles to the distance vector).The product of force x distance remains constant while the distance or the force may vary in relation to the other. So for a given torque if you double the distance you halve the force that is acting at that distance.This is also why smaller wheels have more pushing force; since if the wheel radius (distance from axle) is less, the force acting at the wheel rim is greater.Steam locomotives on railways made use of this relationship. Passenger express locomotives that needed to pull a light load very fast would have large driving wheels, while freight locomotives that needed to pull heavy loads at lower speeds would have smaller driving wheels. The torque acting on the driving wheels broadly speaking would be the same since there were mechanical limits to the steam pressure available, the length of travel of the piston (and hence attachment point to the wheel), and the cross-sectional area of the piston (limited by steam flow, valve gear and such like).

author: Kremmen
author_rank: Super Contributor
timestamp: September 26, 2013, 06:14:34 am
content: IanB's explanation is basically right, but let me expand it a bit.Torqueis a property ofrotating systems. It is a vector quantity, oriented in thedirection of the rotating axis.It has no distance from anything and there is no lever arm associated with the torque vector as such.What you (OP) are pondering and what IanB is effectively explaining, is the transformation of torque from the rotating system to force in the linear system, and back. Physically, attaching a lever arm of length l to a rotating shaft and pushing the arm with force F will induce a torque of Fl in the shaft. The linear force has now been converted to rotating torque. Thus force and torque are duals in different reference coordinates. And just because of the above example, the unit of torque is conveninently expressed as force times length.So, bearing in mind that T=Fl you of course get the force from torque just by dividing by the length of the arm: F=T/l. Your motor produces torque T' as specified above. You can translate that to linear force simply by attaching a wheel to the motor shaft. Now the linear force available at the effective radius r of the wheel is of course F' = T'/r.In practice, and especially if your drivetrain is complex, you need to subtract the drivetrain power losses from the available torque. Sometimes this is simple, sometimes complex especially if there are viscous elements included, such as oil filled gearboxes etc. At least you should allow for the friction force of moving the load, and any gravity effects due to motion other than horizontal. Subtracting the losses will give you the effective torque at the output shaft of the transmission, and from there the maximum load you can move in the first place. Reducing the losses and load to the motor shaft will give you a basis for calculating available torque for acceleration. Bear in mind that inertia opposes acceleration. If you do careful design you should reduce all moments of inertia - the rotating parts of the motor, the transmission, the load - to the motor shaft. Now you can calculate the time to accelerate to a target speed, based on how much extra torque you have after load and losses, to overcome inertia.Edit: typo

author: TimNJ
author_rank: Super Contributor
timestamp: September 26, 2013, 06:44:09 pm
content: Ah thank you both! That makes plenty of sense. It kind of reminds me of Ohm's law in a strange way...but not really.Well I guess that makes torque a cross product of l and F. But I'm only in physics 201, so I'm definitely going to be looking up some equations on how to find out losses and hell, even some kinematics.I had a bit of an "ah-ha!" moment. Thanks for that.