BFL13 wrote:
Not sure if we are talking about the same thing but using the glass half full or half empty way of looking at it and getting mixed up.
:) perhaps - I'm not sure either.
The wire just adds some R to what the charger "sees" as the battery. So the thin wire makes it seem like the battery bank is smaller than it is (lower AH total capacity)
I agree that the additional resistance of the cabling is similar to the additional resistance of a smaller battery bank. If the charger is current limited however, the current will be the same no matter what the size of the bank - correct? No matter how big I make the bank, I get no more current, and as long as I don't make the bank so small that I leave the constant current regime, it will, by definition, remain in the maxed out constant current bulk charge mode.
This has the same effect as increasing the charging rate (initial charging amps wrt battery capacity)
Hmmmmm. I'm not sure I agree - (which also means I'm not necessarily disagreeing). The charging rate "(initial charging amps wrt battery capacity)" is the charge current compared to the capacity. Neither of those changes no matter how much resistance is in the cables, provided that we are in the initial bulk charging constant current mode.
which makes the battery reach Vabs at a lower SOC for same Vabs.
This - more or less - sounds like we agree. Yes, The absorption voltage Vabs is the max voltage for the charger. The charger reaches that voltage earlier when the cables have higher resistance. That's why I said that the effect of long thin high resistance comes at the end of the charge cycle - when the charger has left constant current mode. It leaves that mode earlier. From now on, the charger is running at fixed voltage (I'm assuming no remote voltage sense capability).
So amps will taper sooner in SOC,
Yes. With heavier shorter cables, you would still be in constant current mode, instead of tapering the current.
but you still get a faster charge in comparison with using a lower initial rate and a higher SOC where tapering starts.
I had to read that a few times, but I think I agree.
Once you are at the same SOC and battery voltage, the acceptance rate is the same and charging times will be the same from then on to fully charged.
I see what you are saying, and you are right, it's a glass half empty versus half full. I think you are saying that if we wait until the battery has reached the same SOC, the shape of the curve at the end of the charge cycle will be the same for both cases, so the effect of the high resistance cables shows up at the start. I'm saying that if we start two charge systems and batteries with the same SOC at the same start time and one differs from the other only by the resistance of the charge cables, they will have the same charge current at the beginning and the high resistance system will always be at a lower SOC and lower charge rate at the end (charge current started tapering earlier), so the difference is at the end of the charge cycle, not at the beginning.
Now you might say that since the thin wire makes the battery seem smaller than it really is, that you will take longer to charge it since it is bigger than it seems.
Yes. It takes longer to charge and the effect is seen after the constant current regime when both had the same constant current charging - the charge curve is shifted to the right (shifted in time).