Stalling the BaneBot 775 motors

[AdamHeard]

The power draw would be less than what was calculated before, as he said 20-40% of stall torque, which would mean the voltage is lower as well.

Assuming 20%, that is 17 Amps @ 2.4V, which works our to 40W.

Not as bad a number, but far less than we plan on drawing; Getting that down to just 10% of stall torque bumps you to 10W, a much nicer number.

[Leav]

Originally Posted by AdamHeard

The power draw would be less than what was calculated before, as he said 20-40% of stall torque, which would mean the voltage is lower as well.

Assuming 20%, that is 17 Amps @ 2.4V, which works our to 40W.

Not as bad a number, but far less than we plan on drawing; Getting that down to just 10% of stall torque bumps you to 10W, a much nicer number.

Sounds like a typical engineering trade off, but one I could definitely live with.

[dtengineering]

Try strapping a cooling fan to the motor. A few zap ties and one of the little fans might make a big difference... you could even try making a little duct work to blow it all in one end of the motor... but that is probably overkill.

Our nerf ball shooter in Aim High was an FP that ran about 5-10 amps. It would be on for a fair bit, but not all, of the time during a match, it would be turning... and it would be... not hot, but definitely warm, by the end of a match.

I would suggest that you try running the motor at your predicted loads on a test bench in the shop before you even mount it on the robot. Just lock the shaft and run the motor at your expected current. Actually, since this is a test... start at half your expected current. Hold it there for two minutes. It would help if you had an IR thermometer to track case temperature, but remember that the real heat will be forming in the coils and will take a while to get out to the case. This will be fairly simple to do if you have a clamp-on ammeter, but if you don't, I believe the Jag can be convinced to give you feedback on your current consumption... you may even be able to just send a "give me 5 amps" command to the Jag. The real heating situtation during a match will be quite different, of course... the motor will draw a lot of power as you lift the arm, and then, perhaps before that heat has a chance to be "blown away" by the motor's fan, you'll be stalled... so it may not be just the stall heat that you have to deal with. Of course, then you'll lower the arm and the motor may have a chance to rest and cool, but perhaps you can run it through a few cycles like that over a two minute stretch.

Then keep in mind that you may or may not have time to cool your motor between matches, particularly during practice matches. We've been fortunate to clear tech fairly quickly from time to time and get practice match after practice match after practice match. One year we had to pull out from the practice field because the drive CIMs were starting to overheat! Just think what would have happened to one of the little motors!

Jason

P.S. Another time your motors will be subjected to extended heating and use will be in the post-season while doing demos.

[AdamHeard]

Originally Posted by dtengineering

Try strapping a cooling fan to the motor. A few zap ties and one of the little fans might make a big difference... you could even try making a little duct work to blow it all in one end of the motor... but that is probably overkill.

Our nerf ball shooter in Aim High was an FP that ran about 5-10 amps. It would be on for a fair bit, but not all, of the time during a match, it would be turning... and it would be... not hot, but definitely warm, by the end of a match.

I would suggest that you try running the motor at your predicted loads on a test bench in the shop before you even mount it on the robot. Just lock the shaft and run the motor at your expected current. Actually, since this is a test... start at half your expected current. Hold it there for two minutes. It would help if you had an IR thermometer to track case temperature, but remember that the real heat will be forming in the coils and will take a while to get out to the case. This will be fairly simple to do if you have a clamp-on ammeter, but if you don't, I believe the Jag can be convinced to give you feedback on your current consumption... you may even be able to just send a "give me 5 amps" command to the Jag. The real heating situtation during a match will be quite different, of course... the motor will draw a lot of power as you lift the arm, and then, perhaps before that heat has a chance to be "blown away" by the motor's fan, you'll be stalled... so it may not be just the stall heat that you have to deal with. Of course, then you'll lower the arm and the motor may have a chance to rest and cool, but perhaps you can run it through a few cycles like that over a two minute stretch.

Then keep in mind that you may or may not have time to cool your motor between matches, particularly during practice matches. We've been fortunate to clear tech fairly quickly from time to time and get practice match after practice match after practice match. One year we had to pull out from the practice field because the drive CIMs were starting to overheat! Just think what would have happened to one of the little motors!

Jason

P.S. Another time your motors will be subjected to extended heating and use will be in the post-season while doing demos.

You could do this test by locking the shaft to force stall, and setting the voltage to the proper value. To get 10% of stall, set 10% of 12V.

