Deep, let me see if I can make this a little more understandable. For the sake of this example, let's say it take 10,000 volts to jump a 1/8" spark gap. So you buy a coil capable of producing 40,000 volts. At what voltage does the spark jump that same 1/8" gap? If you said 10,000 volts, you are the winner. Does that make it a little clearer? But lets say the higher voltage fines a pin hold in the spark plug wire leading to the actual spark plug. What do you think is the chances of it jumping through that pin hole to ground?
A really good interesting thing to understand is the typical power pole workers. Each day before going out to work on the very high voltage power lines, they take their insulated gloves and have them tested to verify there are ZERO pin holes in them. Because even one little pin hole spells electrocution to the wearer if contacting high voltage power. JMHO!
Gm280, I really do appreciate you taking the time and try to explain this in layman's terms. However, I have a degree in electrical engineering from one of the top 10 schools on the planet and I did start my career as an electrical engineer designing and building automatic control systems for few years. Granted, this was over 20 years ago and I am a little rusted now after all of these years in the software engineering domain away from EE but I still do remember my basics.
First, saying that in the start (cold) condition requires a lower voltage than in normal operating temperature goes against everything I know about the subject. When the engine is cold and the fuel mixture ratios are not nominal, this will require stronger and longer lasting spark from the ignition system to ignite the mixture and start the engine. That is why we keep cranking and cranking until it finally fires.
Second, under an operating condition that will requires, 20KV to generate a spark, for example, a coil with a maximum potential output of 50KV ? while it will not actually generate this voltage ? it will however maintain the spark 3 times longer than a coil with max potential output of 20KV. In other words, the spark duration will triple while holding everything else constant. Same plug gap, same voltage required to generate the spark, same cylinder pressure, etc. It is not a question of output voltage, this will remain the same. But rather it is a question of stored energy, this however is not the same.
This can actually be proven mathematically but I will not bore the people here with the math.
Longer spark duration equal more efficient burn. Some folks confuse the internal combustion engine with an internal explosion engine which doesn?t actually exist. It combust the charge, it doesn?t explode it. So, claiming that once the spark is presented to the mixture it is all over is treating the situation as if we are designating a bomb while in reality we are initiating a burn of the air/fuel mixture. This burn is not optimal in all conditions and the effeciency of the burn is function of (among so many different things) how much fuel we were able to ignite initially which in turn is a function the spark duration. The longer time the spark will remain the more mixture will ignite which will lead to a cleaner more efficient burn which will translate to more engine efficiency.
Third, when the condition of the spark plug changes (becomes fouled or the gap increases) the higher output coil will still be able to jump this gap sparing us a misfire. Now, I am not advocating for skipping maintenance or ignoring the root cause of the problem, of course the problem needs attention and should be addressed. But $h!t happens and until we get to it and solve the problem, I would appreciate the extra strength that will spare me some misfiring while I am trying to make it back home.
Fourth, it takes only 1000V AC to ionize air under normal atmospheric pressure and make it conduct electricity according to Paschen's Law if I still remember my physics. Your grill lighter can generate a spark with that voltage. So, if your insulator is compromised with a pin hole in the circuit that will be a path of least resistance to discharge the coil, it will not matter if your output is 50KV, 40KV,30KV or 20KV. All of them will use it as a path of least resistance and you will lose the spark.
Finally, and again, I assure you I am not offended by any shape or form by your attempt to explain and simplify. I mentioned my background for two reasons only
- For you to feel free to get more technical if you are so inclined without worrying about this being a foreign language to me.
- So I don?t come across as someone who is arguing without knowledge or just for the heck of it. I was actually determined to just let it go and move on but just couldn?t I guess.
Thanks all of taking the time and share your experience. As always much appreciated.
