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Unread 29-03-2008, 21:33
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Re: Cim breakers tripping?

Quote:
Originally Posted by Hazmatt View Post
Hello. My team was having a little trouble at the West Michigan regional with the main CIM motor's breakers tripping towards the end of the match. My mentor tried to explain that as the victors heated up, resistance would rise causing a higher current draw, but that seems backwards to me.

My thought was that maybe 2 victors weren't enough to push our bot and the CIMs were being over-taxed. The bot weighs 119.2 pounds(without batteries) We are using a two wheel center drive system, 1 motor per wheel using standard toughboxes with direct drive to the wheel (no further reduction after the toughboxes). Would adding another CIM to each wheel remedy this situation? Thanks alot!
Here's the math:

The average current draw on your motors is a function of the torque seen by the motor and the average voltage from the Victor.

The torque seen by the motor depends somewhat on the friction in your drive train but mostly on the force against your wheels.

The coefficient of friction of the kit wheels (on carpet) is about 1.0 and for traction wheels is closer to 1.2. This means that your wheels can produce tractive force of up to (1.0-1.2) * the normal force (the downward force on your wheel, usually due to gravity.) If your 120 lb mass is evenly distributed on four wheels, the normal force on each wheel will be 30 lb and the maximum force on each wheel before slipping can be 30 - 36 lbs (normal force times coeefficient of friction). HOWEVER, with a center wheel drive and depending on how your other wheels are configured, you could have most or all of the mass on your two drive wheels. This raises the maximum force before slipping to 60 - 72 lbs per wheel. This maximum is important because the maximum torque seen by the motor occurs just before the wheels slip.

The force on your wheels and their radius determines the torque. With 6 inch wheels the radius is 3 inches so the torque on the wheel drive shaft will range from 7.5 to 18 ft-lbs. (with 30lbs - 72 lbs of normal force).

The standard toughbox gearbox has a ratio of 12.75. So the torque on your motor before accounting for the losses in transmission is on the order of 0.59 ft-lbs to 1.41 ft-lbs which is 113 oz-in to 270 oz-in. The normal losses in the transmission will be on the order of 15 - 30% so the actual torque as seen by the motor will be in the 133 to 318 oz-in range (with 15% losses). Now if you look at the torque curves for the CIM motor you will see that the expected currents (at 12Volts) at these torque levels are about 50 amps and 120 amps respectively. In fact, if all your weight is on your center wheels you are pretty close to stall on the motors.

If your robot has 1/4 of its weight on each wheel and you are pushing against the wall you will draw at least 50 amps. If 1/2 your weight is on each wheel (due to your CG or the geometry of what you are pushing against) your motors could draw as much as 120 amps each without slipping. Similar forces can be seen during maximum acceleration and reverse braking. You can't actually draw 120 amps from each motor at 12V because the internal resistance of the battery will cause a voltage drop at the battery. (Your actual voltage will be more like 9 Volts under such heavy loads)

As others have pointed out, the 40 amp breakers have a time delay characteristics (and so does the 120 amp breaker). Auto resert breakers typically hold at 100% and trip at around 135% of rated current. However, it can take as long as 20 seconds to trip at 135%. So you can run at about 54 amps for this long. You could typically run at 200% (80 amps) for 2 - 6 seconds. Your popping breakers are a sign that you are probably averaging well above 40 amps for several seconds. The longer you play while pushing the motors above 40 amps average, the closer you are to popping the breaker.

The toughbox is designed for operation with one or two CIM motors with a chain stage of 22/15 to 6 inch wheels. You are missing the 1.5 torque multiplier from the chain drive which is leaving you a bit outside the envelope. The center wheel drive may be compounding the problem because you can end up with nearly all of the weight on these two wheels, especially during accelleration and pushing. To answer your question,if your design allows, a second CIM will halve the torque load and put you well into the pink. With 4 CIM's fused at 40 amps, you do have some risk of popping the main breaker which is not-resettable leading to game-over. This is particularly true if the root cause of your problem is mechanical as discussed below.

Another consideration is your final drive shaft. The above calculations assume a fully working drive train that does not have extraordinary friction due to binding and/or misalignment. If you are driving the wheels directly - is your shaft supported at the far end? If so is the axis through all three bearings precisely aligned? If your shaft is binding (under load) the torque seen by the motor can be considerably higher. This might not show up when you are up on blocks but only when the wheels are loaded. Again, the center drive can compound the issue with greater weight on the shaft.

If your wheels are just hanging off the toughbox shaft this may actually be the root cause of the issues you are having. This might show up if you measure the force required to pull your robot manually (with the victors jumpered for coast). This gives you a measure of the friction through the drive train when the wheels are loaded.

For reference, we are using two CIM's per drive with rear wheel direct drive using a modified toughbox transmission geared for 6.4 to one. This is similar to your configuration except with 1/2 the gear ratio. With two CIM's on each wheel, the torque on each of our CIM's is about the same as yours. Our wheels, however, will spin (by design) before we pop the breakers because our CG is well forward of the drive wheels and our normal force is on the order of 25lbs.