Calculating Lamp Specs?

[Ria]

I'm afraid you lost me by about the 2nd line 

Hopefully Max or someone else with more brain cells than me reads this..!

[Drew]

My apologies in advance for the incredibly long post. For a while now I have been asking periodic questions on LG about the specifics of calculating lamp parameters. Of course many say that it is just impossible, but I have gotten somewhere with the help of a few people. Here's where I am right now:

PLamp: RMS lamp power (W)
VLamp: RMS lamp voltage (V)
ILamp: RMS lamp current (A)
ZBal: Ballast impedance (Ω)
OCV: RMS open-circuit voltage (V)
C: Mystery unknown constant

These variables make up the following equation:

ZBal = (C * √(OCV^2 - VLamp^2)) / ILamp

Which can be rearranged as:

ILamp = (C * √(OCV^2 - VLamp^2)) / ZBal,
and:
C = (ZBal * ILamp) / √(OCV^2 - VLamp^2)

One may think that the value C is power factor, but inputting lamp power factor calculated by PLamp/(Vlamp*ILamp) gives completely false results. Take the following examples:

400W MV:
PF = 400 / (135 * 3.2) = 0.926
ZBal = (0.926 * √(220^2 - 135^2)) / 3.2 = 50.27 ohms
50.27 / 44.5 = 1.13 = 13% off
Says the lamp needs 50.27 ohms of impedance when the specs say 45.5 ohms

55W SOX:
PF = 56 / (109 * 0.59) = 0.87
ZBal = (0.87 * √(480^2 - 109^2)) / 0.59 = 689.3 ohms
689.3 / 775 = 0.89 = 11% off
Says the lamp needs 689.3 ohms of impedance when the specs say 775 ohms

So obviously that isn't gonna work, it gives wrong answers. So we use the rearranged equation to look specifically for this special C constant, which is different for every lamp, however reasonably consistent within technologies. I have found the following average values of C for different technologies:

MV: 0.831
MH: 0.873 (highly variable)
HPS: 0.913
LPS: 0.973
FL: 0.915

So this definitely isn't the power factor. It seems to trend in the opposite direction of power factor, with MV being the lowest and LPS being the highest, which definitely points away from this being power factor. I thought maybe this was just the distortion power factor excluding displacement power factor, but I have no idea.

Anyway, now with these values of C:

400W MV:
C = 0.831
ZBal = (0.831 * √(220^2 - 135^2)) / 3.2 = ohms
45 / 44.5 = 1.01 = 1% off

55W SOX:
C = 0.973
ZBal = (0.973 * √(480^2 - 109^2)) / 0.59 = 770 ohms
770 / 775 = 0.99 = 1% off

So obviously these values give good results, they are averages that were trained on the existing data from the reference circuits. But do they work for circumstances outside of the reference circuits? Well apparently yes, and another LG user helped me prove it with their variable reactor ballast.

Standard specs for 18W SOX-E:
PLamp: 18W
VLamp: 15V
ILamp: .35A
ZBal: 829Ω
OCV: 300V

And we can use the C equation to get:
C: 0.985
which is only a little bit higher than the LPS average of 0.973, but will help us get a more accurate answer.

With these values, using the ballast impedance equation, we can extrapolate what ballast impedance we will need for different OCVs, such as 220, 230, and 240V. We get the following values:
220V: 598.1Ω
230V: 627.2Ω
240V: 656.2Ω

The LG members experimental values with a variable reactor are below. 240V was not able to be determined just due to practical constraints.
220V: 602Ω
230V: 632Ω
Both of these values are within 1% of my predicted values, meaning whatever the values of C means, it does not change significantly with different circuit characteristics.

This makes it possible to predict fairly accurately the necessary ballast impedances and probable lamp currents for lamps running on alternative gear. This seems to be very accurate when the reference circuit is available (so an individualized C can be calculated) but the average C values can be sufficient (within 2-5% usually) when a reference circuit is unavailable.

So this is my question:
What is C? Where does it come from? It is obviously some sort of characteristic of the arc discharge type because it is fairly consistent within technologies, but it is not the total power factor. I was greatly doubting the ability of these equations to work outside of the standard circuits, but apparently they do so quite well (for 18W SOX-E at least).

Also if anyone else has the right equipment, we might be able to prove that these equations work with other discharge technologies.

What do you think? Just thought I would get another set of eyes on this.
Thanks