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CryptoStorm 21-01-2016 14:16
How many CIMs in your drivetrain?
 
I started as a FIRST student in 2004. to my knowledge we've always used 2 CIMS per side on the standard "tank drive" type setup. We used mecanum wheels once which required independent drive of 1 CIM per wheel.

Does anybody have any (scientific) input on using only 1 CIM per side? (2 total)

how about 6 CIM drive?

pfreivald 21-01-2016 14:57
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by CryptoStorm (Post 1527880)
I started as a FIRST student in 2004. to my knowledge we've always used 2 CIMS per side on the standard "tank drive" type setup. We used mecanum wheels once which required independent drive of 1 CIM per wheel.

Does anybody have any (scientific) input on using only 1 CIM per side? (2 total)

how about 6 CIM drive?
We prototyped a 6 CIM drive during the offseason using VexPro's dog shifting dual-speed gearboxes, and it works fantastically.

We also did octocanum for about five years, and could beast through just about everyone...except for 6 CIM drivetrains, despite being geared down about as preposterously low as you could get, and with great traction.

I would strongly recommend not going to one per side--you'll lose every pushing match you get into with a four-CIM drivetrain, and each of those CIMs are going to draw more current, work harder, get hotter, and drain your battery more than four or six. The only advantage is weight, and it's not a good enough tradeoff.

MaGiC_PiKaChU 21-01-2016 15:11
Re: How many CIMs in your drivetrain?
 
has anyone ever done a drivetrain with more than 6 motors? (not including steering for swerve)

Ether 21-01-2016 15:37
Re: How many CIMs in your drivetrain?
 

Here's a quick calculation using MCALC to compare

one CIM drawing 60 amps at 12 volts
vs
two CIMs each drawing 30 amps at 12 volts:

Code:
Motor Calculator  build MCALC_2014d 2/3/2014 1255pm


Select motor:
1)CIM     6)FP0673  d)FP2719  f)FP9012    g)FP9015    h)FP9013 j)MiniCIM
a)am-0912 b)am-0914  7)DensoL  8)DensoR    c)Denso0160 k)BAG    u)UserDefined
2)RS395  3)RS540    4)RS550  e)RS775-12  5)RS775-18  m)RS555  n)am-0915
r)am-2193 s)am-2235  t)RS390  v)RS545      p)am-2161&2194
w)VEX2177hi  z)VEX2177std  1


CIM FR801 001, AM802 001A @ 12.00 volts:

@ free (no load):
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
    0.0  0.000    5310  100.0    2.7        0.0        32.4    0.0

@ stall:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
  343.4  2.425      0    0.0  133.0        0.0      1596.0    0.0

@ max power:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
  171.7  1.212    2655    50.0    67.8      337.1      477.1    41.4

@ max efficiency:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
    42.8  0.302    4648    87.5    18.9      147.1        80.2    64.7

Select input:
1)oz-in  7)watts_in    3)rpm    5)amps  b)rpm&amps  8)eff%
2)Nm    6)watts_out  4)rpm%  9)volts  a)rpm&ozin  m)main menu  x)exit 5


enter amps: 60

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


Select motor:
1)CIM    6)FP0673  d)FP2719  f)FP9012    g)FP9015    h)FP9013 j)MiniCIM
a)am-0912 b)am-0914  7)DensoL  8)DensoR    c)Denso0160 k)BAG    u)UserDefined
2)RS395  3)RS540    4)RS550  e)RS775-12  5)RS775-18  m)RS555  n)am-0915
r)am-2193 s)am-2235  t)RS390  v)RS545      p)am-2161&2194
w)VEX2177hi  z)VEX2177std  u


Enter the following motor specs in the order and units indicated,
separated by spaces or tabs:
volts  stall_ozin      free_rpm        stall_amps      free_amps
12      686.8          5310            266            5.4

UserDefined-> 12.0 686.8  5310 266.0  5.4 @ 12.00 volts:

@ free (no load):
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
    0.0  0.000    5310  100.0    5.4        0.0        64.8    0.0

@ stall:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
  686.8  4.850      0    0.0  266.0        0.0      3192.0    0.0

