How many CIMs in your drivetrain?

[Sperkowsky]

We had 2 added 2 more today. Only problem is we were only able to get one screw in each Cim. We will try more probably making a tool to get in there but, we sadly have run a drive train an entire season using 1 screw per Cim and can verify it works.

[CryptoStorm]

You're specifying X amount CIMs at 60 amps. It is manufacturing a theoretical load that will require 60 amps.

The load is different.

What happens if you specify a load?

[D_Price]

We are going to be running either a 6 CIM drive or a 4 CIM drive depending on our lift mechanism. At this point in the build we are using the 6. 3 on each side.

[ratdude747]

General rule of thumb for motors (at least what I was taught in college): A motor is most efficient when run exactly at the rated power output. More or less will result in a drop in efficiency. Obviously the latter is a better scenario than the latter, but in terms of % efficiency, all else being equal, going overkill isn't an improvement.

That said, it's far easier for a 2 cim drive to get pushed past the rated output, and given that batteries are usually recharged between matches, the time when efficiency (for realistic practical FRC purposes) counts is under full load (read: shoving match), so that's why a 2 cim drive will overheat and (probably) draw more amps in such a scenario.

The reason 6 cim drives lead to brownouts is not a motor issue per se; it's simply that such a drivetrain has more capability than a single battery can provide (at a voltage that the control electronics/VRMs can handle). So, realistically, in a FRC legal robot, you'll never see the full potential of a 6 cim drivetrain. You may see more than a 4 cim, sure, but you'll probably never get to full load simply because the battery can only give so much power (internal resistance is the limiting factor assuming good battery wiring). Hence the need for brownout prevention for such drivetrains.

[Jared Russell]

Originally Posted by ratdude747

A motor is most efficient when run exactly at the rated power output.

Maximum power and maximum efficiency occur at two separate points on the motor curve for brushed DC motors.

Maximum power occurs when P(out)=radial velocity * torque is maximized, which is at half free speed and half stall torque.

Maximum efficiency occurs when P(out) / P(in), or (radial velocity * torque) / (current * voltage) is maximized. This always occurs somewhere between the maximum power point and free speed (it varies from motor to motor depending on free current).

[Ether]

Originally Posted by ratdude747

General rule of thumb for motors (at least what I was taught in college): A motor is most efficient when run exactly at the rated power output.

The max power (rated power) of the DC brushed motors used in FRC occurs at roughly half the free speed.

The max efficiency occurs at a higher speed.

EDIT: Jared beat me to it.

[taharder]

Roughly, max efficiency occurs at 1/6 of stall torque. That means that you are also operating at 5/6 of free speed.

P(out)_eff=(T_stall/6)*(5*w_free/6) = 5/36*T_stall*w_free

P(out)_max = (T_stall/2)*(w_free/2) = 9/36*T_stall*w_free

So for a 300W CIM you should be able to count on 54% P(out)_max or 167W at max efficiency.

Edit: Reviewing motors.vexrobotics.com this is good enough for picking gear ratios but not exactly accurate for bag and 775pro. Of course, picking an initial gear ratio should lead to further testing and measurement. This rule of thumb lets you choose gear ratios in your head and predict mechanism performance on the back of a napkin which is where all the real engineering happens.

[CryptoStorm]

Originally Posted by Ether

The max power (rated power) of the DC brushed motors used in FRC occurs at roughly half the free speed.

The max efficiency occurs at a higher speed.

EDIT: Jared beat me to it.

Isn't that basically what your calculation is based on though? It's 60 amps in both calculations, but the 2 motor setup is putting out less torque at a higher RPM.

I realize that load isn't a constant when you're talking about acceleration, but there are situations where a single motor would be more efficient.

Aside from limiting inrush current you could never *decide* that you're going to apply 60 amps to a running motor who's continuous load only requires 20 amps.

If I had a windows machine I'd mess around with mcalc, looks like a nifty program

[Jared Russell]

Originally Posted by CryptoStorm

Isn't that basically what your calculation is based on though?

Yes, by putting two CIMs together in a gearbox you are basically creating a "super motor" with the same free speed but double the stall torque, stall current, and free current.

Obviously with larger loads the two CIM gearbox will be more efficient (at one CIM's stall load, the single CIM gearbox is stalled and therefore has 0% efficiency, whereas the two CIM gearbox is near its max power point).

At lower loads, the one CIM gearbox will be more efficient (at some tiny load just >0, both gearboxes are running at roughly free speed, but the two CIM gearbox is drawing twice as much current).

[ratdude747]

Originally Posted by Jared Russell

Maximum power and maximum efficiency occur at two separate points on the motor curve for brushed DC motors.

Maximum power occurs when P(out)=radial velocity * torque is maximized, which is at half free speed and half stall torque.

Maximum efficiency occurs when P(out) / P(in), or (radial velocity * torque) / (current * voltage) is maximized. This always occurs somewhere between the maximum power point and free speed (it varies from motor to motor depending on free current).

Originally Posted by Ether

The max power (rated power) of the DC brushed motors used in FRC occurs at roughly half the free speed.

The max efficiency occurs at a higher speed.

EDIT: Jared beat me to it.

Well, obviously I wasn't taught correctly. Then again, the class was using AC motors (both single and 3 phase), and with those, we found such to be true. Maybe inductive (AC) motors behave differently?

(Sorry for the misinformation, I feel kinda dumb here).

[cbale2000]

A lot of interesting discussion (and Math) in this thread.

What are peoples thoughts on using a combination of CIMs and MiniCIMs in drive systems?

My team is planning on using 4 CIMs + 2 MiniCIMs in our drive system for the second time this year (on a pair of 3 CIM ball shifters). Do you loose efficiency doing this due to the differences in the MiniCIM performance? Worth the extra weight?

[Ether]

Originally Posted by CryptoStorm

What happens if you specify a load?

OK let's specify a load: 7.88 Nm @ 363 RPM.

1 CIM solution:
9.73 gear ratio, 46.2 amps @ 12 volts, 255 watts waste heat, 54.1% efficiency

2 CIM solution:
12.75 gear ratio, 38.6 amps @ 12 volts, 164 watts waste heat, 64.7% efficiency