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Motorformler

Elmotorer

[[1]]

$1 inch = 0.0254 m\,$

$1 m/s = 3.6 km/h\,$

$V_{pitch} = RPM * \frac{2 *\pi * Prop pitch_{inch}}{60} * 0.0254 m/{inch} * 3.6 km/h$

$P_{out} = Thrust * V_{pitch}\,$

$Efficiency = \frac {P_{out}}{P_{in}}$

$P_{in} = U * I\,$

Feature Article - Motor Constants, How To Find Them And Use Them by Joachim Bergmeyer [[2]]

Introduction

Have you ever bought a tiny DC motor from a surplus store and asked yourself at which current and voltage it might work best? Do you believe that the well-known KP00 motor is an "amp hog" and eats up the current more quickly than it should? Then read on; you might find some help and a new point of view.

Abstract

This article describes how to measure and calculate the motor constants of small, brushed, DC motors at home with common and relatively cheap tools and instruments. Furthermore, it describes how to make use of them to choose a good point of operation for an unknown (or known) DC motor i.e. the working voltage and current. It also gives some insight to the physical basics. For this purpose, some high school level mathematics will be used.

What Is A Brushed DC Motor?

The motors we are referring to consist of very few parts.

The Stator: In most cases, it is the outer housing of the motor. It is made of a soft (does not remain magnetized) magnetic material, most commonly iron and contains magnets that are arranged to have one north and one south pole. There are also two bearings, either plain plastic or porous brass bushes or (in more expensive motors) ball bearings.

The Rotor: This either consists of an iron core carrying sets of windings (called poles) or is coreless. In the latter case, the windings are wound by a machine and then fixed by resin such that they stand alone without the iron core. The number of poles is most commonly three, and sometimes (in better motors) five. More poles tend to be used only with bigger motors that are too large for small indoor models.

The Commutator: This consists of two halves; one-half fixed to the rotor, the other to the stator. Its purpose is to route the current from the motor terminals to the rotor windings and switch the current in such a way that the windings of the poles generate magnetic forces that cause the rotor to turn indefinitely. This is the part of the brushed DC motor that makes it brushed. It is the brushes that connect the rotating rotor to the stationary stator.

How do voltage, current, torque and rpm relate?

A DC motor converts current into torque and voltage into rpm. Unfortunately, there are losses. Friction in the bearings must be overcome by an idle current I0. The resistance Ri of the rotor and alternator cause voltage losses, and there are also so-called "iron losses" caused by current in the core iron which is generated by the changing magnetic field in the rotor. In this article, the iron losses are ignored because they are also represented by the idle current (and because they are not calculated as easily as the other losses). Besides that, the often-used coreless motors do not have a core and therefore have much less iron losses, which make them more efficient than cored motors.

The Fundamental Motor Equations

The generated voltage of a motor and the rpm have a fixed ratio. It is called the rpm constant :

1. $K{v}=rpm/Ui\,$ The input power of a DC motor is the terminal voltage times the current:

2. $P_{in}=U_{mot}*I_{mot}\,$ The output power from a mechanical point of view is rotor torque times rotating speed (in radians per second):

3. $P_{out} = M * ( 2 * \pi ) / 60\,$ Here arises a problem, as is difficult to measure the torque. You have to build some sort of test fixture and you need a good balance to measure the counteracting forces. Even if you do the former and already have or are willing to buy the latter, it is difficult to get exact readings. However, there is another possibility. We will not measure the torque, but calculate the output power as the input power minus the losses. $U m o t$ is the voltage that we can measure at the motor terminals. $I m o t$ is the motor current. However, not all the voltage counts for the output power, only the voltage generated in the rotor windings according to formula (1) does. a part of the voltage is lost at the inner resistance $R i$ when flows through it. therefore, we have to subtract the voltage loss across the motor resistance , (which is, according to Ohm's law, $R_i * I_{mot} \,$ ) from the battery voltage:

5. $U_i = U_{mot} - R_i * I_{mot} \,$ The motor torque caused by the idle current is needed to compensate for the friction of the bearings and the alternator; we do not see any torque outside the motor from this part of the current so we subtract it from the battery current:

6. $I_{eff} = I_{mot} - I_0\,$ Now we multiply the effective voltage by the effective current and get the output power:

4. $P_{out} = U_i * I_{eff} = ( U_{mot} - R_i * I_{mot} ) * ( I_{mot} - I_0 )\,$ We can now find the output power without measuring torque! The efficiency is the ratio between input and output power:

7. $\eta = \frac{P_{out}}{P_{in}} = ( U_{mot} - R_i * I_{mot} ) * ( \frac{I_{mot} - I_0 }{U_{mot} * I_{mot}} ) \,$ The rpm can be calculated using equation (1) and (5):

8. $rpm = K_v * ( U_{mot} - R_i * I_{mot} ) \,$ Now we have shown that with the help of the motor constants $R_i\,$ , $K_v\,$ and $I_0\,$ , we can calculate all these values including the $rpm\,$ . We just need the motor current and motor terminal voltage, both of which are easy to measure. We need a voltmeter and an ohmmeter. The cheapest and easiest way is to use two multimeters at the same time. These multimeters do not have to be very expensive, but those with digital display are easier to read, and of course, multimeters that are more exact are more expensive than simpler ones. There are also combined instruments available that show voltage, current and input power at the same time; these are a taste of luxury. There are even more interesting values! The most important questions when trying out a new motor are...

How much power can I expect?
At which current will I get maximum power?
Which efficiency can I expect?
At which current will I get maximum efficiency?
Which current should I use?
How fast will the motor turn?

All these questions can be answered by help of the motor constants. First, we should decide about the power source, because that will tell us how much voltage we can use. In general, more voltage leads to more efficiency, more power, more current and, worst, more weight. The latter is because we micro-flyers tend to use the smallest available batteries that provide us with the necessary power. Since energy to weight ratio is worse with smaller cells, we always should try to use less cells first, not smaller cells. The only reason to use smaller cells is when we are down to one LiPoly cell only and this one cell is too heavy! The new star under the power sources is the Kokam Li-Poly cell with 145mAh, giving us about 3.6V under load and up to 1A. About the same can be expected from three Sanyo 50mAh NiCad cells (for a shorter time, of course) or from three GP 120mAh NiMH cells. One of these power sources is generally used in the "1oz-class" (models with an all-up weight of up to ~30g). More voltage needs a DC/DC booster which adds complexity, costs and weight, or more cells, which also adds weight and therefore is used more in the "2oz-class" (all-up weight up to ~60g). (Apart for some pager motor powered models that require a DC-DC and weigh <1oz, Graham) Therefore, depending on the class to which our model belongs, we assume that we have 3.6V or 7.2V and up to 1A.

Maximum Power

A formula for the current at the point of maximum power for a given motor can be derived from formula (4).

10. $I_{Pmax} =\frac{U_{mot}+R_i*I_0}{2R_i}\,$ Power, efficiency, and rpm can be calculated using formulae (4), (7), and (8). Just put in the motor constants, the chosen motor terminal voltage and use $I_{Pmax}\,$ for $I_{mot}\,$ . We should never use more than in any case with any motor because if you pull more than then the efficiency goes down more quickly than the power consumption rises, so we will get less (not more) power. The motor will become very hot, since all the electrical power that is not converted into mechanical power will be converted into heat instead. Of course, this tells nothing about how long the motor will stay alive at this power level. Formula (10) only tells us at which current this particular motor will have the most output power when used with the assumed voltage. Electrical motors generally have no fixed maximum working voltage, so this has to be answered by experience. As stated above, the motor could quit by overheating, by burning the brushes or by destroying the rotor. The coreless motors are more sensitive because their rotor has little mass and thus will heat up very quickly, and since it has no iron core, if the resin that fixes the windings melts then the rotor will expand and rub to the outer housing of the motor, generating a lot of friction. In this way, I myself damaged some tiny coreless pager motors.

