Ohm output on a ls1???
Ohm output on a ls1??? ok i was wondering what the Ohm output is on our motors? anyone know or able to point me in the right direction??? i think i stumbled onto something...good that is.
thanks
brandon
thanks
brandon
Ohms are a measurement of resistance to the flow of electrical current. The higher the resistance measured in ohms the less current will flow through a conductor at a given voltage. Do you want to know the resistance of the plug wires?
no i dont need to what the resistance is.
i know im lacking here in this area... so please be patient w/ me as i try to word it properly w/ out sounding like a complete fool lol.
when motors rev up from idle..... how much current is lost through the wires. say its 13volts at idle and then at 7k its 9volts.....do you know what im talking about or trying to say? am i wording this rite?
i know im lacking here in this area... so please be patient w/ me as i try to word it properly w/ out sounding like a complete fool lol.
when motors rev up from idle..... how much current is lost through the wires. say its 13volts at idle and then at 7k its 9volts.....do you know what im talking about or trying to say? am i wording this rite?
from the way you're asking, i think you might need to get a better understanding of electrical engineering;
ohms (R for resistance) is the unit of electrical resistance;
volts (V) is the unit of electrical potential, or force (electromotive force)
amp (I for current) is the unit of electrical current for the amount of electrical charge flowing past a given point per second.
rule #1: V = I * R
rule #2: Power = V * I
the ignition coil on all car's today run on 14vdc and consume, I think, around 8 amps. That is 112 watts of power, so figure the coil consumes around 100 watts of power.
Each time the coil is triggered to send a spark to the spark plug (along the spark plug wire) that ~100 watts of power going into the coil is converted. Power is not created nor destroyed, only converted, that's conservation of energy. And because it is not a 100% efficient conversion of energy, you lose a lot of that power when it goes to the spark plug. I think only around 25 watts of power make it to the spark plug, what is happening is you're trading current for voltage; really high voltage so there will be a spark across the spark plug gap of 0.060" but it only happens for microseconds and the amount of current (amps) is very small.
The voltage is anywhere from 20000 to 60000 volts, if you want to consider me as a reliable source. That's what I've seen over the past 15 years for ignition coil's specifications, albeit advertised from the manufacturer. But it has been very consistent, you are only going to get so much voltage because the power to the coil has never changed- always been 14 vdc and 5-10 amps. Some coils are advertised as 100000 volt output, and I have tried them and had them give out very easily and very consistently at 3000+ rpms, they overheat. Most coils today, instead of looking like this
have an E shape I think it's called, and look like this below. I believe this includes all coil on plug or coil near plug ignitions, including all chevy LS series of engines. The magentic field is oriented such that it gives a better transfer of energy for a higher voltage on the output (spark).
so, anywhere from 20000 to 60000+ volts going through the spark plug wires, depending on the type of coil.
how many volts you lose through the spark plug wire:
that's an interesting debate because because the circuit from the coil to the spark plug is an open circuit and I'm not aware of any way of accurately measuring it other than what might be claimed by the manufacturer, which i don't trust. So you can't just use V = IR to figure it out. And, it is not simply DC current from the coil the plug, there is a frequency to it along with a backward type of spike from the primary side of the coil. That's why RF or EMI suppression is needed on spark plug wires. That being said, the resistance of the wires probably don't play a large part in the amount of spark energy reaching the spark plug.
Another real area of concern however is how the ignition coil operates, what it's rise time is. That means how long you have to feed it power before it will output it's advertised spark voltage. As engine rpm increases, the coil fires more per unit of time so the time the coil has to build up it's magnetic field is less. If it's too short, you'll get less output voltage = weak spark, and the engine will sputter at high rpm. This is a problem with the traditional style of coil on a V-8 engine where there is one coil being used for 8 cylinders. Now, you have a coil for each spark plug so in a V-8 applications I think the time is lengthened by a factor of 8 (or 6 for 6 cylinder, 4 for 4 cylinder) which helps the coil immensely, it has a longer time at high rpm to generate maximum output voltage.
ohms (R for resistance) is the unit of electrical resistance;
volts (V) is the unit of electrical potential, or force (electromotive force)
amp (I for current) is the unit of electrical current for the amount of electrical charge flowing past a given point per second.
rule #1: V = I * R
rule #2: Power = V * I
the ignition coil on all car's today run on 14vdc and consume, I think, around 8 amps. That is 112 watts of power, so figure the coil consumes around 100 watts of power.
