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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Yeah, it’s accumulated in those 2 or 3 big capacitors on the VESC input and the reason you need an anti-spark switch.
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You mean a map of the motor efficiency like this one?
https://www.eleceng.adelaide.edu.au/students/wiki/projects/index.php/Projects:2015s1-61_Computer_Aided_Measurement_and_Analysis_of_Equal_Efficiency_Characteristics_of_Electrical_Machines
Or
https://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/f2017/apg67_kss223_ejy25/kss223_apg67_ejy25/kss223_apg67_ejy25/index.html
So Y-axis is motor current(torque) and X-axis would be motor speed(BEMF)?
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Yes exactly like that.
And yep you got it right with amps and BEMF.
Good find, i dug for a while and couldn’t find and overlay like that
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Adelaide represent! 
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@Deckoz @professor_shartsis Been letting this thing brew in the back of my mind and I think I have a pretty good understanding of the parameters and their effects now, but then I started to think why does the efficiency go down with different torque/current at some specific speed?
For example in the linked graph @ 2400 RPM, if we’re running the motor with low torque/current or high torque/current it’s efficiency is worse than at around 30 Nm?
Well I found another research paper with the following graphs for IM (induction motor), IPM (interior permanent magnet motor) and SPM(surface permanent magnet motor), which they generated with a similarly spec’d motors of the different types.
So, if I apply my understanding. At low torque/current the motor is essentially operating closer to the no-load state, which is effectively a speed dependent static drag on the motor, which uses in relation a bigger portion of the power with low torque/current for the output power (aka, just to spin the motor).
And then at the higher torque/current we’re getting exponentially increasing copper losses, which then limit the efficiency at high torque/current?
Referenced research paper: http://porto.polito.it/2627142/1/effciency_mapping_ecce_rv2.4.pdf
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Yep copper losses, and the magnetic flux linkage. The iron in the bell’s magnets go through demagnetization and have losses as well depending on the magnet type (n42h n42sh n52sh etc)
The magnets don’t go through demag. If they do they are done. The only losses in mags are Eddy currents. Regardless of type