[Tesseract]

Originally Posted by G-man

You mean sag, right? A DC battery doesn't really have ripple.

If a DC battery has a non-zero impedance and it supplies a load which draws current in pulses (like a motor controller) then it will exhibit both sag and ripple.

Sag is the drop in voltage caused by the source having a non-zero resistance. It is literally the voltage drop from the average, or DC, current being drawn across the internal and external resistance of the battery circuit.

Ripple is the AC equivalent of sag, and is typically a triangular voltage waveform caused by the motor controller drawing current in (capacitor-integrated) pulses from the battery pack interacting with the non-zero impedance of said battery pack.

Originally Posted by G-man

Let's assume for the sake of investigation we had lockout contactors, and separate controllers, properly sized wiring, etc. Under this (theoretical) scenario, would there be an advantage?

Yes, switching losses are partially proportional to voltage (but also to current, switching frequency and switching transition time) so there would be a slight boost in efficiency by rewiring the pack to half the voltage at double the ampacity when the motor is operating at low RPM, especially at high torque. But since this is the least-efficient portion of the motor's operating curve it is best to minimize the time spent at low RPM and high torque.