Each time the coil is triggered to send a spark to the spark plug (along the spark plug wire) that ~100 watts of power going into the coil is converted. Power is not created nor destroyed, only converted, that's conservation of energy. And because it is not a 100% efficient conversion of energy, you lose a lot of that power when it goes to the spark plug. I think only around 25 watts of power make it to the spark plug, what is happening is you're trading current for voltage; really high voltage so there will be a spark across the spark plug gap of 0.060" but it only happens for microseconds and the amount of current (amps) is very small.
The voltage is anywhere from 20000 to 60000 volts, if you want to consider me as a reliable source. That's what I've seen over the past 15 years for ignition coil's specifications, albeit advertised from the manufacturer. But it has been very consistent, you are only going to get so much voltage because the power to the coil has never changed- always been 14 vdc and 5-10 amps. Some coils are advertised as 100000 volt output, and I have tried them and had them give out very easily and very consistently at 3000+ rpms, they overheat. Most coils today, instead of looking like this
have an E shape I think it's called, and look like this below. I believe this includes all coil on plug or coil near plug ignitions, including all chevy LS series of engines. The magentic field is oriented such that it gives a better transfer of energy for a higher voltage on the output (spark).
so, anywhere from 20000 to 60000+ volts going through the spark plug wires, depending on the type of coil.
how many volts you lose through the spark plug wire:
that's an interesting debate because because the circuit from the coil to the spark plug is an open circuit and I'm not aware of any way of accurately measuring it other than what might be claimed by the manufacturer, which i don't trust. So you can't just use V = IR to figure it out. And, it is not simply DC current from the coil the plug, there is a frequency to it along with a backward type of spike from the primary side of the coil. That's why RF or EMI suppression is needed on spark plug wires. That being said, the resistance of the wires probably don't play a large part in the amount of spark energy reaching the spark plug.
Another real area of concern however is how the ignition coil operates, what it's rise time is. That means how long you have to feed it power before it will output it's advertised spark voltage. As engine rpm increases, the coil fires more per unit of time so the time the coil has to build up it's magnetic field is less. If it's too short, you'll get less output voltage = weak spark, and the engine will sputter at high rpm. This is a problem with the traditional style of coil on a V-8 engine where there is one coil being used for 8 cylinders. Now, you have a coil for each spark plug so in a V-8 applications I think the time is lengthened by a factor of 8 (or 6 for 6 cylinder, 4 for 4 cylinder) which helps the coil immensely, it has a longer time at high rpm to generate maximum output voltage.
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if you are talking about available electrical power in the entire car's electrical system, that is provided by the alternator. The alternator is rated at a certain maximum output, typically 100 amp. The voltage is regulated at 14 volts DC. Most alternators become capable of outputting maximum aperage around 2000 engine rpm (the alternator has smaller pulley so it also spins faster). The alternator output is directly linked to how fast it spins, so at a low 600 rpm there isn't enough power to run the headlights, heater, radio with engine running. There will be less than 14 volts, possibly less than 12 volts of the battery- the headlights will be dim and there will be less than ideal power feeding the ignition coil which will result in lower voltage to the spark plugs. But, it's not low enough to make a real significant on engine performance at that 600-800 rpm range. And once at 1000-1500 and higher engine rpm there's enough juice and all ignition coils are running at full voltage/power. does that help?
