Moved to hereHaving read through the patent application I'll try and answer some of these.
OK Phloop, The spark plug wire is connected to the distributor cap where it gets its charge from the coil. The voltage goes into the capacitor where it builds until it is released to the plug in a single short burst. The reason for the ground wire is because that is how capacitors work. In order to drain them, in this case across the spark plug gap you need a closed circuit thus the ground wire from the negative side of the capacitor.
A capacitor is two metal 'plates' with an insulating material (dialectric) between them. In this case the 'capacitor' appears to be formed by wrapping a metal shield around the wire. The wire core is one plate, the silicon insulation is the dielectric and the metal shield is the other plate. By varying the length of the metal shield the capacitance can be changed. The 'capacitor' is effectively connected between a point somewhere along the length of the wire and ground. If you're familiar with points ignition then you'll know about condensors - a condensor is a capacitor, one end to coil, other to ground. Like Sidekicksrock says, an electrical component needs at least two wires (+/-)
1st as the coil voltage is building in the capacitor what is actually increasing? The capacitor will continue to store voltage until it is equalized with the voltage source in this case the coil so 30-40,000 volts. But would the amperage increase over a stock induction ignition?
The capacitor is similar to a
very small battery. What is increasing in the capacitor is the charge. Like I said before the max. voltage across the plug is defined by a number of parameters, on a Zuk its probably no more than 20KV max, maybe 8KV no load.
The coil does not determine the voltage. a 60kV coil can produce a 30kV more than a 30KV coil but with no other changes the max. voltage across the plug will still be around 20kV. The additional current provided is tricky to calculate, you would need to know the effective resistance of the spark plug itself.
The charge available is easy Q=CV where C = capacitance and V = Voltage. Some fairly rough calculations:
Q = 75pF * 20K = 1.5uC
now 1 Amp is 1 C/s so if the discharge time was 1 second then there would be a current of 1.5 millionths of an amp.
Over the stated 400nS then the current is 3.75A average (see below).
Here is another issue, the discharge time will be fixed by the Capacitance and the effective resistance of the lead AND plug. What is the effective resistance of the plug? I don't know but if theres 20KV across it then it must be reasonably high V=IR, if R was zero then so would V
2nd what would be used to trigger the discharge? Would the increased plug gap be enough to delay the spark and allow the capacitor to charge or would there need to be a resistor or some other trigger in line which would open once a certain voltage is reached?
The capacitor will begin to discharge when any point in the circuit has a lower voltage. In this case the voltage across the plug will fall when the spark is formed. Now because of the resitances in the leads the voltage at the coil is slightly higher than the voltage at the capacitor and the voltage at the capacitor is considerably higher than the voltage on the plug, so the current will flow towards the plug.
There are a number of points in their literature that don't seem to be quite right. Even if they do (and I have a lot of doubts) then there is still the issue of whether there really are any gains to be had from a single high energy very short duration spark in an automotive engine. All the research available seems to indicate that a longer spark duration is preferable.
Its also worth noting that a spark with half the energy over twice the time has the same power because the overall power is the
integral of the function.