Quote:
Originally Posted by asid61
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.
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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.