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Whatever you are thinking, you need to remember that very high horsepower LSX engines spinning above 7500 rpm use the stock ignition system. The stock LS1 ignition system is essentially ideal and does not need to be improved.
from the way you're asking, i think you might need to get a better understanding of electrical engineering;
ohms (R for resistance) is the unit of electrical resistance;
volts (V) is the unit of electrical potential, or force (electromotive force)
amp (I for current) is the unit of electrical current for the amount of electrical charge flowing past a given point per second.
rule #1: V = I * R
rule #2: Power = V * I
the ignition coil on all car's today run on 14vdc and consume, I think, around 8 amps. That is 112 watts of power, so figure the coil consumes around 100 watts of power.
Each time the coil is triggered to send a spark to the spark plug (along the spark plug wire) that ~100 watts of power going into the coil is converted. Power is not created nor destroyed, only converted, that's conservation of energy. And because it is not a 100% efficient conversion of energy, you lose a lot of that power when it goes to the spark plug. I think only around 25 watts of power make it to the spark plug, what is happening is you're trading current for voltage; really high voltage so there will be a spark across the spark plug gap of 0.060" but it only happens for microseconds and the amount of current (amps) is very small.
The voltage is anywhere from 20000 to 60000 volts, if you want to consider me as a reliable source. That's what I've seen over the past 15 years for ignition coil's specifications, albeit advertised from the manufacturer. But it has been very consistent, you are only going to get so much voltage because the power to the coil has never changed- always been 14 vdc and 5-10 amps. Some coils are advertised as 100000 volt output, and I have tried them and had them give out very easily and very consistently at 3000+ rpms, they overheat. Most coils today, instead of looking like this
have an E shape I think it's called, and look like this below. I believe this includes all coil on plug or coil near plug ignitions, including all chevy LS series of engines. The magentic field is oriented such that it gives a better transfer of energy for a higher voltage on the output (spark).
so, anywhere from 20000 to 60000+ volts going through the spark plug wires, depending on the type of coil.
how many volts you lose through the spark plug wire:
that's an interesting debate because because the circuit from the coil to the spark plug is an open circuit and I'm not aware of any way of accurately measuring it other than what might be claimed by the manufacturer, which i don't trust. So you can't just use V = IR to figure it out. And, it is not simply DC current from the coil the plug, there is a frequency to it along with a backward type of spike from the primary side of the coil. That's why RF or EMI suppression is needed on spark plug wires. That being said, the resistance of the wires probably don't play a large part in the amount of spark energy reaching the spark plug.
Another real area of concern however is how the ignition coil operates, what it's rise time is. That means how long you have to feed it power before it will output it's advertised spark voltage. As engine rpm increases, the coil fires more per unit of time so the time the coil has to build up it's magnetic field is less. If it's too short, you'll get less output voltage = weak spark, and the engine will sputter at high rpm. This is a problem with the traditional style of coil on a V-8 engine where there is one coil being used for 8 cylinders. Now, you have a coil for each spark plug so in a V-8 applications I think the time is lengthened by a factor of 8 (or 6 for 6 cylinder, 4 for 4 cylinder) which helps the coil immensely, it has a longer time at high rpm to generate maximum output voltage.
ohms (R for resistance) is the unit of electrical resistance;
volts (V) is the unit of electrical potential, or force (electromotive force)
amp (I for current) is the unit of electrical current for the amount of electrical charge flowing past a given point per second.
rule #1: V = I * R
rule #2: Power = V * I
the ignition coil on all car's today run on 14vdc and consume, I think, around 8 amps. That is 112 watts of power, so figure the coil consumes around 100 watts of power.
Each time the coil is triggered to send a spark to the spark plug (along the spark plug wire) that ~100 watts of power going into the coil is converted. Power is not created nor destroyed, only converted, that's conservation of energy. And because it is not a 100% efficient conversion of energy, you lose a lot of that power when it goes to the spark plug. I think only around 25 watts of power make it to the spark plug, what is happening is you're trading current for voltage; really high voltage so there will be a spark across the spark plug gap of 0.060" but it only happens for microseconds and the amount of current (amps) is very small.
