Please un-check this box. If you happen to hit 60k ermp while ridding it will apply the breaks, then your face will meet the road.
In regards to the duty cycle, hit full throttle with the wheels in the air and check the telemetry. Then do the same going half speed or going up a hill. See the results.
looking at the original poster’s data:
39.1v pack voltage * 15.1a battery current = 590.41w electrical
(39.1v pack voltage * 40% duty) * 36.9a motor current = 577.11w electrical
(15.64v effective voltage) * 36.9a motor current = 577.11w electrical
the difference is likely caused by either a combination of iron & esc losses, or possibly decimal rounding inaccuracy of the duty cycle value in the logging app.
Yes. Thanks 
Yeah, I know. You’re totally right. Devin’s math was more like: 40 V , 1 ohm = 40 A = 1600 W 10% duty cycle = 4 V * 400 A = 1600 W BUCKED DOWN! LOGIC! WHY CAN’T YOU EXPLAIN WHERE I’M WRONG! I WILL WRITE 4 ANSWERS TO MY OWN POST!
There are probably 3 topics with 300+ post each by devin about this, and I really really really do not want to revive him. He tends to return as @devin2, @devin4… I’m almost scared of speaking it out. Like saying “Candyman” 3 times…
if the winding resistance in this example is in fact 0.01ohm then it seems correct.
((40v battery * 10% duty) - 0v bemf) / 0.01ohm winding = 400a motor current
(40v battery * 10% duty) * 400a motor current = 1600w electrical
(4v effective) * 400a motor current = 1600w electrical
you can calculate the instant back emf voltage to use in this equation by dividing the instant rpm by the kv… for example 1000rpm / 100kv = 10v back emf voltage
so if we assume the motor is 100kv and presently turning 1000rpm, then:
((40v battery * 35% duty) - 10v bemf) / 0.01ohm winding = 400a motor current
(40v battery * 35% duty) * 400a motor current = 5600w electrical
(14v effective) * 400a motor current = 5600w electrical
5600w electrical / 40v battery = 140a battery current
though again, vescs are fail-safe programmed within the firmware to limit motor current below 120a despite any user settings.
DC of 10% doesn’t push 10 times more current than 100%.
Now you’re trolling, @devin . The motor had 1 ohm at 100%. At 10% DC the motor windings still have one ohm.
^you can estimate the resistance of your motor + vesc by using your throttle with the motor stalled
in this case we have 20a battery current, 78a motor current, 49.7v battery voltage
49.7v battery voltage * 20a battery current = 994w
994w & 78a motor current into the ohm’s law calc (0v bemf - stalled)
this gives roughly 0.163ohm for the motor winding + vesc.
effective voltage for 20a battery current with the motor stalled in this case is ~12.74v effective. (49.7v battery)
the largest source of error in this method is the battery voltage was actually a bit lower than 49.7v while the 20a battery amps were drawn due to voltage sag – so the electrical wattage in this case was in fact a bit lower than 994w while the 20a battery amps were drawn.
Dude, yes. I know.
Last try, then I’m out. I know that you know that the meaning of DC is not that DC of 10% pushes 10 times more current than DC 100%. And that the power at DC 10% is lower than at DC 100%. That’s all I was saying.
I’m too scared of summoning Devin5 to continue.
