Lets talk about "Voltage Sag"... [POLL]

I guessing this is due to the fact you have a wider voltage window, but how can you define Vs = Vw*N, for N is the number of cells (and number of cells in the entire pack, or the number in series?)? That doesn’t make much sense. For example, I run 12s, so my usable windows is 38.4v to 50.4v, so 12v is my usable window for my entire pack. According to your formula, it’s also my sag. But I sag at most 3v under high loads if I’m in the right part of my battery pack (fully charged for example, because that first 2 volts or so drop off quickly). I never sag 12v, so I don’t get what you formula is telling me, other than the range of my batter’s voltage.

Maybe we are talking about different sags… We are talking about the sag under load, not from using the energy out of the cells through normal discharge.

From another forum, I found this:

For instance if the internal resistance is 50 milliohms (0.05 ohms), then the voltage sag on a 6 A load would be 6 x 0.05 = 0.30 V.

Which seems more like the sag we are after. So…

sag = load * resistance

Can anyone confirm or deny this formula? My understanding prior to finding this was that resistance directly affects sag, which is true in this formula.