Quote:
Originally Posted by shorod
I still struggle with this. They would almost need to be measuring the current internal to the spark plug, where virtually no resistance is involved, to accurately measure 500A. Accurately being a key here. Using their numbers then, and Ohm's Law, P=I^2 * R. Solving for R = P / I^2 = 5E6 / 0.25E6, so there needs to be 20 ohms or less total resistance when they are measuring the current. I don't think they're going to get there in any real world application.
Not that any of this really matters. Heck, political candidates stretch the truth (and maybe even tell some lies) all throughout their campaigns and still get elected. Some people are even pleased with the results from certain candidates. Why can't the same go for spark plugs?
I'm not suggesting that these plugs can't make an improvement in some applications. I just like to be able to rationalize the claims a product makes before I purchase it. That being said, I have been known to use some products that I couldn't justify the claims, yet they seemed to work for me....
-Rod
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I was told by the inventor and Chief technical officer the following:
That they do have less than 20ohms resistance!
Impedance is the square root of inductance divided by capacitance, we have 10 nanohenries and 30 Pico farads
So:
Z=ÖL/C = 18.2ohms Ö 10nH/30pF
So assuming the spark occurs at 20kV then:
Current (I) = V/R
1098 amps = 20kV/18.2 ohms
Since the voltage is falling when the current is rising there is a crossing point about ½ way
So then V=10kV and I=500A so peak power is 5MW
We do not use any current sensing device that could affect the spark.
The measurement uses a B-Dot sensor that measures the B field of the arc itself
Yes, we know how to accurately measure 1000A rising in 1nS
James