Motor theory... and practice.

[Olle P]

Investigating what’s behind the “Old Truths” and general advice regarding our typical use of electric motors.

This apply to brushed and brushless motors alike, although for brushed motors I’m aware that some of the mathematical models used only work well within a narrower voltage range. The general conclusions still hold though.
I’ve stuck to dissecting the influence of winding turns and feed voltage, since those are the primary factors that we as users can influence when choosing motor and battery. The rest of the internal motor hardware is simply assumed to be “constant”.

The article is very wordy and there are many formulas presented. I’ve broken this subject into one post for each motor property. Each post, apart from the last ones, is subdivided into a short presentation, Theory, Reality Check and Conclusion.

Contents:

Title                           Post #
Introduction, Legend               1 (This post.)
Motor data explained               2
Kv rating                          3
Torque                             4
Power                              5
Heat and Power efficiency, part 1  6
Heat and Power efficiency, part 2  7
Conclusions                        8

Introduction:
I’ve started this thread as a spin-off from another thread.

Originally Posted by JohnRobHolmes

Once inside the motor coils, the respective voltage and amperage that power is produced from doesn't matter if we keep the amp/turn and copper volume constant. The torque and power can't be changed with a different KV, low KV motors do not produce more torque. They do produce more torque per amp, but they also need higher voltage applied to get it. ...

In theory (and what seems like technically possible) John is correct, as shown below.

My previous statement, to the effect of “More turns => More torque, more efficiency and less Kv” stems from what’s generally accepted as the “truth” and also on the fact that the general (and good) advice is to gear down when switching to a higher Kv motor. I blame only myself for not previously having done the proper research…

Now I’ve spent some time going back to the basic physics of electric motors and compare that to reality to find out what is true.
The physics is simply an applied use of Ohm’s law, Faraday’s law, Kirchhoff’s voltage law and some other basic rules of electronics and mechanics.
For the mathematical models I’ve throughout assumed that all pieces of the motors except the windings are identical and constant. The amount of copper used for the windings is also constant (unless otherwise explicably noted).
“Reality” is represented by well proven general advice and experience as well as manufacturer’s data for the somewhat older range of Turnigy TrackStar motors. I used these motors because of the relatively exhaustive amount of data given and I run the 17.5T version myself. The full range isn’t directly comparable without caution since the motors up to 8.5T have an internal fan that the others don’t and some data lack precision. The 4.5T version can be found here.
I’m aware it’s not sufficient to have a select very few motors to make perfect conclusions set in stone, but it’s far better than having only the theories.
Some results are intriguing, to say the least…

Legend
Physics and maths go hand in hand when doing the theoretical study, so here is a legend for terms used later. Notice that there’s a difference between upper and lower case letters:
N = Number of winding turns.
R = Electrical resistance in the winding. Unit: Ω
f = Motor angular velocity, used as a variable. Unit: rad/s
f0 = Angular velocity at no load for a given voltage.
Kv = f0*π / 30*voltage. Unit: rpm/V (For brushed motors this is more voltage dependent than for BL motors. I use it as a constant though.)
V = Feed voltage, when used as a variable. Unit: V
V0 = The feed voltage, a constant.
VE = Electromotive Force. Unit: V
I = The electrical current through the motor. A variable. Unit: A
I0 = The no load current at a given voltage.
T = The torque. Unit: Nm
Ts = The stall torque.
T0 = The torque required to overcome motor friction.
P = Power. Can be electrical or mechanical. Unit: W
Pin = Electrical power drawn by the motor.
Pout = Mechanical power given off the motor axle.
Ph = Power generated as heat within the motor due to winding resistance.
Pm = Power generated as heat within the motor due to mechanical friction.
L = Load. A value from 0 to 1, often given as a percentage.
E = Power efficiency.
k[index] = used for various constants depending on the motor design

[Olle P]

Motor data explained

There are eight motors (older generation TrackStar) used for the “reality checks”.
These have the windings 3.5T, 4.5T, 5.5T, 6.5T, 8.5T, 10.5T, 13.5T and 17.5T.
The motors up to and including 8.5T have integral fans mounted on the rotors, which can be expected to add some loss of output torque. The results from this study show that in practice the fans have very little overall influence though.

What data is presented?

• These are two-pole sensored brushless inrunners of 540-size. Good to know but mostly irrelevant.
• Winding turns: Not much to say about that.
• Kv-value: An essential piece of data.
• Max voltage: Good to know.
• Maximum current: Not how much the motor is willing to draw, but how much it can handle at a constant load without overheating at some unspecified level of cooling. ... or so I thought...
• Maximum power: Not how much power the motor can deliver but the power drawn by the motor while fed the “maximum current” at the “maximum voltage”.
• Winding resistance: Required for the mathematical models used.
• No load current: My analysis indicates that it is valid for a 7.4V feed, not the maximum voltage. Also essential for some of the mathematical models.
• Physical dimensions: Irrelevant for this discussion.

