# Control Tower - February 1999

### Jim Bourke presents Part 6 of "Understanding Electric Power Systems"

Death of a Friend

As you may already be aware, Andy Fok, the owner of Unbeatenpath Imports, died of a massive heart failure in his home on January 21, 1999. This unexpected death was especially tragic as Andy, age 36, left behind his wife, Darlene, and his three young boys: Anthony, age 6, Vincent, age 4, and Ryan, age 2. In addition, Andy and Darlene are expecting a fourth son who is due in April.

Andy founded UPI 3 years ago in an attempt to bring high-quality models into the US for the first time. Andy was a visionary who saw a need for imported hobby goods and worked feverishly to create the business ties necessary to pull things off.

Andy was a long time sponsor of this webzine and a friend to many people in the hobby. He will be missed.

Please direct cards of support to Andy's family at 6088 Mary Lane Court, Oconomowoc, WI 53066.

I am told that UPI will continue business but I have no specific knowledge of what form that will take. Please do not email me with questions regarding UPI orders. I promise to announce any information as I receive it.

Understanding Electric Power Systems - Part 6

This month I present part 6 of my series on electric power systems.  Please refer to my September 1998 column if you are looking to catch up on missed information.

Recap

In the January, 1999 edition of this series I completed the simplified motor model by presenting motor losses and limitations.

Here are the answers to last month's Q&A:

 Q&A Joe's Super Thermal 2000 R/C electric sailplane now has a new motor that will let it climb out vertically to 500 feet.  Joe's sailplane weighs 5 lbs.  How much work is accomplished during the vertical climb?  How much power is the motor producing if the climb takes 10 seconds? 2500 foot pounds of work are expended when lifting a 5 lb object to 500 feet.  250 foot-pounds per second of power are used to accomplish this task in 10 seconds, which is just under .5 horsepower.   Suppose I have a motor with a Kv of 1500.  What is the Kt value? Recall that: Kt x Kv = 1352 Which means that: Kt = 1352 / Kv Kt = 1352 / 1500 Kt = .901 Therefore, the Kt value must be .901 inch ounces per amp.   Consider the following motors: Motor1: Kv = 4000, Io = .2 amps, Rm = .150 Ohms Motor2: Kv = 2000, Io = 2 amps, Rm = .050 Ohms Which motor is best for an indoor flyer that will draw only 2 amps on 6 cells?   What will be the power out and efficiency of the chosen motor? Before running the numbers on both motors, it might help to examine the two options and use a little bit of logic to make the selection.  You should notice that Motor number 2 has an Io of 2 amps.  Therefore our simplified motor model will predict that this motor will not produce any torque at 2 amps.  For that reason, it cannot be used in such a low-current application.  The following calculations will demonstrate this. Motor 1: Pout = (Vin - (Iin * Rm)) * (Iin - Io) Pout = (6 - (2 * .150)) * (2 - .2) Pout = (6 - .3) * 1.8 Pout = 5.7 * 1.8 Pout = 10.26 watts Eff = Pout / Pin Eff = 10.26 / (Vin * Iin) Eff = 10.26 / (6 * 2) Eff = 10.26 / 12 Eff = .855 Eff = 85.5 % Motor 2: Pout = (Vin - (Iin * Rm)) * (Iin - Io) Pout = (6 - (2 * .050)) * (2 - 2) Pout = (6 - .1) * 0 Pout = 0 Eff = Pout / Pin Eff = 0 / Pin Eff = 0% So it seems that Motor number 1 is definitely the better choice.   Which motor is best for a sport plane that will draw 30 amps on 10 cells?  What will be the power out and efficiency of the chosen motor? For this problem we are using the same motors as above.  However, the Io of motor 2 will not be as important because the current is much higher.  In this case the armature resistance is the primary determinant of which motor is better for our application. Motor 1: Pout = (Vin - (Iin * Rm)) * (Iin - Io) Pout = (10 - (30 * .150)) * (30 - .2) Pout = (10 - 4.5) * 29.8 Pout = 5.5 * 29.8 Pout = 163.9 Eff = Pout / Pin Eff = 163.9 / (Vin * Iin) Eff = 163.9 / (10 * 30) Eff = 163.9 / 300 Eff = .546 Eff = 54.6 % Motor 2: Pout = (Vin - (Iin * Rm)) * (Iin - Io) Pout = (10 - (30 * .050)) * (30 - 2) Pout = (10 - 1.5) * 28 Pout = 8.5 * 28 Pout = 238 Eff = Pout / Pin Eff = 238 / 300 Eff = .793 Eff = 79.3% The point of this exercise is that different motors are suitable for different applications.   Suppose I have a motor with an RPM limit of 30,000.  With a certain battery pack and a 50 amp current draw, this motor is able to spin a 12x8 propeller at 29,000 RPM.   If I remove the propeller and rev up the motor, what will happen? There is a potential that the RPM will exceed the motor's limitation of 30,000, causing damage to the motor.   Look on the internet at the following websites: www.maxcim.com,   www.aveox.com, www.astroflight.com.  All three of these websites list motors with their constants.  Compare the constants, price, and weight of the various motors.  Do you see any trends? You may or may not have noticed the following things: The higher the number of winds, the lower the Kv. Within the same brand of motor, different armatures produce different Kt and Rm values that are roughly proportional to each other.  In other words, as the Kt increases so does the Rm. As a motor gets heavier, its ability to withstand current increases.  High torque limits are typically found on large motors. Within a given size of motors, such as the Aveox 1406 series, the total amount of torque that can be created is nearly a constant.  There is a difference in the torque per amp, but not the torque per watt. All of these relationships will be explained in a future column.  For now just consider them.

