Replace 'L' by 'Y' in this figure, like the next figure.
The most simple motor considered in this page. This motor has 2 pole magnets and 2 turn coil(M=2). ----
Here, we consider only a steady state for the motor revolution (i.e. constant rpm condition). You might have learned the Lorentz force at high school. Now again, you may read the text book for physics. The Lorentz force is written as follows.
F=ma=q(E+v*B)+A -----(0)
This force F has a vector, which has three components.
---------------------------------------------------
Fz=ma'=q(E+v*B)+A1 -----(0a)
First, the up-down directional component Fz of force vector F in the figure is related to the electron flow dynamics along the coil string. Easily speaking, this equation indicates the generalized Ohm's law along the coil circuit. The electron dynamics corresponds to the electric current. For the component,
m: total electron mass flowing along the coil string, a': acceleration rate of the electron ----- (easily speaking, time variation of current). ----- hence, a'=dv'/dt; v' is the electron speed. q: total electron charge in the coil, E: electric field . ----- hence, L*E=V where V is battery voltage and L is total coil length. v: revolution speed (at the coil's edge). ----- hence, v=S*W where S is the radius of the coil and W is motor rpm. B: magnetic field generated in the magnet. A1: coil resistance force, which prevents the electron flow. ----- A1=Rm*I*q/L where Rm is the coil resistance, I(=q*v') ----- is current and L is total coil length.Now, The acceleration rate "a" is zero because the drawing current is constant in the steady state revolution. It means that the electron is not accelerated and also not deccelerated.
Additionally, the total length L is simply M*(2Y+4S) for the motor shown in the figure. Now, M is the coil turn number. Finally, eq.(0) becomes
V=Ke*W+Rm*I ----(1)
where Ke=L*S*B.
------- Coil resistance Rm -------
Coil string seems to be a conducting material which has no resistance. However, it is not ture. Exactly, every coil string has a small resistance. Since Rm value is generally very small, the measurement is not so easy.
Under no-magnetic field condition, the Ohm's law is written as V=I*Rm, as well-known. Now, the Rm value in eq.(1) is the total resistance along whole of the coil string. Always, less resistance is better for the motor performance but it is limited in some points. For example, thicker string has less resistance but the thickness of the string is limited in the motor size. For example, too thick string cannot be inserted in a small motor, In usual, it depends on the coil turn number M, i.e. total string length L. Also, shorter string has less resistance Rm but such a change modifies the motor performance (Kt, KV, and so on) due to the change of L, as shown above.
----------------------------------------------
----------------------------------------------
Ft=ma=q(E+2*M*v'*B)+A2 -----(0b)
Second, the rotational component Ft(=theta) of force vector F creates the force driving the motor. For this component, the definition of afore-mentioned values must be modified as follows.
m: mass of the rev. part (coil rotor), a: acceleration rate of the motor, --- hence a=dv/dt where v is the revolution speed. q: total electron charge flowing in the coil, E: electric field, =0 for this component, v': electron speed, hence q*v'=I, B: magnetic field generated in magnet, M: coil turn number, A2: force for the output torque. this is propotional to T/S, where the output torque T and rev. part's radius S are used. Note that torque is not force. Between these values, the rev. radius S is always required to be related.The second term of the right hand side is being multiplied by a factor M*2. It means that the Lorentz drive force (q*v'*B=I*B) is enhanced by the coil turn number M. In addition, the left and right parts in the coil totally give the factor 2. Hence, it finally becomes M*2.
Now, because of no-acceleration "a=0", we can put Ft=0. Also, "E=0" can be set. Subsequently, we replace T by T+To where the new T is the output torque and To is no-load torque. To can be measured under no-load condition. Finally, eq.(0b) becomes
T=Kt*(I-Io) ----(2)
where Kt=2*M*Y*S*B and To=Kt*Io.
------- no-load current Io ------
Io is called as no-load current. It is measured in the no-load condition (no-gear, no-propella and no-main rotor). Now, in an appropriate voltage range, Io is assumed to be constant. Io is directly related to the motor efficiency. So, it is very important. The measurment itself is easy, if you have a volt meter and a tach (rpm) meter. However, depending on the load history of the magnets and the voltage, Io may somtimes vary, so that the measurent may be not so stable.