I think it’s a shame that the newer motor designs are not presented with this much data!

[Olle P]

First property: Kv rating

The “old truth” is that “when the number of turns go down the Kv goes up”. Is it true?

Motor torque (required to overcome any internal friction and accelerate the rotor) is produced by the current running through the coils. The current is driven by the feed voltage and limited by the winding resistance. If these were the only factors involved the current would be constant and very high, with the motor accelerating until it broke down. Another factor is also limiting the current and this is the Electromotive Force (EMF) (a voltage).

Theory:
The current is thus defined by (V – VE)/R.
Faraday’s law says that the EMF (at a given motor speed) is proportional to the number of turns.
VE = kF*N*f
At no load f = f0 = Kv*V0*π/30
VE = kF*N*Kv*V0*π/30 =>
Kv = VE*30 / (kF*N*V0*π)
Set VE ≈ V0 so VE/V0 ≈ 1 and we get Kv ≈ 30/kF*N*π
It’s obvious the theory states that if N goes down Kv goes up.

Reality Check:
With lower N-value the Kv does go up.
The theory says 30/(π*Kv*N) should be constant (kF). For the actual motor range it’s close, with kF = 0.000286±0.000013.

Conclusion: The old truth holds true!

[Olle P]

Second property: Torque
Old truth: ”More winding turns gives more torque.”
Standard, well proven, advice: “When changing to a motor with less turns, gear down [to reduce the required torque].”

Theory:
The (total) torque depends on the current and the number of turns.
T = kT*N*I
Maximum current (and thus torque) is given at stall, when EMF=0V.
Then Ts = kT*N*V0/R
With the assumption of “constant copper” follows that if you double N the wire cross section is halved (doubling R) and the length of wire doubled (doubling R once more).
Thus R = kR*N^2
Giving
Ts = (kT*N*V0) / (kR*N^2) = (kT*V0) / (kR*N)
Look at that! Torque is proportional to 1/N. Say what? That’s the exact opposite of the old truth!
There must be something more to this…

So what happens if the motor manufacturers cheap out and use the same wire for all winds?
Then R is proportionate to N (wire length) so
Ts = (kT*N*V0) / (kR*N) = kT*V0/kR
Stall torque becomes a constant relative to N. It’s still not enough to verify, or even to explain, the old truth.

Reality Check:
Is the wire gauge constant or optimized?
In the motor range R correlates much better to N^2 (0.200±0.035) than it does to N (increase with N), so it’s optimized using available standard gauges.
From that follows that for a given voltage and load percentage a low turn motor will provide more torque than a high turn motor, opposing the old truth.

Intermediate Conclusion:
One conclusion is that high turn motors provide more torque at the same current, not at the same voltage. They need more voltage to reach that same current.
Higher torque (with low turn motors) allows for higher gearing, so one should expect broad recommendations to gear up when changing to a motor with fewer turns but instead the advice is to gear down. WHY?
I will return to this issue later on, since I think it’s related to the following properties…

[Olle P]

Third property: Power
The “old truth” state that “with a decrease in turns the power goes up”.

Theory:
Mechanical power: P = T*f
I’ve already proven that Kv increase with lowered N, and shown that Ts also seems to increase with lowered N.

Neglecting the internal friction(*) to make the mathematical formulas cleaner:
For a given load L
T = Ts*L = (kT*V0*L) / (kR*N)
and
f = f0*(1-L) where f0 = kF*V0 / N
Thus
P(out) = (kT*V0/N)*L * (kF*V0/N)*(1-L) =
= kT*kF*V0^2*L*(1-L) / N^2

From this we can read that:
P is proportional to 1/N^2.
P is proportional to L*(1-L). (It’s 0 when L is 0 or 1 and there’s a maximum when L = 0.5.)
P is proportional to V0^2.

Reality Check:
For a car speed is a function of power. It’s well known that R/C race cars go faster if they get a motor with less turns.
It’s also well known that they go faster with higher voltage.

Conclusion:
The old truth holds true!

(*) The power lost due to internal friction is typically magnitudes smaller than the output power.

[Olle P]

Fourth and fifth properties: Heat and Power Efficiency, part 1
The “old truth” state that “with a decrease in turns the power efficiency goes down”.
John Holmes use to say “Volt up, gear down” [to get less heat (and more efficiency?)].

Theory:
When running a motor of any kind the power fed to the motor (Pin) exceeds the mechanical power delivered by the motor (Pout). The difference comes out mostly as heat. (Combustion engines also have other losses.)
The relation between output and input is called Power efficiency (E) and is defined as E = Pout/Pin

The heat generated is something we generally don’t want but also can’t avoid.
From the definition of efficiency follows that for a given power out you get less heat if the efficiency is better.

