Thanks for correcting the link and I appreciate your proof-reading.
(But, just to let you know, I did understand the forward-slash convention as I typically use something similar with parentheses and slashes.)
Anyways, I was eyeballing the equations I posted earlier and wondered, "Hmmm... what proportion of current would the batteries approach if the load current became really big(i.e., the load resistance approached zero)?".
Equations:
And as R_l goes to zero, you're left with I_a = V_a/R_a and I_b = V_b/R_b. Taking the ratio I_b/I_a and you get (R_a/Rb)*(V_b/V_a). If the voltages of the two batteries are nearly equal(Which I'm assuming we'd want them to be close so the boost pack can start delivering when the capacity pack drops in voltage), then V_a~=V_b and I_b/I_a~= R_a/R_b as (V_b/V_a) ~= 1. So, basically, you can calculate the ratio the currents of each battery approach based on their internal resistances.
For example, a 100 mOhm booster pack in parallel with a 200 mOhm capacity pack would have a current nearly twice as much as the capacity pack at high currents. A 50 mOhm booster pack in parallel with a 200 mOhm capacity pack would have a currently near 4 times as the capacity pack at high currents. If you know the booster pack will supply roughly x times as much as the capacity pack at high currents and you know the maximum current draw of the load, I think you can easily calculate how much current the capacity pack will draw at this max current draw. For example, if you knew the booster pack would approach 4 times as much current output as the capacity pack(the booster has 4 times less internal resistance) at large currents and you knew the controller would only draw 100 amps at max, you'd then know the capacity pack would roughly approach 20 amps at max and the booster pack would give approximately 80 amps(In real life, probably a little more but not much more) since it would provide nearly 4 times as much. As the booster pack depletes, though, the (V_b/V_a)~=1 becomes more errant as the booster pack's voltage drops and so you'd expect the capacity pack to start increasing its current output. If the booster discharges to the point where the booster cuts out, then the capacity pack would be hit with large current demands in which case some means to throttle the current limit on the controller would be needed. Luckily, my scooter has a home-made supplemental controller with a programmable current limiter so I'd be able to easily send the "booster pack turned off" status to the supplemental circuitry which would then change the current limit on the fly as needed. I don't see this option being available to many current users(on the fly current limit changing for the motor's controller), however... so that might be a kink with this idea. But for my use, and whomever may have their own similar solution, it should be doable.
Now that I've done that, I think my ping battery has an internal resistance of around .2 ohms. A 10s2p pack of a123s would have (10s*10mOhm = .1 ohms... .1ohm/2 = .05 ohms) which is about 4 times less resistance so I might expect it to contribute about 4 times as much or more current than the ping when the booster is enabled. Is that... acceptable? Let's see, at 80-100 expected amps, the ping would contribute around 17-20 amps or a little less which sounds nearly ideal since I was planning on limiting its current to 20 amps anyhow. So, wooohoooooo! This is a cheap, exceedingly easy way to implement the booster! No extra current limiting necessary! Yeehaw!
However, this all awaits real world testing, so let's let those results come in first before popping open the champagne.
