Re: mallory unilite hookup confusion
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Re: mallory unilite hookup confusion



I recant everything I just said, ......its probably a slightly reduced diameter wire in the BR which causes the resistance increase and subsequent voltage drop
  Neal


On Fri, Oct 18, 2013 at 7:49 PM, neal zimmerman <neal.zimmerman@xxxxxxxxx> wrote:
Yes, I reviewed  my original post and I was wrong. I forgot to say that in start the resistor is indeed bypassed, then as Joe says you switch over to a different path which goes THRU the BR and that's where the magic happens.To correct my last post I should have said that heat buildup causes resistance and consequently voltage drop. That would be a more correct way to say it.
  Neal


On Fri, Oct 18, 2013 at 7:43 PM, neal zimmerman <neal.zimmerman@xxxxxxxxx> wrote:
yes, but the resistance/voltage drop is caused by the heat buildup in the BR.  When the car is warmed up that BR is warm
  Neal


On Fri, Oct 18, 2013 at 6:43 PM, Joe Bitterman <furyrestored@xxxxxxxxx> wrote:
You get a full twelve volts while cranking the engine because there is a feed from the starter relay which feeds the points12 volts in the cranking position of the ignition switch. As the motor starts and you release the key to the run position the wire from the starter relay no longer is putting juice to the distributor and it now gets it's reduced volts from the ballast resistor in the run position

Sent from my iPhone

On Oct 18, 2013, at 9:12 PM, neal zimmerman <neal.zimmerman@xxxxxxxxx> wrote:

To my way of thinking, electricity likes a certain temperature range, not too cold, not too hot. If I can remember from school resistance is caused by one of 3 things ( 2 of which I can only remember right now LOL) One is small wires, and the other is temperature extremes. #3 might be length of wire... longer wire, equals more resistance. The little coil in the ballast resistor is cold at first so you get full 12 volts to coil. As it heats up it causes resistance which starts to drop the volts. When I think about this it is similar in a vague way to how the electric chokes work too, except with the choke its a bi-metallic spring which starts to tweak upon heating and opens the choke plate.
  Neal


On Fri, Oct 18, 2013 at 4:12 PM, Paul L. http://1962to1965mopar.ornocar.org/ml-lennemann63.html <pjlenn@xxxxxxxxx> wrote:
To my way of thinking the resistor doesn't care which end is which.  It will pass voltage and reduce it either way.  The heat is the result of the voltage drop being converted to heat energy.  Just like an element on a stove top.  A diode (like in an alternator will only pas voltage in one direction. 

Paul L.


On Friday, October 18, 2013 11:57:15 AM UTC-5, neal wrote:
 OK, it's getting close to fire up on the barracuda, . I wanted to double check how the previous owner had hooked up the Mallory Unilite set up, and  ( surprise, surprise!) he did it wrong according to the mallory tech sheets. Amazing it ran. Anyway they say to hook in  the red wire from their distributor to the 12 volt side of ballast resistor (with a little picture of a key and ignition switch), but according to my wiring diagrams, the dark blue wire from one side of resistor and the brown wire  on the other side of resistor both run to the ignition switch. The dark blue goes to "1" on ign. switch and brown goes to "2" on ign switch.
So which wire is the 12 volt side of resistor?? I have a feeling its the blue, but not sure.

  BTW is there a front door and  a back door to a ballast resistor or does it not care??
  Thanks , Neal Zimmerman Eugene Oregon

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