Joining 2 dif. motors

[Joe Johnson]

Originally Posted by Matt Adams

Now with the above illustration.. don't think of motors as people rowing a boat, where if one is rowing harder the boat will turn. Think of the motors as a bicycle with two riders. If one works harder, that's fine. But if someone's legs just can't physically move fast enough, it's bad news![/font]


Matt[/color]

I generally agree with what Matt has to say, but I want to point out that the free speed of a motor is not like the speed of sound in which there is some major discontinuity in the behaviour of the motor. A motor does not have an opinion about what speed it spins at. In particular, it has not qualms about going faster than its free speed.

What I think folks need to keep in mind is that the speed/torque line does not end at the Y-intercept, nor the X-intercept. If you couple 2 motors together, they are going to run with some (fixed) relationship between their speeds. SO... at each given speed, the torques are going to ADD. They add even if you are opperating at a speed higher than the free speed -- BUT, the motor running faster than its free speed is providing NEGATIVE torque at that point.

As Matt said, this is not a 100% tragedy. The motor does not suddenly blow up like a it was in a cheap action movie. Just as Matt said, it hurts the effeciency (because, just like the speed/torque curve, the effeciency parabola does not end either -- you have NEGATIVE effeciencies for that one motor in this range -- which only makes sense, you are putting still putting positive electrical energy in and you are also adding work via the shaft i.e. getting negative energy out). If you opperate in this range very long, you will smoke your motor but that is more of a gradual thing not an explosive process.

Another thing to think about is that up to this point, we have only been comparing motors as if there is only ONE speed/torque curve. This is not true. There is a family of speed/torque curves based on the voltage applied to the motor. It is possible to give more or less voltage to one motor or the other to reduce some of these problems - it is not a perfect solution, but it can be a way to help the motors share the load a bit more effectively.

Joe J.

[eugenebrooks]

[/font][/color]
Input welcome, could a EE out there explain back EMF for me a little better?
Matt[/color][/quote]

If one motor is over driven by the other, it becomes a generator,
in effect and saps power from the drive train to backfeed electrical
power in into the robots electrical system. This is easily demonstrated
by turning a motor by hand and watching the leds on robot control system.
You will see them light up.

If the over driven motor is sapping power from the mechanical drive
train, it is taking up power that would otherwise move the robot forward,
hence the loss in efficiency.

[Matt Adams]

Originally Posted by eugenebrooks

If one motor is over driven by the other, it becomes a generator, in effect and saps power from the drive train to backfeed electrical
power in into the robots electrical system. This is easily demonstrated
by turning a motor by hand and watching the leds on robot control system.
You will see them light up.

If the over driven motor is sapping power from the mechanical drive
train, it is taking up power that would otherwise move the robot forward,
hence the loss in efficiency.

Since I asked this question, I've taken a class at Purdue that dove into DC motor information. What you're saying is mostly correct. At the time, I was specificly looking for how to find the back-emf constant, that is, relating the amount of power loss quanitatively to RPM. I've come to discover that these coefficients are usually listed within motor specs from the manufacturer, and is essentially the torque constant for most of the motors that we'd see in FIRST.

Thanks for answering the question, though.

Matt

[pryoplasm]

ok, a few things, i dont know the exact stuff for what my team did this year because i specialized on the electrical stuff rather than the mechanical side, but i can give a rough idea of what we did

we designed and made our own gearboxes, with different reductions for the drill and the CIMs. not sure what the exact ratio is. also, a heads up, unless the drills/CIMs are mounted in the same direction (ie. both facing left) then one will have to run in reverse, which isnt the same as the high speed for the forward drill.

also, if you only want to use 2 motors for the entire drivetrain, its possible for tankdrive. we did that, oh, maybe 3 years ago, the game with the ramp that could tip back and forth.., used 2 drills in what i can only incorrectly describe as a figure 8 looking thing, with 2 drills in the middle of each side, 2 chains going to the front and back wheels. we also had a sweet transmission that could pop a pin and slide around in neutral, but that belongs in a different thread.

hope this helped

[Ryan F.]

