Re: ignition and calculations

[cookiedan@domain.elided (Daniel Nees)]

John,
    O.K. first off let me apoligise for not continuing this yesterday. I
had a lot to do and not enough time to write length disortations. 
   Second. I was wrong on the size of the ballast resistor. I measured
mine with my DMM and it read 2ohms.
   Since this was started last week I don't recall where we where headed
on this line, so; I'll pick-up where I thought we left off. 
   So, since I was wrong all my earlier figures are off. We now have a
start circuit of 12V, which is still correct. That is feed off the
starter solinoid. Probably off the S terminal of the solinoid. We have a
run circuit of 3.5 ohms resistance. (2ohms ballast resister + 1.5ohms
coil) which yields E/R=I 3.43Amps. Now we know in a series circuit that
the current is constant in every component. Therefore the voltage drop
across the ballast resistor will be I*R=E or 3.43A*2ohms (ballast
resistor)= 6.86V, and the Voltage drop across the coil will be
3.43A*1.5ohms=5.145V. The total Voltage in the circuit will be 6.86V+
5.14V=12V. So, my run circuit is supplying 5.145V to the coil. (exactly
what I had said) Now the current in the start circuit, just for kicks,
is E/R=I, or; 12V / 1.5ohms=8Amps. 
   So, going back to my transformer/coil ratio of 1667:1 (This is from
using a 20,000V coil with a 12V primary. 20,000/12=1667:1), at run, with
a points ignition, the secondary is putting out 8,568 Volts. Now a
electronic ignition like the Gold box holley or Prestolite use full
voltage run circuits, therefore; the run circuit output of the coil will
be max output of that coil. Supposing the stock coil is a 20,000V
output, then every spark will be 20,000V. (In a perfect world with no
heat in the coil.)
   With that I contend then that I was right inmy statement that the
primary circuit in run was only 5Volts and that was to protect the
points. The math I had earlyer in the week was faulted because I used my
memory, not my written formulas. That is why I carry a copy of 'Ugly's'
in my toolbox at work. If it isn't written down somewhere I will forget
it.
   O.K. John, Howard, John, your shot.

Dan Nees
cookiedan@domain.elided

   1979 Scout II 345, Auto, 3.07's, Trak-Lok. One Ugly and Trusty Truck
covered in mud! Thinking about throwing a couple of big buckets of mud
on it just before I get to the fairgrounds!

1971 Scout II 304, D30/44, D20, power steering.

http://members.tripod.com/~IHCaholic/scoutindex.html


---- Begin included message ----

• To: "Daniel Nees" <cookiedan@domain.elided>
• Subject: Re: ignition and calculations
• From: John Hofstetter <hofs@domain.elided>
• Date: Sun, 30 Aug 98 20:47:10 -0700
• cc: <ihc@domain.elided>, "Thomas J. Harais" <tjhemh@domain.elided>
• Content-Type: text/plain; charset="US-ASCII"

>So, 12V*2ohms=24V really is a
>statement of what it would take to get through the 2ohm resistor. The
>formula is for Voltage drop. E*R=Ed. 

Danny,

It looks to me as if your formula would give you compensation for voltage 
drop caused by the resistor being in the circuit. In other words, 24 
volts running through a 2 ohm resistor yields the same current as 12V 
with a resistance of 1 ohm. I understand your reference to choosing wire 
size and such, but can't see the relationship to the ballast resistor, 
since we don't compensate for the ballast resistor by sending more 
voltage through it. 

??????????
John H.

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