The power “lost” in an electric motor is two-fold:

• Heat as a result of wire resistance.
  Ph = R*I^2
• Heat as a result of friction.
  Pm = T0*f

We know also that Pin = Pout+Pm+Ph

At no load:
Pin = V0*I0
Pout = 0W
Ph = R*I0^2
Pm = T0*f0 = kT*N*I0*f0

Put together: V0*I0 = R*I0^2 + kT*N*I0*f0
(Divide by I0.) V0 = R*I0 + kT*N*f0
V0-R*I0 = kT*N*f0
We know from post 2 that V0-R*I0 = VE at no load and that
VE = kF*N*f0, so now we get
kF*N*f0 = kT*N*f0, which is true only if kF ≡ kT
Let’s verify by dimensional analysis:
kF = VE / (N*f) corresponding to [V]/([1]*([1]/[s]) = [Vs]
kT = T / (N*I) corresponding to [Nm]/([1]*[A]) = [Nm/A]
At first glance [Vs] ≠ [Nm/A] but then I realize that
1Nm=1J=1Ws=1VAs so 1Vs = 1Nm/A and therefore kF ≡ kT.

Now we can also calculate T0:
V0*I0 = R*I0^2 + T0*f0
T0 = (I0*(V0-R*I0))/f0

Reality check #1:
(From here on the “reality” gets a bit shady, since it’s all a matter of applying the mathematical models on the given motor data, no actual measurements to verify that the models are correct.)
What’s the calculated value of T0 for the motors?
T0 = 16±6 mNm
Quite a spread, relatively, but the input data (especially I0) isn’t very high precision either…

Reality check #2:
What’s the heat generated in the windings?
At 7.4V and no load:
R*I0^2 = 0.36±0.15 W, so quite a spread, again probably due to poor precision in input data.
At maximum rated current:
R * Imax^2 = 15.4±1.0 W, Principally the same all over the range.
At 7.4V and stall:
R*Is^2 = V0^2/R = V0^2 / kR*N^2
For the 3.5T motor the heat generated at stall is a whopping 26.08 kW while for the 17.5T motor it’s a more moderate 1.07 kW.

Reality check #3:
What’s the mechanical heat generated at no load?
We’ve already stated that T0 is roughly the same for all motors whereas f0 is the inverse function of N, so lower turns means more heat due to mechanical friction.
At 7.4V for the 3.5T motor it’s 95.8 W and for the 17.5T motor it’s just 14.6W.
With load added the speed goes down and with that the internal heat generated by the friction (assuming the friction itself stays the same).

Intermediate conclusion:
At a constant voltage you get more heat from a motor with less turns. No surprise there…

Power efficiency
The heat goes up with less turns, but we already know that so does the output power as well.
Power efficiency (E) is the relationship between output and input power.
E = Pout / Pin
Pin = V0*I
Pout = T*f
I = (Is-I0)L + I0
T = (Ts-T0)*L
f = f0*(1-L)
Ts = Is*N*kT
T0 = I0*N*kT
f0 = V0/(N*kF)
kF = kT
E = (Ts-T0)*L*f0*(1-L) / (V0*((Is-I0)*L+I0)) = V0*N*kT*(Is-I0)*L*(1-L) / (V0*N*kF*((Is-I0)*L+I0)) =
= (Is-I0)*L*(1-L) / ((Is-I0)*L+I0)
Power efficiency doesn’t seem directly related to N although N does have some influence on I0.
Solving E’(L) = 0 says E reaches maximum when L = ((Is*I0)^½-I0) / (Is-I0)

Reality Check: #4
How does L for maximum E vary with N?
Optimum L goes up with N, from 5.73% for 3.5T to 10.82% for 13.5T. (The 17.5T motor as well as the 4.5T motor deviate a little from the trend.)
Looking at the numbers reveals a good correlation between optimum L and the square root of N.

So what’s the efficiency at optimum load?
At optimum load the efficiency is also somewhat dependent on N. Most efficient is the 3.5T motor at 88.6% and least efficient the 13.5T motor at 78.3%.
Higher turn motor is NOT more efficient!

Intermediate Conclusions:
It’s quite obvious that higher turn (low Kv) motors are not more efficient as such!
So let’s explore further to see where the “old truth” comes from…

Theoretical models applied to the motors:
Comparing the motors at 7.4V feed and the maximum rated current.
The 17.5T motor operates at its maximum efficiency. 10.5%load, 79% efficiency, delivering 99.4W power on the axle and 26.4W heat.
The 3.5T motor at the other end of the scale is far from its optimum load, at 2.0% load instead of 5.7%. Still the efficiency is 83%, the power on the axle is 508W and the heat generated is 107W.
This tells me the maximum current rating has little to do with the heat generated! (Actually, more current does mean more heat, but my point is that the “extra” heat doesn’t come from the winding resistance, and there’s also a difference in output while generating that heat. The value for maximum current doesn’t seem to correlate to the risk of motor overheating though.)