Making use of our motor model

Over the last several months we have examined electric motors in considerable detail. Starting with a mythical, lossless, perfectly efficient, Ideal Motor we learned the basics of how electric motors work. Over time the addition of new information has caused our understanding of electric motors to grow more complex, hopefully at a proper tempo.

Last month we completed our Simplified Motor Model, which allows us to predict the power out of an electric motor, given the amount of power coming in. Previously, we had learned how to compute the power absorbed by a given propeller. If we put all of this knowledge together, it seems trivial to determine how much current a motor will draw given a certain number of cells and a certain propeller. Unfortunately, this kind of operation is more difficult to carry out than it might seem at first glance, therefore this month’s column is devoted to the subject of solving common electric flight problems.

Review

Recall that the power absorbed by a propeller (Pa) can be computed using the following formula:

Pa = Kp * (RPM ^ 3) * (D ^ 4) * P

Where RPM is specified in thousands and diameter and pitch of the propeller are specified in feet. The Kp value is the "propeller constant" which is nothing more than a value used to fudge the result depending on what brand of propeller we are considering. For the general case we can use a Kp of 1.25 and be somewhere in the ballpark.

Recall also that the power out of an electric motor (Po) differs from the power in due to two different losses: the torque loss and the RPM loss. The power out can be computed by:

Po = Vout * Iout
Po = (Vin – (Iin * Rm)) * (Iin – Io)

Where V is the input voltage, Iin is the input current, Rm is armature resistance, and Io is the no-load current.

Solving for Iin

The goal of this section, given our simplified motor model, is to:

Take a given motor, propeller, and battery pack and compute the current drawn by the motor at full throttle.

The complication we must consider is that we cannot know the power absorbed by the propeller unless we have the RPM. But we cannot have the RPM until we know the losses due to the current draw. And, naturally, we cannot have the current until we know how much power the propeller will absorb.  Now you can see why I started this series with Ideal Motors!

The first step in solving this problem is to examine the Pa formula and modify it a bit. Lets create a new constant called Ka which represents the power absorption factor of the propeller alone.  This step will allow us to make the formula a little bit easier to understand.