1: Basic equations for motors:
As shown in the above section,
applying for the simple motor structure shown above,
eq.(0) is re-written as the following two equations in each component.
In the motor physics, these two equations are the most important.
V=Ke*W+Rm*I ----(1)
T=Kt*(I-Io) ----(2)
Ke=L*B*S(=1/KV) -----------(1')
According to eq.(1),
when the current I is assumed to be very low
(note that this is no-load condition),
we can see that the motor runs at KV(=1/Ke) rpm per 1V.
---- KV (=1/Ke) and the motor size -----
According to eq.(1'),
the motor constant Ke (=1/KV) is propotional to the magnetic field
intensity B generated in motor magnet. In general,
Neodyum magnet has much stronger magnetic field B
than that of Ferrite magnet used widely in 540class brushed motors
for RC car. As shown in eq.(1'), stronger magnet naturally
tends to give lower KV value.
In addition, as the coil turn number decreases,
the coil string length L also decreases, and therefore, KV increases.
Also, as the motor radius S decreases, KV increases. And, vice versa.
As a result, such a low (High) KV motor is low (high) rpm motor
and tends to be powerful (powerless).
Kt=2*M*S*Y*B ------(2').
This is called as torque constant.
---- Kt and the motor size ----
According to eq.(2'), the Kt value is propotional
to the field intensity B and coil string dimension = M*S*Y.
------ the relation of Kt and Ke ---------
Now, the total coil length L is written as
M*(2*Y+4*S), following the figures shown above.
If Y is much larger than S, we can assume L=2*M*Y.
This is a rough approximation. It means an ideal no-core coil !!!
At the time, Kt=Ke is obtained. In real motors,
these values is not the same but close.
In fact, in AVEOX web page
(for the specification table of brushless motor), the Ke/Kt values for
every brushless motor in 14xx/xY series are always almost constant.
It suggests that those motors have the same basic structure
and component for the coil, motor can and magnet (Neodyum).
In the following discussions, let us put Kt=Ke for simplicity.
In addition, we introduce the following equations.
Output power ; Pout=T*W ----(3)
Input power ; Pin=I*V ----(4)
The efficiency rate ; Q=Pout/Pin ----(5)
Combining eqs.(1)-(5), we obtain the output power curve and
efficiency rate curve for each motor.
The motor performances (such as, W, I, Pin(=I*V) and Pout)
are determined by four motor constants; Rm, Io, Kt and KV (=1/Ke).
Now, let us think that
the motor axis rpm W is determinted by current I and voltage V.
Instead of current I, the torque T may be considered.
The torque T is determined by load conditions, such as prop. size and
so on.
motor axis rpm: W=(V-Rm*I)/Ke --------(6) important !
=(V-Rm*((T/Kt)+Io)/Ke
e-current: I=(T/Kt)+Io -------- (7)
Input power: Pin=I*V=((T/Kt)+Io)*V ----(8)
Output power: Pout=(I-Io)*(V-Rm*I)*(Kt/Ke) -----------(9) important !
=T*(V-Rm*T/Kt-Rm*Io)/Ke
Hence,
Efficiency rate: Q=Pout/Pin*100 (%)
Loss power : Ploss= Pin-Pout
Let us put Kt=Ke for simplicity.
Figure 2a presents the output power Pout of Hacker
B40-8L and 12L for a constant voltage (8.4V).
These motors have different coil turn numbers (M=8 and 12),
so that Rm, Io, Kt and Ke are different,
as listed in motor table .
The 45deg. oblique dash line is the ideal Pout line for 100% efficiency rate.
The loss power Ploss is obtained by (I*V-Pout).
The abscissa is the current I which is propotoinal to the input power
Pin=I*V.
--- Basic discussions ----
These output power curves Pout are always parabolic for every motor.
Note that the current range (hence, range of Pin)
shown in the figure is amazingly high.
The max output power Pmax(=0.5((V-Rm*Io)**2)/Rm)
is obtained at I=0.5(Io+V/Rm). For 8L, Pmax is about 1.7kW and
for 12L, Pmax is about 0.8kW. Can you believe the values ?
