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Hi again pro's …


On my brother's 302, we installed a pertronix I in the distributor with a flame-thrower coil (1.5 ohms).

As pertronix recommand, a 12v to the + side of the coil, (no more resistance wire) with the red wire of the module also at the +side of the coil.
Black module wire to the - side of the coil.

This setup looks like it cured a part throttle bog, but the coil is pretty hot, with my infrared temp gauge I get 172deg
after an 1/2 hour drive in the city. Whit my original resistance wire to +side of coil, but the red module wire to a 12v, my temp is 140, but I have
my part throttle bug again. Could it be my sparkplug gap ? Should I gap them .035" or .045" or ??

I sent an e-mail to pertronix 6 days ago and still no feedback.

Altought pertronix recommand 8 amps MAX to the coil, so with what I've learned is: 12v/1.5 ohms (coil) =8 amps
But with charging system I can see 13.5 volts at 2000rpm at the coil, so we are at 9 amps min.

So do you think that coil will last long, and does somebody as the same wiring that's actually running for years without problem ?

Thanks a lot.
 

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Pertronix has had their head up their respective you-know-whats for a while. They don't put this on the Ignitor I instructions, but very plainly on the Flamethrower Coil instructions....Lord knows because it's more applicable to the Ignitor....

Do NOT remove the ballast resistor or resistance wire if the primary resistance is lower than specified or if you are using the stock coil.

You can purchase a ballast resistor in a ceramic housing, bolt it to the coil mounting bracket, move the wires from the coil to the input side of the ballast resistor and a new short wire from the other side to the coil+. This way your coil will have correct resistance and your Ignitor will have 12V. The "down" side is that you won't have 12V to the coil during cranking. To do this you'll have to find the BROWN wire from the solenoid "I" terminal, trace it along to the firewall plug, snip it, and run an extension also to the coil+.

Make sense?
 

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Pertronix has had their head up their respective you-know-whats for a while.
This sentence makes me happy. Can't say why exactly but I'm going outside to piddle with cars in a good mood now.
 

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Ohm's law states for X amount of residence, doubling the voltage quadruples the amount of watts. Multiple watts by 3.4 will give you BTU. A typical coil with with a impeadence of ~ 1.5 ohms plus a primary resistance of ~ 1.5 ohms gives you ~3 ohms impeadence. That's working with ~7 volts. So you've doubled the coil voltage and cut the total primary impeadence to 1.5 ohms, no wonder it's running hot
 

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no wonder it's running hot
Ok Tom, so what is hot? My coil is mounted to the cylinder head on the front of an engine running around 190º. Since I have both pre and post resister wires available at the coil, the post resister wire end is taped up and sitting on the intake, I could swap them in a few minutes and see if it makes any difference in how the car runs. If I swapped to the Motorcraft DG470 coil would that make the resistance / impedance issues go away?

Not trying to be an a$$ but am confused like many others
 

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So I took my own advice and did some wire swapping. The Pertronix 1 and the FiTech 12v keyed ignition remained at pre resister voltage while only the flame-thrower coil went to the factory resister side of the wiring. Car fired right up and acted exactly as before right up to the 6400 rpm redline. Since the Pertronix 1 is designed to work with the resister side anyway I might try to swap the Pertronix to the resisted side of things to see if they're is any loss of performance.

I was only able to do about 20 miles of driving before the local constabulary was in my rear view mirror. One of the benefits of living near a county commissioner!
 

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Ohm's law states for X amount of residence, doubling the voltage quadruples the amount of watts. Multiple watts by 3.4 will give you BTU. A typical coil with with a impeadence of ~ 1.5 ohms plus a primary resistance of ~ 1.5 ohms gives you ~3 ohms impeadence. That's working with ~7 volts. So you've doubled the coil voltage and cut the total primary impeadence to 1.5 ohms, no wonder it's running hot
Tom : Just so you don't lead someone astray, your formulae are incorrect.