The voltage is anywhere from 20000 to 60000 volts, if you want to consider me as a reliable source. That's what I've seen over the past 15 years for ignition coil's specifications, albeit advertised from the manufacturer. But it has been very consistent, you are only going to get so much voltage because the power to the coil has never changed- always been 14 vdc and 5-10 amps. Some coils are advertised as 100000 volt output, and I have tried them and had them give out very easily and very consistently at 3000+ rpms, they overheat. Most coils today, instead of looking like this
have an E shape I think it's called, and look like this below. I believe this includes all coil on plug or coil near plug ignitions, including all chevy LS series of engines. The magentic field is oriented such that it gives a better transfer of energy for a higher voltage on the output (spark).
so, anywhere from 20000 to 60000+ volts going through the spark plug wires, depending on the type of coil.
how many volts you lose through the spark plug wire:
that's an interesting debate because because the circuit from the coil to the spark plug is an open circuit and I'm not aware of any way of accurately measuring it other than what might be claimed by the manufacturer, which i don't trust. So you can't just use V = IR to figure it out. And, it is not simply DC current from the coil the plug, there is a frequency to it along with a backward type of spike from the primary side of the coil. That's why RF or EMI suppression is needed on spark plug wires. That being said, the resistance of the wires probably don't play a large part in the amount of spark energy reaching the spark plug.
Another real area of concern however is how the ignition coil operates, what it's rise time is. That means how long you have to feed it power before it will output it's advertised spark voltage. As engine rpm increases, the coil fires more per unit of time so the time the coil has to build up it's magnetic field is less. If it's too short, you'll get less output voltage = weak spark, and the engine will sputter at high rpm. This is a problem with the traditional style of coil on a V-8 engine where there is one coil being used for 8 cylinders. Now, you have a coil for each spark plug so in a V-8 applications I think the time is lengthened by a factor of 8 (or 6 for 6 cylinder, 4 for 4 cylinder) which helps the coil immensely, it has a longer time at high rpm to generate maximum output voltage.
With the coil you probably don't have that much loss though. Some constant loss to keep the core useable and some variable depending on output voltage.
Maybe 60% absolute max.
I would also say the current is more important than the voltage at this point. I think the rule for an airgap which has high resistance is something on the order of 3000 V/mm. That might be why I have heard people talking about aftermarket coils with more current but not talking about Vo which we can't readily change Vin anyways. Why change the voltage?? The spark gap is relatively fixed especially when hotrodders like cold plugs.
EG: I have heard that MSD blasters for our cars have 3x the current of stock. I have heard that Top Fuels have 1amp per cylinder. Stuff like that.
I think it is mostly the heavy wire gauge and packing requirements that keep us from getting any more out of a stock vehicle.
from the way you're asking, i think you might need to get a better understanding of electrical engineering;
ohms (R for resistance) is the unit of electrical resistance;
volts (V) is the unit of electrical potential, or force (electromotive force)
amp (I for current) is the unit of electrical current for the amount of electrical charge flowing past a given point per second.
rule #1: V = I * R
rule #2: Power = V * I
the ignition coil on all car's today run on 14vdc and consume, I think, around 8 amps. That is 112 watts of power, so figure the coil consumes around 100 watts of power.
Each time the coil is triggered to send a spark to the spark plug (along the spark plug wire) that ~100 watts of power going into the coil is converted. Power is not created nor destroyed, only converted, that's conservation of energy. And because it is not a 100% efficient conversion of energy, you lose a lot of that power when it goes to the spark plug. I think only around 25 watts of power make it to the spark plug, what is happening is you're trading current for voltage; really high voltage so there will be a spark across the spark plug gap of 0.060" but it only happens for microseconds and the amount of current (amps) is very small.
