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-   -   Battery Leads and Resistance (http://www.chiefdelphi.com/forums/showthread.php?t=119062)

Mr V 10-09-2013 13:39

Re: Battery Leads and Resistance
 
Quote:

Originally Posted by cgmv123 (Post 1290753)
.002 ohms (.004 ohms accounting for both + and -) can definitely impact performance. At 100 amps, you lose .4 volts. At 532 amps (4 CIMs at stall), you lose 2.12 volts. That's just accounting for the resistance of the wire, there are other components that cause voltage drop.

The calculation was based on the fact that they extended the battery location aprox 2.5' so 5' of additional wire.

The 532 amp figure is irrelevant as the battery can not produce that kind of current. I don't have the time to look it up right now but IIRC the max current the battery can produce is 300a or so. When it is under a 300a load the voltage measured across the terminals will be much less than the open circuit voltage, probably in the range of ~10v or so.

The 100a figure is more realistic as to what will be seen in the real world on a regular basis. So the real world voltage additional drop seen at the power distribution board from extending the wires 2.5' will be in the neighborhood of .2v, or less than 2%, which is not something that you will notice in regular play. If the motors are operating at anything less than 98% throttle a very slight increase in throttle input will compensate for that additional voltage drop.

Which is why I said in the real world the difference will be so small that it is not really noticeable.

Al Skierkiewicz 10-09-2013 13:44

Re: Battery Leads and Resistance
 
Mr. V,
The battery can supply 600+ amps fully charged but the terminal voltage will be less than 12 volts due to drop across the internal resistance. CIM motors will draw 131 amps stall (start) current each. While your mileage may vary, a four CIM drive under most conditions will draw 400 amps when starting.


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