This is true !!! In real use, we are using these motors, at most,
under 0.3-0.5kW. If you used the motor around Pmax, the large loss power
should immediately burn the motor up. In this repect,
please overwrite the efficiency rate curve by yourself in this figure.
Then, check that the max eff. point is located
in the very low power range (= left side) of the figure.
It is the power range for real use.
In addition, note that 'V/Rm = Imax'
corresponds to the right edge (foot) points
of each motor's Pout curve (parabola curve).
For 8L, Imax is about 0.8kA which is amazingly high !
This corresponds to the most dangerous situation.
Because, the motor generates no output power,
and hence, the motor is stopping !!!
In other words, the motor drive axis is rigidly fixed,
so that it means zero rpm. At the time,
the whole input power is converted to the heatloss !!!
There is no way in which motor does not burn up. Dangerous !
In Figure 2a, the most important point is that
the shape (size) of the Pout curves
is basically determined only by the Rm value.
Accordingly, the motor's powerfulness (=Pmax)
is also determined only by the Rm value.
Because, in this figure, the no-load current Io(=1-4A) for every motor
is plotted to be very lose to zero (rather, Io is almost zero !!!, now).
Also, note that Ke and Kt are meaningless for the Pout curves,
because of the assumption Kt=Ke.
Hence, in general, lower Rm motor is more powerful.
However, "Powerful motor" is not necessarily "a good motor".
As shown in motor table ,
such a low Rm motor has large Io value, which generates
larger Lo-loss. When you choose a motor suitable for your RC-model,
Io and Rm must be well-balanced for the RC-model.
------- Pout curve is modified by the input voltage -------
If the voltage is changed to 11V (=10cells) from 8.4V (=7-8cells),
the Pout curves also change, as shown in the next figure.
Comparing This Figure 2b with Figure2a,
every motor becomes more powerful than that of 8.4V.
In fact, Pout curves are shifted to higher power side and enlarged
(swelling). Also, Pmax for each motor became higher.
Using equations (8) and (9),
the power loss generated in the motor is described by
Pin - Pout = Rm*I*I + Io*V -Io*I*Rm ----- (10).
Most of the loss power is heat loss (The other is almost
dissipated to sound noise).
Too large loss power rapidly heats the motor.
Especially, when the magnet is strongly heated,
the magnet power may be lost forever. The motor die.
Figure 2c shows the Ploss curves
for hacker B40-6L, 8L, 10L, 12L and 16L.
The voltage is constantly 8,4V.
Comparing Figure 2a with this figure, note that
this current range is shifted to lower side.
This figure shows the power range for real use.
In fact, in usual e-heli, current 10-30A flows,
as shown in my many motor data pages
for ECO8, EP Concept SR and so on.
If you only do hovering, 10-15A will be sufficient.
Lower loss power is anyway better for every meaning.
Which motor do you choose in your heli ?
According to Figure2c, 10L seems to be the best in 10-30A,
because of the low loss power.
Also, 8L is not so bad ?.... The loss power is around 10-20W.
At this point, you can imagine the 20W loss power because
it is similar to the wattage of your solder tool.
How long time dose it take until the solder tool gets hot ?
5min ? or 10min ? If the loss power was 10W, longer time is required
to get hot. By the way, how big is the motor, comparing with the solder tool ?
Different size will results in the difference of the heat capacity.
..... Large motor will permit long run time without overheating.
Evidently, 6L generates too high loss power, so that the motor
is out of our choise (7-8cells heli use). While,
if you like mild flight (= lower power flight = long duration flight),
you may goto 12L or 13L..... Those motors generate
very low loss power at less than I=15A which
is sufficient for hovering.
Note that the voltage is constantly 8.4V, now.
If the voltage is changed, these curves change.
Also, note the voltage drop due to the inner resistance of battery.
In fact, exactly speaking, 8.4V corresponds to 8cells rather than 7cells.
---- Additional important comment ---
Accordingly, 10L seems to be the best for 7-8cells heli (around 8.4V).
However, note that this motor theory is not a perfect theory.
In fact, if you like acro flights, 8L or 9L may be better than 10L,
for 7-8cells heli.
While, if you like mild flights, 10L or 11L..... will be better.