[Olle P]

Fourth and fifth properties: Heat and Power Efficiency, part 2
This far I’ve compared the circumstances with same voltage and same load.
What about same output power? The power is what we want and use, so how does different motors behave while delivering the same power?

Case 1: Same maximum power, delivered at 50% load.
With the 17.5T motor running at 14.8V and 50% load it will deliver as much as 1,038W on the axle (slightly less power than the heat generated).
The motors with less windings need much less voltage to deliver the same power at 50% load. The 3.5T motor needs only 2.96V and will consume 73W less from the ESC while doing it!
Comparing the currents we find the 17.5T motor requires 146A while the 3.5T will need a full 707A.
Comparing the torque delivered at 50% load on these voltages we find it to be roughly the same across the spectrum, which isn’t surprising since the motors are also running at the same speed.
From an efficiency point of view it’s therefore in theory slightly better, or at least just as good, to use a motor with fewer turns running the same gearing on lower voltage.

Compare model to what’s feasible:
In real life there are some limiting factors that come into play.
Delivering 146A at 14.8V is hard, but not impossible using a 4S LiPo able to deliver 70C or more peak. There are ESCs able to handle it as well.
707A at 3V is much worse. Not many ESCs can handle that current, and it takes a battery pack typically of 1S 10Ah with 70C peak rating to do it (and matching cables and connectors).

Case 2: Same stall torque.
How much heat is developed at (the same) stall torque?
Ts = kT*N*Is so Is = Ts/(kT*N)
Pstall = R*Is^2 = kR*N^2*Ts^2 / (kT*N)^2 = kR*Ts^2 / kT^2
The power (heat) at stall is independent of N, varies with actual torque (squared).

Some more theory:
Let’s look at the other end of the envelope. Motors need some voltage above 0 to start in the first place because the (stall) torque needs to overcome the friction at rest.
Assume the required torque is Tx, same for all motors of same type.
Then we know that Tx = kT*N*Istart = kT*N*Vstart/R = (kT*N*Vstart) / (kR*N^2) = (kT*Vstart) / (kR*N) =>
=> Vstart = Tx*kR*N/kT
Vstart is thus proportional to N, so the relative difference between minimum speed and maximum speed stays roughly the same even if the low turn motor is run at a lower voltage.

Volt up, gear down?
Is there some merit to this advice?
Starting from the situation above, with the 17.5T motor fed 7.4V at 10.5% load (maximum rated current). The input is 125.8W and the output is 99.4W.
If we double the voltage while keeping the output power the same the load is reduced to 2.4% (by means of gearing down). Then the current drops from 17A to 10.9A but the input power increase to 161.0W, which means the heat generated is more than doubled!
Why doesn’t it help to volt up and gear down? Because the motor was running at peak efficiency. Gearing down reduced the efficiency and thus increased the power loss (read: heat).
A motor geared to operate at loads above peak efficiency will benefit from gearing down.
The current, which in itself might be the limiting factor, is reduced by using higher voltage while keeping the load constant.
Running RC cars one “issue” is that the load is far from constant. During hard acceleration and (typical for crawlers) when the wheels get stuck in tight spots there’s a very high load.

So to reduce current: Volt up, gear down!
To increase efficiency (reduce heat): Optimize the gearing and voltage, not necessarily by gearing down.

Other observations
The relationship between Is and Imax at a given voltage is much bigger for low turn motors. If we to that add that the heat generated is proportional to U^2/R it’s easy to understand that at high loads a low turn (low resistance) motor will heat up much faster than a high turn (high resistance) motor.

Heat as the limiting factor
How do we maximize the output power while keeping the generated heat at a constant level?
That’s done by maximizing the efficiency!
Ph + Pm = Pin * (1-E)

Theory and reasoning
We’ve seen that low turn motors provide the most efficiency for a given maximum load.
The “cost” for this efficiency comes in the form of a high current and a requirement for reduced voltage.
Low voltage makes it difficult to get a perfect match in the choice of battery.
High current at that low voltage makes the resistance in cables and ESC a concern.

What numbers are we talking about?
Say we want 50W continuous output from the 3.5T motor.
Using the optimum load at 5.7% we need a voltage of 1.40V (equivalent of a single, fully charged, NiMH cell) and the motor will draw 40.5A (56.7W).
The 17.5T motor also providing 50W at optimum load will require 5.29V, 12.1A (64.2W)
The difference in heat is thus considerable, with the 17.5T motor generating more than twice as much heat as the 3.5T motor. (14.2W vs 6.7W)
The difference is that at stall with these voltages is quite different though. The 3.5T motor will draw 933W and the 17.5T motor only 546W, so it’s almost the opposite.
Evaluating further we find that the low turn motors run faster at optimum load (and thus require lower gearing) than the high turn motors while providing the same power. That’s probably where the “higher Kv => less gearing” stem from.

