How many CIMs in your drivetrain?

[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.

[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

[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

[Andrew Schreiber]

Originally Posted by Ether

CIM FR801 001, AM802 001A @ 12.00 volts:
oz-in Nm rpm rpm% amps watts out watts heat eff%
151.0 1.066 2975 56.0 60.0 332.2 387.8 46.1

Probably not the right place to mention this but the thread on MCalc is actually closed.

In your MCalc32 equation sheet you say power_out = speed * torque. Given your Speed (2975) and the torque (1.066) I'm confused how you get 332.2 W out.

Edit: admittedly the only reason I noticed this was I don't have a windows machine and liked the concept of MCalc so started a Go version. (I also needed to learn Go for a project at work and this seemed like a simple program)

[Ether]

Originally Posted by CryptoStorm

Given your Speed (2975) and the torque (1.066) I'm confused how you get 332.2 W out.

Convert everything to SI units to do the computations.

Convert 2975 rpm to 2975/60*2pi = 311.5 radians per second

311.5 rad/sec * 1.066 Nm = 332 Nm/sec = 332 joules/sec = 332 watts

[Andrew Schreiber]

Originally Posted by Ether

Convert everything to SI units to do the computations.

Convert 2975 rpm to 2975/60*2pi = 311.5 radians per second

311.5 rad/sec * 1.066 Nm = 332 Nm/sec = 332 joules/sec = 332 watts

Ah, that's the part I was missing. Thank You. Now I'm only off by ~0.01 which I mostly attribute to using the motors.vex.com numbers which might be different.

[Ether]

Originally Posted by Andrew Schreiber

Probably not the right place to mention this but the thread on MCalc is actually closed.

I didn't close it. That happened a while back when the server crashed or got updated or re-hosted or something. I'll ask Brandon to re-open it.

[James Juncker]

So wouldn't it be more efficient to power all wheels independently with CIMs so that they have less wasted energy due to slipping or one wheel getting less traction?

[Ether]

Originally Posted by James Juncker

So wouldn't it be more efficient to power all wheels independently with CIMs so that they have less wasted energy due to slipping or one wheel getting less traction?

Try laying out your logic in a bit more detail and see if you come to the same conclusion.

[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?

[MrRiedemanJACC]

Originally Posted by Ether

Take a careful look at what I posted:



332+388 = 720 total watts consumed (one CIM)

447+273 = 720 total watts consumed (two CIMs)

They both consume the same power @ 12 volts @ 60 amps, but you get more output mechanical power and less waste heat with two CIMs.

So you don't need as many amps with the 2 CIMs to get the same output power as 1 CIM.

So if I may add to that and continue the idea with a 3 CIM gearbox and 60 amps total load on the the three motors, you use MCALC at 20 amps per CIM and get this:
CIM FR801 001, AM802 001A @ 12.00 volts
oz-in=45.6, Nm=0.322, rpm=86.7, amps=20, watts out = 155.3, watts heat is 84.7 and eff% = 64.7


So I'm seeing the three CIMs each drawing 20 amps at 12 volts:
465.9 watts output mechanical power
254.1 watts waste heat

vs these two choices that Ether's calculations from before.

one CIM drawing 60 amps at 12 volts:
332 watts output mechanical power
388 watts waste heat

two CIMs each drawing 30 amps at 12 volts:
447 watts output mechanical power
273 watts waste heat

So not as much of a gain in efficiency improvement as I had hoped, but the package is still capable of much more.