@ max power:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
  343.4  2.425    2655    50.0  135.7      674.2      954.2    41.4

@ max efficiency:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
    85.6  0.605    4648    87.5    37.9      294.3      160.4    64.7

Select input:
1)oz-in  7)watts_in    3)rpm    5)amps  b)rpm&amps  8)eff%
2)Nm    6)watts_out  4)rpm%  9)volts  a)rpm&ozin  m)main menu  x)exit 5


enter amps: 60

UserDefined-> 12.0 686.8  5310 266.0  5.4 @ 12.00 volts:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
  143.9  1.016    4197    79.0    60.0      446.6      273.4    62.0


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

GeeTwo 21-01-2016 16:44
Re: How many CIMs in your drivetrain?
 
We did a 2 CIM drivetrain last year, but Recycle Rush was definitely a special case; you didn't want to accelerate quickly or drive very fast. We're starting with 4 this year, and if we change it, we'll be going to six.

Quote:
Originally Posted by MaGiC_PiKaChU (Post 1527899)
has anyone ever done a drivetrain with more than 6 motors? (not including steering for swerve)
I've read about people considering 4 CIM + 4 min-CIM designs for Mecanum and 4-wheel Killough, but was unable to find any examples that had been built and played.

dellagd 21-01-2016 17:29
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by GeeTwo (Post 1527955)
I've read about people considering 4 CIM + 4 min-CIM designs for Mecanum and 4-wheel Killough, but was unable to find any examples that had been built and played.
Here you go

That year we ran 4 CIMs + 4 Mini-CIMs in our shifting 8in Mecanum Drivetrain. Looking back, mecanum probably wasn't the best choice that year with the heavy defense (:ahh:), but we sure tried our hardest to make it work. For the most part it did, and we were able to handle many defensive situations with the great traction of the extremely wide wheels, but the cons far outweighed the benefit.

With 8 CIM-motors, we had the potential to run 320 Amps of current into our drivetrain before any of the individual 40A breakers would trip. Even operating well below that, there was incredible risk for popping the main breaker, not to mention browning out components, both of which happened in matches on multiple occasions. In high gear, any defensive situation was almost always a brown-out and radio reboot on the field. In low gear, we had maybe 10 seconds before a brown-out. With the current sensing capabilities of the new PDP, careful software could allow a team to more safely run this kind of setup, but back then, it was far more risk than the team could effectively manage or I would ever recommend.

CryptoStorm 21-01-2016 22:16
Re: How many CIMs in your drivetrain?
 
We talked about using 6 cims for a all terrain off season robot. I've got a couple 36v motors I'm going to donate though.

The team is considering 2 CIM drive this year. I'm not recommending that, especially with tank treads. I think they will change their minds when they see how poorly it performs after the chassis is built. Hopefully we can drive it this weekend.

Maybe I'll be proven wrong though..

CryptoStorm 21-01-2016 22:56
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by Ether (Post 1527908)

Here's a quick calculation using MCALC to compare

one CIM drawing 60 amps at 12 volts
vs
two CIMs each drawing 30 amps at 12 volts:

Code:
Motor Calculator  build MCALC_2014d 2/3/2014 1255pm


Select motor:
1)CIM     6)FP0673  d)FP2719  f)FP9012    g)FP9015    h)FP9013 j)MiniCIM
a)am-0912 b)am-0914  7)DensoL  8)DensoR    c)Denso0160 k)BAG    u)UserDefined
2)RS395  3)RS540    4)RS550  e)RS775-12  5)RS775-18  m)RS555  n)am-0915
r)am-2193 s)am-2235  t)RS390  v)RS545      p)am-2161&2194
w)VEX2177hi  z)VEX2177std  1


CIM FR801 001, AM802 001A @ 12.00 volts:

@ free (no load):
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
    0.0  0.000    5310  100.0    2.7        0.0        32.4    0.0

@ stall:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
  343.4  2.425      0    0.0  133.0        0.0      1596.0    0.0

@ max power:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
  171.7  1.212    2655    50.0    67.8      337.1      477.1    41.4