Ohm's Law is V = IR, where V = voltage, I = amps and R = resistance.

Watts = V x A. Thus if you double the voltage, with the same amps, it will only double the watts, not quadruple them.

Bob
 

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Tom : Just so you don't lead someone astray, your formulae are incorrect.

Ohm's Law is V = IR, where V = voltage, I = amps and R = resistance.

Watts = V x A. Thus if you double the voltage, with the same amps, it will only double the watts, not quadruple them.

Bob
Bob, if you double the voltage with the same resistance the amperage will not be the same. If I have 12 volts at 6 ohms, amperage is 2 amps, 2x2x6=24 watts. 24 volts at 6 ohms is 4 amps, 4x4x6=96 watts
 

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Ok Tom, so what is hot? My coil is mounted to the cylinder head on the front of an engine running around 190º. Since I have both pre and post resister wires available at the coil, the post resister wire end is taped up and sitting on the intake, I could swap them in a few minutes and see if it makes any difference in how the car runs. If I swapped to the Motorcraft DG470 coil would that make the resistance / impedance issues go away?

Not trying to be an a$$ but am confused like many others
No worries. Here's the thing. A coil is nothing more then a transformer. It has a turns ratio so it one side has 10 turns of wire to the other side of 1 turn, you have a 10:1 turns ratio and putting 12 volt to the 1 turn side, the 10 turn side will produce 120 volts. The rub is that a coil or transformer will not work on DC. It will work with AC or pulsating DC. To produce current mechanically you need 3 things, conductor, magnetic field and the conductor cutting the magnetic field. So you can either move the wire or move the magnetic field. Take a look at a wire with 12 volts applied from the battery. As soon as you remove the 12 volts, the magnetic field collapses, you now have the 3 things needed to produce power. Yes, it will fight the power being re applied such as when the points close to energize the coil. The faster the AC frequency or the pulsating DC the power being generated from the collapsing field opposes the incoming power more and more. You'll eventually will reach a point where the coil will simply stop working. Up to that point you will reach a point of diminishing returns or basically full saturation of the coil. A good example is a mechanical voltage regulator. They had 2 relays, 1 wound with a very fine wire that regulated voltage and 1 wound with a heavier wire for amperage. The fine wire with many wraps of wire would get saturated quickly and limit voltage while the heavier winding would have to have a very heavy draw of amperage to work. I worked in a Chevy dealer parts dept. Ignition coils were listed for manual or automatic transmission. The automatic coils were designed to produce their hottest spark at the governor shift point. I would like to clarify one point about turns ratio and ignition coils. That 20,000 volts isn't so much the turns ratio but rather the suddenly collapsing magnetic field that produces that 20,000 volts. You're also producing a sizeable spark on the primary side of the coil that back feeds into the car's electrical system. That's why there's a condenser aka capacitor which absorbs that spark.

Basically any coil is going to produce a spark but randomly installing any coil may not produce desired results. Personally I don't believe applying 12 volts to a 6 volt coil is really going to make much if any difference in spark output due to the magnetic component. Systems like factory HEI use a low resistance coil which means the gauge of the primary wire is heavier allowing more amperage because the HEI can better manage dwell time and really saturates the coil with a lot of energy. It's like a arc welder to static electricity. Again these get limited to that point where the coil will stop working efficiently. This is where CD ignitions come in. They will operate their coils on a higher voltage. the CD box may send the coil 24 volts which means they can get more wattage for the amperage compared to 12 volts.

For DC the terms are resistance and wattage. For AC, due to the magnetic component instead of resistance, it's impeadence which is represented by the letter "Z" and instead of wattage it's "VA" or Volt Amps.
 

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Discussion Starter #11 (Edited)
Thanks guys very instructive, but still have some questions :


My old no name coil as "12V" stamped on it along with this: "EXTERNAL RESISTOR REQUIRED".
I did checked it with my multimeter and it has exactly 1.5 Ohms.


So I assume that it was only to prevent the points from burning right ?
And if I'm right, I can use that coil with my pertronix with no resistance wire right ?




Next one...
If I run pertronix with a flame-thrower coil and full 12 volts at the + side of the coil with NO resistance wire, like I said in my first post, running
at 2000 rpm it's more 13.5V, so if my math is right, that circuit will draw 13.5v/1.5ohms= 9 amps, that is more than the 8amps MAX.
that pertronix recommand, and surely not good for a long period of time. So to be safe if I can find a ballast resistor of 0.3 ohms,
I'll get 13.5/(1.5ohms coil + 0.3ohms ballast) = 7.5amps at max voltage.


If someone can approve my math I'll shop for a 0.3 ballast but I don't think there's one 0.3 ohms.


And do you think it will affect performance if a coil is getting 9, 10, or 12 volts at the + terminal ??


Thanks a lot
 

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Thanks guys very instructive, but still have some questions :


My old no name coil as "12V" stamped on it along with this: "EXTERNAL RESISTOR REQUIRED".
I did checked it with my multimeter and it has exactly 1.5 Ohms.

So I assume that it was only to prevent the points from burning right ?
And if I'm right, I can use that coil with my pertronix with no resistance wire right ?

Next one...
If I run pertronix with a flame-thrower coil and full 12 volts at the + side of the coil with NO resistance wire, like I said in my first post, running
at 2000 rpm it's more 13.5V, so if my math is right, that circuit will draw 13.5v/1.5ohms= 9 amps, that is more than the 8amps MAX.
that pertronix recommand, and surely not good for a long period of time. So to be safe if I can find a ballast resistor of 0.3 ohms,
I'll get 13.5/(1.5ohms coil + 0.3ohms ballast) = 7.5amps at max voltage.


If someone can approve my math I'll shop for a 0.3 ballast but I don't think there's one 0.3 ohms.


And do you think it will affect performance if a coil is getting 9, 10, or 12 volts at the + terminal ??


Thanks a lot
The wiring of the stock coil might not be able to handle the higher currents whereas, presumably, the Flamethrower was designed specifically to allow for direct full battery voltage per their documentation.

Your math is correct but here are two possible explanations:
(1) When they spec'ed it they may have spec'ed it backwards... "12v battery divided by 1.5 ohms.... 8 amps... put that in the spec" (BTW: Where did you find that number?)
(2) The overall resistance when the Flamethrower is used with an ignitor1 (vs points) is possibly different. The ignitor1 turns on a transistor to allow current through the primary (vs points which are effectively a switch). The ignitor1 transistor likely has some "on-resistance" that would then be in series with coil resistance and, thus, reduce the overall primary current. Unfortunately, I cannot find any spec for this "on resistance".
 

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Could it be my sparkplug gap ? Should I gap them .035" or .045" or ??
Do you know what your spark plug gap is? Mine is at .036 and yesterday after posting on this thread I got bored and swapped my flamethrower coil to the factory resister wire from full voltage and had no observable loss in performance and certainly no bog as you describe.
 

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Thanks guys very instructive, but still have some questions :


My old no name coil as "12V" stamped on it along with this: "EXTERNAL RESISTOR REQUIRED".
I did checked it with my multimeter and it has exactly 1.5 Ohms.


So I assume that it was only to prevent the points from burning right ?
And if I'm right, I can use that coil with my pertronix with no resistance wire right ?




Next one...
If I run pertronix with a flame-thrower coil and full 12 volts at the + side of the coil with NO resistance wire, like I said in my first post, running
at 2000 rpm it's more 13.5V, so if my math is right, that circuit will draw 13.5v/1.5ohms= 9 amps, that is more than the 8amps MAX.
that pertronix recommand, and surely not good for a long period of time. So to be safe if I can find a ballast resistor of 0.3 ohms,
I'll get 13.5/(1.5ohms coil + 0.3ohms ballast) = 7.5amps at max voltage.


