Ignition only

[Shannen at grolen.com]

I'll scan some pics when I get to my pc, if no one else comes through.

DI>Ya, well meaning and good but.
DI>Your trying to expalin things as a DCv,  and it's just not that way.
DI>This has to "balance" on both sides, doesn't apply.

DI>Does any one know of a on-line picture or scan of a scope pattern of a plug
DI>firing?.
DI>If we all had some common reference maybe I could better explain what the
DI>scope is actually displaying and then folks might get a better grasp of
DI>things and we can get off of this DC silliness
DI>Bruce




DI>From: "James Ballenger" <vtjballeng at yifan.net>
DI>> I have a non-EE explanation.  Based on Ohms law, which should apply
DI>> here.  Let's assume the coil has the ability to only put out a set
DI>amperage
DI>> per fire and that amperage has to go across the spark plug on each side.
DI>> Use V=IR.  The cylinder with higher pressure will have a higher voltage
DI>for
DI>> the same amperage.  The cylinder with the lower pressure will have a lower
DI>> voltage for the same amperage.
DI>> As an fictional example:
DI>> Combustion side.  40,000V = 10000ohms*4amps
DI>> Waste side 5,000V = 1,250ohms*4amps
DI>> Same current passing through both of them... it is just like a
DI>> series circuit with the voltage source in the center as I see it.  If this
DI>> is wrong, please tell me why.
DI>> James Ballenger

DI>> If the
DI>> >combustion end has a high potential, and the waste end has a
DI>> >low potential
DI>> >across the gaps, wouldn't the Amperage be inversely
DI>> >proportional in it's
DI>> >respective cylinder, to the other one? I'm just a Refrigeration
DI>> >Tech/Industrial Electrician here, not an EE. Please, if I'm incorrect,
DI>> >correct me and enlighten us all. Thanks
DI>> >Scott Creech

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