Found some ignition resistance....

[BUBBAS IGNITION]

Moving some swap meet parts around this morning and tripped over these two resistor blocks. The one (small one) is a off shore replacement (growing green stuff already) and the other one with the relays is a orginal ford i bought at a swap meet.

They both show .5 ohms !!

Using ohms law and a new skip coil ( .6 ohms) total resistance would be 1.1 ohms and using a six volt battery charging at 7.2 volts the system would see 6.5 amps .

No wonder these systems have some much drama . A compare 12 volt system would use approx 4 amps !!

Then add some more with the green stuff growing from the rice dirt on the off shore parts and it would cause even more problems!!!!!

[ford38v8]

Bubba, with your talking about volts, amps and ohms, I'm lost. Please translate in terms that I can recognize: I've always been under the impression that a Ford resistor will drop voltage by about 2 volts, and that number will vary up or down depending on the charging rate at the moment, to provide a steady voltage to the coil. Am I right in this theory, and how does this relate to the findings you describe today?

Thanks!

[flatheadmurre]

You forget it´s a ballast resistor...when it heats up the resistance increase.
Hook it up and measure voltage drop over it will raise when heated up.

[ford38v8]

Originally Posted by flatheadmurre

You forget it´s a ballast resistor...when it heats up the resistance increase.
Hook it up and measure voltage drop over it will raise when heated up.

Yes, that's what I"m talking about. Available incoming voltage means nothing until the circuit is closed at the points, when the true outgoing voltage can be measured, right?

[BUBBAS IGNITION]

Originally Posted by ford38v8

Bubba, with your talking about volts, amps and ohms, I'm lost. Please translate in terms that I can recognize: I've always been under the impression that a Ford resistor will drop voltage by about 2 volts, and that number will vary up or down depending on the charging rate at the moment, to provide a steady voltage to the coil. Am I right in this theory, and how does this relate to the findings you describe today?

Thanks!

Using the above 6 volt example . Volt Drop is amps x resistance : so 6.5 amps x .5 ohms would equal a voltage drop of 3.25 volts..........then 6.5 ampsx .6 ohms would = 3.9 volts 3.9 plus 3.25 = approx 7 volts..... with the remaining voltage dropped with wiring ( special) and connectors etc....

These rules change with charging voltages and heat ( like flatheadmature) mentioned but will be very close to the actual voltages used in the system....

[Jack E/NJ]

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Then if you also factor in point dwell, heat dissipation, rpms, local flora & fauna, plate tectonics and the phases of the moon, you soon must conclude that these ignitions simply won't work. 8^) Jack E/NJ

[flatheadmurre]

If i got what´s going on right that resistor is a real clever little thing and works like a hot start.
Cold it´s .5ohm but after a while with current going through it the resistance raises a considerable amount dropping the current flow down to running condition.
So instead of having a bypass from the starter fot hot start like in a modern system that resistor does all the work.

[4dFord/SC]

So, does resistance cause heat, or vice-versa?

[Jack E/NJ]

I'd guess yes. Jack E/NJ

[BUBBAS IGNITION]

Originally Posted by 4dFord/SC

So, does resistance cause heat, or vice-versa?

Yes resistance does cause heat. However low resistance causes higher current and higher current = even more heat.....

[Lawson Cox]

Sorru fellows. I FLUNKED GREEK. lol

[JSeery]

You can talk voltage, but it is really the current (amps) at the points that matters. I would think that 6 1/2 amps at the points would make for a short lifespan!

[quickchange34]

Well said Bubba . Your tech is right on !