[Gdeaver]

We plan to provide assist to our arm. Gas shock or cylinder to take the load off the arm motor. Constant high temps on the motor can weaken the motors magnets cutting power output and increasing the amp draw.

[Al Skierkiewicz]

Fox and Leav,
I believe your analysis is correct. Assuming that the pulse width is sufficiently long enough to over come rise time caused by the inductance/resistance of the motor, full power is being applied during the pulse. i.e. 12 volts * 86 amps=1032 watts. That power is being dissipated only 20% of the time so the average power is 213.6 watts over the full period of the PWM. This does not account for other losses. That amounts to a lot of heat over a two minute match. Please remember that the current is also heating the circuit breaker. Since the breaker is a heat sensitive device, it may trip well before the motor hits the fail point. As the breaker resets fairly quickly, that adds more to equation than simply the motor and the more a breaker resets, the hotter it becomes. Also if you are using a Jaguar, you may trigger a fault condition there due to high temperature or sustained high current which shuts the Jaguar down for 3.5 seconds.
Leav, to answer your other question, fans are allowed if they follow all the other electrical rules for wire and breakers.

[Jon Stratis]

FWIW, we've done something similar in the past...

In 2008, we used a FP motor to lift the trackball on an elevator. Once at the top, we had to stall the motor in order to keep the elevator from drifting down. While it worked and we never blew that motor (after all, we only needed to stall it up there for a few seconds at a time), it did get very hot - hot enough we put a shroud around it to prevent people from getting burned by accidentally brushing it.

Contrast that with 2009, where we used a Banebots motor (I don't recall the exact one) to power a conveyor belt. In our second competition, we burned a few of them out from balls getting stuck and stalling the motor. Not one of our finest moments!

[thefro526]

Originally Posted by Leav

I'm reading the 2011 manual to try and figure out if cooling fans are allowed.

Leav, we're planning on using a fan to cool each of our 775's as well. There are 6 Small Fans, and 2 Large fans included in the KOP This year, and since they were included in the KOP, according to <R45 - A> they are legal.

<R45> Motors specifically permitted on 2011 FRC ROBOTS include:
A. all motors, actuators, and servos listed in the 2011 KOP Checklist,

The Fans I'm talking about can be found on page 12 of the KOP Checklist Rev B.

[fox46]

I believe your analysis is correct. Assuming that the pulse width is sufficiently long enough to over come rise time caused by the inductance/resistance of the motor, full power is being applied during the pulse. i.e. 12 volts * 86 amps=1032 watts. That power is being dissipated only 20% of the time so the average power is 213.6 watts over the full period of the PWM. This does not account for other losses. That amounts to a lot of heat over a two minute match. Please remember that the current is also heating the circuit breaker. Since the breaker is a heat sensitive device, it may trip well before the motor hits the fail point. As the breaker resets fairly quickly, that adds more to equation than simply the motor and the more a breaker resets, the hotter it becomes. Also if you are using a Jaguar, you may trigger a fault condition there due to high temperature or sustained high current which shuts the Jaguar down for 3.5 seconds.
Leav, to answer your other question, fans are allowed if they follow all the other electrical rules for wire and breakers.

Thanks Al. Good call on the circuit breakers. When I was reading up on chopper style speed controls last night the inductance/resistance of the motor was where things started to get messy, differential equations began popping up and my desire to keep reading was all of a sudden non-existent lol. I don't believe the rise/fall time would have a significant effect on current consumption but its definitely a factor, especially at low power settings when the chopping frequency is high.

[Ether]

Originally Posted by Al Skierkiewicz

Fox and Leav,
I believe your analysis is correct. Assuming that the pulse width is sufficiently long enough to over come rise time caused by the inductance/resistance of the motor,

I think that assumption is false (for the Jag/RS775 combination we are discussing here) and therefore the cited analysis is not correct. But I have been unable to locate an inductance spec for the RS775 so I can't state that for certain.

Al, what is your basis for assuming this is true for the Jaguar/RS775 combination? Have you measured the RS775 locked-rotor inductance? If so, would you please share that information with the CD community. If not, make an educated ballpark guess what you think the inductance is, and I will run a simulation model and post the results here.

[Al Skierkiewicz]

Ether,
I am basing the statement on the experience of rise time vs inductance with any motor. Since the Jaguar is switching at 15kHz, current through any motor may not be able to rise to full stall current during a short pulse. I am simply suggesting that an estimate of power dissipated can be made if one ignores rise time for this discussion.

