paper: 4 CIM versus 6 CIM theoretical calculations

[asid61]

Originally Posted by GeeTwo

A great overview of the continued advantages of 6 CIMs over 4 - provided that you watch the total current!

Assuming 200A for the drive train is simplistic, but a good way to compare apples to apples. No matter what the battery state is, you will have a certain limited current you can draw from it before beginning the brownout process.

The first (and so far only) technical error was the sentence that spans from the first page to the second. It should read something like:

This does not change any of the high-level conclusions.

Another thing to consider if planning a drive train with current limiting is whether using some other motor will be more effective to your strategy than CIMs. A cursory review of the mini-CIM finds it to be inferior in most cases, as it has lower efficiency than the CIM at any given current draw, but there are probably cases where it is the proper choice, due to weight considerations or specific needs. If you're going to do proper current/speed and/or thermal monitoring, you may want to add the 775 pro or other low thermal mass motor to your considerations.

Thank you!
Whoops, I didn't notice that error. Rest assured that the calculations were done the correct way (and not the way stated in the paper!) I'll change that and re-upload soon.
I was considering a 10 775pro drivetrain just before I left for the mountains on Friday, and will do some calcs for those too soon.

Originally Posted by MichaelBick

In practice, my guess is that you can draw way more current than 200 amps. Just based off of my team's 2016 drivetrain, which was 4 CIM with no voltage ramping, we experienced no brownouts even going from full forwards throttle to full backwards throttle. If we assume that our robot was traction limited below 40% of our max speed (a conservative assumption), we were drawing over 300 amps max.

Another cool aspect to current limiting is changing the limits in real-time based on each subsystem's need. For example, in a 4 CIM "hard right" turn -- 0% power on the right drive side, 100% on the left -- you may only be drawing 50% of what the battery is able to provide. If you have 1.5 or 2 times the power and current limit, you can double the current limit on the left side so that you fully utilize the battery. This same idea can be used to change the current limits to better follow a motion profile or prioritize superstructure motions.

One last important item to keep in mind is the main breaker. A high power drivetrain with a current limit pulls the maximum current for more time than an equivalent lower power drivetrain. Because the main breaker is heat triggered, it is sensitive to prolonged periods over the 120 amp breaker limit, and therefore high power drivetrains are more likely to blackout than lower power drivetrains.

Given that the talon SRX now supports current PID, a lot of doors have been opened for teams to find performance improvements with current limiting.

You're definitely correct, 200 amps is really conservative estimate. I couldn't dig up any current draw graphs from this year (where we ran 4 CIM and had a few brownouts before adding a center drop), but I hope to get some more concrete data to apply to make a more optimized curve. That being said, the overall gains in efficiency and the output power still exist with a higher current draw, and I think that due to the nature of the 6 CIMs that actually might drag the gap out more.

The main reason to use 200 amps was to allow me to do just a comparison between 4 CIM and 6 CIM given other constant variables.

I'm not too worried about the main breaker. It was a problem in 2014, but main breaker blows these days are extremely rare IMO, just because brownouts happen well before blows. Early in the season we browned out many times in practice, but never faced a breaker blow.

I forgot that the SRX can do that! That makes this way easier. In that case using the function in my paper isn't really necessary because you can just PID the current directly. That being said, coming up with a separate algorithm that takes into account battery voltage drop and max current vs. time would probably be better.

Originally Posted by Aboudy Dairi

Towards the end of the appendix A is a sentence that starts with "One final thought: ..." Is that supposed to carry on to the last paragraph somehow or is it just incomplete?

Great paper, this was a good introduction to 6 CIM drives for me in terms of the math. The idea of current limiting sounds pretty cool and is definitely something that could be implemented with the Talon SRXs to tune performance and prevent brownouts.

Oh geez, that was supposed to say:
"One final thought: Because a 6 CIM drive hits 200 amps at almost 4,000 rpm, where as a 4 CIM drive hits 200 amps at 3,300 rpm, the 6 CIM drive will be able to utilize all 200 amps for acceleration closer to the free speed of the drivetrain. A 4 CIM drive might only be able to get 200 amps worth of torque up to 12fps in a 16fps free speed drivetrain, whereas the 6 CIM drive can get up to 14fps before dropping under 200 amps current draw."
That makes the gap between 4 CIM and 6 CIM larger.
Thank you!

Originally Posted by ThaddeusMaximus

1. Mechanical power output is not everything. You need to have useful power; proper gearing. However, yes, having more mechanical power output won't hurt.

2. How do you magically draw a hard limit of 200 (or whatever) amps without effecting voltage input to the motors? Though you may be "limited" to 200 amps, when you're drawing this much current, voltage dips due to both surface depletion (?) and voltage drop in the loads to the motors. This shouldn't be ignored. When you drop away from the 12V, the motor curve you use up there is no longer valid.

Look into using kV and kT values to model things and take this voltage drop into account.

