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Old 09-21-2010, 02:53 AM
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Harvey Harvey is offline
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Join Date: Jul 2009
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Hi Erik,

A typical ignition is wired this way:


Battery(+) -------> IGN(+)[Primary Winding] IGN(-) -------> [Contact Points and Condensor] ------> Battery(-)

Battery(-) -------->[Secondary Winding]--------->[Distributor Center Post]----->[Distributor Rotor]---------[High Tension Cylinder Wires]------>[Spark Plug]------->Battery(-)

From this diagram two things should be noted:
1. The condensor sits across the points to minimize burning the contacts by arcing

2. The energy for the secondary path is provided solely by the field collapse of the primary winding. This should be obvious because the secondary path starts with the Battery(-) and ends with the Battery(-). Therefore all the energy is supplied magnetically in the ignition coil.

In your case you are melting wires - presumably the high voltage wires from your explanation. Wires do not melt due to voltage, but instead due to internal heating caused by forcing current through them beyond what they can tolerate. The engineering rule of thumb for copper wire is 700 circular mills of wire mass per ampere. If this is exceeded, the wire will heat up.

This is where the voltage applies: The current is the result of Voltage being applied across a resistance - in this case it would appear to be that resistance of your wires. Using #13AWG (note that British wire gauge is different here) you get exactly 2.00 Ohms per thousand feet of wire. The circular mils of that wire (5180) only supports up to 7.4 Amps without heating the wire. This means that you can only safely have 14.8V applied for every 1000 feet of wire. The formula is E = I x R where E is electromotive force (voltage) and I is intensity (current) and R is resistance.

Using a distributor is a great way to switch High Voltage, but it will not prevent the burning of wires. If you intend to use short lengths of wire that have High Voltage on them then you should have a load between them and ground to limit the current. A spark plug is a good load because the spark itself has a reasonably high resistance. But you will still want to measure the current through your secondary circuit to keep it below the rated values.

I hope that helps
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