Joining 2 dif. motors

[K.Shaw]

Team 237 would like to know how to join the Chiaphua and drill motors into one for the drive system. Can anyone help us on this feat and tell us in exact detail how you can do this? Pictures, diagrams, anything that can help us out post it here or email it to me at kevin185@optonline.net . Thanks alot!

p.s. even joining any other 2 motors post it here also thanks

[GregTheGreat]

Originally Posted by K.Shaw

Team 237 would like to know how to join the Chiaphua and drill motors into one for the drive system. Can anyone help us on this feat and tell us in exact detail how you can do this? Pictures, diagrams, anything that can help us out post it here or email it to me at kevin185@optonline.net . Thanks alot!

p.s. even joining any other 2 motors post it here also thanks

Are you saying use a chiaphua and a drill for each of the 4 wheels? or like 2 sets of that in the back and FP's in the front?

If you can describe exactly what you are trying to do I will look through all my downloaded stuff and archives to see if I can find something like it.

-Greg The Great

[K.Shaw]

ok cool umm we are trying to use tank steering with a drill and chiaphua motor on each side of the robot. meaning 1 chiaphua and drill motor on one side and aanother set on the other side.

[Matt Adams]

Originally Posted by K.Shaw

ok cool umm we are trying to use tank steering with a drill and chiaphua motor on each side of the robot. meaning 1 chiaphua and drill motor on one side and aanother set on the other side.

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

[Rickertsen2]

I avn't read the whitepaper listed above, it might say what i'm about to say but anyway... Don't match the moters at free speed. I have seen many teams do this. This is the quickest way i know to mismatch them. Match them under some typical load. Further matching can be accomplished through software.

[K.Shaw]

thanks for the info guys keep replying to this post! we need everyones input

[Cory]

Search... there are 1000000000000000000 threads on this topic (well, not that many, but there are certainly more than I care to count) download a white paper such as the Technokat's transmission, or Team 116's, or Team 716's They all offer complete prints and hints/tips etc.

[Matt Adams]

Originally Posted by Rickertsen2

I avn't read the whitepaper listed above, it might say what i'm about to say but anyway... Don't match the moters at free speed. I have seen many teams do this. This is the quickest way i know to mismatch them. Match them under some typical load. Further matching can be accomplished through software.

From the best of my understanding, nothing is wrong with matching at free speeds. If you don't, at higher RPMs you will have back EMF occur, reducing efficiency if you have one motor trying to go faster than the other's free speed.

Take a look at the following figure for a Drill in low gear and Atwood coupled together at the drill's free speed:




Let's say that you have 200 in-lbs of torque applied to the shaft. In this case, the motors coupled at free speed will spin at about 225 RPM. The atwood motor will be contributing more torque to the system than the drill. I do not see a problem with this. I admit that there's uneven load sharing between the motors... but there is nothing fundamentally wrong with this from a mechanics standpoint.

I think that some people believe that if motors aren't sharing equal load, then there's a "problem" or the motors are "fighting" each other. This is not the case, and isn't "bad." You don't have an even current draw from each motor.. but your motors aren't being ruined, and your efficiency is constant. They're pretty happy. One is working harder, but they don't get jealous of each other! The only time problems arise is if there's a situation with little torque, and they're coupled at a non-free speed. One motor (higher free speed) says, "Woo hoo! let's go really really fast!" and the other says, "But my max speed is slower than yours. I'll try to do what I can to keep up, but we won't be able to go as fast as you combined!"

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!

Now, you could optimize your motors such that they're intersecting at a low torque that your robot could never achieve due to friction, slightly below the free speed. This could help have the amount of torque (and current) distributed between the motors be more even. But having the load sharing imperfect is not a "bad" thing if you're careful about analyzing the situation to make sure the load isn't so uneven that a breaker on one motor will trip before the other can effectively contribute.

Let's take a look at this:




NOTICE: As you can see, the free speed of a combined motors in low torque conditions (to the left of their intersection point) I made to be the geometric mean of the two speeds. This is not completely correct, (neither is the fact that these show all linear relationships between torque and RPM, this is simplified). There are coefficients that determines the ratio between the two motors free speed which sets what the RPM the motors will have at low torque values. However, the RPM in low torque conditions is NOT the fastest motor's speed, nor is it the slowest. The reason for this is unique to DC motors, and refers to the back-EMF, which occurs when a DC motor is spinning faster above it's free speed. You run this risk of this occuring if you do not match them at free speed.

