amps at points 37 ford flathead

[37 ford firetruck]

New to the site. Trying to locate the number of amps acceptable for the points in a factory distributor with a 6 volt positive ground. 85 horse flathead.

[scooder]

What are you trying to work out?
Martin.

[Terry,OH]

About 4.5 Amps You can put more current through but the life on the points is much less.

[JSeery]

Even that is a little high, 3.0 to 3.5 would be even better.

[37 ford firetruck]

The responses are much appreciated. I have "inherited" a 1937 ford firetruck Model 79 with the original flathead. It's been off the road for 20 years and was just limping along at that time from having last been touched in the mid 70's. I'm learning to respect the engineering from 80+ years ago. The original coil on the three bolt diver helmet distributor had been replaced with a coil adapter kit many years ago. The wiring had dry rotted and I have purchased a new adapter kit, new 6 volt external oil filled coil with 1.2 ohms resistance, new OEM Ford points, and cleaned and oiled the distributor. I am trying to calculate the resistance needed to get strong spark as well as save my points. What I would like to do...I have a new Cole Hersey ignition switch with a spring return start position and would like to run straight 6-6.2 volts to the start position to be held while using the original start button on the dash. The spring would return the ignition to run and have the proper resistance to the coil (so I will have two wires to the coil). Ok, new to the flathead ignition mentality, so I would like know if my plan is realistically functional. If so, I need to calculate the balance of the resistance needed along with the 1.2 ohms already offered from the coil. My calculations show I would need a .6 ohm resistor on my "run" wire with 7 volts running. Thoughts out there?

[Terry,OH]

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Using your coil of 1.2 ohm and an original Ford ignition resistor of 0.5 ohm gives you 1.7 ohm total. With 6.7 volts that gives you 3.9A Sounds like good numbers to me.

[JSeery]

If you need an after market resistor here are some numbers. The ones I have part numbers for are higher resistance:

NAPA* Echlin* ICR11 (1.35 Ohms)
NAPA Echlin ICR23 (1.20 Ohms)
NAPA Echlin ICR34 (1.40 Ohms)
Lucas* 3BR (1.3 - 1.4 Ohms)
Mopar* DCC-4529795 (1.40 Ohms)
Accel* ACC-150250 (1.35 Ohms)
Standard* RU-4 (1.35 Ohms)
Standard RU-23 (1.20 ohms)
Standard RU-37 (1.40 Ohms)

If you can find an original Ford resistor that would work fine, most of the replacement ones are junk.

[MGG]

Do you know anything about the Echlin ICR40? It looks like the originals in appearance.

[JSeery]

Originally Posted by MGG

Do you know anything about the Echlin ICR40? It looks like the originals in appearance.

It does look like the originals, the issue will be what the core inside the wire is made from. It needs to be ceramic. And I do not know anything about them, hope someone on the Barn has had some experience with them.

Note: Found this thread - http://www.fordbarn.com/forum/showthread.php?t=71654
https://fordbarn.com/forum/showthread.php?t=176624

[MGG]

Please be patient with me as I am an electrical neophyte. If I put 6.3 volts from the battery into the resistor and it has a 0.35 ohm rating, does that mean that 0.65 (1.0-0.35) x 6.3 = 4.1 volts comes out of it? If not, what is the correct calculation? Thank You

[Terry,OH]

MGG: Your example is not practical, since you do not have a second resistance. Lets add the ignition coil mentioned above in post #5 of 1.2 Ohm and your simple series circuit of 6.3V input and the second resistor of 0.35 Ohm. In this circuit the current is the same through each device and the circuit total resistance is additive (These principles are basic laws of electricity). I(current)=E(voltage)/R(Total resistance) or I=6.3/1.55 so I=4.06 Amp. The voltage dropped at the resistor is E=IR or 4.06 X 0.35 so the resistor drops 1.42V and the voltage at the coil 4.06 X 1.2 or 4.87 V. Just one more idea here, to check this work. The sum of all the voltage drops in a simple series circuit is equal to the voltage input (another basic law) so if you add the voltage drop at the resistor 1.42 & the drop at the coil 4.87 you have the input voltage of 6.3V

[G.M.]

First of all I would have an original coil rebuilt by Skip Haney in Florida.
Use an original resistor and an IH-200 condenser. Now you have a pretty
much original system and is easy to trouble shoot. You have to start with
a fully charged 6 volt battery which is 6.3 after sitting a few hours after
being charged. The voltage of the coil wire to ground will have two
voltages, close to 6.2 when points are open and in the range of 4.0
points closed. You can tell when points are open or closed by the voltage
at the coil wire. Jog starter with the voltmeter connected from coil
wire to ground and you can see it change. DON"T leave the ignition
switch on more then 2 minutes when testing, if points are closed
the coil windings are shorted to ground, the winding get real hot and
melt the coil. You need at least 3.5 volts to the coil, points closed or
it will be hard to start. If battery is fully charged and you have low
voltage at the coil check ground, wire connections, switch and resistor.
You will see where the voltage lose occurs. G.M.