Maximum Efficiency

The motor current at the point of maximum efficiency is:

11. $I_{\eta max} = \sqrt { \frac {I_0 * U_{mot}}{R_i}}\,$ The derivation can again be seen here. We also have a nice formula to calculate the maximum efficiency directly (we could also put into formula (7) for the same result):

12. $\eta_{max}=(1-\sqrt{\frac{I_0 * U_{mot}}{R_i}})^2\,$ The motor current at full throttle should never be less than because the output power would be very low and the efficiency would be bad. So, we have some nice guidelines on what we can expect from a motor and which current we should try to reach by choosing prop and gear ratio.

The Rules

Use at least $I_{\eta max}\,$ . If this is already too much power for your model, use a smaller motor.
Never use more than $I_{Pmax}\,$ . Otherwise, your motor will burn and the output power will be low.
It is a good rule of thumb to use about 70-90% of $I_{Pmax}\,$ for static current at full throttle for the small motors we normally use.

The output power, rpm and of course the efficiency can again be calculated using equations (4), (7) and (8) in any point of operation, so we will know anything that we need.

But how do we get the constants?

We get the constants by measuring values that are relatively easy to get, and then use some special formulas.

Firstly, we measure the idle current $I 0$ . This is dependent on the motor terminal voltage, so we measure it at about 2/3 of the operating voltage (means: battery voltage) that we want to use later. This is because it will more or less be the rpm level at which the motor will run later with the best compromise of efficiency and power. Measure the idle current without having a gearbox attached to the motor.

Then we choose two different loads that we believe to be near the upper and lower border of the operating range of the motor. In the simplest case, these are the two props that we think are ok but which we do not know which fits better. It does not matter whether the props are directly driven or geared, but in the geared case, of course we have to multiply the measured rpm with the chosen gear ratio.

We take two sets of data. $I_{mot1}\,$ , $U_{mot1}\,$ and $rpm1\,$ with the first prop, $I_{mot2}\,$ , $U_{mot2}\,$ and $r p m 2$ and with the second prop. We should take the voltage directly at the motor terminals to avoid a measuring error. All values should be taken at the same moment, or it might be easier to use a battery with big capacity because the voltage will then not drop quickly during the measurements.

If we find later that the measured currents are about 30-60% of $I_{Pmax}\,$ for the lighter load and about 80-100% of $I_{Pmax}\,$ for the heavier load then we guessed right about the two different test loads. If the readings are much different from that then we should choose some other loads (means: props) and redo all our measurements and calculations. Otherwise, our results might be less precise than they could be.

The motor resistance can be calculated as:

(9-6) $R_i = \frac{rpm_2 * U_{mot1} - rpm_1 * U_{mot2}}{rpm_2*I_{mot1} - rpm_1*I_{mot2}}\,$

...and the speed constant $K_v\,$ will be:

(8-4) $K_v = \frac{rpm1}{U_{mot1}-R_i * I_{mot1}}\,$

Examples

The Kp00 Motor

I have measured the following data at this little workhorse and got the following results.

Measurement with load 1 (bigger prop): U1 2.10 V I1 0.66 A n1 13900 rpm

Measurement with load 2 (smaller prop): U2 2.08 V I2 0.62 A n2 14450 rpm

Measured idle current at 2.1V: I0 0.10 A

We put our values into the formulas and get the following motor constants:

(9-6) Ri = 1.559 Ohm

(8-4) kv = 12980 rpm/V

Assuming an average discharge voltage of 3.3V for one Kokam 145mAh Li-Poly cell, we get the following predicted data:

(10) = 1,108 A

This fits well to the maximum current of the Li-Poly cell. We don't want to go all the way up to the maximum power anyway. (Very Good!)

(4) = 1.59 W

This is quite a lot of power. A rule of thumb says that the INPUT power of a sports plane should be 50W/kg, So we could expect to be able to power models up to 3.3V*1.1A/50W*kg = 72g. This isn't a micro flyer anymore, right? Right! I would recommend this motor for models in the 1oz-class (around 30g maximum). (good again)

(7) = 43%

This is a quite high efficiency at the maximum power point of a so small motor (50% is the theoretical maximum value here!), also very good.

