Kv * turns = constant?

ProxRB

10 mW
Joined
Jun 24, 2015
Messages
22
Quick question, for a given motor geometry, is Kv proportional to 1/turns?

Corollary to that question, for a given motor geometry since motor_internal_resistance = constant * turns^2, motor_internal_resistance = constant / Kv^2. Max_motor_current = constant * Max_motor_power = constant * motor_internal_resistance = constant / Kv^2. So there is a HUGE penalty to use unnecessarily low Kv motors?
 
First, I'll define my variables just for clarity's sake.
Kv: rpm/volt
Kt: torque/amp
Rm: motor phase resistance
P: power
I: current
V: voltage
w: angular velocity
T: torque

Yes Kv is proportional to the inverse of turns. A 200Kv, 4 turn motor will be 100Kv if wound with 8 turns.

Rm is proportional to turns^2, so yes Rm is proportional to 1/Kv^2.

The problem with your next statement is that P=I^2*R, so I=sqrt(P/R), so I is proportional to 1/sqrt(R) and also proportional to Kv.

The other thing you may be missing is that the goal is motor torque/power not motor current. We can make power with volts or amps it doesn't matter. Kv is proportional to the inverse of Kt, so if we double Kv to get more current handling we halve Kt which means we end up with the same amount of torque.

tl;dr Changing Kv doesn't change any of the performance characteristics of your motor, it only changes what combination of P=IV you use to get to that same performance.
 
What if we also double the gear ratio in addition to doubling the kv? We get the same speed, torque, and power for the same voltage and current input. But we also have 1/4 the resistance, so effectively double the current capability, and double the max torque?
 
Short answer - no. When you double the the KV you also get half the KT and get half of the torque for the same current. However, when doubling the reduction ratio you will get only half the heat.

Long answer - read this :twisted: :
https://endless-sphere.com/forums/viewtopic.php?f=2&t=64907

Avner.
 
Wow that thread is gigantic! I'm having a lot of trouble following the discussion. From what I can tell, liveforphysics agrees with me?

Let's use pronghorn's notation and speak in math terms. Sorry, it is my weakness here, but I think I have difficulty understanding the English. Equations though, I can understand them.

Adding one variable to the notation:
Q_max = maximum allowable rate of heat generation, which is a function of the max temp of winding insulators, the de-magnetization temperature of the magnets, the surface area of the stator over which to dissipate the heat, the color of the stator for radiative cooling, the desired service life of the bike etc etc. Practically speaking, this is an unknown value, but constant for a given make/model of motor regardless of how it is wound.

Starting from these relationships
1) Q_max = I_max^2 * R_m
2) P_max_mech = w_max * T_max
3) w_max = Kv * V_battery
4) T_max = Kt * I_max
5) R_m = constant / Kv^2 (assuming constant copper fill)


Now algebra:
I_max = sqrt(Q_max / R_m)

P_max_mech = Kv * V_battery * Kt * I_max
P_max_mech = V_battery * I_max
P_max_mech = V_battery * sqrt(Q_max / R_max)
P_max_mech = V_battery * sqrt(Q_max / (c / Kv^2 ))
P_max_mech = Kv * V_battery * sqrt(Q_max / c)

So in general, to maximize the system's power, pick the highest Kv you practically can. At some point you wont be able to fit the gearing to convert the power from high w, low T to a more useful low w, high T at the wheels, so you can't pick an ARBITRARILY high Kv. But as a general rule of thumb, you can err on the high side I think.
 
Every Turn Count for the same motor has the exact same power, torque, speed.
Some sellers promote their same motors as Speed or Torque, THAT IS THE MYTH!

Lower Turn Count motors have a Higher RPM/VOLT, which means they go faster for the same voltage, say 60V.
Higher Turn Count motors have a LOWER rpm/volt, which means they go slower for the same voltage, say 60V.

Lower Turn Count motors have larger (and shorter) winding wire which can handle more amps for a longer time.
That winding wire has LESS resistance.

Higher Turn Count motors have thinner (and longer) winding wire which handles WAY LESS amps, or can handle the same amps as a Lower Turn Count motor, but for WAY LESS time.
That winding wire has MORE resistance.

It is all a balancing act, like a teeter-totter. You can play around with the Motor Simulator in the Tools section and select two examples of a low and high Turn Count motor in the MXUS category say 4503 vs 4506, that is a 45H motor with 3T vs 6T.

This link does another motor as an example.
https://endless-sphere.com/forums/viewtopic.php?f=2&t=64907&p=985052&hilit=justin#p984783
 
Markz, ProxRB is not messing up the turn count thing. He's just recognizing that spinning faster and gearing down to the same speed results in more torque capability.

ProxRB, you don't necessarily need to increase Kv to increase speed. You could also increase your battery voltage. Besides being able to fit gearing, motors also have speed dependent losses, so there are limits to how fast a motor can spin. Given a certain heat rejection ability for a motor, you want to balance these speed based losses and torque based (copper resistance) losses.

This thread here has some good info:
https://endless-sphere.com/forums/viewtopic.php?f=30&t=16376

I would start reading in the Motor Technology section for more info on this stuff.
 
thepronghorn said:
This thread here has some good info:
https://endless-sphere.com/forums/viewtopic.php?f=30&t=16376

I am not commenting on any ones post, it was more a refresher for myself. I know there is more info to add to my MYTH BUSTER statement in my previous post. However, as stated: Like going to a smaller wheel diameter gives you more torque, and less heat, and that torque which is instant from zero on up. But that also slows the speed down. Again its all a balancing act to what are our goals for our eeeerides. I know for a low turn count motor in 26" takes a longer time to get up to speed then having more torque by the gearing of going to smaller wheel. Some like the feel and response of lots of torque, for traffic riding, or hill climbing.

There is another angle to this all, its what components do I have right now on hand, what do I need to buy, what terrain will I be riding most of the time. This all impacts what we are going to build. Best not to dive into it without knowing.

This is all neither here nor there just take it as ramblings or a refresher.
 
ProxRB said:
So in general, to maximize the system's power, pick the highest Kv you practically can. At some point you wont be able to fit the gearing to convert the power from high w, low T to a more useful low w, high T at the wheels

You also need to consider the inefficiency of gearing. Real-world (suboptimal) gearing is approximately 90% efficient (10% loss) per stage and the maximum practical reduction you can achieve per stage is likely to be around only 5:1 to 10:1.
 
Punx0r said:
ProxRB said:
So in general, to maximize the system's power, pick the highest Kv you practically can. At some point you wont be able to fit the gearing to convert the power from high w, low T to a more useful low w, high T at the wheels

You also need to consider the inefficiency of gearing. Real-world (suboptimal) gearing is approximately 90% efficient (10% loss) per stage and the maximum practical reduction you can achieve per stage is likely to be around only 5:1 to 10:1.

It's also dependent upon whether the motor is already run at it's rpm limits. eg with the different Kv winds available with Astro's motor models, their test reports all show an identical power rating for a given model motor regardless of the Kv of the wind. It's really only with DD hubmotors that higher Kv makes the motor capable of more power, which is due to 2 factors:
1. DD hubmotors aren't run anywhere near their physical rpm limits.
2. We have practical limits on voltage due controller availability.

As long as you're flexible in system voltage, then the Kv of a specific motor isn't important. The power of a motor is determined by it's steel and magnetic circuit along with heat dissipation. The amount of copper fit into the windings is what's important, not how many turns are wrapped around the teeth. Assuming the same copper fill, the Kv only determines the combination of voltage and current to get exactly the same result with another identical motor wound to a different Kv.
 
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