Pa = Kp * (RPM ^ 3) * D^4 * P
Ka = Kp * D^4 * P
Pa = Ka * RPM ^ 3

It should stand to reason that all of the power coming out of the motor will be absorbed by the propeller.  Therefore, the calculated power absorption of the propeller and the the calculated power out of the motor must be equal.

Po = Pa
Vout * Iout = Ka * RPM^3

Next, we can arrange our formula so that all of the constants involving the motor are on the left side of the equation and all of the constants involving the propeller are on the right side.

Vout * Iout = Ka * RPM ^3
Vout * Iout / RPM^3 = Ka

Now we can substitute for RPM because the RPM is a product of the motor's Kv value and the motor voltage (minus losses).

Vout * Iout / RPM^3 = Ka
RPM = Vout * Kv
Vout * Iout / (Vout * Kv)^3 = Ka
Iout / (Vout ^2 * Kv ^3) = Ka

Now we can substitue back in the values of Iout, Vout, and Ka to see our formula in all of its glory:

Iout / (Vout ^2 * Kv ^3) = Ka
Iout = Iin - Io
Vout = Vin - (Iin * Rm)
(Iin – Io) / ((Vin – Iin * Rm)^2 * Kv ^3) = Kp * D^4 * P

This gives us equation 6-1:

Equation 6-1: (Iin – Io) / ((Vin – Iin * Rm)^2 * Kv ^3) = Kp * D^4 * P

That is as far as we can comfortably go in an attempt to simplify our formula. Now our task is to plug in all of our constants and solve for the remaining value. In this case we are attempting to solve for current (Iin). The steps to perform this task are as follows:

1. Select a motor, propeller, and voltage.
2. Compute the Ka value (the right hand side of the equation) using the propeller constants and the motor's Kv value.
3. Choose an acceptable starting guess for the input current.
4. Calculate the expression on the left hand side of equation 6-1.
5. If the computed value is higher than the Ka, choose a new input current that is lower than the previous choice.  If the computed value is lower than the Ka, choose a higher input current.
6. Repeat steps 4 and 5 until you are satisfied with your answer.

Lets go ahead and compute an example:

Step 1: Select a motor, propeller, and voltage:

• Motor = Aveox 1409/3Y (Kv = 1333, Rm = .040, Io = 1.6 Amps)
• Propeller = 9x6 APC (Kp = 1.11)
• Input voltage = 10 volts

Step 2: Compute the Ka of the chosen prop:

Ka = Kp * D^4 * P
Ka = 1.11 * .75^4 * .5
Ka = 1.11 * .316 *.5
Ka = .175

Step 3: Choose a starting value for Iin

Rather than just taking a stab in the dark, lets use some logic to choose the current.   Lets start off using the current value predicted by the Ideal Motor Model, on the assumption that it will be fairly close:

I = Pa / V
I = Ka * RPM^3 / V
I = Ka * (V * Kv)^3 / V
I = Ka * V^2 * Kv^3
I = .175 * 100 * 2.37
I = 41.5 amps

Step 4: Calculate the value on the left hand side of equation 6-1.

(Iin – Io) / ((Vin – Iin * Rm)^2 * Kv ^3)
(41.5 - 1.6) / ((10 - 41.5 * .040) ^2 * 1.333^3)
39.9 / ((10 - 1.666)^2 * 2.37)
39.9 / (8.334^2 * 2.37)
39.9 / (69.45 * 2.37)
39.9 / 164.6
.242

Step 5: Compare the two values and adjust the current as appropriate.

I note that .242 is higher than the Ka value of .175 so I will adjust the current down to 30 amps.

Step 6: Repeat steps 4 and 5 until satisfied with the result.

(30 - 1.6) / ((10 - 30 * .040)^2 * 1.333^3)
28.4 / ((10 - 1.2)^2 * 2.37)
28.4 / (8.8^2 * 2.37)
28.4 / (77.44 * 2.37)
28.4 / 183.5
.155

So 30 amps is too low.  Next we can try 35 amps:

(35 - 1.6) / ((10 - 35 * .040)^2 * 1.333^3)
.191

Since 35 amps is too high we can now try 32.5 amps:

(32.5 - 1.6) / ((10 - 32.5 * .040)^2 * 1.333^3)
.172

Now lets try 33 amps:

(33 - 1.6) / ((10 - 33*.040)^2 * 1.333^3)
.175

At this point we can stop searching because we've found the answer.  We just predicted that an Aveox 1409/3Y will draw 33 amps on 10 cells with an APC 9x6 propeller.