[Ether]

Originally Posted by Al Skierkiewicz

I am simply suggesting that an estimate of power dissipated can be made if one ignores rise time for this discussion.

If you "ignore rise time" and assume that "full power is being applied during the pulse" you will get a very wrong answer if your assumption is wrong.

I do not know what the inductance of the RS775 is, but even if it is as small as 20 micro Henries, then power consumption at 20% duty cycle (with a Jag and locked rotor) will be roughly 40 watts, not 213.6 watts.

If anyone reading this has access to the necessary test equipment and would be willing to measure locked-rotor inductance of the RS775 and post the result here, I will run a simulation and post a graphic of the current waveform (and power calculation).

[fox46]

For the purpose of this discussion I think we are getting too far into the theory behind the component. It would be a very difficult piece of hardware to work with if it only output 4% power when you tell it to give 20%. For the purpose of this discussion and our use, I am quite sure that the Jaguar compensates for the exponential effects of Inductance/switching frequency/rise time and changes its pulse frequency so that the output power varies linearly with the input instructions. In this case, an instruction of 20% output would result in an intuitive 20% power (W) output and thus ~200W.

If the jaguar input controlled the switching frequency- linearly, then I think Ether is right. After one of our team members hooked up a jaguar backwards today, I opened it up to take a peek inside. The output has a capacitor on it- I believe this would effectively "smooth" the output power and generate a physically lower voltage/amperage. With the capacitance/inductance/resistance and switching frequency you effectively have an RLC circuit. Relative to the switching frequency, output would indeed change exponentially just like Ether's calculation.

I believe we are just thinking of two different parameters of which we are commanding 20% thereof- My thinking is that the input commands control the power output wheras Ether's thinking is that the input commands control the switching frequency. I'm going to stick with the former so tell Regis my final answer is ~200W!

If you know what the frequency is (someone with an oscilliscope?), we know what the percieved output voltage, current and resistance is at full power, I can take a look at the guts of the Jag and see what the capacitance is. Can we not deduce the inductance?

[Ether]

For the purpose of this discussion I think we are getting too far into the theory behind the component.

This is pretty basic AC circuit theory, and it has a profound effect (40 watts versus 200 watts) on the answer you get, so it's worth understanding.

I am quite sure that the Jaguar compensates for the exponential effects of Inductance/switching frequency/rise time and changes its pulse frequency so that the output power varies linearly with the input instructions.

Sorry, it doesn't work this way. First, the output frequency does not change at all. It is a constant 15000 Hz. Only the duty cycle changes. The duty cycle varies linearly with input command (unless, of course, you are running closed loop, but that's another story entirely).

In this case, an instruction of 20% output would result in an intuitive 20% power (W) output and thus ~200W.

It doesn't work that way. A 20% command gives a 20% duty cycle.

If the jaguar input controlled the switching frequency- linearly, then I think Ether is right.

The Jag does not control switching frequency. The frequency is held constant at 15000 Hz, which gives a pulse width of 66.7 microseconds. What the Jag changes is the "on" portion of that 66.7 usec.

After one of our team members hooked up a jaguar backwards today, I opened it up to take a peek inside. The output has a capacitor on it- I believe this would effectively "smooth" the output power and generate a physically lower voltage/amperage. With the capacitance/inductance/resistance and switching frequency you effectively have an RLC circuit. Relative to the switching frequency, output would indeed change exponentially just like Ether's calculation.

Not sure what you mean by "exponentially" here. The effective voltage changes linearly with input command, and the effective current also changes linearly. The effective power thus changes as a square function.

I believe we are just thinking of two different parameters of which we are commanding 20% thereof- My thinking is that the input commands control the power output

They do, but not linearly.

wheras Ether's thinking is that the input commands control the switching frequency.

No, it controls the duty cycle, and thus the effective voltage and current.

I'm going to stick with the former so tell Regis my final answer is ~200W!

Are you a betting man?

If you know what the frequency is

It's 15000 Hz.

Can we not deduce the inductance?

If you knew how many windings, and the core material magnetic properties, you could make a rough calculation.

[AustinSchuh]

Originally Posted by fox46

... The output has a capacitor on it ...

http://www.luminarymicro.com/index.p...8&Itemid=59 1

This user manual for the black Jaguar has the schematic. I'm not sure which capacitor you were seeing, but there is no capacitor on the output of the Jaguar. There is a bootstrap capacitor in the circuit, but will not generate the effect you are describing.