Quote:
Originally Posted by inkling16
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.
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Check out the datasheet. The capacity of the battery is much different when the current is different. The battery is rated for 18 amp hours, which it can achieve when used in low current situations, but when used in FRC situations, the capacity is likely closer to 7 amp hours.
I don't know exactly why this is true, but I'd be willing to bet that the chemical reaction isn't quite as effective/efficient when it happens really quickly.