Feeling left out.

[John Hofstetter <hofs@domain.elided>]

>From: n9ads@domain.elided (Howard R Pletcher)
>Subject: Re: ignition and calculations
>
>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.

Hey Dan,
How come I didn't get this post? I feel picked on, left out, neglected, 
and so on.
John H.