The voltage is anywhere from 20000 to 60000 volts, if you want to consider me as a reliable source. That's what I've seen over the past 15 years for ignition coil's specifications, albeit advertised from the manufacturer. But it has been very consistent, you are only going to get so much voltage because the power to the coil has never changed- always been 14 vdc and 5-10 amps. Some coils are advertised as 100000 volt output, and I have tried them and had them give out very easily and very consistently at 3000+ rpms, they overheat. Most coils today, instead of looking like this
have an E shape I think it's called, and look like this below. I believe this includes all coil on plug or coil near plug ignitions, including all chevy LS series of engines. The magentic field is oriented such that it gives a better transfer of energy for a higher voltage on the output (spark).
so, anywhere from 20000 to 60000+ volts going through the spark plug wires, depending on the type of coil.
how many volts you lose through the spark plug wire:
that's an interesting debate because because the circuit from the coil to the spark plug is an open circuit and I'm not aware of any way of accurately measuring it other than what might be claimed by the manufacturer, which i don't trust. So you can't just use V = IR to figure it out. And, it is not simply DC current from the coil the plug, there is a frequency to it along with a backward type of spike from the primary side of the coil. That's why RF or EMI suppression is needed on spark plug wires. That being said, the resistance of the wires probably don't play a large part in the amount of spark energy reaching the spark plug.
Another real area of concern however is how the ignition coil operates, what it's rise time is. That means how long you have to feed it power before it will output it's advertised spark voltage. As engine rpm increases, the coil fires more per unit of time so the time the coil has to build up it's magnetic field is less. If it's too short, you'll get less output voltage = weak spark, and the engine will sputter at high rpm. This is a problem with the traditional style of coil on a V-8 engine where there is one coil being used for 8 cylinders. Now, you have a coil for each spark plug so in a V-8 applications I think the time is lengthened by a factor of 8 (or 6 for 6 cylinder, 4 for 4 cylinder) which helps the coil immensely, it has a longer time at high rpm to generate maximum output voltage.
ohms (R for resistance) is the unit of electrical resistance;
volts (V) is the unit of electrical potential, or force (electromotive force)
amp (I for current) is the unit of electrical current for the amount of electrical charge flowing past a given point per second.
rule #1: V = I * R
rule #2: Power = V * I
the ignition coil on all car's today run on 14vdc and consume, I think, around 8 amps. That is 112 watts of power, so figure the coil consumes around 100 watts of power.
Each time the coil is triggered to send a spark to the spark plug (along the spark plug wire) that ~100 watts of power going into the coil is converted. Power is not created nor destroyed, only converted, that's conservation of energy. And because it is not a 100% efficient conversion of energy, you lose a lot of that power when it goes to the spark plug. I think only around 25 watts of power make it to the spark plug, what is happening is you're trading current for voltage; really high voltage so there will be a spark across the spark plug gap of 0.060" but it only happens for microseconds and the amount of current (amps) is very small.
The voltage is anywhere from 20000 to 60000 volts, if you want to consider me as a reliable source. That's what I've seen over the past 15 years for ignition coil's specifications, albeit advertised from the manufacturer. But it has been very consistent, you are only going to get so much voltage because the power to the coil has never changed- always been 14 vdc and 5-10 amps. Some coils are advertised as 100000 volt output, and I have tried them and had them give out very easily and very consistently at 3000+ rpms, they overheat. Most coils today, instead of looking like this
have an E shape I think it's called, and look like this below. I believe this includes all coil on plug or coil near plug ignitions, including all chevy LS series of engines. The magentic field is oriented such that it gives a better transfer of energy for a higher voltage on the output (spark).
so, anywhere from 20000 to 60000+ volts going through the spark plug wires, depending on the type of coil.
how many volts you lose through the spark plug wire:
that's an interesting debate because because the circuit from the coil to the spark plug is an open circuit and I'm not aware of any way of accurately measuring it other than what might be claimed by the manufacturer, which i don't trust. So you can't just use V = IR to figure it out. And, it is not simply DC current from the coil the plug, there is a frequency to it along with a backward type of spike from the primary side of the coil. That's why RF or EMI suppression is needed on spark plug wires. That being said, the resistance of the wires probably don't play a large part in the amount of spark energy reaching the spark plug.
Another real area of concern however is how the ignition coil operates, what it's rise time is. That means how long you have to feed it power before it will output it's advertised spark voltage. As engine rpm increases, the coil fires more per unit of time so the time the coil has to build up it's magnetic field is less. If it's too short, you'll get less output voltage = weak spark, and the engine will sputter at high rpm. This is a problem with the traditional style of coil on a V-8 engine where there is one coil being used for 8 cylinders. Now, you have a coil for each spark plug so in a V-8 applications I think the time is lengthened by a factor of 8 (or 6 for 6 cylinder, 4 for 4 cylinder) which helps the coil immensely, it has a longer time at high rpm to generate maximum output voltage.