Heres the simple method we used that worked just fine. We used the drill motors with the FIRST supplied grearboxes, and just stuck those on the front. For the CIM's we designed a simple gearbox with a relatively simular output speed, and just chained the two drive shafts together. All the gearbox consisted of was a 12 tooth gear right on the CIM moror, that lead to a 60 tooth on an idler shaft which also contained a 16 tooth gear. That 16 tooth gear was then meshed with a 70 tooth gear on the output shaft. It turned out running well for us, but the CIM's might have been running a little too fast, so if I used a concept like this again, I would change the gears around a little to acquire a slower output speed.

Here's a crude diogram of out Gearbox

CIM MOTOR ---12 tooth gear
.......................60 tooth gear----------16 tooth gear
.................................................. ......70 tooth gear--wheel!

Simple is good.

[Matt Adams]

Originally Posted by rforystek

It turned out running well for us, but the CIM's might have been running a little too fast, so if I used a concept like this again, I would change the gears around a little to acquire a slower output speed.

Here's a crude diogram of out Gearbox

CIM MOTOR ---12 tooth gear
.......................60 tooth gear----------16 tooth gear
.................................................. ......70 tooth gear--wheel!

In general, I've found it much more efficient to combine a pair of motors, and then run this combined set out via some sprockets to the wheels. This ensures that one wheel won't ever try to run a bit faster than the others. If you try to run them individually, your motor torque and speed outputs vary as the speed of your robot changes. Fortunately, the Chips and Drills happen to have roughly the same trip points when matched at free speed, which is why I'd imagine a lot of people don't notice it.

Matt

[Tytus Gerrish]

bosch 307 Watts @37.4A
CIM 300 Watts @40 (estamate)

300+300=600
600W=one gearbox

600+600=1200
1200W 2gearboxes

1 HorsePower=746 Watts

1200/746.7=1.6

1.6 Horsepower

[lupjohn]

I noted in your PPT workup that you mention back EMF. I have tried to get this concept across to several teams that this problem means that one motor is actually reducing the power available to drive the robot since that translates to a "braking" effect on the power train. I have not measured this effect but what does it do to the drivers on the motor controllers. These devices are current sources and not current sinks unless I have my theory wrong. Anyway thanks for the great work up. I just need more time to translate this for my students. LRU

Originally Posted by Matt Adams

Oh boy.. do I have something for you.

As part of the FIRST college course that I taught this semester, I have some rather comprehensize data on combining both the chips and drills... from a number crunching standpoint, it's very complete.

There aren't any sprocket sizes, CAD drawings, or the like.. but theory wise.. this is (fundamentally) very strong material. I will be publishing a white paper along these lines soon on www.boilerinvasion.org.

Here's the PowerPoint. If you have questions, please ask. If someone has corrections, comments, or concerns, please ask.

http://web.ics.purdue.edu/~mjadams/motorsinfirst.ppt

Matt

[Matt Leese]

Originally Posted by lupjohn

I noted in your PPT workup that you mention back EMF. I have tried to get this concept across to several teams that this problem means that one motor is actually reducing the power available to drive the robot since that translates to a "braking" effect on the power train. I have not measured this effect but what does it do to the drivers on the motor controllers. These devices are current sources and not current sinks unless I have my theory wrong. Anyway thanks for the great work up. I just need more time to translate this for my students. LRU

I don't quite follow what you meant in your original post but I'll try to respond as best I can. I'm not quite sure what you mean by the one motor reducing power available to the power train.

The back EMF generated by a motor is caused by the spinning magnet field of the motor. The back EMF is proporitional to the rotational velocity of the motor. The higher the back EMF, the less current that the motor is drawing.

I think what you're worried about for the speed controllers is driving a current back into them. The speed controllers which are based around an H-bridge basically just use transistor switches. They do not use transistors as current sources. When a current is driven back through them, they will pass it much like they'd pass a current the other direction (particularly since they're based on FETs). This just means that the current is going back to the battery. There's no problem with sourcing current into the speed controllers.

Matt