Re: Let's discuss Packard 6V electrical systems and the change to 12V systems

Posted by gone1951 On 2008/6/21 13:28:02

Looks like I should have read all the postings in this discussion befor I wrote the following but I'll leave it in anyway.

E= IR, IR Drop. The applied voltage has little to do with the voltage drop over the length of a wire. What is important is the I R drop. The loss along a wire is calculated by multiplying the current in amps times the resistance per foot in ohms. the result is the voltage drop per foot. Multiply the voltage drop per foot by the length in feet of the wire and you get the over all voltage loss. As the wire diameter goes up the resistance goes down and the voltage drop goes down. If you make the wire size smaller the resistance goes up and the voltage drop goes up.

P=IE The power needed to turn the motor over with the starter motor doesn't change just because you changed the voltage. The power is the power. It is the measure of the mechanical force it takes to cause the motor to turn over fast enough to start. I is the current in amps and E is the voltage in volts. You can see that to develop the same power with 6 volts as opposed to 12 the current has to be twice as much. This necessitates the need for the larger wire size for 6 volt systems as opposed to 12 volt systems.

I too think that the switch from 6 volt electrical systems to 12 volt systems was for economic reasons. With a 12 volt system all the wiring can be accomplished with much smaller wire including the internal wire of the starter and generator.

Another note: I have owned dozens of cars with 6 volt systems in my day and had no particular problems. The voltage was not really an issue. If you maintain the engine in good condition as well as the battery and charging system you will never have a problem just because the battery has 3 cells instead of 6. Most of the starting problems are a result of low compression or starter problems or a poorly maintained charging system. The carburetor is a good place to look to correct a hard starting condition. Take care of it and it will take care of you.

Bob