The battery current limit & how it affects hill climbing speed

I wondered how different battery current limit settings affect hill climbing ability-- so I did a comparison of 30a vs 60a battery current limits per motor (both 100a motor current limit) with (2) 73kv hub motors at 46v with 84mm tires…

the grade chosen for the comparison is a 31.5% slope which is equivalent to the steepest street in san francisco. the minimum force required in both pounds and watts to maintain speed or accelerate on the 31.5% slope is shown on the purple line, bottom left chart (pounds) and the green line, top middle chart (watts). the vehicle thrust in pounds and watts is shown in the red line, bottom left chart (pounds), and red line top middle chart (mechanical watts).

conclusions:

19mph top speed up 31.5% slope with 100a/30a motor/battery current limits 27mph top speed up 31.5% slope with 100a/60a motor/battery current limits

(the top speed is assessed by where the red line dips below the purple line on the bottom left chart)

Super interesting man! Now, this is true when the battery amps are the bottleneck. On my single drive build, wheel and belt traction are the bottleneck, so I’ve found that on single 6374 30 battery amps performs almost as well as 60 amps and prevents my wheel and belt from slipping. I weigh 180 pounds.

Do you have numbers on motor and VESC temperatures from both?

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this is all predicted.

motor numbers used: 46v battery, (2) 73kv hub motors with 0.07ohm resistance and 84mm tires, 200lbs rider + board, 0.75 wind drag coefficient, 0.6m^2 frontal area

motor current is calculated from:

(effective pwm voltage - back emf voltage) / winding resistance = motor current

back emf voltage is:

motor rpm / kv = back emf voltage

electrical watts is:

motor current * effective pwm voltage = watts electrical

torque in newton meters is calculated with:

(60/(2 * pi * kv)) * motor current

mechanical power in watts is:

torque newton meters * angular speed in radians per second

slope force in newtons is calculated with:

mass in kg * gravity acceleration (9.8m/s^2) × sin(slope angle in degrees)

wind force in newtons is:

(1÷2) * 1.225kg/m^3 air density * front area in m^2 * (meters per second * meters per second ) * standing human drag coefficient

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Ahhhhhhhhh I see.

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