At this point, note that the real power response and
the acceleration performance are not considered
in this present steady state motor theory. Generally,
if you like powerful and vivid flights rather than highly efficient
flights (= long and mild flight), slightly less coil turn motor
(= high KV motor) may be better.
Figure 2d shows the loss power of Hacker B40-8L, 10L, 9S and 11S.
B40-xxS(120g) series motors are smaller than B40-xxL(150g).
Rm values for 8L and 9S are similar. Also, 10L and 11S are similar.
Hence, these power curves (8L vs 9S and 10L vs 11S) are similar, respectively.
While, the loss power of B40-xxL is always a little lower than that of
B40-xxS. It is originated in the difference of Io values.
It means that larger motors (B40-xxL)
tend to have lower Io than smaller motor (-xxS), so that
such a motor has higher efficiency rate.
The detail of the loss power is further discussed below.
.
-------- Compare the Loss power curve ------
Look the next figure. This figure was also calculated from
the motor constants(KV,Io,Rm (note Rm=Ri))
by using the equations shown above.
These two curves show the loss power for
Hacker B40-9L(9turn) and B40-11L(11turn).
For less than 20A, B40-11L generates lower loss power than B40-9L.
While, for more than 25A, B40-9L is better.
The white triangles shown in this figure are respectively
pointing the max efficiency points for these two motors.
It is important that the loss power uniformly increases
as the current increases.
In other words, the minimum loss power is NOT established at the
max efficiency point. Rather, the loss power acceleratedly increases beyond
the max efficiency point. This feature is common for every motor.
In addition, the difference between these loss power curves is, at most,
about 5W. Compared to the total loss power (20-30W), the differece is
not so large. While,
the difference between Aveox1406/2y and 3y will reach 10W. Be careful !!!
This figure shows more easily those features.
You can recognize that the feature is the same as that of Figure 2c.
The variation (a)==>>(b)==>>(c) shows the variation of the coil turn number
from "less coil turn motor" to "more coil turn motor".
If you use the motor in the low power range (left side of the figure),
motor (c) gives low loss power so that you can get long flight duration
and less overheat. Oppositely, if you like high power flight,
motor (a) will be good because the motor generates the lowest loss
in the high power region. In the intermediate power range,
motor (b) becomes the best choise.
In Sec.4, we recognized that the powerful motor(= low Rm motor)
is not necessarily a good motor.
Because, lower Rm motor tends to have large Io.
Hence, we must balance Rm-loss and Io-loss.
In Sec.5, we have known how to find a low loss motor
(in the output power range required) for your RC-model.
In this section, the technique shown in Sec.5 is theoretically
explained in term of balance of Rm-loss and Io-loss.
----6-1: Rm-loss power related to coil resistance Rm -----
In eq.(10), the Rm-loss (=Rm*I*I) is the so-called Joule heat
generated in the coil's copper string.
As the e-current "I" increases, this loss power increases.
When we keep the output power (=I*V) to be constant,
lower input voltage V results in higher e-current I.
In such a case, in order to suppress this loss power Rm*I*I,
lower Rm motors must be selected. Vice versa. As a result, in general,
motor which has smaller (higher) Rm value is suitable for lower
(higher) voltage use.
As wellknown, when the coil string is thicker,
the Rm value becomes less. It is good to suppress the Rm-loss
but the motor becomes fat and heavy.
If you want to keep the motor size,
you may reduce the coil turn number M
in order to insert the thick coil into the same size motor can.
However, unfortunately, such a less turn coil motor (less Rm motor)
has larger Io, leading to large Io-loss power mentioned in the next.
-----6-2: Io-loss power related to no-load e-current Io ----
The Io-loss power (=Io*V) is related to
every loss power out of the coil resistance heat.
Especially, this loss power strongly heats the motor magnet.
This heating is more serious than heating of the coil,
since the hard heating erases the magnet power, forever.
When the input voltage V is low, this loss power Io*V is also low.
When the motor has more coil turn number M (or the motor is larger),
the motor has lower Io value. Therefore,
lower Io motor must be selected for higher voltage RC-models.
Vice versa.
-----6-3: The balance of the Rm and Io loss ------
Finally, low-Rm motors tends to have high-Io value, and vice versa.