If you spend some time with an analysis program to validate these results, please remember that we are using 10 volts in our example, not 10 cells.  This may throw the numbers off a bit depending on how sophisticated the program is.

Solving for the propeller

To solve for the propeller is a little bit different.  We can't quite solve for the propeller the same way we do current because a propeller is actually specified by three distinct constants: Kp, diameter, and pitch.  Instead, we can solve for Ka and then use some reasoning to choose the propeller that matches the result.  According to our model there are an infinite number of propellers available to match a given Ka, but we will use some common sense to narrow that down to just a few.

Recall equation 6-1:

(Iin – Io) / ((Vin – Iin * Rm)^2 * Kv ^3) = Kp * D^4 * P

To solve for the propeller, we follow these steps:

1. Select a motor, voltage and current draw.
2. Compute the left hand side of equation 6-1.
3. Choose a propeller type to determine the Kp value, or use 1.25 for a generic propeller.
4. Divide the value determined in step two by the Kp value.
5. If solving for diameter, choose a pitch and divide the result of step 4 by the chosen pitch  If solving for pitch, choose a diameter and divide the result of step 4 by the chosen diameter.  Alternatively, solve by diameter to pitch ratio.

The following example will be helpful:

Step 1: Select a motor, voltage, and current draw

• Motor = Aveox 1409/3Y (Kv = 1333, Rm = .040, Io = 1.6 Amps)
• Input voltage = 10 volts
• Current Draw = 25 amps

Step 2: Compute the left hand side of equation 6-1

(Iin - Io) / ((Vin - Iin * Rm)^2 * Kv^3)
(25-1.6) / ((10 - 25*.040)^2 * 1.333^3)
23.4 / ((10 - 1)^2 * 2.37)
23.4 / (9^2 * 2.37)
23.4 / (81 * 2.37)
23.4 / 192.0
.122

Step 3: Choose a propeller type

Lets choose an APC propeller with a constant of 1.11.

Step 4: Divide the value from Step 2 by the Kp of the propeller chosen in step 3.

.122 / 1.11
.110

Step 5:

There are three options: solve for diameter, solve for pitch, or solve by ratio.   Lets examine all three:

Option 1: Solve for diameter

In this option we must choose a pitch.  Note that according to the formulas I can choose any pitch I want but, in reality, the ratio of diameter to pitch must be kept between 2:1 and 3:2 for real-world efficiency concerns.  I'll choose a pitch of 5 inches, or .417 feet.

D^4 * P = .110
D^4 * .417 = .110
D^4 = .264
D = .716

Therefore the diameter must be .716 feet or (roughly) 8.5 inches to draw 25 amps from an Aveox 1409/3Y motor on 10 cells with a 5 inch pitch propeller.

Option 2: Solve for pitch

This time we should choose a diamter and solve for pitch.  Lets use 7.5 inches (.625 feet) for the diameter and see what kind of pitch we need.

D^4 * P = .110
.625^4 * P = .110
.153 * P = .110
P = .719

Therefore our 7.5 inch diameter propeller must have a pitch of roughly 8.5 inches (.719 feet) to draw 25 amps.

Option 3: Solve by ratio

To solve by ratio we must express the propeller's diameter in terms of pitch.   This is usually the best way to solve for a propeller as 99% of all useful propellers have a diameter to pitch ratio between 1:1 and 2:1.  Propellers with pitches that are greater than their diameters are usually not good choices for model airplanes, but can work in certain circumstances.  Propellers with diameters that are more than 2 times their pitch are almost always very bad choices for model airplanes and should not be considered as viable candidates.

