[Caleb Sykes]

Originally Posted by Jared

I still don't see this working well for what you're trying to accomplish. Your value of "E available" is not really quantifiable for the battery. If you read the datasheet, you'll see that the battery's capacity (which affects E_available) varies greatly with current draw.

Basically, this means that your effective energy consumed is proportional not only to the energy you're actually using, but the rate at which you're using this energy (or some weird function of the rate).

I guess I still don't fully understand why we wouldn't be able to easily calculate Eavailable assuming we account for the internal resistance of the battery. We should be able to integrate the power over time to get the energy. Something like Eavailable(t) = Emax-(time integral of (V(t')*I(t') + I(t')^2*Rinternal) from t'=0 through t'=t) where V(t) is the terminal voltage of the battery at time t, I(t) is the current supplied by the battery at time t, and Rinternal is the internal resistance of the battery.

This should be all we need to calculate Eavailable, since the initial energy of the battery either has to turn into electric energy that moves through the circuit or turn into heat due to the internal resistance of the battery. I don't see anywhere else that the energy of the battery can go. The only thing I might not be considering here would be that Rinternal might not be a constant, but a function of current and/or temperature. Does anyone know if this is the case? Because if so, that could explain why the battery loses charge more quickly than expected at higher currents.