I run a Hyper9 with 5 Tesla modules and would not recommend running lower voltage on that motor.

I'd also caution you to consider voltage sag when anticipating your range for this system. The internal resistance of my pack is about 45mOhm, which means at wide open throttle (650A) I'm getting 32.5V of sag! My pack charge cutoff is 125V and my discharge limit is 100V. This means even on a full charge I'm forcing the battery below safe discharge voltage at full power.

What about at lower power? Well I use about 25kW to maintain freeway speeds. So at say, 20% SoC (105V) that'd be about 240A, which gives 12V of sag. So I can't maintain freeway speeds at 20% SoC. In fact, I can't sustain much better than 30mph below about 30% SoC, due to the voltage sag.

This is a limitation of lower voltage systems which I don't see discussed much and did not realize myself before I built my conversion. With a higher system voltage, the required current for a given power is lower, which means less voltage sag. Plus as system voltage goes up so does the spread between charge and discharge voltage, which means the (already smaller) voltage sag is an even smaller proportion of the usable voltage range.

But that's only true if the resistance is due to wire and wiring devices and connections, not internal resistance of the cells.

If the same cells were arranged with half as many cells in each parallel group, and so twice as many of those groups in series, they would have the same energy capacity and same power capability, but at twice the overall battery voltage. The internal resistance of each of the groups of parallel cells would be twice as high, and with twice as many groups in series the total internal resistance of the battery would be four times as high.

- modules as-is:
- total internal resistance of five modules: 45 mΩ
- internal resistance of one modules: 45/5 = 9 mΩ
- internal resistance per cell group: 9/6 = 1.5 mΩ

- doubled voltage version:
- internal resistance per cell group: 1.5 * 2 = 3 mΩ
- total internal resistance of battery (60 groups): 180 mΩ

Fortunately, for the same power requirement the current would be half (simply power = voltage * current) - that's 325 A at wide open, and 120 A at 25 kW and 210 V. As a result, the voltage drop for the whole battery would be only twice the present situation - that's about 60 V at wide open and 24 V at 25 kW. Twice as much voltage drop sounds bad, but it's a drop from a starting voltage twice as high, so the sag as a fraction of battery voltage is the same (12/105 = 11.4% and 24/210 = 11.4%). The sag is the same proportion of battery voltage and the same proportion of the usable voltage range, for the same power requirement.

Higher battery voltage is good to reduce the size of conductors and wiring devices required, but doesn't matter to voltage sag due to internal resistance. If those conductors and wiring devices are sized to suit the current there is no benefit in voltage sag of higher system voltage; if conductor size is limited then there is an advantage to reducing the amount of current squeezed through it.

Voltage sag is reduced by using a larger battery, relative to the power demand, regardless of series and parallel configuration. As they say "there ain't no free lunch" - you can't fix battery sag by shuffling connections within the battery.

The benefits of higher operating voltage are in the motor and inverter, not the battery.