Re: ignition and calculations

[n9ads@domain.elided (Howard R Pletcher)]

You'll get no argument from me this time.  I just thought that perhaps
you were trying out a new set of formulas so you could write your own
Electromagnetism textbook.  :-)

Howard Pletcher
Howteron Products Scout Parts

On Tue, 1 Sep 1998 21:39:53 -0400 (EDT) cookiedan@domain.elided (Daniel Nees)
writes:
>
>
>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
>
>
>--WebTV-Mail-281292317-429
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>Message-Id: <199808310343.UAA17468@domain.elided>
>Subject: Re: ignition and calculations
>Date: Sun, 30 Aug 98 20:47:10 -0700
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>From: John Hofstetter <hofs@domain.elided>
>To: "Daniel Nees" <cookiedan@domain.elided>
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>>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.
>
>--WebTV-Mail-281292317-429--
>
>

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