Gearing is important, definitely. I did address that a bit in my paper, but as a general rule based on my experiences I would just run ~15fps on a 4CIM and ~17fps on a 6 CIM. That lets you take advantage of the extra efficiency of the 6 CIM while still getting a bit more acceleration.
My understanding of the motor curves was that the current and RPM decrease proportionally to the proportion of 12v that you are running at. I.E. if you run a CIM at 6v, it will have a free speed of 2,640rpm and a stall current of 66 amps. What do you mean by "voltage drop in the loads to the motors"? Like losses in the wires?
I was supposed to make a new equation that took into account voltage drop and used kT and kV to make a better comparison while I was in the mountains, but I didn't get around to it then. If you have any current draw vs. brownout graphs or anything like that I would appreciate it.

Originally Posted by thatprogrammer

I took a decently quick read and had a few questions.
1. I may be misunderstanding what you are stating here:

Are you saying that a 4 CIM drive should be limited to 38% per a motor in order to keep your robot below 200amps of power usage? If so, I think that the 200 Amp number is actually pretty far from the limit of the power system of a modern FRC robot; we have run 4 CIM drives without any voltage ramping and never had any issues with brownouts.*

2. Would limiting the current draw of a motor be similar to limiting voltage? I.e. would limiting max current have a similar effect to limiting a CIM's output to .5 (In terms of limiting the speed at which the CIM will approach maximum acceleration?)

3. Why not simply control the Max RPM of the CIM in code rather than using a current PID?

*I suspect that part of this has been our somewhat low final RPMS from our gearbox reductions. These have both resulted in very short acceleration periods.

Thanks in advance for the answers!

1. You're right that 200 amps is pretty low. But as a number to use to compare the two drives I would say it's fine. I'm working on a more optimized model, so if you have any current data before brownout I would appreciate it.

2. Michael got it.

3. Current PID (or maximizing current draw without browning out) will let you get the most out of your drivetrain in terms of acceleration and safety. This is important if you want to run a drivetrain at high speeds and still not suffer from decreased acceleration and brownouts.

Originally Posted by MichaelBick

He is saying that @ 0 RPMs you need to provide 38% of full voltage to draw 200 amps total. At higher RPMs you will need to provide more voltage to draw the same 200 amps. Similarly you need to provide even less voltage if the robot is traveling backwards initially.



Halving voltage halves no-load speed, halves torque, and results in 25% mechanical power. On the other hand, limiting current limits torque and thereby limits acceleration. This by itself is very helpful, even without the power increase that is mentioned in the paper, because it causes robots to be much more stable.



If you only control velocity, you will still draw high amounts of current at stall. For example, a 4 CIM drive will draw 524 amps at stall, and 1048 amps going from full forward to full reverse. This is what causes brownouts.

Shoot, I forgot about the power decreasing with different voltages and the like. That actually screws up a big portion of my paper, so I need to go back and redo a lot of calculations. The basic conclusions should be similar, however.

Originally Posted by marshall

I'm a little busy to check this but I'd like to know how these calculations line up with the results in this paper:

https://www.chiefdelphi.com/forums/s...d.php?t=131790

I'm going through it right now, and due to the mechanical power and such changing I can't make a judgement right now. I'll come back to it once I fix my stuff. That being said, the issue with the 234 paper is that it gears very low and assumes you are going long distances, so it's not a perfect comparison. The plots for voltage and current will come in very handy, come to think of it.

[InFlight]

The lead acid batteries used by FIRST are about 17 Amp-Hours. They are certainly capable of discharging in excess of 300 amps for a few seconds.

The roboRIO brown out control is based on a voltage threshold of 6.3 Volts. And thus has 5.7 volts of margin with a fully charged battery.

You could reasonably model this situation as a 12 V voltage source with 0.05 Ohms of resistance to the power distribution board; and 0.3 Ohms of resistance through each of motor controls (Talon SRX) and wiring.

A CIM motor has a stated Stall current of 131 amps at 12 volts. For the instantaneous stall current, you could model the equivalent resistance as R=V/I = 12/131=0.09 Ohms

So each CIM motor circuit would have a Series Resistance of 0.09+0.3=0.39 Ohms.

Parallel resistances are totaled as 1/Rtotal = 1/R1 + 1/R2 +…+ 1/Rn

Four CIMs have a combined parallel resistance of 0.0975 Ohms.
Six CIMS have a combined parallel resistance of 0.065 Ohms

Four CIMS
The Four CIM total Resistance is 0.05+0.0975=0.1475 Ohms

Current:
I=V/R=12V/0.1475=81.4 Amps

The voltage drop at the RoboRio is
V=I*R=0.05*81.4= 4.07 Volts
The roboRIO has 1.63 Volts of margin for a brown out.
12-6.3-4.07=1.63

Six CIMS
The Six CIM total Resistance is 0.05+0.065=0.0115 Ohms

Current:
I=V/R=12V/0.115=104.3 Amps

The voltage drop at the RoboRio is
V=I*R=0.05*104.3= 5.21 Volts
The roboRIO has 0.49 Volts of margin for a brown out.
12-6.3-5.21=0.49


What happens later in the match when the Battery voltage has drooped down to 10.5 Volts?

