Re: Adding FP Motor to '05 KOP Transmission
 

There is nothing magic about matching free speeds of motors driving the same gear.

The combined motors acts very much like a new motor with a new stall torque and a new free speed.

The combined motors add in is way*:

Stall Torque12 = TS1 + TS2

FreeSpeed12 = (TS1 + TS2) * (K1*K2) / (K1+K2)

where
TS1 = Stall Torque 1
TS1 = Stall Torque 1

K1 = (FreeSpeed Motor 1) / (Stall Torque 1) <<< NOT "MOTOR CONSTANT" K
K2 = (FreeSpeed Motor 2) / (Stall Torque 2) <<< NOT "MOTOR CONSTANT" K

This new motor has a new speed torque curve, efficiency curve, etc.

Things do not blow up if you drive the slower motor faster than its free speed. In this case, the slower motor adds a net negative torque to the sum, but what of that? The combined torque would be less than the torque of the faster motor in this range, but again, what of that?

As long as things are not TOO out of wack all is well.

In this case, the FP with the 3.7 gearbox (FP/AM) acts like a 4500 RPM freespeed, 1.3N-m stall motor. The CIM is a 5300RPM/2.4N-m motor.

Above 4500 RPM, the FP/AM is a net drag. How much of a drag? Not too much, the combined Freespeed of the motors will be 5000 RPM (the FP spinning at 18.5K) and at that point the FP/AM will be providing -.14N-m of torque while the CIM will be providing +.14N-m of torque. This is not a disaster!!!

Some of you may be asking, Yeah, I believe you that it is not a disaster but I added a motor and I got LESS torque. True enough, but did you add to the extra motor to get more torque when the robot is zipping down the field or when it was struggling to turn because you're using 4WD with wheels in the corners and no trick wheels? Of course it was for the torque at LOAD.

When the torque on your combined motor gets above .23N-m (which is only at 6% of the combined stall torque) the FP is pulling its own weight!

When you get to 50% of the combined Stall torque (1.9N-m) you're chugging right along at 2400 RPM putting 480Watts of power to your robot.

If you loaded that CIM alone to that same 1.9N-m torque, your robot would be creeping along at only 1100RPM and adding only 220Watts to your robot.

I know which one I would like to be driving...

By the way, in my opinion, the BEST reason, the MOST IMPORTANT reason to have multiple motor drives are not for reasons of Torque but for reasons of CURRENT. If you do a current analysis you will find that the current to accelerate your robot or to turn it is much reduced AND you get two (2) 40 Amp fuses to share this reduced current. BOTTOM LINE: you get more safety margin until your breakers trip -- now that is something that comes in handy when you go start playing hard ball defense ;-)


Joe J.

*One way to think about how motors in parallel combine is that when you look at the graphs, the stall torques are simply added together and the SLOPES (the values I called K above) add like resistors in parrallel:
(1/K12) = (1/K1) + (1/K2)
Solving for K12:
K12 = (K1*K2)/(K1+K2)
Then you see that
N12 = TS12 * K12
Where N12 = Freespeed of the combined motor, TS12 is the combined Stall Torque and K12 is the slope of speed/torque curve for the combined motor.

P.S. This analysis assumes that you are driving the same voltage to both motors. You do not have to do this. You could provide more or less voltage to the various motors in order to make one work harder or less harder than the other. There are advantages to this strategy. But they are small and will have to wait for another day. JJ

Re: Adding FP Motor to '05 KOP Transmission
 

Thank you very much Joe, I am very new to this motor matching buisness and your help is much appriciated. I'll be playing with JVNs spreadsheets and different motor combos later on in the week. Thanks again. :D

Re: Adding FP Motor to '05 KOP Transmission
 

Correct.

From my understanding, these were originally designed to work with older FP motors (03 or 04?) that were running around 20,00 RPM. You are correct in this instance.

You seem to share a pretty concern about 'running into a problem here' with regard to not maching motors at free speed. You do not need to match free speeds to get worthwhile contributions from DC motors.

