Original poster: Terry Fritz <teslalist@xxxxxxxxxxxxxxxxxxxxxxx>
Hi Steve,
At 05:47 AM 7/2/2005, you wrote:
I was thinking of leaving R3 constant for now and finding the power at a given point in time and use the Freau formula to give streamer length. Something like:
L(t) = 2.4 x SQRT( LoadEnergy(t) )
That sounds a bit funny to me. The way I understand it, the instantaneous streamer length isn't a function of the instantaneous power, but of how much energy has been burnt in the streamer load between the start of the burst and "now". That would be:
EnergyToNow=sum from t=start of burst to t=now(voltage across R3*current through it*size of timestep in seconds)
Load energy is actually the integration of energy per dt so it does that...
I kind of lost track of what LoadEnergy is in your code since you sometimes use "power" and "energy" interchangeably too :-P
You should probably add in the energy that was burnt in previous bangs, weighted with an exponential decay function of some kind, to account for how the channel cools and deionizes between bangs. This would help accuracy at bps much different to 120. I suggest using exponential averaging for that.
So the streamer length calculation done at each timestep is:
StreamerLength=Some_Constant*sqrt(EnergyToNow(1-DecayFactor))+(EnergyFromLastBang*DecayFactor));
Some_Constant is chosen so it turns out the same answer as Freau's formula at 120bps. That would be 1.7 * (1/120) * (1/39.xxx inches in a meter)
After calculating a whole bang you would update EnergyFromLastBang-
EnergyFromLastBang=(EnergyToNow(1-DecayFactor)+(EnergyFromLastBang*DecayFactor));
You choose DecayFactor according to the BPS and the thermal dynamics of air ::) Or fudge it so sparks start to coalesce around 70-100bps as we observe in experiments. If DecayFactor=0 then it reduces to the method you're using just now.
The disadvantage is that you now have to model several bangs, enough to reach a steady state. You can probably modify the code somehow to include the assumption that this bang was preceded by a large number of bangs just like it, to get the answer in one shot. But I don't know how to do that.
Cheers,
Terry
Steve Conner