A real life example:
Apart from my crawler I’m also running a drift car.
The typical setup for a drifter is:

• 2S battery.
• 9.5T motor (±1T)
• Gearing close to 8:1

My setup is:

• 4S battery
• 17.5T motor (the TrackStar)
• Gearing 4.25:1.

Reported power consumption with the typical setup is 60W (estimated for driving indoors on carpet).
Power consumption with my setup is 30W (measured while driving outdoor on asphalt).
Why does my setup use roughly half the power of the typical setup when my motor is supposed to be slightly less efficient?
There are a number potential flaws in the comparison:

• One of the consumption levels is just an estimate by the driver, not actually measured.
• The driving conditions are different. I have no idea which surface requires more power.
• The cars and motors are different, but there if anything I think the “typical” car is better than mine, using a more expensive (efficient design?) motor and generally optimized chassis and suspension.
• Driving style differs. I’m a rookie and the other driver is a seasoned veteran.

Can the sum of errors make up for all of the difference? I find it unlikely…
To me one really interesting factor is the difference in gearing. I decided to use the higher gearing based on the assumption that my motor has more torque, but in reality (based on the torque findings above) the difference should be negligible!
My motor should thus see a higher load and my car should behave quite differently than the typical.
I haven’t tried the typical setup, so I can’t comment on any differences, but I have no problems drifting with my setup! (And I rarely need to use more than 1/3 throttle doing it.)
Perhaps my motor is running closer to peak efficiency? But even then the difference should be much smaller. (And the “typical” setup adjusted to use a more efficient gearing.)

[Olle P]

Conclusions
Creating this article has really rocked my world regarding motor properties!
Things that I took for granted have been proven wrong. There were many “eureka moments” along the way, mixed with dead ends and failures (in the model design process).
Do take note that everything below relates to the MOTOR only, unless otherwise explicitly expressed. What happens in ESC and battery is NOT examined!

Here’s a recap of what’s been found:
“When the number of winding turns go down the Kv goes up.”
That’s True!
“When the number of winding turns go down the output power goes up.”
That’s True! (With the voltage kept constant.)
“When the number of turn goes up, so does the torque.”
Wrong for constant voltage! (True for same current though.)
True also if you keep the output power the same and adjust gearing and voltage so that the motor operates near its maximum efficiency.
The maximum efficiency is at a higher load for high turn motors, and therefore it’s better to have them geared a little higher, not because “they have more torque”.
“When the number of turns go up, so does the efficiency.”
Wrong! If anything the efficiency goes down with more turns. (At least while the load is <50%.)
“Volt up, gear down!”
Good advice if the objective is to reduce the motor current. Not good for improving motor efficiency.
To increase efficiency you’re probably better off doing the opposite…

Further findings:

• The no load current is a result of bearing (and other) friction and can therefore not be estimated without actual test data. It can be said that it’s higher for low turn motors though.
• For a given torque the power loss due to wire resistance does NOT depend on the number of turns. Any difference in total power loss is due to difference in speed (for a fixed general design).

Based on these motor findings the general advice should be:

• ALWAYS use very low turn motors. (The lower the better.)
• Adjust voltage and gearing to fit the requirements for power and torque.

It’s obvious that these advice can’t be applied in practice though.

Some limiting factors are:

• The ability of battery and ESC to handle large currents (and/or high switching speeds, in the case of BL motors).
• The ability of ESC and other electronics to use a relatively low voltage (<2V) feed.
• The motor being exposed to peak loads way above “optimum”. (Resulting in very high currents.)
• Not many battery options for very low voltage.

How would this be implemented onto my crawler?
As mentioned I’ve been running the 17.5T motor on 4S in my LCC (a shafty with worm drives) for a while. I had some trouble with the low speed characteristics, but that might have been a timing issue.(*) Otherwise I felt the power and torque to be satisfying.
Adopting the findings above I should:

1. Switch to the 3.5T motor.
2. Use 1S LiPo.
3. Limit the ESC output to 82% (3.0V nominal).
4. Gearing could be reduced a little, but I was running it below optimum just to get the low speed behavior better so no change required.

What problems arise?

• The ESC (standard MMP) can’t handle that low input voltage.
• The Servo and Rx want higher voltage as well.
  For the Rx I can use a voltage booster, but the servo will need a separate feed.
• Finding a suitable sized 1S LiPo will be difficult.