We will leave the CIM resistance unchanged, even though they are now quite hot. Same with the wiring resistances.


Four CIMS
Current:
I=V/R=10.5V/0.1475=71.2 Amps

The voltage drop at the RoboRio is
V=I*R=0.05*71.2= 3.5 Volts
The roboRIO has 0.7 Volts of margin for a brown out.
10.5-6.3-3.5=0.7



Six CIMS
Current:
I=V/R=10.5V/0.115=91.3 Amps

The voltage drop at the RoboRio is
V=I*R=0.05*91.3= 4.57 Volts
The roboRIO is -0.37 Volts below the brownout voltage.
10.5-6.3-4.57=-0.37

Summary:
Standard four CIM Drives consistently maintain a 1 Volt higher brown out margin than six CIM drives. While a Six total CIM Drive can accelerate 150% faster than a Four CIM drive, it does come at considerable more risk of brownout event. Each team will have to balance drivetrain performance vs. robot reliability for the entire match.

[asid61]

Originally Posted by InFlight

The lead acid batteries used by FIRST are about 17 Amp-Hours. They are certainly capable of discharging in excess of 300 amps for a few seconds.

The roboRIO brown out control is based on a voltage threshold of 6.3 Volts. And thus has 5.7 volts of margin with a fully charged battery.

You could reasonably model this situation as a 12 V voltage source with 0.05 Ohms of resistance to the power distribution board; and 0.3 Ohms of resistance through each of motor controls (Talon SRX) and wiring.

A CIM motor has a stated Stall current of 131 amps at 12 volts. For the instantaneous stall current, you could model the equivalent resistance as R=V/I = 12/131=0.09 Ohms

So each CIM motor circuit would have a Series Resistance of 0.09+0.3=0.39 Ohms.

Parallel resistances are totaled as 1/Rtotal = 1/R1 + 1/R2 +…+ 1/Rn

Four CIMs have a combined parallel resistance of 0.0975 Ohms.
Six CIMS have a combined parallel resistance of 0.065 Ohms

Four CIMS
The Four CIM total Resistance is 0.05+0.0975=0.1475 Ohms

Current:
I=V/R=12V/0.1475=81.4 Amps

The voltage drop at the RoboRio is
V=I*R=60 amps =0.05*81.4= 4.07 Volts
The roboRIO has 1.63 Volts of margin for a brown out.
12-6.3-4.07=1.63

Six CIMS
The Six CIM total Resistance is 0.05+0.065=0.0115 Ohms

Current:
I=V/R=12V/0.115=104.3 Amps

The voltage drop at the RoboRio is
V=I*R=60 amps =0.05*104.3= 5.21 Volts
The roboRIO has 0.49 Volts of margin for a brown out.
12-6.3-5.21=0.49


What happens later in the match when the Battery voltage has drooped down to 10.5 Volts?

We will leave the CIM resistance unchanged, even though they are now quite hot. Same with the wiring resistances.


Four CIMS
Current:
I=V/R=10.5V/0.1475=71.2 Amps

The voltage drop at the RoboRio is
V=I*R=60 amps =0.05*71.2= 3.5 Volts
The roboRIO has 0.7 Volts of margin for a brown out.
10.5-6.3-3.5=0.7



Six CIMS
Current:
I=V/R=10.5V/0.115=91.3 Amps

The voltage drop at the RoboRio is
V=I*R=60 amps =0.05*91.3= 4.57 Volts
The roboRIO is-.37 Volts below the brownout voltage.
10.5-6.3-3.5=-0.37

Summary:
Standard four CIM Drives consistently maintain a 1 Volt higher brown out margin than six CIM drives. While a Six total CIM Drive can accelerate 150% faster than a Four CIM drive, it does come at considerable more risk of brownout event. Each team will have to balance drivetrain performance vs. robot reliability for the entire match.

That's some nice math you got there. I'm not sure if your model is perfect however. Are you assuming a stalled CIM? A CIM's resistance becomes less important as it speeds up, although I assume that you are doing the math for starting from a standstill.
A possible solution to this could be just basic speed ramping.

[InFlight]

Every forum needs a lurking electrical engineer.

This was just at the initial state of powering the motors, with no initial velocity.

My resistance values for the wiring resistances are just reasonable guesses. The actual amperage values could be larger or smaller, but the trends with the brownout margins would be consistent.

The other note is the motors windings only get a fraction of the full stall current; because the battery and wiring system resistance can't deliver that much current.

[AustinSchuh]

Originally Posted by asid61

That's some nice math you got there. I'm not sure if your model is perfect however. Are you assuming a stalled CIM? A CIM's resistance becomes less important as it speeds up, although I assume that you are doing the math for starting from a standstill.
A possible solution to this could be just basic speed ramping.

Also, what happens if you go from full forwards to full reverse? Theoretically, you get 2x stall current. Practically speaking, you'll start saturating magnetic fields at some point, so it gets hard to model. Stall is a good starting point.