coil connections

[russcc]

Rumbleseat advised in "coil connections" that in a positive grnd system, the ignition feed goes the negative terminal on the coil for best performance. I have 12 V, negative ground system, so I assume that the connections are reversed. The ignition feed goes through an external resistor. I have 2 coils, the terminals are marked differently. #1, the terminals are marked + & - . #2, coil is marked dist & bat. Please see photos of the 2 coils. On the + & 1 marked one I assume, (again), that the ignition feed goes to the + terminal. Thank you Fordbaners.

[russcc]

Photos.

[koates]

Hi there, You say you have a 12 volt (negative ground) system so I presume these original (looking) 6 volt Ford coils have been rewound to operate on 12 volts, otherwise you cant use them on 12 volts. For a negative ground system coil # 1 is connected negative terminal to distributer points and positive terminal to ignition switch power through the resistor. Coil # 2 for negative ground system connect "DIST" terminal to ignition switch power through the resistor. "BAT" terminal connects to distributer points. Its the other way around on both coils for Positive ground system. Regards, Kevin.

[russcc]

Hi Kevin. Thank you for your response. The coils are original 6 volt, and there is a resistor before the coil to drop it to 6 volts or less. I get it on how to hook up both coils on negative ground system, I appreciate your clarification on that, as I may have made the question a little complicated.

[Terry,OH]

russcc: On a 6 volt system, Ford uses an in line o.5 ohm resistor to drop the voltage to the coil to about 4 volts. The resistor also limits the current through the coil and the distributor points to about 3 Amp. For you to use the Ford 4 volt coil on a 12 V system the resistor would have to drop about 9 volts (9+4=13V) and still pass the 3 Amps. The wattage size of the resistor would be 9 X 3 = 27. W but you do not want to have the resistor to operate at it's limits so double the wattage of the resistor to about 50.W That is one BIG resistor that would put out a LOT of heat, plus the price is $$$. Kevin Koates is correct have the coil reworked by Skip to 12V.

[david.skinner]

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I'm interested in why the spark would be better with one coil direction rather than the other? I can understand the logic of connecting the Bat terminal to the distributor but why does the polarity of the spark matter?
Regards
David

[JSeery]

Yes it matters, it is how the coil is designed. They will work connected backwards but only at about 80% I believe. There are test you can do to assure the spark is polarized correctly, a lead pencil test.

"Why do we worry about coil polarity? Because the spark plugs do care which way the electrons are flowing in the high tension circuit. The spark plug has a thermally insulated center electrode (surrounded by ceramic). With engine running the center electrode runs substantially hotter than the exposed end electrode. Design of the ceramic insulator determines how hot the center electrode will run, leading to the designation of hotter or colder spark plugs. As electrons go, they love to jump away from a hot surface and fly toward a colder surface, so it is easier to drive them from hot to cold rather than from cold to hot. End result is a difference of 15 to 30 percent in voltage required to make spark "initially" jump the gap on the plug depending on which way it is going. So the spark plug prefers to see a voltage potential that is negative on the center electrode and positive on the end electrode for the very first hop of the spark. Oddly enough, this has nothing to do with polarity of the vehicle electrical system, but it is influenced by the common connection inside the ignition coil."

https://www.youtube.com/watch?v=9eAHeKDelF0

[flatheadmurre]

To take full advantage of the ford coil system you need an airwound resistor that is allowed to heat up for additional resistance giving you a hotstart function with less currentflow at running for less heat.
Using a ceramic resistor will not give you this.

[david.skinner]

Awesome answer guys. I'm an electrical engineer and I didn't know that. Many thanks