still the question is where to cut the line between 12S and reducing gearing ratio?
100kv motor a - 900rpm: (10v battery - 9v bemf) / 0.1ohm = 10a motor current
100kv motor b - 9900rpm: (100v battery - 99v bemf) / 0.1ohm = 10a motor current
same motor current so both have the same torque, same copper loss & same ohmic heating. motor b has 11 times greater angular speed than motor a, and the same torque so motor b is producing 11 times as much mechanical power as motor a, but copper loss is the same. which is more efficient?
what about more motor kV? beside gearing ratio?
this problem is still questionable for me
For a given motor geometry with the same amount of copper used in the winding, the motor performance will be the same no matter how you wind it, more turns of thinner wires or less with thick ones, the only thing that changes is that the maximum efficiency and peak torque occur at different voltage/current combination
This is due to the max current density( A/mm^2) being the same for different windings, more than that and you start to saturate the laminations and the efficiency drops a lot
What you need to do is to pair the winding with the available voltage you have for driving the motor, this way you can keep the flux at max (max torque production)
how to determine this?
see, i can easily go 12S4P but why should i if i can do a 10S5P with a higher kV and get the same result in speed… thats what i am thinking right now
I’m just gonna say…
12s4p… 48 cells, was averaging 21 miles of range on a 30q
Pshaw Evo, 13s4p, 52 cells, averages 27 miles, a third more milage almost, for 4 more cells…
My Evo 13s5p, 65 cells, - only 17 more cells not half more, I average 43 miles a charge… double the range…
Both 190kv 6374
Math that…and tell me higher voltage is less efficient. I’ll wait patiently lol.
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I just did some research and i came to this conclusion:
I read all time about the fact that motors with a higher kv have lower torque, which is definitely wrong. So here are some facts about these kind of motors:
kv means RPM per Volt, which we all know. the reciprocal is km, were km = 9.55/kv, unit Nm/A.
km is the unit that determines how much torque (Nm) per 1A the motor produces. motors with less kv do produce more Nm per 1A than motors with higher kv
which means: motors with high kv need more amps to produce the same amount of torque
this is why high kv motors need more amps than motors with less kv. this is for every brushless motor regardless of the power output - 10W or 10kW. a motor with more poles does not have higher torque than a motor with less poles (same voltage, rpm and amp)
example:
Kontronik Pyro 700-45, amp consumption 13A, 10 Poles, 0,3Nm approx. 16000 rpm. outrunner
Lehner 2230/40, amp consumption 13A, 2 Poles, 0,3Nm approx. 18000 rpm. inrunner
used with 10S 36V nominal
Torque cant be compared with the count of poles, torque originates from the air gap between rotor and stator. the distance for the leverage condition between rotor and stator is greater when it comes to outrunners which means more torque. (in case of winding construction)
straight to @Deckoz
i never said high voltage is less efficient - the question hereby is:
Maxing out Voltage VS. Reducing gearing ratio VS. upping kv - the most efficient of those combinations when it comes to: Speed - Range - Torque
see, if i would go for 12S, and want 50 kph topspeed: should i lower my kv and get a higher gearing ratio OR should i up my kv and lower my gearing reduction.
this calculations should be done with the maximum providable power of the vesc or the battery…
what you write sounds good but it’s kv and kt are reciprocal. km shows the relationship of torque to heat which would show what motor is more efficient. as @Pedrodemio was saying and you have to pair the battery with the kv. you don’t want a no-load speed that’s far off in the distance that you don’t go to…then you’ll be inefficient. why exactly?..and is it inefficiency in the motor or just the esc in such a situation where youre forced into less than highest duty cycle.
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Km is wrong in the concept, it is torque for a given current divided by the square root of the current for this given torque squared times winding resistance, in other words Km = T/((R*I^2)^1/2)
Comparing motors using Kt or Kv is deceiving. Yes, you increase the the torque for a given current, but not for the absolute torque the motor can produce
That’s called the Ni constancy, if you increase the number of turns, you use thinner wires, so the absolute current must go down, the works the other way, if you lower the number of turns you get a higher Kv but the wire must get thicker to maintain the slot fill, so you can pump more current and get the same torque, you can’t change N without changing i
Back to Km, the formal definition is air gap flux density times rotor inside radius ( for outrunners) times square root of wire volume in a slot all divided by square root of resistivity, all this is independent of number of turns, Kv or Kt
A given motor geometry with the same cooper fill will have the same performance
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km is the same:
100kv - 10 turns - 1 cross section - 0.1ohm 60/(2 * pi * 100kv)=0.09549296585513720146133 newton meters torque per motor amp (4v battery - 0v bemf) / 0.1ohm = 40a motor & battery current at 0rpm 4v * 40a = 160w electrical 40a * 0.09549296585513720146133nm/a = 3.819718634205488058453nm 3.819718634205488058453nm / sqrt(160w electrical) = 0.3019752726269222102173km
200kv - 5 turns - 2 cross section - 0.025ohm 60/(2 * pi * 200kv)=0.04774648292756860073067 newton meters torque per motor amp (4v battery - 0v bemf) / 0.025 ohm = 160a motor & battery current at 0rpm 4v * 160a = 640w electrical 160a * 0.04774648292756860073067nm/a = 7.639437268410976116907nm 7.639437268410976116907nm / sqrt(640w electrical) = 0.3019752726269222102173km
^km (torque per square root of watt) is the same
1:1 gearing - 10mph ground speed 100kv - 10 turns - 1 cross section - 0.1ohm 1000rpm = 104.719755 radians per second (11v battery - 10v bemf) / 0.1ohm = 10a motor/battery current torque per amp: 60/(2 * pi * KV) = 0.095492nm/a 10a * 0.095492nm/a = 0.95492newton meters torque 11v battery * 10a current = 110w electrical 0.95492newton meters torque * 104.719755 radians per second = 99.99w mechanical copper loss: 10a * 10a * 0.1ohm = 10w wheel torque at 10mph @ 110w electrical = 0.95492nm * 1 gear reduction = 0.95 newton meters summary: 10mph ground speed, 0.95nm wheel torque, 99.99w mechanical, 110w electrical, 10w copper loss
2.150:1 gearing - 10mph ground speed 200kv - 5 turns - 2 cross section - 0.025ohm 2150rpm = 225.14747 radians per second (11v battery - 10.75v bemf) / 0.025ohm = 10a motor/battery current torque per amp: 60/(2 * pi * KV) = 0.047746nm/a 10a * 0.047746nm/a = 0.47746 newton meters torque 11v battery * 10a current = 110w electrical 0.47746 newton meters torque * 225.14747 radians per second = 107.49w mechanical copper loss: 10a * 10a * 0.025ohm = 2.5w wheel torque at 10mph @ 110w electrical = 0.47746nm * 2.15 gear reduction = 1.02 newton meters summary: 10mph ground speed, 1.02nm wheel torque, 107.5w mechanical, 110w electrical, 2.5w copper loss
^same ground speed, same voltage battery, same battery current and motor current, different gearing & the higher kv appears more efficient… it would seem if the 100kv is at constant speed @ 10mph from the load, with the same ground speed & load the 200kv is still accelerating because there is more torque at the wheel…
yeah i meant kt, had to translate from german, my bad.
@professor_shartsis yep, that is now the calculation for different amounts of kv, it seems that staying with higher kv would be the better idea.
but what about the difference between:
Therefor given:
10S5P = Cont. Discharge approx. 100 amp 12S4P = Cont. Discharge approx. 80 amps
staying at low kv and adding more voltage (+2S) (higher current delivery without losses to heat)
VS.
staying at 10S and adding more kv (higher max. discharge current, but losses to heat e.g.)
since it is clear the continuous discharge current will never be over 60 amps, but what about the maximum current provided when accelerating with:
a) higher kv, lower voltage, greater amp supply from the battery b) lower kv, higher voltage, lesser amp supply from the battery
when it comes to efficiency?
it seems in my eyes that staying at 190kv, going 12S and 36/15 gearing seems the best and efficient scenario or does anyone recommend another option ?
But is it an inefficiency in the motor w a poorly paired kv n voltage and if so why?
I’ve seenthe same motor w different windings and kv having different stated lower output with the lowest kv stated as least powerful and it seems when u add many many turns, and are trading speed for torque output you lose some of the max potential power. So have the same copper and iron but lesser performance. Just how a hub motor has to be bigger for same power compared to the motor w a pulley. I think only way to more efficiency as compared to what u last wrote would be to lower the top speed. Or reduce performance by decreasing amp output to the motor. Probably the motor amps. Why are motor amps considered more inefficient than battery amps,or are they not?
the target is to reach 50 kph with the most efficient setup, not to be more efficient driving at slower speeds.
it doesnt seem that you lose torque while going to 12S, while you will definitely lose torque with higher kv if not delivering more amps to your motor - which can be limited by your battery settings.
well, why should i power a motor with 60a if my battery only can deliver 40 continuously?
slowly it seems i can’t really get a correct answer here in this thread either if i should go with 12S or adding kv.
60a “motor amps” continuosly to the motor draws varying amounts of “battery amps” depending on the present angular speed of the motor… less “battery amps” for 60a motor amps at low rpm and more battery amps at higher rpm for the same constant 60a “motor” amps… 60a motor amps will draw nearly 60a battery amps when you are very close to the peak power output of your motor when considering the motor amp limit you’ve selected.
also, my earlier equations showing higher efficiency with higher kv assumes 2 things… that the “km” is identical with both kvs, and the gearing is modified with the higher kv… if you had higher kv but lower km, the higher kv might be less efficient, even if you change the gearing.
so going with 12S?
More eff if uadjust top speed w gearing to what it was at 10s. But assuming u dowqnt that top speed, if u gear it more for lower top speed even more eff
i just calc’d it, i need about 32 teeth wheel pulley and about 15 -16 teeth motor pulley… this sounds to me like a big torque loss if i only could add up 2S more for more rpm at higher speeds
a motor running with 10S instead of 12S has roughly 80% as much peak mechanical power and rpm with the same motor amp (torque) limit… so onto my question…
at 10s how can he reach the same top speed with a motor delivering only ~80% as much peak mechanical power and rpm but the same amount of copper loss and torque?
gearing can increase torque at the proportional expense of rpm, or increase rpm at the proportional expense of torque. mechanical power is torque in newton meters * angular speed in radians per second. changing the gearing doesn’t change the peak mechanical power output of the motor, unlike changing the voltage or motor amp limit.
Yea. Les power so less speed.
12s bro no bs. don’t understand why this is even a discussion!
ride a 10s set up and ride a 12s setup…
10s is for ladies who cant handle the power 
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