I saw some of the nicely done drive trains such as Woburn as the image posted by the Technokats in a different message. I was wondering how one could go about figuring out the gear ratios for connecting two different motors to a singular drive train. My thought was that find the ratio of RPM’s and then match gears with the reverse ratio. Is that correct? For example the drill motor has 300RPM (not true) and the fisherprice motor has 100RPM, would you put a gear ration of 1:3 on the two motors, where the gear on the fisher price motor is 3 times greater than that on the drill motor?
Just remember - any amp that goes through the FP or Chiphua motors is an amp that could be better spent in the Drill Motor. Multiple motors on one axis gains you the right to put multiple speed controllers, and therefore multiple fuses on one axis. That’s it. Does that have value? Yes. Is it worth the weight of the Chiphuas and the associated gear trains? That’s up to your team.
The question of whether to put multiple motors in parallel is not a simple one to answer.
I don’t not agree with those who say that you must always match free speeds of the motor or you must always match stall torques or those that say that an amp through a non-drill motor is an amp better spent going through the drill motor (sorry Ryan, but I have to call 'em as I see 'em).
This is a very complex issue.
Let me try to simply it. The two motors geared in parallel (by which I mean that the speeds of the motors are related through a fixed gear ratio) essentially function as a single motor. The speed torque curve is a straight line, just as it was for the individual motors. How many points determine a line, 2 of course. In order to understand the behaviour of the 2 motors system you just need to figure out how the system reacts at two points.
To get the two points, use the method of superposition. Calculate the speeds and the torques of the output shaft for the motors individually. To calculate the behaviour of your system, set the speeds of the two motors equal and then add the two torques
For me, the easiest two points to calculate are at stall point and at the free speed of the motor that makes the output shaft rotate slowest.
Drill motor (20,000 RPM free, .7 N-m stall) with ratio 5:1 at .8 efficient
Chiaphau (5500 RPM free, 2.2 N-m stall) with ratio 2:1 at .9 efficeint
Note that I have not matched either the free speed or the stall torque – why should I?
Also note that I have put an efficiency of the Chiaphua’s gearbox higher because I suppose that a 2:1 can be done in 1 gearstage while a 5:1 may take 2 gear stages depending on space requirements – and because I want to make a point to folks that effeciencies are something you should think about as you do your calculations
Output Free Speed 20,000/5 = 4,000 RPM
Output Stall Torque .75.8 = 2.8 N-m
Speed-Torque Curve for output shaft for this motor alone:
Speed = (-4,000/2.8) Torque + 4,000 (simple y=ax + b algebra 1 stuff)
Solving for Torque:
Torque = (-2.8/4,000) Speed + 2.8
Output Free Speed 5,500/2 = 2,250RPM
Output Stall Torque 2.22.9 = 4.0 N-m
Speed-Torque Curve for output shaft for this motor alone:
Speed = (-2,250/4.0) Torque + 2,250
Solving for Torque:
Torque = (-4.0/2,250) Speed + 4.0
Now let’s do the Superposition:
Speed = 0
Torque = 2.8 + 4.0 = 6.8 N-m
Free Speed of motor that drives output slowest (in this case the Chiaphua):
Speed: 2,250 RPM
Torque = Torque from Chiaphua + Torque from Drill
Torque from Chiaphau = 0 (this is the definition of “free speed”)
Torque from Drill = (-2.8/4,000) * 2,250 + 2.8 = 1.2 N-m
Torque = 1.2 N-m
With these two speed torque points (0,6.8) & (2,250,1.2) you can draw the whole speed-torque curve.
Alternatively you could just write the equations out after doing a bit of algebra:
TorqueOut = (-2.8/4,000 - 4.0/2,250) * SpeedOut + 2.8 + 4.0
TorqueOut = (- 6.8/2,750) SpeedOut + 6.8
I wrote the equation the way I did so that it was obvious what the stall torque is (6.8N-m) and what the free speed is (2,750 RPM).
This gearbox of two motors acts just like a SINGLE motor with a free speed of 2,750 RPM and a stall torque of 6.8 N-m.
Pretty simple, right? Well, not really.
To see what this means, lets suppose that you wanted to run at half stall load, 3.4 N-m. In that case the output shaft would be spinning at 1,400 RPM, you would be getting about 470 Watts of mechanical power out of your gearbox. Where is that power coming from? Let’s see:
At 1,400 RPM, the Drill is providing:
Torque = (2.8/4,000)*1,400 + 2.8 = 1.8 N-m
This means that the drill is providing roughly 250 Watts of mechanical power
Similarly, the Chiaphua is providing 1.5 N-m and 210 Watts (there are some rounding errors creeping in to my calculations but the idea is sound).
This seems pretty good, they are sharing the load pretty well.
But now suppose that I want to run at 1 N-m load? The speed would then be 2,350 RPM. At that speed, the Drill is contributing 1.2N-m of torque and 282 Watts of power while the Chiaphua is contributing (-0.2) N-m of torque and (-47) Watts of power (yes is it providing NEGATIVE torque and NEGATIVE power, which is expected, you are asking the motor to go faster than its freespeed, that requires something to drive it faster than it would like to go under no-load, in this case, that something is the drill motor!)