Hence, when we keep a constant power I*V,
low-Rm motors are recommended for low voltage and high e-current.
Inversely, low-Io motors are recommended for high voltage
and low e-current.
----- 6-4: Loss power permitted in each model ----
Anyway, we must enoughly suppress the loss power in flight.
For the purpose, you must choose the motor suitable for
your flight conditions which depend on
the power range and flight style.
For instance, like F5B motor gliders, the combination of
high power (500-2000W) and short duration (5-20sec) is safely used.
Inversely, like mild trainer airplanes,
the combo of low power(50-100W) and longer duration(10-15min)
will be also safe. In such a case, we can expect the sufficient
cooling effect.
Evidently, the combo of high power and long duration is dangerous
for the overheat. Use sufficient heavy motor
which has low-Io and low-Rm values.
As discussed so far, KV value is basically
not related to the motor performance, itself.
The KV value is only used for the determination of the gear ratio.
Of course, the gear ratio depends on each RC-models and your flight styles.
In order to determine the gear ratio, firstly, you must determine the prop.
rpm or main rotor rpm, which you require in your RC-model.
It must be done out of this motor theory.
Finally, the gear ratio can be roughly estimated to be KV*V*0.9/(your rpm),
where the battery voltage V is roughly set to be 1.1V*(cell number).
The factor 0.9 is called as load rate for Neodyum magnet motor
under the safe loading.
If the motor has Ferrite magnet, 0.7-0.75 must be used instead of 0.9.
Again, note that this calculaion is independent of the motor performance.
It means that you must check if the loss power is safely low,
as shown in Sec.5 and 6 (otherwise, check the heat level in real flights).
If the loss power is sufficiently low for the motor,
the motor should not burn up. But, if it not so,
the motor will die, due to the overheating.
In such a case, change the coil turn number or battery voltage
on the basis of the motor performance.
Otherwise, goto a larger motor which is tougher and has higher
efficiency rate.
In this page, I explained how to determine the motor performance
with the values Rm, Io, Kt, KV(=1/Ke).
The heat power loss discussed here is mainly
classified into the following two terms.
1; One of the heatloss originates in the copper string resistance
in the motor coil. This is related to the value Rm (Rm-loss).
2; Another heatloss originates in the magnetic effects caused
in the magnet. This is related to the value Io (Io-loss).
However, in this page, we ignored some important loss power mechanisms.
Firstly, we assumed that the weight balance of motor rotor is
perfectly ideal. But, in general, manufacturer's products have
an amount of the balance error. Even slightly unbalanced
rotor largely prevents the revolution, especially for higher rpm.
This suppresses the efficiency rate presented in this page.
Second, as the motor is heated by the continuous run,
the coil resistance Rm increases, therefore leading to the rapid increase
of the Rm-loss. In this respect, even
slightly large initial loss power leads to
the rapid acceleration of the heating by the continuos motor run.
Third, the motor considered here has no switching mechanism (circuit) !
It is not real.
It means that there is (no brush and) no controller (ESC) in this discussion.
But, as you know, the rotational switching timing largely influences
the motor performance. This is also strongly related to
the efficiency rate.
I don't know how much does these ignored loss power suppress the
efficiency rate given in this page.
V: input voltage (NiCd battery),
Ke: (=1/KV),
KV: noloading motor (rpm / 1 volt)
W: motor axis rpm,
Rm: resistance of motor coil,
I: electric current,
T: axis torque,
Kt: torque cofficient,
Io: no loading e-current
Eq.(1) presents the electron dynamics in coil strings of the motor.
Easily speaking, this is related to the generalized Ohm's law.
As afore-mentioned, a motor constant is derived as follows.
Eq.(2) presents the rotational force balance of the motor coil,
which is also based on the Lorenz force equation (0).
The next relation shows a motor constant Kt.
2: Additional equations:
3: Power and Efficiency rate curves :
4: Output power curves : 
Figure 2a. 
Figure2b.
5: Loss power curves : 
Figure2c.
----- Heavy motor tends to be highly efficient.---- 
Figure2d.
6: How is the loss power influenced by Rm and Io ?
7: The gear ratio is determined by the KV value :
8: Summary and Appendix: