@whaddys 60.5 tire revs per sec since the gear ratio would be 3.83 (it’s the 3rd # from the bottom in the upper chart)… the motor would be putting out it’s peak mechanical power with that motor current limit at 13925rpm (14th # from the bottom in the upper chart)
to get those rpms, you

look at the battery current limit per motor (32a)

divide 32a by the max duty .95 to get the motor current at peak mechanical power

I^2R = W to get the copper loss in watts at peak mechanical power (in this case 0.05ohm was chosen for R)

multiply the battery current limit by the battery voltage to get the electrical watts at peak mechanical power

subtract the copper loss watts from the electrical watts to get the peak mechanical power per motor

calculate the torque per motor amp KT = 60/(2 * pi * 220kv)

multiply the motor current at peak mechanical power by the KT to get the motor torque at peak mechanical power

divide the peak mechanical power in watts by the motor torque at peak mechanical power to get the motor angular velocity in radians per second at peak mechanical power

convert the radians per second at peak mechanical power to rpm

divide the motor rpm at peak mechanical power by the gear ratio to get the wheel rotations per minute, then divide by 60 seconds to get the wheel rotations per second
the trickiest part of the entire equation is starting with a slope, vehicle mass, frontal area, drag coefficient, fluid density of air, peak mechanical power and then converting that to max possible meters per second which is done at the 8th number from the bottom in the upper chart… in this case that’s slightly simpler since 0 slope was chosen.
i previously described how that’s done here: