OK Batteroo,

Since you doctorates depend on undergrads to do your work for you, here it is:

Apple Trackpad (this will be very similar to the keyboard since both are bluetooth):

0% indication voltage: 2.10V / 1.05V per cell

Functional cutoff voltage: 1.85 / or 0.925V per cell

Before you and your "engineer" go doing handsprings down the hallway because 0.925V is higher than the 0.8V specified by Duracell, I refer you to the same graph Dave did below, where you will see that 0.925 is very nearly depleted. If you need an exact % of area under the curve of energy remaining, I'll have that in a few days, as I need to do that for a project anyway. By counting squares, I'm estimating the remaining charge is less than 5%. Please show me the math of how you plan to extrapolate that <5% into 800% better battery life.

Oh, and your argument about the voltage drop under load? Bullshit. Peak current is about 0.050 mA, and the measured voltage drop against fresh AA cells under that load is approximately 0.02V per cell. 0.04V one way or another does not meaningfully change reality here. Will running electric motors have a higher voltage drop due to the current draw? Of course they will. But since your front page advertising is directed squarely towards modern electronics, I'm going to point this out to you (as if it hadn't been done enough already): virtually all well-designed electronics use up their battery capacity nowadays.

In other words, Apple is already extracting a very practical amount of energy from their batteries and leaving enough headroom to help prevent leakage due to over-discharge. As a former Apple employee, it's surprising to me you don't know that. Hmm...