Second option is to use 2S LiPo which will eliminate the problems above but require the ESC output to be limited to 41%. That might in turn provide problems in the ESC as well.
How much current will I need to handle?
At around optimum load there won’t be much difference in current, so that’s not a problem. However, the stall current at 3.0V is 1.4kA for the 3.5T motor, up from 0.3kA of the 17.5T motor. That might be a problem, not only for the MMP but also for the battery. Experience says that the full stall current won’t be used, but 1/3 of it is a reasonable maximum, so 475A from the ESC and 195A from the battery must be manageable.
At least HobbyKing doesn’t have a suitable (2S, ~1.5Ah, 130C peak discharge) battery in their assortment.
At the end there will need to be a compromise to find out a working combination of parts that actually exist …
… and since I’m now using a completely different motor it’s not an actual issue to start with!

Where do I go from here?
From the above findings and conclusions I realize that the next “black hole” in my knowledge is ESC performance.
How (in)efficient is an ESC depending on variations in current, voltage and (motor) frequencies?
I can make a few guesses, but they’re by no means “educated” guesses…
Unfortunately I don’t even know where to start looking for the type of knowledge needed, so anyone that can give me a nudge in the right direction is welcome to do so.
________________

(*) Lately I’ve found indications that the markings on the sensor timing ring are way off. The markings indicate that timing can be adjusted between 15 and 45 degrees (with 30 degrees being the default factory setting), so I set it to “15 degrees”. The real numbers might very well be -15 to +15 degrees, so I ran it at -15 degrees actual timing (if the indication is true).

[wouterb]

Wow, what a post. Read 3 wks ago, still trying to get my head around all the equations. ;)

I'm I right in concluding that there is no such thing as to much KVs? (Assuming rest of the system can handle the amps).
A high KV motor will draw a lot of amps trailing uphill at half trottle, right?

In practice; I'm looking for a motor for a crawler/trailer rig. General opinion is 3000 to 3500KV (sensored) is the way to go. Following your lecture a 4250KV (sensored) motor will be just as good. (Adapted ESC/batteries/wires and geared lower). Is this assumption correct in your book?

[Jtroutt19]

This is...FANTASTIC! Thank you for going through the trouble to document your findings. This is invaluable information and should be made a sticky!

[Olle P]

Originally Posted by wouterb

Read 3 wks ago, still trying to get my head around all the equations.

You're still ahead of me. It took me about twice as long to figure it all out in the first place.

Originally Posted by wouterb

... General opinion is 3000 to 3500KV (sensored) is the way to go. Following your lecture a 4250KV (sensored) motor will be just as good. (Adapted ESC/batteries/wires and geared lower). Is this assumption correct in your book?

Short version: Almost, but NOT geared lower, and not as good as a 7,000 Kv.
The torque (gearing) is really a key finding here!
The speed/torque line for a 7,000Kv motor on 1S is (as good as) identical to the line of a 3,500Kv motor on 2S. And since the "faster" motor is also more efficient I see no reason at all to gear down warranted by the motor...

There's a HUGE disclamer to go with that since I can't actually recommend such a set-up (7,000Kv) though.

3,500 Kv is recommended with 2S LiPo (or possibly 3S, if you want lots of power).
A same line 4,250 Kv motor running the same gearing should perform the same on 2S as the other on 3S, while pulling about 50% more current.
If the switch is done using the same ESC, the ESC can be expected to double its generated heat due to resistance. Other effects on losses due to the higher current I just don't know.

[Olle P]

Originally Posted by Jtroutt19

This is...FANTASTIC! Thank you for going through the trouble to document your findings. This is invaluable information and should be made a sticky!

Thank you for taking the time to notice!

What good would this be without documentation?
Spreading the findings for a peer review (as well as to educate some less schooled) is just as important as doing the research in the first place, if not more important.

[JohnRobHolmes]

Thank you for the comprehensive theory examination, it is quite thorough!



When comparing two motor winds for efficiency, the easiest way to compare apples to apples is by keeping the motor rpm and input load constant. With the way you are examining, you are getting better efficiency with higher KV and it is a correct and thoroughly inspected comparison (I think you got heat right, don't have time to run through all the numbers). As you state, the load is always changing which makes it even more of an unknown. On the side of nearing no load, the motors drop to zero efficiency so you might find more juicy info towards the heavily loaded wattage.

Other observations
The relationship between Is and Imax at a given voltage is much bigger for low turn motors. If we to that add that the heat generated is proportional to the difference squared it’s easy to understand that at high loads a low turn motor will heat up much faster than a low turn motor.

