Ignition only

[James Ballenger]

Bruce,
	When I posted my message, I said "If this is wrong, please tell me
why."  You simply told me I am wrong, with no explanation as to why you
think I am wrong.  Please offer some constructive criticism and tell me
exactly why this is wrong.  Simply telling me or anyone else they are wrong
and further degrading their opinions by calling them silliness and referring
to a need for a better grasp on the concept does little to help this thread
or anyone's understanding of the subject.  Please tell us why this is DC
silliness, not just that it is.  

James Ballenger



>-----Original Message-----
>From: owner-diy_efi at diy-efi.org [mailto:owner-diy_efi at diy-efi.org]On
>Behalf Of Bruce Plecan
>Sent: Friday, November 10, 2000 4:29 PM
>To: diy_efi at diy-efi.org
>Subject: Re: Ignition only
>
>
>
>Ya, well meaning and good but.
>Your trying to expalin things as a DCv,  and it's just not that way.
>This has to "balance" on both sides, doesn't apply.
>
>Does any one know of a on-line picture or scan of a scope 
>pattern of a plug
>firing?.
>If we all had some common reference maybe I could better 
>explain what the
>scope is actually displaying and then folks might get a better grasp of
>things and we can get off of this DC silliness
>Bruce
>
>
>
>
>From: "James Ballenger" <vtjballeng at yifan.net>
>> I have a non-EE explanation.  Based on Ohms law, which should apply
>> here.  Let's assume the coil has the ability to only put out a set
>amperage
>> per fire and that amperage has to go across the spark plug 
>on each side.
>> Use V=IR.  The cylinder with higher pressure will have a 
>higher voltage
>for
>> the same amperage.  The cylinder with the lower pressure 
>will have a lower
>> voltage for the same amperage.
>> As an fictional example:
>> Combustion side.  40,000V = 10000ohms*4amps
>> Waste side 5,000V = 1,250ohms*4amps
>> Same current passing through both of them... it is just like a
>> series circuit with the voltage source in the center as I 
>see it.  If this
>> is wrong, please tell me why.
>> James Ballenger
>
>> If the
>> >combustion end has a high potential, and the waste end has a
>> >low potential
>> >across the gaps, wouldn't the Amperage be inversely
>> >proportional in it's
>> >respective cylinder, to the other one? I'm just a Refrigeration
>> >Tech/Industrial Electrician here, not an EE. Please, if I'm 
>incorrect,
>> >correct me and enlighten us all. Thanks
>> >Scott Creech
>
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