Would adding a Fisher Price motor, run through an AndyMark Planetary reducer, put enough strain on the existing KOP transmission to the point where it would not work, or to the point where the efficiency of the system is so low it wouldn’t be worth using?
If you can answer this question with the math and/or know-how to back it up, could you also please explain the principles used to find this answer, or direct me to some further reading on this topic?
Thanks alot,
Tom
What do you mean adding? You want your robot to run on two FPs and two CIMs? That would be perfectly fine. However, as I recall the 6V FP running at 12V going through the AM planetary will give you a no lload speed of something like 6500 rpm where the CIMs are closer down to 5500 rpm. Quite a difference, but you could just pwm limit the FPs perhaps. Also, I think the shaft on the AM planetary is slightly longer than the CIM so you may have to trim it. Other than that, I’d say you’re good to go.
The gear reduction in the AM planetry is 3.67:1. The no load speed of the FP I use is 20,000 (I dont think this is the 6V, because my team replaced all of ours with 2004 FPs). The gear reduction on the FP motor brings its speed down to about 5500 RPM which is a good mate speed with the CIM, of course as long as you use the same motor pinion.
However, what are you trying to do exactly? As Sanddrag said, a drive train with 2 CIMs and 2 FPs running through the KOP? If you use the 2004 FPs (the ones that dont burn out) you will actually be getting less power, so there is no reason why it wouldn’t work. If you are using the 2005 FPs (the ones that burn out rather easily) it will be a little more power than a CIM, about 60 watts per drive train. However, if you use the '05 you also have to limit its speed as Sanddrag said, to match its speed with the CIM, so you might not be getting any more power than a regular CIM. Also, you have to take into account the inefficiency of the AM planetary reduction.
If you are using two 2004 FPs and two CIMs for your drive train, you should be absolutely fine.
Hope that helps.
That’s a great idea - just make sure to use the 12V type of FP, either from 2004 or the ones you could’ve exchanged for at your first 2005 competition. The AM planetary outputs about the same speed as the CIM if you use those 12V FPs, and the great thing is that they also use the same output shaft and mounting holes as the CIM as well. Also note that the type of FP has been different in each of the past 4 years (maybe more?); in 2002 it was a Johnson FP (~274 Watt) with a 20 tooth pinion and a free speed of 20,000. In 2003 it was a Mabuchi FP (~150 Watt) with a 24 tooth pinion and with a free speed of 15,000. Then in 2004 it was a Johnson FP (~263 Watt) with a 19 tooth pinion and a free speed of 15,700. Last year was the 6V debacle which we hope doesn’t happen again.
All the above data is thanks to Paul C, Joe J, Andy B, and others on CD.
Our 6V motors didn’t mind one bit running at a full 12V. One of them even lost its internal fand blades and it still didn’t even get warm. It is all in how you use them. I hate such generalizations as (for example) “the 6V FPs burn out, you should use the 12V ones.” We had great success with the 6V ones and enjoy every Watt of power we have over everyone else.
You are absolutely correct, because we customized a gearbox for the 6V FPs we had like 900:1 reduction to actuate our arm, and then when we got to our competition to switch out our FP’s, they had a different number of teeth on them, and without the resources to swap the pinions, we stayed with our 6 V FPs and they held up just fine. However the AM Planetary is not enough reduction for the 6V motors running at 12 V.
I 'm banking on them not giving us 6V motors again.
We joined the fisher price with gearbox to the KOP transmission last year and used it for our arm shoulder joint. No issues at all.
But back to your question. I have to point you to some greater minds.
and
The KOP transmission was designed to take the torque loading in the arm configuration. If you stay below the max torque at the output stage in the arm configuration, then you can use the FP motor to drive the KOP transmission.
We used the globe motor with one KOP gearbox to drive our arm last year and it worked fine. The KOP transmission manual has the output torque curve for the arm configuration in the Appendix. If you stay below that maximum output torque, then you should be good to go.
-Paul
Thanks for the replies everyone.
Just to clear things up, I was talking about modifying the back plate of the transmission to accept a 3 motor input. (2 CIM, 1 FP)
Oh, NOW you tell us. 
Those gears are 1/2" face width if I’m not mistaken. IMHO, shouldn’t be a problem as long as it is done with care and precision.
Tom,
Again, refer to my post above. Do the math. If you are at, or near the limit on the chart in the manual, then you will be fine. My initial numbers tell me you should be O.K using the one stage option.
-Paul
Okay, I’ll chime in with a question.
Looking at the KOP FP motors that my team received the part number on the can is 00968-9003.
searches through FIRST specs
Those motors apparently have a free speed of 16,320 RPM at 12.000V. Thats all well and good but when you apply that to the andymark planetary gearset with a reduction of 3.67:1. That yields a free speed of 4446.87 RPM. Okay so what, right? Well being as the andymark planetary gear set fits up in the same mounting dimensions as the CIM motor I think most people (myself included) assume that the speed would match up. The CIM motor free wheels at 5310 RPM; that’s over an 800 RPM difference from the FP/AM. Clearly you cannot just say replace one of the CIMs in the KOP transmission with a FP/AM combo and have it work. Unless maybe the andymark gear set was designed for a different FP motor, perhaps one that spun about 20k RPM. If you run that number through the gearbox you get 5449.59 RPM, much closer to the CIM. I don’t remember which of the ten or so different FP motors made 20k RPM (maybe 2004?). Anyway I guess I was just seeking some input on this, because I can see some people (myself included) possibly running into a problem here.
Sure you can – I think. There is no absolute reason they have to both have the same free speed. Through much of the match both motors will be running outside their most efficient range anyways. As I understand the wonderful world of DC motors, the free speed differential won’t make any difference, as long as you arent’ trying to make that FP motor run past its maximum free speed (something you can insure in software). You could even argue that with two motors of slightly different torque characteristics on each transmission that you might get an aggregate torque curve that is better over a wider range of speeds.
I encourage someone who actually knows what they are talking about to correct me if I’m wrong (which is entirely likely – I’m really more of a software and program management kind of mentor).
Mark and I are looking into this. This lower rpm FP motor is definitely a suprise to us. We need to do some tests and report back soon. My initial reaction is that the FP motor will still be a boost of torque, but with a lower efficiency compared to a FP motor with a 20,000 free speed rpm.
Andy
Thanks for the input Rick and thanks for looking into the problem Andy. Always nice to see people standing behind their products. 
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 = (K1K2)/(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
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.
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:
http://www.playlemonland.com/first/fpcimmatch.gif
http://www.playlemonland.com/first/fpcimmatch2.gif
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
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.
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?