If someone can approve my math I'll shop for a 0.3 ballast but I don't think there's one 0.3 ohms.


And do you think it will affect performance if a coil is getting 9, 10, or 12 volts at the + terminal ??


Thanks a lot

Did you read my previous post? Run a ballast resistor. Your math is wrong. Impeadence for the coil is going to be 2πfxl. As the rpm or frequency changes so will the impeadence. Gap the plugs to factory specks. The coil is going to produce only so much power regardless of how much you try to increase primary voltage.
 

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Did you read my previous post? Run a ballast resistor. Your math is wrong. Impeadence for the coil is going to be 2πfxl. As the rpm or frequency changes so will the impeadence. Gap the plugs to factory specks. The coil is going to produce only so much power regardless of how much you try to increase primary voltage.
I think Patbel is trying to figure out why he has a bog with the resister and not at full voltage. Agreed, fresh plugs at the factory gap is step one.
 

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I think Patbel is trying to figure out why he has a bog with the resister and not at full voltage. Agreed, fresh plugs at the factory gap is step one.
A low voltage spark isn't going to cause a bog. If anything, it's going to cause a misfire and a bunch of popping and snorting as the spark breaks down due to cylinder pressure. A bog is typically a FUEL issue and, if not, almost certainly a spark TIMING issue.
 

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A low voltage spark isn't going to cause a bog. If anything, it's going to cause a misfire and a bunch of popping and snorting as the spark breaks down due to cylinder pressure. A bog is typically a FUEL issue and, if not, almost certainly a spark TIMING issue.
It seems something else is going on. I must admit I was surprised when I lost no performance when going from full, to resister voltage, on my flamethrower coil. But I'm currently employed as a cabinet maker. A man's got to know his limitations.
 

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Discussion Starter #18
Do you know what your spark plug gap is? Mine is at .036 and yesterday after posting on this thread I got bored and swapped my flamethrower coil to the factory resister wire from full voltage and had no observable loss in performance and certainly no bog as you describe.


Thanks for your answer Nailbender, the plugs were at .044", so I gapped them to.035", use the flame-thrower coil with stock wire (with resistance),
and also notice that my accelerator pump cam was on #2 on throttle shaft, switched it to #1. I've learned that #2 is for cars idling at 1000 rpm and more,
#2 will retard the cam about 5 degrees to compensate for the already rotated throttle shaft. Now it's running good !! I did not switch back to #2 to see if
the bug was related to that, anyway it's running good.

Thanks
 

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Discussion Starter #19
Did you read my previous post? Run a ballast resistor. Your math is wrong. Impeadence for the coil is going to be 2πfxl. As the rpm or frequency changes so will the impeadence. Gap the plugs to factory specks. The coil is going to produce only so much power regardless of how much you try to increase primary voltage.

Thanks Huskinhano,

you mean the resistance will change with rpm (voltage change) ?

Yes I did gap the plug to stock spec and it's ok !

Thanks again
 

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Discussion Starter #20
The wiring of the stock coil might not be able to handle the higher currents whereas, presumably, the Flamethrower was designed specifically to allow for direct full battery voltage per their documentation.

Your math is correct but here are two possible explanations:
(1) When they spec'ed it they may have spec'ed it backwards... "12v battery divided by 1.5 ohms.... 8 amps... put that in the spec" (BTW: Where did you find that number?)
(2) The overall resistance when the Flamethrower is used with an ignitor1 (vs points) is possibly different. The ignitor1 turns on a transistor to allow current through the primary (vs points which are effectively a switch). The ignitor1 transistor likely has some "on-resistance" that would then be in series with coil resistance and, thus, reduce the overall primary current. Unfortunately, I cannot find any spec for this "on resistance".
I saw these number on that page……….
 

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