The common thought process is, "If I have one motor that has a free speed of X and one that has a free speed of 2X, and I don't match their speeds, then one will 'work too hard' " This is not correct.

What I think is most important to realize is that if you have a robot sitting around, and you put a full 12 volts to those motors, the initial speed is 0.

It's more important, in my opinion, to match your motors current load at zero speed, a speed where your machine is often, rather than at free (max) speed, where your robot is driving VERY SELDOM, if ever.

I'm a visual guy, and JVN is a spreadsheet guy, so when my ambition to compare plot something combines with John's ambition to make an awesome spreadsheet you get:





I know when I was first leaning about combining DC motors, I feel really comfortable with the 2nd graph, but the idea of the first one really bothered me. "What happens when the motors spin at 5500 RPM?!" was the terrifying question.

I'll let others on these forums answer this question if anyone is really concerned about it, but the short answer is that when a motor turns beyond its rated free speed, you start developing back emf, or a back voltage, that theoretically will actually cause an torque that needs to be overcome by the motor whose free speed is higher. I'm sure that others can clarify or correct my statements.

If you look at the first graph, if we combined a pair of motors at those ratios and then attached them to a shaft which needed a total of around 4.4 Nm of torque, many would say, "Well, each motor can contribute about 2.2 newton meters at 0 RPM, so I'd guess the speed the shaft would turn would be really slow (right around zero). I'd agree with them.

However, I know I personally got confused about what would happen if I needed say 3 Nm of torque. I'd divide he 3 Nm between the two motor and look and see that at 1.5 Nm, each motor travels at a different speed. I thought this was a doomsday situation. It's not!

The way you solve the above 'puzzle' is by finding out at which speed the sum of the torques yield 3 Nm. In this situation, it would be about 1200 RPM. The CIM would contribute roughly 1.75 NM, while the FP would contribute 1.25 Nm of torque. The contributions are different, and that's okay (though some might debate that it's not optimal for certain situations. I'm sure they'll contribute their insight.)

See also:
Combining 2 Dif. Motors
Shifting Gears

Just some thoughts,

Matt

Re: Adding FP Motor to '05 KOP Transmission
 

I am very old (to motor matchings and to sun risings) or at least I feel it.

I did not mean to be condescending. I am sorry if I came off grumpier than needed.

The take away point is to get the motor speeds matched about as well as you can and don't worry about it -- worry about the rest of the robot design.

FIRST is not an optimization problem, except in the sense that FIRST force syou to find the optimal answer to this question: What is the robot your team (with your team's people and resources) can complete in TIME that will give your team the best chances of success?

There is no sense optimizing other parts of the robot design process if you blow your answer the above questioin.

Joe J.

Re: Adding FP Motor to '05 KOP Transmission
 

Joe, your response was neither condescending nor pretentious, and I’m sorry if I made you feel that way. These responses are exactly what I was looking for. Thanks everyone and especially Matt and Joe, if anyone has more pertinent info please don't hesitate to chime in, I really appreciate it.

Oh and Tom, sorry for the thread hijack. :o

EDIT: Matt which spreadsheet did you use to plot those motor curves?

Re: Adding FP Motor to '05 KOP Transmission
 

Joe and Matt...

Thank you both very much for the analysis above. I learned something from both of these posts.

All,

Adding to their input, here is a general view:

Whenever a motor is being driven over its free speed, it begins to act as a generator. So, in this case, if the AM Planetary motor gets over 4,400 rpm, then the FP motor is a generator. This would only happen if the CIM Motor was also running at 4,400 rpm or above. Since the CIM motor is providing power to a mechanical system (drive base, conveyor, etc.), there are losses in that system. Most FIRST drive bases, if they have low losses, don't operate at over 80% efficieny (there is usually a 20% loss from the motor to the desired output).

My point is that CIM motors don't ever run above 80% of their free speed. Most of the time, they are running at a much lower rpm, as was said above.

In my opinion, coupling a CIM motor with a 2006 FP motor (and a 3.67:1 reduction gearbox such as the AM Planetary figures to be a workable solution.

Andy B.