Another thing is the secondary output is dictated by secondary resistance. The higher the resistance, the higher the output.
Lastly, the bestest way to view just whats going on, is to use an oscilloscope and secondary inductive pickup to measure the secondary voltage. You can also use an inductive amp probe and view the coil charge and saturation on the primary side.
If anyone is interested, I have a bunch of scope screen shots of all kinds of cool ignition waveforms. If anyone wants to see any, post up and I'll post some tomorrow.
Thats a pretty good overview. A few things to consider. The inductive voltage spike is caused by the primary coil collapsing it magnetic field and discharging through the secondary side of the coil. This is common when any saturated coil collapses. Its commonly seen in fuel injectors too. Or AC compressor coils (The reason they have diodes)
Another thing is the secondary output is dictated by secondary resistance. The higher the resistance, the higher the output.
Lastly, the bestest way to view just whats going on, is to use an oscilloscope and secondary inductive pickup to measure the secondary voltage. You can also use an inductive amp probe and view the coil charge and saturation on the primary side.
If anyone is interested, I have a bunch of scope screen shots of all kinds of cool ignition waveforms. If anyone wants to see any, post up and I'll post some tomorrow.
Another thing is the secondary output is dictated by secondary resistance. The higher the resistance, the higher the output.
Lastly, the bestest way to view just whats going on, is to use an oscilloscope and secondary inductive pickup to measure the secondary voltage. You can also use an inductive amp probe and view the coil charge and saturation on the primary side.
If anyone is interested, I have a bunch of scope screen shots of all kinds of cool ignition waveforms. If anyone wants to see any, post up and I'll post some tomorrow.
The new Oscopes with floppy or flash drives are a little too expensive for me! Or most Oscopes at all.
1FMF..... Thanks for the explanation...
ok heres the thing... a buddy of mine made a type of booster (increased spark) for a spark wire for a 2 stroke. and it made more power. thats what i was really asking about. if there is any small power to be had from doing something like this to our wires. or is this something that wont work.
-brandon
ok heres the thing... a buddy of mine made a type of booster (increased spark) for a spark wire for a 2 stroke. and it made more power. thats what i was really asking about. if there is any small power to be had from doing something like this to our wires. or is this something that wont work.
-brandon
but if you have your headlights, heater, radio, and everything on at idle then the alternator at low rpms probably isn't spinning fast enough to maintain 14v output. It can be down around 12v until you get the rpms up around 1500-2000 where the alternator can produce the needed power. In this case you would be correct in that your volt gauge on the dash would increase, which is normal, and technically you would then get increased spark energy out of the coil since the electrical system had been lower.
Now if with just the car running and all accessories off you see your volt gauge around 12-ish at idle and it climbs to 14 when you rev the motor then you might have an alternator problem, or other electrical system problem.
ok heres the thing... a buddy of mine made a type of booster (increased spark) for a spark wire for a 2 stroke. and it made more power. thats what i was really asking about. if there is any small power to be had from doing something like this to our wires. or is this something that wont work.
there's really no performance to be had on any auto ignition system year 2000 or later. Especially any coil per cylinder type setup, where there's 1 coil per cylinder instead of 1 coil for all 8 cylinders. Like was said the LS7 is running to 7k rpms or something on the stock ignition system no problem. And once you get a hot enough spark, any more isn't going to do much for performance. It's the combustion process of air/fuel that does the real work, not the spark.
check this out
http://www.chevyhiperformance.com/te...ics/index.html
that is a little misleading/incorrect. Normally your alternator should output 14v at all rpm's so by revving the engine there should not be any voltage change and you'll be getting the same spark energy to the plugs;
but if you have your headlights, heater, radio, and everything on at idle then the alternator at low rpms probably isn't spinning fast enough to maintain 14v output. It can be down around 12v until you get the rpms up around 1500-2000 where the alternator can produce the needed power. In this case you would be correct in that your volt gauge on the dash would increase, which is normal, and technically you would then get increased spark energy out of the coil since the electrical system had been lower.
