Re: ignition and calculations

[John Hofstetter <hofs@domain.elided>]

>I pulled out my IH shop manual and my Chiltons Truck and Van repair for
>1971-1978 books. The IH manual says the resistor is there to protect the
>points from having to pass 12V under normal run conditions. The by-pass
>is there to allow TWELVE VOLTS to the coil at start up especially in
>cold weather. My Chiltons (I'm sure this is in the IH manual to I just
>didn't look for it until I opened the Chitons) says that the external
>resister is a .43 Ohms resister. If you do the math then you find that
>V*Ohms=lost V or what is left after the resistance has been powered.
>Therefore 12*.43=5.6V That is where the 5V at run comes from.
>
>Dan Nees
>cookiedan@domain.elided

Danny,
I'm a little hazy here and Howard hasn't given me any recent lessons on 
Ohms Law, but if that formula that you list here works, and putting in 
the numbers in the same places you put them, then with a 2 ohm resistor 
you get 12V*2=lostV or what is left after the resistance has been 
powered. Therfore 12*2=24V. That doesn't make any sense, Dan. I wish it 
did, because then instead of spending big money on my inverter, I'd run 
my 12V trailer voltage through a 10 ohm resistor and I'd have 120 volts.
According to the Ohms Law that Howard has insisted that I heed if I'm 
going to talk electricity with him, whatever happens to the voltage 
depends on the amperage drawn through the resistor.

Some givens: The current remains the same in every part of a series 
circuit.
             The voltage across several resistors in series is the sum of 
the voltages across 
                 the separate resistors. 

I looked in my Motor Manual and the only diagram I could find with 
ballast resistor in it was for an electronic ignition module and in that 
diagram the resistor that was .43 ohms wasn't the ballast resistor. It 
showed the ballast at .68 ohms. I measured an old ballast resistor I have 
in reserve and got .7 Ohms. My IHC coil shows a resistance of 1.5 Ohms. 
Adding the two resistances I get 2.2 Ohms. 12V/2.2 = 5.4 amps. At the 
ballast resistor I get 5.4 amps * .7ohms = 3.8 volts used by that 
resistor. Because we need a total of 12V to be accounted for the coil 
must use 8.2 V. note: we also could have calculated that voltage drop by 
multiplying the 1.5 * 5.4 = 8.2V.
                                   Another way
                                In the formula below
                             (the voltage total here must = 12)
                                       3.8V used  8.2V used    
           12V and 5.4 amps in  --->   .7 ohms + 1.5 ohms --->  5.4 amps 
still. 

Voltage after ballast  12 - 3.8 = 8.2V

Whether or not, this calculation is correct, I like the result because it 
is pretty much consistent with the measurements I've done in the past. I 
hope the engineers on the digest will explain this more simply, or if I'm 
wrong, correctly, than I've been able to do it. Actually, Danny, even 
though I couldn't agree with your result, I liked your much simpler 
calculation a lot better than mine, and besides that I could do away with 
my inverter.

John H.
 

John Hofstetter  "Ol'Saline's Web Site" www.goldrush.com/~hofs
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