The heat generated from any kv of motor is proportional to the load and rpm, and does not depend on the wind directly(except brushed motors- in practice, faster than 27t three slot or 13t five slot there is noticeable torque loss because of amp induced brush losses). Although heat is square of amps, the copper cross section of each change in turn is also an exponential function. So they balance and efficiency does not change for a given load @ rpm UNLESS copper cross section has been altered. You do have a typo in this observation but the examination of heat can be shelved until cross section is added to the mix too. By fixing voltage and changing KV, you are changing the rpm which will change the efficiency curve by default. You have found that spinning a motor faster is more efficient for a load, same as increasing voltage with a fixed kv

Theory and reasoning
We’ve seen that low turn motors provide the most efficiency for a given maximum load.
The “cost” for this efficiency comes in the form of a high current and a requirement for reduced voltage.
Low voltage makes it difficult to get a perfect match in the choice of battery.
High current at that low voltage makes the resistance in cables and ESC a concern.

The cost of the higher amps and copper for higher KV systems is rather small, an ounce or two more wire goes a long way. It does tax the FETs and copper harder, as there is a difference between the battery amperage and motor phase amperage. So it becomes more of a balancing act of a few factors.

#1 The system must provide the wheelspeed you need at gearing the vehicle can provide.
#2 You are constrained by the voltage choices and volume of a battery.
#3 You are constrained by the voltage limits of the hardware, FETs and power capacitors rating with a buffer for ripple voltage.
#4 The motor has a maximum rpm rating of rotor and iron. 400hz or less commutation rate for m19 silicon steel.


The motor controller works better at low amps than high, so using the highest voltage possible for every watt keeps it cool. For any electrical motor: the faster you can spin the motor for a given load, the higher the efficiency (assuming its still doing a lot more work than the no load losses and rpms are within spec!). So this gives us a few points to work from.



Here are my personal guidelines / tastes when I select a system:
-High performance Motor KV X voltage should net between 30,000 and 50,000 rpms for inrunners and brushed, and 15,000+ rpms for outrunners. 3300kv on 3s for example. For performance brushed, 35t to 27t three slot or 16 to 11t five slot running 3s. Brushed motors with regular copper or silver brushes don't like above 15v. To use 5s and 6s, higher resistance brushes would be needed to reduce arcing.
-Gearing to taste or torque or heat requirements. If you can't gear slow enough, you should have gotten lower KV.
-Higher weight vehicles can use higher voltages and lower KVs to reduce ESC temperature, 1750kv on 6s would meet my RPM and wheelspeed needs.




The original goal behind my "Volt Up, Gear Down" slogan was to help folks get better low speed control and drag brakes. Over time the average motor speed crept up and up, from 55t motors on 7.2v to folks that throw 3500kv motors on 3s into a stock wraith with 20/80 gears and still crawl it at the same spots (me ) . It's all in the sake of torture testing, I promise!



Great tech and depth to this thread, bravo sir!

[drzoo2]

Thank you putting this together. I look forward in my spare time going through it in detail.

I'm a little lax in electric motor theory and this looks like a great read.....

Subbed!

[Olle P]

Originally Posted by JohnRobHolmes

When comparing two motor winds for efficiency, the easiest way to compare apples to apples is by keeping the motor rpm and input load constant.

Might be true, but I opted to use constant output since that's what we want from our rigs.

Originally Posted by JohnRobHolmes

With the way you are examining, you are getting better efficiency with higher KV and it is a correct and thoroughly inspected comparison... As you state, the load is always changing which makes it even more of an unknown.

After posting this article I've come to the conclusion that variations in load (during use) doesn't matter to the overall conclusions!

Originally Posted by JohnRobHolmes

On the side of nearing no load, the motors drop to zero efficiency so you might find more juicy info towards the heavily loaded wattage.

See post #7, "Case 2".

Originally Posted by JohnRobHolmes

The heat generated from any kv of motor is proportional to the load and rpm, ... You do have a typo in this observation ...

My "observation" is very poorly expressed, good catch, and will be re-written!

What I really meant was:
1. "Relationship" refer to Imax/Is, which is considerably smaller for a low turn motor.
2. Then, with the same voltage and very high load (~stall) the heat developed in the windings gets much higher.
(3. With the same voltage the power output, and therefore generated heat, at moderate loads is also higher. In this "observation" that part is omitted and I can't recall why. Most of the article has been rewritten multiple times before posting, so it might have slipped by.)

Originally Posted by JohnRobHolmes

... It does tax the FETs and copper harder, as there is a difference between the battery amperage and motor phase amperage. So it becomes more of a balancing act of a few factors.

#1 The system must provide the wheelspeed you need at gearing the vehicle can provide.
#2 You are constrained by the voltage choices and volume of a battery.
#3 You are constrained by the voltage limits of the hardware, FETs and power capacitors rating with a buffer for ripple voltage.
#4 The motor has a maximum rpm rating of rotor and iron. 400hz or less commutation rate for m19 silicon steel.

