Miller foot pedal question

Alright, I know I'm doing the "Thread Necro" move and I apologize. I'm attempting the same thing as the OP but I don't want to cut the original Miller potentiometer and JB weld the shaft. Reason for this is because work just upgraded to a wireless miller pedal and was going to throw out the old wired pedal. I was lucky enough to snag it before the trash man came. I'd like to keep it as "original" as possible in the unlikely event it gets called back.
This got me thinking, the pot is just used as a voltage divider. The only reason the resistor would have to be 47k and not the miller 1k is the current involved.
Does anyone know the "nitty gritty" of the electrical signals and circuitry involved here?

The main differences will be the resolution and slope of the divider and how the rest of the circuit is loaded. The wrong resistance will change the linearity of the pedal. The Miller pot is nothing special, just get another long shaft pot to replace it and leave the Miller pot intact, in case you want to return it to stock. You can also get pot shaft extensions if you want.