Now if with just the car running and all accessories off you see your volt gauge around 12-ish at idle and it climbs to 14 when you rev the motor then you might have an alternator problem, or other electrical system problem.
without seeing it I can only guess at what it really is. And what was the ignition system originally on the 2-stroke?
there's really no performance to be had on any auto ignition system year 2000 or later. Especially any coil per cylinder type setup, where there's 1 coil per cylinder instead of 1 coil for all 8 cylinders. Like was said the LS7 is running to 7k rpms or something on the stock ignition system no problem. And once you get a hot enough spark, any more isn't going to do much for performance. It's the combustion process of air/fuel that does the real work, not the spark.
check this out
http://www.chevyhiperformance.com/te...ics/index.html
but if you have your headlights, heater, radio, and everything on at idle then the alternator at low rpms probably isn't spinning fast enough to maintain 14v output. It can be down around 12v until you get the rpms up around 1500-2000 where the alternator can produce the needed power. In this case you would be correct in that your volt gauge on the dash would increase, which is normal, and technically you would then get increased spark energy out of the coil since the electrical system had been lower.
Now if with just the car running and all accessories off you see your volt gauge around 12-ish at idle and it climbs to 14 when you rev the motor then you might have an alternator problem, or other electrical system problem.
without seeing it I can only guess at what it really is. And what was the ignition system originally on the 2-stroke?
there's really no performance to be had on any auto ignition system year 2000 or later. Especially any coil per cylinder type setup, where there's 1 coil per cylinder instead of 1 coil for all 8 cylinders. Like was said the LS7 is running to 7k rpms or something on the stock ignition system no problem. And once you get a hot enough spark, any more isn't going to do much for performance. It's the combustion process of air/fuel that does the real work, not the spark.
check this out
http://www.chevyhiperformance.com/te...ics/index.html
That was a very cool article. Thanks for the read. I think I am sold on CD systems now.
A random figure from the article...
I did some further digging around and found this article about how to make your own CD system:
http://www.uoguelph.ca/~antoon/circ/...s/zappers.html
Who is in? JK.
thanks again my dad was saying that as well... that in late model cars that the coils and wires are real efficient as compared to years ago.
what ever my buddy made into the spark plug wire, i gained 3 secs on my lap times, w/ no other changes....ill have to get incontact w/ him about it and get all the details about his project.
heres the engine im talking about...
http://www.yamaha-motor.com/sport/pr...64/0/home.aspx
thanks again
brandon
what ever my buddy made into the spark plug wire, i gained 3 secs on my lap times, w/ no other changes....ill have to get incontact w/ him about it and get all the details about his project.
heres the engine im talking about...
http://www.yamaha-motor.com/sport/pr...64/0/home.aspx
thanks again
brandon
1FMF..... Thanks for the explanation...
ok heres the thing... a buddy of mine made a type of booster (increased spark) for a spark wire for a 2 stroke. and it made more power. thats what i was really asking about. if there is any small power to be had from doing something like this to our wires. or is this something that wont work.
-brandon
ok heres the thing... a buddy of mine made a type of booster (increased spark) for a spark wire for a 2 stroke. and it made more power. thats what i was really asking about. if there is any small power to be had from doing something like this to our wires. or is this something that wont work.
-brandon
Toward the back of the GMHTP Magazine is an ad that tell you if you buy their product it will increase your coil voltage to 18 volts. On LSX motors you got to buy 2 one for each side of the engine. And they only claim to make 7 additional Hp after retuning. Now if V= I * R and your coil/plug wire/sparkplug is a consistant value then its easy to see how you could increase the current (I) by increasing the Voltage (V). But would spending $ on it be worth 7 hp? I don't think so.
Last edited by NC98Z; Jan 31, 2008 at 03:06 PM. Reason: add