i thought the v6 was 18v, if so and you want to drop it to 12v you could add a resistor in series (as you probably know )

You can use ohms law to calculate, but a quick method ... Vdrop req divided by I (current) required = R

then work out the power rating, v x I =Pw... but a hidden red herring is not to work out the power rating after the volt drop, but the supply voltage to be safe

readings will vary as the head motor becomes under load with carpet resistance.

eg 6v drop at 1 amp = 6 ohms ? P=18w

I am a little rusty with formulae so i will apologise now for any errors

I am building a fishing rod ring wrapper ( i do have a commercial one too ) i used a sewing machine motor but even with a variable rheostat i couldn't get motor speed down low enough ( my fault for getting the gearbox calcs wrong ) so i split the motor open and revealed a couple of 1 ohm resistors in series with the brushes, i increased these to 10 ohms 1 watt and got the min motor speed perfect.

Another point i noticed, run a v6 on the bench, whilst on remove the head - the cyclone motor still runs - then reattach the head. head wont run until switched off and back on.