Thanks for all the help dude. I just can’t seem to get my head around it. I might come back to it a few weeks from now but I gotta focus on some work. 
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if you run that formula for a given battery voltage, back EMF, electrical wattage and resistance, and then you decrease the resistance, you will see the motor current increase, and that’s why you can feel more torque with a bigger, lower resistance, same kv motor.
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I got stuck on this for a bit because I was thinking how can motor_current be independent of battery amps (2nd equation) …
but I think i got it …
- the battery amps are in the wattage_electrical (3rd equation)
- I had been thinking a fixed voltage. voltage is not fixed it will sag.
- either that or vesc will hit battery current limits and reduce duty_cycle.
Lets say you know the electrical wattage (battery amps * battery volts), motor rpm, kv, bemf_v, and winding resistance, it will be hard using those equations to deduce the motor current and duty cycle that makes all of them valid (except by trial and error or rearranging them)…
that’s where this one comes in handy:
(ps winding resistance is ohms not mOhms)
The motor amps are separate from the battery amps because they are at a significantly different (lower) voltage than the battery voltage, and the duty cycle determines how much lower. If you apply the full battery voltage to the motor at low speeds (100% duty) you would get way, way, way too many amps and too much power.
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If you look at the form of this equation, it’s simply ohms law, where I is the motor current, V is the voltage to the motor, and R is the resistance in Ohms—
I=V/R
but we have 2 different corrections to the voltage… the duty cycle reduces the battery voltage by the duty cycle %, so when the motor isn’t turning, we get:
I=(V*duty%)/R
next when the motor starts turning, there’s a further correction to the voltage due to the back emf generated by the moving magnets which subtracts from the applied voltage giving:
I=((V*duty%)-BEMF_V)/R
where BEMF_V = motor_rpm/kv
To calculate the electrical wattage, it isn’t the battery voltage times the motor current, but the reduced voltage to the motor factoring the duty cycle (but not the bemf) times the motor current, so the electrical wattage is the battery voltage times duty cycle times motor current (in bldc):
(Bat_V * duty%) * motor_current = electrical_wattage
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I totally get this.
that said, motor current is not not completely independent of battery current because once you move to higher rpm you are battery amp limited. which is still in these equations, i was just trying to illustrate my initial confusion that it wasn’t there and subsequent understanding that it is.
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which gets us back to this… with the 18s option, there’s no good reason to lower the motor current setting, because at low speeds you can do 30a motor per motor without exceeding 20a battery per motor, so the 18s system gets 50% more torque at the wheel at low speeds when geared for the same top speed, without using any extra motor current, without exceeding the lower battery current limit, and without generating any extra waste heat.
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in a 6374 or higher package what would be the optimal KV for the current lot of available esk8 motors for 18S to not mess up most ESCs (ubox, 100D, etc) or motors?
I’d just pick a kv that keeps the motor under around 10000rpm to avoid losses from spinning it too fast, and then gear it for your favorite top speed and choose the lowest resistance you can fit and afford.
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18s? that’s awfully high i thought 12s was reasonable. if doing regen its going to have voltage spikes and need to be lower than if not.
no one is getting anywhere close to losses from the motor spinning being substantial enough to be a limit unless youre doing some weird extreme gearing. even at 10,000 rpm top speed, depending how you ride, the iron losses arent going to overtake the copper but likely the motor or maybe the bearings arent designed for such speed and maybe fly apart. spin it up to 10,000 rpm on the bench and see what the losses are if you can safely
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18S have withe Highe KV moto
More Torqe
More efficiency withe Low resistance moto (you limit the amps withe you smart VESC)
More Speed 
You can Lower Amps fore the same Power.
-You have better stable battery Volt fore stable Power all situations.
-Longer Battery Life.
Wouldn’t you need taller gearing to get more torque out of a hv setup using high kv motors? increasing voltage alone dosnt net you more torque by itself right?
as I understand it:
10kv
1:1 gearing
Max 30mph (no load) too speed
Vs
100kv
10:1
30mph
They will have the same torque at the wheel for any given amps based on the kv and gearing.
The low kv slower system will have greater resistive/copper loss, be more inefficient, get hotter (possibly too hot), less range.
So ideally u spin as fast as you can for best efficiency…until you finally hit a balance with iron losses (eddy n hysteresis).
Once u get beyond I think like 80,000erpm (vesc4.12 I think only good to 60k) losses…
http://vedder.se/2014/10/chosing-the-right-bldc-motor-and-battery-setup-for-an-electric-skateboard/
But I wish he explained why he chose 60k erpm and I’ve heard the same but it is more complex than how the iron losses were produced up till that speed:
https://cdn.shopify.com/s/files/1/0076/7098/8859/files/8072_new_chart.pdf?159
But then again I see u can easily get to 8000rpm on a board:
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You have withe all Motos More Torqe withe more Volts all times.
This seems to run on PWM and do insane speeds only limited by its ability to not fall apart. Maybe needs sensors. So fast w a motor with magnets and would have big eddy current brake.
: simple
Vs complex:
High speed no magnets. Maybe most efficient at high speed but inefficient at high torque.
I heard @longhairedboy was making a switched reluctance motor board for someone and it used a 3ft diameter pulley and jackshaft for ultimate efficiency. i think it was going to do like 800k rpm or something. that guy is always pushing the limits.
Vesc loses tracking because the FOC isn’t implemented right.
It seems like the most efficient system is the one with the most aggressive gearing and an erpm that doesnt cause your control loops to become unstable.
This is why people like inrunners, low pole count and you can get the winding resistance really low.
A motor’s figure of merit is determined by Kv times the square root of its line-to-neutral winding resistance. Lower is better.
However, higher kv motors have a higher saturation current.
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You can this Limitet withe VESC i se no problem.
I think in the future you can limit the AMPS over RPM this is best way handle this.
Higher the gear ratio you increase the losses within the gears same with more gears stages. The most efficient system is a balance and a compromise.
Larger motors=larger bearings=larger surface area witch adds friction there always a point that that the benefits out way the cost it’s a case of finding that balance.
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Every Mech E tells me that real gear losses are closer to 2% per stage rather than 10%.
Since copper losses dominate just about everything else, this break even point is much more on the motor than on the transmission.
Especially since most boards need high torque at low speeds rather than the opposite. Copper losses in Watts are proportional to the square of requested torque.
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