However, will you ever have motors truly reach free speed in a robot? Nope. So can you adjust the ratios by a little so that they're share the toque more evenly? Of course. Even matching free speeds is impossible since the free speed varies from motor to motor of the same type.

However, keep in mind that if you have low torque conditions that yield RPMs above the slower motor's free speed, you loose efficiency.

And this is going to bring up one heck of a hail storm... I know it.

Please, let's keep this at a technical level, and be as specific if you can. Disagree with facts. Graphs help out. Hack away at mine if you wish. Just keep it legit and not hunches. We'll settle this.


Input welcome, could a EE out there explain back EMF for me a little better?
Matt

[Rob Colatutto]

Originally Posted by Matt Adams

Let's say that you have 200 in-lbs of torque applied to the shaft. In this case, the motors coupled at free speed will spin at about 225 RPM. The atwood motor will be contributing more torque to the system than the drill. I do not see a problem with this. I admit that there's uneven load sharing between the motors... but there is nothing fundamentally wrong with this from a mechanics standpoint.

In fact, sometimes you will want to do something similar to this. For 2002 my team designed the gearbox so the bosch would be moving the robot almost to the point where the atwood was not adding any power under normal movement to keep current draw down. Then when we had goals or if we were pushing another robot we'd slow down about .5ft/s but then the torque would increase greatly while not drawing too much extra current. I wouldn't suggest doing something with that big of a difference in normal load rpm's of the shafts but if you wanted to keep current draw down on normal conditions you could do something similar. Also, I would suggest combining the two motors together as soon as possible in the gearbox to use less gears and keep weight down.

[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

[Matt Adams]

Originally Posted by Rob Colatutto

Then when we had goals or if we were pushing another robot we'd slow down about .5ft/s but then the torque would increase greatly while not drawing too much extra current. I wouldn't suggest doing something with that big of a difference in normal load rpm's of the shafts but if you wanted to keep current draw down on normal conditions you could do something similar.

Rob-

How did you "slow down"? Via the joystick or by some sort of a shifting device? I'd have to do some thinking about it, but initially I'd say that that just "slowing down" via a joystick wouldn't let one motor kick in more torque than the other since the amount of torque needed would be constant at any moving speed, and presumably, the joystick would control both motor's input voltage.

Matt

[Pat Roche]

Sorry if i repeat what someone else is saying....im short of time and didnt read the entire post...but you can simply use the drill motor alone(no clutch) and gear the two into a 1:1 ratio. From this point you can decide upon the rest of your gearing, etc.

If my memory is correct:
Drill no clutch:about 19000
Chip:about 5000
so its close to a 4 to 1 ratio
this means that the gear on the chip needs to be 4x bigger that the drill for a 1:1 ratio.

Hope this was helpful


-Pat

[SarahB]

Originally Posted by T134guy

If my memory is correct:
Drill no clutch:about 19000
Chip:about 5000
so its close to a 4 to 1 ratio
this means that the gear on the chip needs to be 4x bigger that the drill for a 1:1 ratio.

Hope this was helpful
-Pat

Bosch normal load speed is 19670 and the Atwood normal load speed is 4967

[Rob Colatutto]

Originally Posted by Matt Adams

Rob-

How did you "slow down"? Via the joystick or by some sort of a shifting device? I'd have to do some thinking about it, but initially I'd say that that just "slowing down" via a joystick wouldn't let one motor kick in more torque than the other since the amount of torque needed would be constant at any moving speed, and presumably, the joystick would control both motor's input voltage.

Matt

There was no option to slow down, the robot slowed itself down when the additional load of the goals was added to the robot. The drills couldn't fully handle the extra 362 pounds and so the motor slowed down and then the Atwood picked up for it, drawing less current than running the Atwood at a slower rpm. We also had a 'software shifter' which took the joystick feedback and divided it by 1.5 so the operator could switch the speed down when latching onto goals.

[Matt Adams]

Originally Posted by SarahB

Bosch normal load speed is 19670 and the Atwood normal load speed is 4967

This is incorrect.

Atwood at normal load (from 2003 spec sheet) is 4378 RPM.
Free speed is 5,500.

Free speed on Bosch is 19,670 RPM as Sarah stated for normal load speed.

Just wanted to make sure that the right data is out there.

Check it out here:
http://www.usfirst.org/robotics/2003/specsheets.htm


Matt