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05-16-2016, 10:34 PM | #1 |
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amps at points 37 ford flathead
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.
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05-17-2016, 01:49 AM | #2 |
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Re: amps at points 37 ford flathead
What are you trying to work out?
Martin. |
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05-17-2016, 05:09 AM | #3 |
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Re: amps at points 37 ford flathead
About 4.5 Amps You can put more current through but the life on the points is much less.
Last edited by Terry,OH; 05-17-2016 at 05:48 AM. |
05-17-2016, 07:15 AM | #4 |
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Re: amps at points 37 ford flathead
Even that is a little high, 3.0 to 3.5 would be even better.
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05-17-2016, 07:56 AM | #5 |
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Re: amps at points 37 ford flathead
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?
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05-17-2016, 08:48 AM | #6 |
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Re: amps at points 37 ford flathead
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05-17-2016, 08:58 AM | #7 |
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Re: amps at points 37 ford flathead
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. |
05-17-2016, 03:48 PM | #8 |
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Re: amps at points 37 ford flathead
Do you know anything about the Echlin ICR40? It looks like the originals in appearance.
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05-17-2016, 04:10 PM | #9 | |
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Re: amps at points 37 ford flathead
Quote:
Note: Found this thread - http://www.fordbarn.com/forum/showthread.php?t=71654 https://fordbarn.com/forum/showthread.php?t=176624 Last edited by JSeery; 05-17-2016 at 04:23 PM. |
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05-17-2016, 05:49 PM | #10 |
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Re: amps at points 37 ford flathead
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
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05-18-2016, 06:21 AM | #11 |
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Re: amps at points 37 ford flathead
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
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05-18-2016, 07:56 AM | #12 |
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Re: amps at points 37 ford flathead
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.
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