Note that it is not the fault of the Chiaphua motor, we could have just as easily picked ratios that made the drill contribute negative power at lower loads/higher speeds.
I think that perhaps this discussion should be condensed into a white paper because it is coming up more and more frequently and it is always a source of confusion and mis-information.
I hope this has shed a bit of light on the subject for now at least.
I am off the the Buckeye Regional… …wish me luck.
A similar analysis of motor current and motor effeciency (by which I mean the ratio of the mechanical power out divided by the electrical power in) would make it clear that it is incorrect to assert that current is somehow mis-directed if not sent through the drill motor.
Generally speaking, motors are more effecient when lightly loaded. Given that you have multiple motors of approximately balanced power ratings (namely the drill, the Chiaphua, and the Fisher-Price), for a given load (i.e. torque at the output shaft), you can choose gear ratios that would make the torque on the motors be 1/Nth the torque it would see if it was doing all the work on its own.
This means that instead of running at 3/4th of the stall load for the drill for instance, if you are the TechnoKats, you run the drill, the Chiaphua, and the Fisher-Price each motor at 1/4th of their stall loads.
Let’s be clear about how big a deal this is: At 3/4th stall, the drill is about 25% efficient – it turns 3/4ths of the electrical power put into it into HEAT!. At 1/4th stall, the drill motor is over 70% efficient – that means that it generates about 1/3rd the heat per unit of mechanical power you need at the output shaft – believe me, this is a good thing.
I don’t want to belabor the point, but think about it, the Chiaphua’s and the Fisher-Price motors are going to have similar gains in efficiency. AND THESE EFFECIENCIES DON"T MULTILPY as most cases they do so basically your drive system is going be 3 times as good at turning electical power into mechanical power. This means that you are generating 1/3rd the heat for a give amount of power you need. AND… …HERE IS THE EVEN MORE IMPORTANT THING… you have THREE motors not ONE to dissipate that heat.
Bottom line: 3 motor systems like the Technokat’s drive are going to have much longer battery lives and have a much easier time staying cool than those that use only one motor per side of their robot.
Dr. Joe -
I am not understanding your use of gearbox efficiencies at the motor stall point. Gear efficiency is a function of kinetic friction. But, if the gears are not moving - the definition of stall, yes? - then there is zero kinetic friction. The efficiency should be 1.0?
But, if you are referring to electrical efficiency, then it is zero at stall, because there is no mechanical output - yes? That would not generate any useful equations, though, would it?
Or, are you simply putting the efficiencies into the equations at that point so they will propagate through the rest of the equations derived from it?
You are absolutely correct about this topic deserving a “white paper”. It would be fascinating and useful information. But first I think we will need a thorough discussion of the basics of motor testing, and the four graphs - torque vs speed, torque vs current, torque vs power, and torque vs efficiency.
There is some good information on this site : http://www.micromo.com/03application_notes.asp
See what you think of it - might be a good starting point. Of course, I am a mechanical engineer, so maybe this looks better to me than it will to you.
The effeciency term in the equations for the stall torques are a catch all for a ton of losses that occur at the interfaces of the parts: The losses at the bearing, the losses as one gear tooth meshes with another, the losses due to squishing grease out of the way, etc.
For normal spur gears, with reasonably well designed bearings, the effeciencies should be in the mid to high 90’s. As a practical matter, with one of a kind gears running in one of a kind gearboxes designed under insane time pressures, I think it is wise to use 90%. It should be conservative, but you will thank your stars for the extra torque when you need it.
P.S. I agree, that the motor curves you discussed are very important for FIRST folks to understand. DrJoesMotorCalc.xls in the whitepapers section of these very forums will calculate all four of them for you. All you need are 4 pieces of information: FreeSpeed, FreeCurrent, StallTorque, & StallCurrent for any one voltage and the spreadsheet will give you all 4 charts at whatever voltage your gearbox designin’ heart desires!
Why not gear down the drill motor to match the slower Chiaphua motor at (no load)free speed? Then you would not have the problem of negative power and negative torque at higher speeds.
I hate to beat a dead horse, as this thread is quite old, but I’ve been thinking about this situation recently, and I have a question.
In the situation presented where the Chiaphau is contributing NEGATIVE work to the system, would adding an Overrun Bearing/Coupling help in any way?
This would allow the Chiaphau to contribue ZERO work to the system in the case where the free speed of the system was higher than than the free speed of the Chiaphau. As the torque requirements increase, and speeds decrease, the Overrun Bearing would allow the Chiaphau to contribue POSITIVE work.
In effect, you would only be adding the Chiaphau motor to the equation in situations where it’s really needed.
I was thinking of exactly such a system a while ago (with a one-way bearing), but the fatal problem with such a system is that it only works in one direction - and FIRST robots almost always have drivetrains that have to go equally in either direction. There may be a way to make it work with some complicated gearing and multiple one-way bearings, but if so, I don’t know what it is.
I seem to have forgotten that fact that the motors need to run both ways.
I wish there was a delete post function.