Not much help else where, plug wire resistence..
#1
Not much help else where, plug wire resistence..
Im checking my plug wire resistance, most are around 920ohms on a 10.5inch wire from end to end (connector on the spark plug side is probably 1/2inch in so so say 10in. of conductor) i have one wire that is around 1200ohms, do these need to be replaced or are they good?
#5
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I checked a set of Taylor wires at 160 ohms for a stock legnth LS-1 wire. Then I checked a stock wire at 730 ohms. Stock legnth wires are about 9", so it sounds like your wires, with an extra 1 1/2" of conductor are about on par with a stock wire. Your 1200 ohm wire seems to have a problem somewhere. Most likley a poor connection at the boot. I'm not sure which kind of wires you bought, but they are pretty high resistance wise. I would expect a little better from a performance wire. Hence the $40 Taylors only having 160 ohm resistance.
#6
What does the lower resistence actually help? I mean are there performance gains or do they just maintain a spark better in higher rpms? If i can justify spending $80 on plugs (NGK's) and wires (MSD's) i'll do it, i just think i might be losing spark up top causing a missfire, but i dont know for sure, it doesnt feel bad, but my dyno #'s are low and my plug/wire setup is 2 years old now
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I read an article where Granatelli's wires were used, they claim to have zero ohms. they did increase hp. I'll look for the mag and post which one. I do know less ohms means more spark cause you are not lossing it by resistance. thats why the lower the ohms the better. I'll look for the magazine/article.
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Originally Posted by cowboysfan
I read an article where Granatelli's wires were used, they claim to have zero ohms. they did increase hp. I'll look for the mag and post which one. I do know less ohms means more spark cause you are not lossing it by resistance. thats why the lower the ohms the better. I'll look for the magazine/article.
fuerzaws6
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it is physically impossible to have zero ohms- but you can come close. they are probobly rating it per inch or something. Zero is impossible.
the higher your resistance, the less current flows through the wire. at high cylinder pressures (peak torque) this is bad. It can also cause incomplete burns of fuel etc.
Summit had a bunch of wires for $45 shipped.
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Resistance is basically "friction" for electricity. Since voltage drops over an impedance the resulting loss is heat. Lower the resistance of the wires and the heat loss at the wires will decrease. That means it has to dissipate somewhere else... the plug.
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Its electrical engergy, not heat energy. if less is converted to heat, then less is converted to heat. nothing "has" to happen at the plug. There isnt a constant amount of heat that needs to go somewhere.
not that it matters, the heat of combustion will make the spark heat negligable
not that it matters, the heat of combustion will make the spark heat negligable
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Energy *has* to go somewhere. Not all of the engergy is released at the plug's electrode therefore it must release somewhere. That's why you go through plugs faster with lower Ohm wires.
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its conservation of energy, not conservation of heat energy.
if high R wires make X units of heat in the wire, why does the plug wont be x units hotter.
if high R wires make X units of heat in the wire, why does the plug wont be x units hotter.
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Conservation of engery doesn't limit conversion. ie. Electric to heat. eg. lightbulb. Your coils are still producing the same amount of power. You just dropped the resistance in the wire therefore the voltage drop is reduced (at the wire). That means you increased the energy at the plug (good thing) which will cause the electrodes to wear faster (bad thing).
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Originally Posted by technical
Conservation of engery doesn't limit conversion. ie. Electric to heat. eg. lightbulb. Your coils are still producing the same amount of power. You just dropped the resistance in the wire therefore the voltage drop is reduced (at the wire). That means you increased the energy at the plug (good thing) which will cause the electrodes to wear faster (bad thing).
I dont see how it means "Since there is less resistance in the wires, the heat must dissipate somewhere else"
although more heat will be at the plug, i dont see why the heat from what would be in the wires goes into the plug.
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Because the voltage drop is now greater at the plug and there is less of a drop through the wires.
Think of a lamp. There is no break in the circuit but the lightbulb's filament isn't a great conductor. I gives off heat because of the resistance. Now skin the power cord to the point it becomes resistive, the lightbulb will dim. The high Ohm plug wires are like the skinned power cord. Drop the resistance and the lightbulb (spark plug) glows brighter, but at the expense of longevity. The ignition "circuit" actually has a break.. the plug itself between the electrodes. This is where the last voltage drop occurs. If that is the largest delta in the circuit, then the it will also be the hottest and the first to wear out.
Sorry if I don't explain myself well.
Think of a lamp. There is no break in the circuit but the lightbulb's filament isn't a great conductor. I gives off heat because of the resistance. Now skin the power cord to the point it becomes resistive, the lightbulb will dim. The high Ohm plug wires are like the skinned power cord. Drop the resistance and the lightbulb (spark plug) glows brighter, but at the expense of longevity. The ignition "circuit" actually has a break.. the plug itself between the electrodes. This is where the last voltage drop occurs. If that is the largest delta in the circuit, then the it will also be the hottest and the first to wear out.
Sorry if I don't explain myself well.
Last edited by technical; 11-22-2005 at 02:38 PM.
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I think i misunderstood what you were saying (or at least I hope!)
lets say the wire with high R gained X values of heat
and the low R gained zero
the plug with wire high R gains heat Y
with low R, it wont be Y+X, just greater than Y.
I was reading your original statement to mean X +Y
lets say the wire with high R gained X values of heat
and the low R gained zero
the plug with wire high R gains heat Y
with low R, it wont be Y+X, just greater than Y.
I was reading your original statement to mean X +Y
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Originally Posted by treyZ28
that company is so full of crap, i wouldn't buy a word of what they say.
it is physically impossible to have zero ohms- but you can come close. they are probobly rating it per inch or something. Zero is impossible.
it is physically impossible to have zero ohms- but you can come close. they are probobly rating it per inch or something. Zero is impossible.
superconductors our plug wires are not.
but yeah, if they said zero ohms, they are full of it.