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Unread 02-04-2007, 15:22
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Al Skierkiewicz Al Skierkiewicz is offline
Broadcast Eng/Chief Robot Inspector
AKA: Big Al WFFA 2005
FRC #0111 (WildStang)
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Re: Robot doesn't turn in high gear

Quote:
Originally Posted by newton418 View Post
When designing the drive train, I figured we could run each small CIM at 40 amps (since they are on a 40 amp breaker); however, I was informed that the power distribution block is rated for 85 amps (I'm not exactly sure, but around this number). This means we weren't getting the current, and thus the torque, we had planned on. Perhaps you guys are facing a similar problem.
This is another one of the myths of electrical design. The Rockwell blocks are designed for continuous current of 85 amps per block. The 40 amp breaker will function with up to 600% of it's current trip rating for several seconds. Neither of these devices limit the current to that of their ratings. But there are limiting factors in the wiring of the robot. Things to consider are the series resistance of the wire, terminals, Victors, and distribution. A simple rule of thumb I use is the "wire/foot" or WF. It is simple to equate series resistance with the resistance of a foot of #10 wire which is .001 ohms. Using this as a rule of thumb, then the battery has 11 WF internally, the #6 feed wiring has 0.5 WF per foot of wire (you must include both the postivie and negative wire) while the #12 has almost 2 WF per foot, the Victors have 4 WF of series resistance and you can have between 1 and 3 WF for every crimp if not done with proper tooling, not tight on the block or screw terminal or not fully crimped. So, how does this affect your design? At 100 amps of current (roughly the stall current of either Chalups motors or the old FP motors) each wire foot will drop 0.1 volt. Just take a look at the circuitry that feeds your motor, add up the resistance using this method and multiply by 0.1 to find the voltage drop to your motor.
Ex. 4 ft #6 wire, breaker panel, one victor, 4ft of #12 from fuse panel to Victor to motor, 20 connections= (4x0.5)+1+4+8+20=35x0.1=3.5 volts drop in all losses at stall. This translates to a loss of about 1/3 of the available current or about a loss of almost 100 oz.in. of torque and a drastic change in the efficiency.
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Good Luck All. Learn something new, everyday!
Al
WB9UVJ
www.wildstang.org
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Storming the Tower since 1996.