Mecanum power draw issue when weighted down

We have a mecanum drive configured with AM 6 inch mecanum wheels, AM CIMple Box 4.67:1 gearboxes, 1 CIM per gearbox, 1 encoder per wheel wired directly to a JAG, 1 JAG per wheel using the CAN bus and in speed mode with PID control. The system was running great when it weighed 40 lbs (just the chassis, drive train and electronics). When we loaded it to approximately 100 lbs, JAGs began browning out randomly and we don’t think we can reliably get through a 2 minute match. We are consuming so much power that batteries barely make it 3 minutes. After much CD research, disassembling & testing motors and a mess of other things, we’re looking at going to a different gearbox with closer to an 11:1 gear ratio because the CIMs supposedly draw less current at higher RPMs. Is that worth the time, effort and money? Just to be clear, even loaded, the bot drives well. It just consumes so much power that JAGs start failing after 1-3 minutes. We were thrilled with our first mecanum drive at week 2. Now (after weighing it down) we are a bit panicy to say the least. Thanks in advance.

You don’t need a whole new gearbox, but you really need to add to your reduction via sprockets or anything else, because you are geared way too fast if you’re direct driven off those gearboxes.


For designing motors into any system I first go to JVN’s Design calculator. Downloadable on the white paper section.

Under 1 speed drive I inputted then numbers and the result:

Firstly notice your speed is 24ft/sec which is beyond anything you could ever need, most teams in the single speed area gear for about 14-15fps. Then notice your amp draw at over 200 per motor under load… remember the jaq is rated for 40 continuous.

Now if you go down to say a 11:1 ratio, which i would do, you will loose speed (but still have more then enough) but gain toque which will give you more pushing power and also with a higher reduction you will use less amps due to the fact your motors are working less.

Now you can do this with a chain reduction, as stated above instead of buying 4 new gearboxes.

Best of luck.

Increase your reduction, significantly. With such a little reduction and 6" wheels, you’re putting a large load on the motors, and they’re drawing a lot of current to compensate. As a result, you’re browning out your control system and/or tripping your 40A breakers (which may be the root cause of you describing your jaguars as “failing”). Additionally, with a large load on the motors, they will spin a slower rate (which is the root of the “CIMs supposedly draw less current at higher RPMs” comment).

I’d highly suggest taking a look at the CIM motor curves and reading some of the great material out there on designing a drive system for FRC.

We’ve run mecanum on 9:1, 10.71:1, 12:1, and 12.75:1 reductions. Even at 9:1, the power draw started to get high enough to make us nervous (please don’t ask for specifics–this was five years ago and I don’t have that kind of memory!)

Our octocanum last year was 10.71:1 on the mecanum wheels and that worked exactly as well as we wanted it to.

So as others said: Gear it down. My suggestion would be a 2:1 (or slightly higher) reduction from what you’ve already got–either by buying new gearboxes (which will be expensive and probably require a redesign, if they’re even in stock) or using belts/chain (which will be much cheaper and definitely require a redesign).

You must have entered some numbers incorrectly.

The stall current for a CIM at 12 volts is only 133 amps.

Or perhaps you did not mean “per motor”.

You could always add 4 more cims/mini cims to your 4 gearboxes to increase efficiency. I’d still recommend a higher gear ratio too.

Could you please post your calculations or test data?

I worked it out assuming the drive running around 45A and it doesn’t look like it will be more efficient. You’d need 2 CIMS running at about 23.8A to cover for the 1 at 45A.

Ether’s objections aside, for the cost of four more motors and four more speed controllers, you can just buy new gear boxes.

all values at 12 volts unless otherwise specified:

1 CIM  at 45 amps         =  295.6 watts @ 3586 rpm

2 CIM  at 38 amps (total) =  **295.6 watts** @ 4645 rpm

2 CIM  at **45 amps** (total) =  347.6 watts @ 4503 rpm

2 CIM  at 45 amps (total) =  283.2 watts @ **3586 rpm**  at 9.97 volts

all calcs are approximate. # of digits does not imply real-world accuracy

I worked out my calculations based on the torque, at 45 Amps I got 111 oz inch which equates to 2 CIMS at 23.8A each

That gives a misleading answer because you are equating the torque at different speeds for 1 CIM vs 2:

1 CIM at 45 amps = 111.5 ozin @ 3586 rpm = 295.6 watts

2 CIMs at 45 amps total (22.5 amps each) = 104.4 ozin @ 4503 rpm = 347.6 watts


2 CIMs at 47.6 amps total (23.8 each) = 111.2 ozin @ 4450 rpm = 366 watts

1 CIM at 47.6 amps = 118.3 ozin @ 3480 rpm = 304.5 watts

Well i thought i was trying to answer a question about the power consumption under a certain load with 1 motor vs 2 motors which is why i used the torque values even though they provide different mechanical power values.