Placebo effect. 
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Maybe we should split this topic?
some of this may be useful to someone else one day 
kind of de-railed this build thread
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Agreed sorry for clutter
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Agreed, very good info in this discussion. Don’t know if you guys knew it but @Hummie was one of the first if not the first to push the limits of motor amps when Vesc 4 was still relatively new. He convinced me to try it and I found a substantial increase in power just going from 60a to 80a.
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an esc works by a similar principle to a dc to dc buck converter in continuous mode which can also lower the effective voltage while increasing the amps. the battery is periodically connected and then disconnected from the winding at a frequency on the order of 25khz. During the times the winding is disconnected from the battery, the winding is shorted to itself which allows the current generated during the brief on times to continue circulating during the off times. since the current continues to flow in the winding during the off times, the “motor amps” are numerically higher than the “battery amps.” the battery may see 10a @ 50v, 10% of the time which averages to 1a, while the motor effectively sees 10a @ 5v the entire time. In reality the current rises in the motor from, say, 9.5a to 10.5a during the 10% on times, then decays from 10.5a to 9.5a during the off times.
as the back emf voltage produced by the motor increases with increasing speed, it is necessary to increase the on time (duty cycle) of the circuit to maintain the same number of motor amps, which draws more and more battery amps.
Pack V * duty cycle % = V effective
(V effective - V bemf)/Resistance=Motor Current
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The vesc does adjust motor amps… Most of us use current control, which the primary function is amp adjustment, Just it is adjusting amps and rampung duty cycle to change motor voltage to generate more wattage If it tried to throw as many amps at the motor as it can…say 80 amps motor max
If you hit 80a motor max at 10% duty 50v * 10% duty = 5v * 80a = 400w
We can only produce 400w on 10% duty of 50v at 80 amps, but the load(your weight + demand of acceleration or speed) is requiring more wattage.
The 80 motor amps will stay 80 motor amps and increase duty cycle to output more wattage, until
- The load decreases(stopped acceleration)
- motor gets into it’s efficiency torque rpms where it generates torque with very little amps above x rpm
- The combination of motor amps, and duty cycle makes a wattage/current higher then max battery
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50v battery * 10% duty = 5v effective volts (5v effective - 0v bemf) / 0.0625ohm winding = 80a motor amps 5v * 80a motor amps = 400w electrical watts 80a motor * 80a motor * 0.0625ohm = 400w copper loss 400w / 50v battery = 8a battery amps
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Exactly
I was just trying to keep the maths simple for everyone. As we could bring in BEMF, and load and heat saturation…but…yea 
Because it is more “powerful”, what we fell is is not power, but acceleration, and it’s directly proportional to torque and that’s proportional to current
So if the duty cycle is low enough that you are not hitting the battery limit, that’s means low speed since duty cycle and speed generally have a linear relation, you are getting more torque if you motor off the line until the battery current limit is hit comparing the 80A to 100A
The more you increase the motor current the more torque you get until a given speed, but if you maintain the same battery limit, this speed that you can use max torque gets lower as you keep increasing motor max
This if of course ignoring motor saturation or it breaking itself apart
As example, my board can theoretically climb a 27% incline, but only at a max of 16 km/h since that’s the speed that the motors can keep the max current of 80A without hitting the 30A battery limit
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Can you clarify/explain what you mean with the boldened text?
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Most motors have a torque curve that could likely be provided by manufacturer. Example
Let me know if this doesn’t make sense…
As side note this is pronounced on single drives. Struggle along til say 16mph then out of no where it feels like turbo boost because the motor has gotten into it’s efficiency range… So 0-16 is moderate acceleration, but 16-24 feels like a rocket…
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KT (torque per amp) can be calculated directly from KV:
KT (newton meters torque per motor amp) = 60/(2 * pi * kv)
0.0954nm/a = 60/(2 * pi * 100kv)
therefore 80a motor current * 0.0954nm/a = 7.632 newton meters torque @ 100kv
7.632 newton meters torque * angular speed of the motor in radians per second = mechanical watts
(1000rpm * 2 * pi)/60 = 104.719 rad/sec
7.632 newton meters torque * 104.719 radians per sec angular speed= 799.22w mechanical power
799.22w mechanical power / 1000w electrical power = 79.92% electrical to mechanical efficiency
the difference between the electrical power and mechanical power appears as ohmic heating in the motor windings.
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This part is the one I have issue with. What exactly do you mean by with this? Does it generate more torque with little amps @ high rpm? Isn’t torque theoretically (not counting losses or other un-ideal effects) directly related to motor current?
Am I to understand that the motor generated more torque with lower motor current at higher RPM?
EDIT: I also looked up that same graph as you linked, before asking the question, because I still couldn’t understand what you meant by this specific part of the sentence.
EDIT2: Or do you mean “it generates very little torque with very little amps above x rpm”
If volts is speed and current is throughput, and load is equivalent to watts
If the load stays the same, the voltage goes up(duty cycle) and amps goes down.
Typically you won’t hold max motor amps up until the max duty cycle equal to max battery amps… As the motor is “unloading” in a sense as voltage increases… as the load is staying the same(board + your weight @ throttle %)
Higher amperage is affected by resistance more then voltage. So at low voltage, high amps. More watts are wasted as heat. where at higher voltage lower amperage, less watts are wasted as heat and translate into mechanical power
@professor_shartsis has a good example above to calculate efficiency.
So you meant that in the scenario of the motor current being restricted by battery current?
AKA, if you have hit the battery max and continue speeding up, then motor amps will start to decrease to keep the same power?
i’m sorry DeckyBro you stood up and now you have to endure… props mate
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What’s VESC are you using? If it’s 4.12 or any variant with just two shunts the boost is because noise in the current measurement that makes the duty cycle jump to 95%
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Ah no I mean more so efficiency in combination with duty as efficiency and torque curves change with duty(motor voltage)
What ends up really happening is you have many voltages and many curves, if you were to plot a graphs with the various curves on many voltages(duty) you would end up some something that looks similar to surge graph for turbochargers…but for bldc
Basically same electrical power over varying voltages results in different efficiency
Anyway I cannot find any images where people have overlaid the graphs at different voltages(duty) from the same bldc. but it would looks similar to pump/compressor graphs
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