I know, I know, another annoying question about calculations... 

Last time we talked about this, I mentioned this equation that was devised:
ILamp = (PF * √(VOC^2 - VLamp^2)) / ZBal
Where ILamp is RMS lamp current, PF is distortion power factor, VOC is supply voltage, VLamp is RMS lamp voltage, and ZBal is ballast impedance. I use this equation to determine the power factor:
PF = PLamp / (VLamp * ILamp)
Where PLamp is wattage. This is the definition of power factor, is it not? I am pretty sure it is. But something doesn't quite check out here. Here is an example, I am using 35W S68 North American HPS, but this same phenomena happens to most other lamps:
PF = 35 / (55 * .83)
PF = 0.77
Cool. So the distortion power factor is 0.77. So let's plug it in and see if we can get the right lamp current with all the other specs being known:
ILamp = (0.77 * √(120^2 - 55^2)) / 116.5
ILamp = .7A
So the current is .7A. But no, that is false, it is supposed to be .83A, that is an error of over 15%. I think the power factor is the problem. A while back I calculated average power factors for different technologies (hps, mv, etc) using the more complex of the two equations (not the definition of power factor, the other one). Let's try my calculated HPS average power factor of 0.913. Keep in mind that this value is averaged from across all HPS lamps of which I have specs.
ILamp = (0.913 * √(120^2 - 55^2)) / 116.5
ILamp = .84A
Well I'll be darned! That new magical PF value gave a result that is within 2% of the actual value. Very nice!

So power factor is the problem. Either the way I am calculating PF is wrong, or the way I am using PF in the big equation is wrong. Which one is it? And thank you all for your continued mathematical help, I think we are on to something here.

Check your equation. Decreasing the power factor must increase the current, ie. when calculating the current you must divide by the power factor, not multiply by it



Also, the formula with Zbal is a bit abridged, but that has smaller effect on the precision of the outcome

Z = sqrt( X^2 + R^2 ), where X is the inductive impedance, R is the resistance representing the losses (most of it is plain DC resistance of the winding wire, but there is some more to account for the core)

So the ballast itself does have its own angle between Z and X, which depends on the ballast efficiency, and does not have anything to do with the lamp PF or the complete circuit PF

When calculating, the Vdrop of the ballast consists of Vdrop(X) and Vdrop(R), which together form Vdrop(Z). However, Vdrop(R) is the one which is mostly in phase with the Vdrop of the lamp arc

So, the complete Vdrop sum would be something like :

Vline = sqrt( Vdrop(X)^2 + ( Vdrop(R) + Varc )^2 )

@Ash
Thanks for the comment!

So the PF should be in the denominator is what you are saying? Let's try it out with the 35W HPS again with the calculated PF for this particular lamp (.77):
ILamp = √(VOC^2 - VLamp^2) / (ZBal * PF)
ILamp = √(120^2 - 55^2) / (116.5 * .77)
ILamp = 1.19
Well that definitely increased the current, but way too much. That is more than double our previous 15% error. Obviously I am not doing this right lol, do I have to do something with squaring/square rooting in the denominator?

I did do some research on the inductive vs resistive impedance of the ballast, but incorporating that into the equation would probably require measuring the inductive impedance and resistive impedance of each ballast you are doing that for. I have no idea how to measure the leakage inductance of a coil while excluding the non-leakage inductance non-destructively, so maybe leaving that out would be easier. I'll consider adding that if the results we get at the end aren't close enough for our purposes.