[Jack E/NJ]

And now we all anxiously await Bubba's answer to 4dford's vice-versa part of the question, does heat cause resistance? 8^) I'm gonna guess yes & no. Jack E/NJ

[flatheadmurre]

I´m a lousy teacher but i take a stab at it feel free to straighten it out...some references to water in ().
Heat is powerloss.
To get it you need current (flow) and resistance (restriction of flow).
The restriction of flow eats up a bit of the voltage (pressure) thats why we can measure voltage over components having rsistance in them.
By multiplying the current and voltage present in a component we get the dissipation.
In the scenario with 6.5a and a .5 ohm resistor we will have 3.25v over the resistor the heat dissiation will be 6.5A x 3.25v = 21w of heat.

[big job]

Bubba, I still have the same old question. I have serviced a zillion all makes of cars
trucks busses when they ruled the roads. Not one used a resistor, just Ford. Question is.
why? For example find me a resistor, say in a 1936 Plymouth think I'll eat it lol. After
high school I worked for my uncles bus Co. we had a few Flxible Clippers Buick power
all 12Vt Pos ground. They had those big beehive coils and condenser like a C flashlight
battery - again no resistor even with 12 volts against the 12vt rule today, but, thats how
it came from the factory. Had a WC28 White big Mustang 6 12vt neg grd. no resistor.
I just see any reason, OR, is this just a Ford thing? I totally know FoMoCo 1956 up have
a coil resistor triggered by the solonoid to shoot 12vts to start but then, some do not have resistor, don't know why but they ran & ran. Can't see what duel points have to do with it.
Worked on a Chrysler Marine X265 flathead Mallory duel points - no resistor 12vts too and
no resisted coil either. Oh well..................sam

[Kurt in NJ]

If you take the cap off the top of a beehive coil there is a spotlight bulb in there for a ballast---at least the 2 I have do

[51 MERC-CT]

Originally Posted by big job

Bubba, I still have the same old question. I have serviced a zillion all makes of cars
trucks busses when they ruled the roads. Not one used a resistor, just Ford. Question is.
why? For example find me a resistor, say in a 1936 Plymouth think I'll eat it lol. After
high school I worked for my uncles bus Co. we had a few Flxible Clippers Buick power
all 12Vt Pos ground. They had those big beehive coils and condenser like a C flashlight
battery - again no resistor even with 12 volts against the 12vt rule today, but, thats how
it came from the factory. Had a WC28 White big Mustang 6 12vt neg grd. no resistor.
I just see any reason, OR, is this just a Ford thing? I totally know FoMoCo 1956 up have
a coil resistor triggered by the solonoid to shoot 12vts to start but then, some do not have resistor, don't know why but they ran & ran. Can't see what duel points have to do with it.
Worked on a Chrysler Marine X265 flathead Mallory duel points - no resistor 12vts too and
no resisted coil either. Oh well..................sam

Suspect that the reason you never saw a stand alone resistor in some Fords or other vehicles is because they use a resistor wire going to the coil.

[JSeery]

It is going to depend on the coil, some coils have resistance built into them. There are advantages to an external resistor, but internal resistance coils work. There are also various types of resistors used including resistance wire. Here are some examples of ballast resistor used by GM and Chrysler. The first is a 1953 thru 1972 GM and the second is 1956 thru 1959 & 1964 thru 1971 Chrysler. A 6 volt ignition can run with a 1.5 ohm coil without an external ballast resistor, this is what Ford did during some of the later flathead years. 6 volts divided by 1.5 ohms equals 4 amps, about what you want. The earlier flatheads used a ballast resistor because of the advantages an external ballast resistor provided but this was later dropped when they went to the tin can coils, a cost cutting move I would guess.

[rotorwrench]

The old Ford coils were decent technology for the era but would be considered a poor design by todays standards. The old designs simply couldn't take much heat and still function reliably for a long period. The fix was the ballast resistor to control the current that reaches the coil's primary winding. The voltage is just the carrier but it is affected by the "load" of a resistor that uses some of the current available in the circuit. The voltage drop is just one of the results of that current consumption. The ignition coil benefits from a more stable current flow to the primary coil while in operation. It still makes for a decent enough spark at the secondary and the coil lasts longer.