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Very complex question...
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.
Example time:
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
Drill alone:
Output Free Speed 20,000/5 = 4,000 RPM
Output Stall Torque .7*5*.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
Chiaphua alone:
Output Free Speed 5,500/2 = 2,250RPM
Output Stall Torque 2.2*2*.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:
Stall Condition:
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.
Joe J.
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