However, a deeply discharged battery will draw more current initially
Depends - it might not.
A battery that's very deeply discharged, i.e,. dead, will also have a very high resistance and won't draw much at first. After it charges up some the resistance will drop and
then it'll draw more until it gets close to full when the resistance rises again and the current flow drops off.
The point in writing this is that, as stated the voltage drop is a function of current and the resistance of length of the conductor. Well the 15' length of wire is static, but the current is not; it's variable. To put this another way, as the battery becomes charged there will be less and less voltage drop, maybe a couple hundredths of a volt difference.
Yea, exactly.
I think it's easy to overlook that voltage drop as regards feeding a load from a battery behaves differently than voltage drop as regards charging a battery.
For instance supplying a 1200w load from a 12v battery would give you 100a current flow and X voltage drop. However, as the battery voltage drops down, the amperage flowing goes up to supply the same 1200w. If the battery gets down to 10.5v, the amp flow would be closer to 115a. So when figuring voltage drop for LOADS you would need to use the worse case numbers. You'd need to plug 10.5v, 115a into the voltage drop calculator to find out what size wire you really need to supply that 1200w load.
In other words, with LOADS, over time, the amperage flow FROM the battery increases, and so does the voltage drop.
But it's the OPPOSITE effect when CHARGING a battery. When charging, over time, the amp flow decreases, and so does the voltage drop. By the time the battery has reached a surface charge equal to the bus voltage, there isn't enough voltage drop to notice.
And even if there is theoretically a voltage drop, what effect would it have on charging? None really.
Let's say the battery is at 11.5v. That will hold the bus voltage down to 11.5v. The voltage regulator will see that and switch on the alternator and keep it on till the bus voltage reaches whatever the set point is (14.4v or 14.11v or whatever).
But the bus voltage won't reach that point until the battery surface charge reaches that point.
As long as the alternator is turned on, the POTENTIAL voltage of the bus is going to be higher than the actual voltage of the battery (thus negating the effect of a voltage drop), until the battery surface charge reaches equality with the bus voltage, and then there isn't any voltage drop anyway.
A smaller wire can make it take longer for the battery to reach a surface charge equal to the bus voltage, but that's about it.
And from that point on, it's just a trickle/absorb effect until the battery chemistry is finally fully activated.