(8) = 20405 rpm

We can only use small props or have to use big gear ratios, since this motor turns fast! The micro scale builders will enjoy this, as most micro scale flyers will be able to use a prop of about scale diameter! If you are searching for efficiency of the prop then you will have to use a big gear ratio. We have all the possibilities, good again.

(11) = 0.46 A

(12) = 61%

(4) = 0,93 W

(8) = 33524 rpm

Now comes the excellent part. The point of maximum efficiency for this motor at this voltage is still within the area of useable power! Yes, this motor likes amps, but it gives us a lot of torque back for it also! Lighter models (around 20-25g) will be able to stay in the air with this power, and a motor this small performing with an efficiency of more than 60% is outstanding! Yes, we will have to use a high gear ratio to make use of this high rpm.

My conclusion is that the KP00 motor is a very efficient and powerful one if you use a light load (=small prop or high gearing) and let him turn as fast as he wants to. The useable current range reaches from 0.5A to 1A at 3.3V, which is exactly what we want for micro scale models around 1 ounce total weight. My Guided Mite did loops! Yes, I love this little powerhouse! :-)

The N-20 Lv Motor

It is available from many surplus sources; mine are from Allelectronics.

I've measured the following data with this motor and got the following results:

Measurement with load 1 (bigger prop): U1 3.19 V I1 0.84 A n1 7900 rpm

Measurement with load 2 (smaller prop): U2 3.36 V I2 0.52 A n2 15700 rpm

Measured idle current: I0 0.12 A

We get the following motor constants:

(9-6) Ri = 2.592 Ohm (8-4) kv = 7803 rpm/V

I've used this motor with a DC/DC booster that happened to deliver 5.2V. We get the following predicted data:

(10) = 1,063 A

This fits well to the maximum current that my DC/DC booster can deliver. (good)

(4) = 2.3 W

This is more power than from the KP00. The rule of thumb (for bigger models) says 5.2V*1.06A/50W*kg = 110g. I would doubt that this motor would carry around a model that heavy, but it works at least for up to 60g. (good again)

(7) = 42%

The efficiency of this motor is slightly lower, but still good.

(8) = 19075 rpm

We still have all the possibilities. (good again)

(11) = 0.491 A (12) = 57% (4) = 1,456 W (8) = 30653 rpm

As said above, at 5.2V this motor is not as efficient as the KP00 at 3.3V, but is still good. It is heavier, but it will carry around models that are heavier also.

The N-20 LV performs well at a minimum voltage of 5V. It works notably less efficiently from one Li-Poly cell without booster. It is therefore probably not ideal for the very small models. However, it works well in my models around 2oz. In addition, it is cheap! So, if one ever quits (no one did yet for me), take the next one! It will probably quit sooner if working from two Li-Poly cells, but this should be a quite efficient drive system if you keep the current low enough (say at about 1A maximum).

Mega AC 16/15/7

Mega 16/15/7

7Cell No Prop 11850 rmp 0.5A

$K_v=11850/7*1,3\,=1302$

$R_i=\frac{U_1*rpm_2-U_2*rpm_1}{I_1*rpm_2-I_2*rpm_1}=\frac{8,1*8580-11,4*6920}{11,4*6920-8,1*8580}=0,65\Omega\,$

Your Own Motor:

You can download a blank form that can be printed out and filled in with your measured values here. You will need Adobe Acrobat Viewer to read it.

Conclusion

I wanted to show that these calculations are no black magic and that you don't have to grope in the dark when you ask yourself how to make the best use of the tiny motor that you have recently found somewhere. Some predictions are calculated easily, and it is even possible to develop the mathematics yourself.

Bibliography

The following books helped me to understand the theory of DC motors.

K. Gieck, Technische Formelsammlung, Gieck Verlag 1981, Heilbronn, Germany
Dipl.-Ing. L. Retzbach, Ratgeber Elektroflug, Neckar-Verlag 1999, Villingen-Schwenningen, Germany

Explosionsmotorer

$P = K * Pitch * Diam^4 * RPM^3\,$

Om P är i enheten W och Pitch och Diameter är i tum, så är $K = 5.3 * 10^{-15}\,$

Motor Formulas