I agree that this is where things get interesting in transforming theory to practice.
#1 is a matter of motor power. It must be able to deliver enough of it for a sufficiently long duration without overheating.
#2 Goes without saying.
#3 The low voltage limit is also (more?) interesting. At how low voltage will the control circuit function properly?
#4 This is the part where I have absolutely no knowledge.

Originally Posted by JohnRobHolmes

Here are my personal guidelines / tastes when I select a system:
... higher voltages and lower KVs to reduce ESC temperature, ...

That's what I've figured out too.
While sticking within the mechanical (speed) limitations for the motor optimizing system efficiency is a matter of balancing the power losses in motor, ESC and battery.
I've never experienced having an ESC or battery heating up due to use, so it's probable that my motors are "off" using too little current...

Originally Posted by JohnRobHolmes

Great tech and depth to this thread, bravo sir!

Thanks! Coming from you it's an extra honour.

[Jtroutt19]

So I got a question.

First when I got back into the R/C hobby a couple years ago the whole volt up gear down popped into my head. Running higher voltages is more efficient correct? That is what I always thought. My limited but ever expanding knowledge of electrical/electronic theory told me this was true. That being said why then not make every esc to where is can handle 4s-6s or even higher? And why do you see 1s ESC's?

The limitations of an ESC being able to handle higher voltages are in the voltage limit of the IC's on the board correct? As well as the programming of the MCU. So what changes in design of brushless motors allows them to handle higher voltages?

[EeePee]

There's no point and 1S controllers are for 12th scale pan car racing. Where they also run single digit motor winds and weigh next to nothing.

[JohnRobHolmes]

Originally Posted by Jtroutt19

The limitations of an ESC being able to handle higher voltages are in the voltage limit of the IC's on the board correct? As well as the programming of the MCU. So what changes in design of brushless motors allows them to handle higher voltages?

Its the power FETs and capacitors that limit the voltage, the ICs do have limits too but other hardware influences this. So while the IC might have a 15v max on a pin, a voltage divider will be used to bring a 25v voltage signal down to the range of the IC.

[JohnRobHolmes]

Originally Posted by Olle P

Might be true, but I opted to use constant output since that's what we want from our rigs.

Whether constant output or constant input, I'm just pointing out you got the result of higher KV or lower voltage being more efficient because you didn't fix the wattage or wheelspeed in certain examples. It becomes a comparison with multiple changes at once. As long as we aren't running into switching losses (spinning past 400hz eRPM) or no-load conditions (not enough load compared to no load current) we can always safely assume a faster spinning motor with more gear reduction is the better choice. It is a basic function of how electric motors work, the efficiency is always lower at a lower rpm but fixed load.


A simplified example, I have a 17.5 turn motor and want to compare 2s or 3s use before buying batteries. The reason I keep insisting on a fixed wheelspeed or fixed motor RPM during the comparison is because it mimics the needs in real world application. I want XXX speed, now how can I get there the best way using the gearing, voltage, and KV choices in front of me.


For 500w at XXX wheelspeed, the motor will require 150% more current using 2s than 3s when we adjust gearing so that final wheelspeed is unchanged. The faster spinning motor will always yield higher efficiency when gearing can offset the change. Volt up, then gear it down to the needed wheelspeed.


The same example using different KV of motor and fixed voltage is the same result. This is just as common in the real world. I have a bunch of 3s batteries, now which motor KV should I buy to give me the best performance at my needed speed within the gearing limits of my vehicle. The faster spinning system at our intended wattage will have better efficiency. KV up, then gear it back down to the needed wheelspeed.


The third way to look at this would be a fixed motor speed and geardown, but using different voltage and KV combinations to get there. Inside the motor the efficiency is not changed if copper fill is unaltered. Outside the motor we get efficiency gains using higher voltage and lower KV, and the constraints become voltage limits of hardware and shapes of batteries. In general, I would consider this a moot issue on the whole because we are forced into certain voltages ranges because of the choices in the market.

KV up, Gear down isn't as catchy as Volt up, Gear down, but they are technically the same method of using higher motor speed to net better copper efficiency. The added bonus of this method is that peak torque at the wheel improves and the wheelspeed is more constant under load changes, just like using a higher ratio transmission and faster spinning engine in a full size vehicle.

[Olle P]

Originally Posted by JohnRobHolmes

The heat generated from any kv of motor is proportional to the load and rpm, and does not depend on the wind directly ... Although heat is square of amps, the copper cross section of each change in turn is also an exponential function. So they balance and efficiency does not change for a given load...

My only flaw is marked in red. Should read "high" (and does so now).

The key is that at stall the power drawn is equal to U^2/R, with U being a constant in this example. A low wind motor does have considerably lower resistance, as you point out, and therefore will generate more heat.

The cross section isn't exponential though, but linear, and so is the length of the wire.
The resistance, which is cross section times length, is exponential.