So I was thinking about main breaker trips with 6 cim drive trains, and I was thinking about how to avoid that.
But because torque is directly proportional to current, and the motors are all cims, for a given current (say, 120a) wouldn’t the toruqe output be the same regardless of the number of cims?
Basically, if the amount of current was limited per cim, through the 40a breakers only, then 6 cim drive would have more max power. But if the power of the system is limited to 120a total, then how is it advantageous to use 6 cims instead of 4?
Right now I think it’s because the breakers need a few seconds to trip and efficiency gains, but that seems like a small gain to me for the extra space and weight.
Why do you believe that the system is limited to 120A? The main breaker is guaranteed NOT to trip until the current is ABOVE 120A. Even at 240A that main breaker will not trip in under 45 seconds.
Also, a 6 CIM drivetrain pushing the same as a 4 CIM drivetrain will draw less current per CIM, thereby possibly preventing the 40A snap action breakers from tripping.
The real difference between 6 CIM and 4 CIM drivetrains is acceleration. Assuming both drivetrains are geared for the same max speed, the 6 CIM drivetrain will accelerate faster. If the both drivetrains are stalled, the 6 CIM drivetrain will pop the main breaker faster as well. Such is tradeoffs in engineering.
The 120a Breaker is capable of instantaneous currents of over 720 amps before tripping (see attachment), and 240 amps for a considerable amount of time (for a FRC match). While not breaking the easier-to-trip 40 amp fuses, you could run much higher currents than a 4 CIM drive, which would be limited to 160 amps, for some seconds.
Just a word of caution: If you stall a CIM it pulls ~133A. If you stall a 6 CIM drive, it’ll pull somewhere between 400-500% of the rated current for the main breaker (current being limited to the batteries internal resistance, I don’t have that number in front of me right now). That will trip the main breaker in 1-6 seconds (as per the chart above). The 40A snap action breakers will be at ~ 250% of rated current and may take longer than the main breaker to pop (again I don’t have the spec sheet in front of me) and that’s not accounting for any other electrical load you have at that moment.
Basically, ensure your 6 CIM drivetrain is never in stall (make sure it’s traction limited, not torque limited).
If each CIM were drawing 133A, a 6 CIM drive would pull 800 amps.
In addition to frying all 6 motor controllers, the plastic PDB would be molten as well. Your wires would be scalding to the touch, and there would be a large quantity of gaseous substance emitting from the robot. Basically, yes it is a bad thing, but no, it will practically never happen. This is really not something to worry about.
We ran a 6 CIM VexPro ballshifter running 3" wheels, and geared to 5.5fps and 15.5fps. In addition to bulldozing everyone we faced, we never blew a breaker once. That being said, we did have a “ramp up function” in the code. Basically, if you jammed the joystick into full forward, the code would automatically and linearly increase the PWM “throttle” very quickly (read: like 1/2 second) from 0 to 100%. Even when throwing the robot from full forward into full reverse, we never had a breaker-blowing situation. The acceleration time is negligible compared to instant-100% because the ramp up is so quick and you have the comfort of knowing whatever you could possibly do to that joystick will not disable the drivetrain. The ramp up has the advantage of suppressing the instantaneous current draw into a slightly longer duration but lower peak draw.
When testing this year, we found that a new breaker can take 6 stalled CIMs and a compressor (~250 amps) for about 25 seconds before tripping. Note that this is to a large extent cumulative over the course of a match, because these breakers are thermal and they don’t dump heat instantly (or even particularly quickly); that is, if you stall for 20 seconds early on, you’re going to be in real trouble if you get into another pushing match 30 seconds later.
Quick question, how would you stall a 6 cim drivetrain in normal usage? The wheels would lift at some point, even driving into a wall, right?
Of course, we would still build in current limitations and preferably shift.
Get into a pushing match with a robot, if the traction of your wheels exceeds the torque created by your robot then the motors will be in stall. (Assuming all wheels are on the ground)
Even with 2 or 4 wheels off the ground, it’s possible for your wheel traction to still be greater than the torque of a 6CIM drivebase (However, unlikely)
Keep in mind that in a 6wd configuration, there are only 4 wheels in contact with the ground at any given time. It remains the same for 8wd, with only 4 wheels ever actively touching the ground (assuming center two wheels are dropped)
Fun fact: 6 CIMs at stall consume a total of roughly 9576 watts of electrical power, which is over 9000!
Seriously though, a robot with 6 CIMs under “normal load” output (excluding friction) a total of 2.6 Horsepower, while a 4 Cim drive does just 1.75. At exactly 40 Amps per motor the numbers are (theoretically) 3.86hp and 2.57hp, respectively.
And as everybody knows, power/weight ratio is the most important figure. :rolleyes:
With a 125lb robot (including battery and bumpers) geared 6.1:1 with 4’’ HiGrip wheels, we stalled 6CIMs when simultaneously running the compressor on a not-quite-full battery, resulting in a current draw of ~250 amps.
With the compressor off and the battery completely full, the same robot spun the wheels and pulled ~180 amps.
They stall at significantly less than that when you’re running them off of a real battery whose effective voltage drops significantly when you start drawing lots of current from it.
It’s the same reason that you can stall them at all in the first place; if you naively just calculate from the stated stall torque of the motor you’ll find that your robot would have to weigh an ungodly amount to stall them at any gearing you’d see on an FRC bot.
True true. I thought about that but did not account for it.
As the resistance outside the battery decreases and approaches near zero (v=ir so 12=133r so r= ~.1ohms) (the motor stalls), the internal resistance inside the battery becomes a significant percentage of the overall resistance, so the potential difference beyond the two terminals is proportionally less than it were when the internal resistance was a small fraction of the overall resistance. Another point for Ampere’s law.