@ max efficiency:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
    42.8  0.302    4648    87.5    18.9      147.1        80.2    64.7

Select input:
1)oz-in  7)watts_in    3)rpm    5)amps  b)rpm&amps  8)eff%
2)Nm    6)watts_out  4)rpm%  9)volts  a)rpm&ozin  m)main menu  x)exit 5


enter amps: 60

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


Select motor:
1)CIM    6)FP0673  d)FP2719  f)FP9012    g)FP9015    h)FP9013 j)MiniCIM
a)am-0912 b)am-0914  7)DensoL  8)DensoR    c)Denso0160 k)BAG    u)UserDefined
2)RS395  3)RS540    4)RS550  e)RS775-12  5)RS775-18  m)RS555  n)am-0915
r)am-2193 s)am-2235  t)RS390  v)RS545      p)am-2161&2194
w)VEX2177hi  z)VEX2177std  u


Enter the following motor specs in the order and units indicated,
separated by spaces or tabs:
volts  stall_ozin      free_rpm        stall_amps      free_amps
12      686.8          5310            266            5.4

UserDefined-> 12.0 686.8  5310 266.0  5.4 @ 12.00 volts:

@ free (no load):
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
    0.0  0.000    5310  100.0    5.4        0.0        64.8    0.0

@ stall:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
  686.8  4.850      0    0.0  266.0        0.0      3192.0    0.0

@ max power:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
  343.4  2.425    2655    50.0  135.7      674.2      954.2    41.4

@ max efficiency:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
    85.6  0.605    4648    87.5    37.9      294.3      160.4    64.7

Select input:
1)oz-in  7)watts_in    3)rpm    5)amps  b)rpm&amps  8)eff%
2)Nm    6)watts_out  4)rpm%  9)volts  a)rpm&ozin  m)main menu  x)exit 5


enter amps: 60

UserDefined-> 12.0 686.8  5310 266.0  5.4 @ 12.00 volts:
  oz-in      Nm    rpm    rpm%    amps  watts out  watts heat    eff%
  143.9  1.016    4197    79.0    60.0      446.6      273.4    62.0


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
Wouldnt a 1/4 hp motor and a 1/2 hp motor consume the same amount of power if the continuous load only required 1/8 hp. Amperage being based on load.

I know that 2 cims will be capable of more torque, I'm just wondering how noticeable it will be from the perspective of the driver.

Ether 21-01-2016 23:18
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by CryptoStorm (Post 1528144)
Wouldnt a 1/4 hp motor and a 1/2 hp motor consume the same amount of power
Take a careful look at what I posted:

Quote:
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
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.

Sperkowsky 21-01-2016 23:19
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 21-01-2016 23:56
Re: How many CIMs in your drivetrain?
 
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 22-01-2016 00:00
Re: How many CIMs in your drivetrain?
 
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 22-01-2016 00:50
Re: How many CIMs in your drivetrain?
 
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 22-01-2016 00:59
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by ratdude747 (Post 1528184)
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 22-01-2016 01:05
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by ratdude747 (Post 1528184)
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 22-01-2016 01:44
Re: How many CIMs in your drivetrain?
 
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 22-01-2016 02:03
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by CryptoStorm (Post 1528170)
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 22-01-2016 06:23
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by Ether (Post 1528186)
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 22-01-2016 11:47
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by Ether (Post 1527908)

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 22-01-2016 13:29
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by CryptoStorm (Post 1528144)
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 22-01-2016 13:31
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by Ether (Post 1528455)
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 22-01-2016 13:35
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by Andrew Schreiber (Post 1528392)
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 22-01-2016 14:44
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 22-01-2016 14:51
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by James Juncker (Post 1528492)
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 22-01-2016 14:56
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by CryptoStorm (Post 1528235)
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 27-01-2016 00:07
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by Jared Russell (Post 1528185)
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).
Quote:
Originally Posted by Ether (Post 1528186)
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 27-01-2016 00:14
Re: How many CIMs in your drivetrain?
 
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 29-01-2016 21:38
Re: How many CIMs in your drivetrain?
 
Quote:
Originally Posted by Ether (Post 1528153)
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.


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