Re: Warp Factor III is now Alive!
If the resistance is 0.1 you are in trouble. The average resistance would be half that meaning 0.05 so the losses in a single connector block will be 3000^2 * 0.05 = 450000 watts of waste heat in every connector block. (Note: It is impossible to have this much resistance and still get to 3000 amps. If the resistance was actually 0.1 ohms per cell then you would be limited to a short circuit current of only 37 amps average) The resistance better be closer to 0.0001 which would give 900 watts of waste heat per side or 1800 watts per cell. Still too much as this is more than a space heater on every cell.
You need to run current from one side of your connector block to the other and measure the voltage drop from a point different than the connection. Divide the voltage drop by the current to get the resistance. Example. You pass 50 amps through the block and read a voltage of 0.005 then the resistance will be 0.0001 ohms. You may have to pass quite a lot of current if your meter doesnt have a millivolt range. The effective resistance will be about half of this because of the way the cells are paralleled.
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Doug Ingraham
Rapid City, SD
1985 Mazda RX-7 GSL (1400+ EV miles)
Now on the road but still under construction.
Last edited by dougingraham; 07-13-2012 at 03:31 PM.
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