We'll examine three examples of solving by ratio.  The first example will attempt to find a propeller with a diameter to pitch ratio of 2:1, like a 12x6, 8x4, etc:

D=2 * P
D^4 * P = .110
(2 * P)^4 * P = .110
2^4 * P^4 * P = .110
16 * P^5 = .110
P^5 = 1.76
P =  .370
P = roughly 4.5 inches

D = 2 * P
D = 2 * 4.5
D = roughly 9 inches

result is a 9x4.5 propeller

Now lets see what propeller we get if we select a 1.5:1 ratio, like a 12x8, 8x6, etc.:

D = 1.5 * P
D^4 * P = .110
(1.5 * P)^4 * P = .110
1.5^4 * P^4 * P = .110
5.06 * P^5 = .110
P^5 = .022
P = .465
P = roughly 5.5 inches

D = 1.5 * P
D = 1.5 * 5.5
D = 8.25

result: 8.25 x 5.5 propeller

Now lets select a 1:1 ratio (like a 12x12, 8x8, etc) and see what we get:

D = P
D^4 * P = .110
P^4 * P = .110
P^5 = .110
P = .643
P = roughly 7.75 inches

result: 7.75 x 7.75 propeller

We have now determined 5 different APC propellers that will draw 25 amps from an Aveox 1409/3Y on 10 cells.  They are the 8.5x5, 7.5x8.5, 9x4.5, 8.25 x 5.5, and 7.75x7.75.

Here is where some common sense comes in.  I am not sure if any of the above propellers are actually produced by APC.  We will have to look at some close matches and test them with a motor to find the right one.  Based on what we've computed, we might try out a 9x5,  an 8x6, and an 8x7.  Undoubtedly, we would find that our calculations are off a little bit once we did some real world tests but this would be a good place to start.

Quick and Dirty Propeller Guesses

Sometimes we don't need to know exactly which propeller we need to draw an exacting amount of current through our motor.  Sometimes we just want to have a good starting point for experimentation.  For this we can avoid all of the work in the above and just use a variation of the Ideal Motor Model.  We'll start off with formula 6-1 and throw out the things we don't need:

(Iin – Io) / ((Vin – Iin * Rm)^2 * Kv ^3) = Kp * D^4 * P

First off, lets ignore any losses, which gives us:

Iin / (Vin^2 * Kv^3) = Kp * D^4 * P

Our goal is to have a quick and dirty propeller selection so lets make two compromises right now.  First, lets get rid of the Kp factor.  Second, lets specify the pitch to be equal to the diameter.  These two changes cancel each other out somewhat as the Kp is typically somewhere between 1.1 and 1.3 while the pitch is typically 50 to 75% of the diameter.  The net effect is that our result will be a little bit conservative, which is a good thing if we want our first rough guess to be safe for our motor.  After we toss out those terms we can solve for diameter and rearrange the formula for easy calculation:

Iin / (Vin^2 * Kv^3) = D^4 * D
Iin / (Vin^2 * Kv^3) = D^5

D^5 = I / (V^2 * Kv^3)

I can do this calculation in a couple of seconds on a calculator and it works pretty well.

Example 1:

I need to draw 12 amps from a 6V Speed 400 motor on 7 cells.  The Kv of this motor is 2700.  What propeller should I try?

D^5 = 12 / (7^2 * 2.7^3)
D^5 = 12 / 964
D^5 = .0124
D = .416 feet or 5 inches

So our quick and dirty formula says a 5x5 propeller will be a good place to start.   That might encourage us to try a 6x3 or 6x4 as well.

If you go through the effort of using our more complex calculation you will come up with a prediction of 10 amps or so for a 5x5 propeller.  The quick and dirty formula works!

Example 2:

I need to draw 35 amps from an Astro 25G motor on 16 cells.  The Kv of this motor (including gearbox) is 595.  What propeller should I try?

D^5 = 35 / (16^2 * .595^3)
D^5 = 35 / (256 * .2106)
D^5 = .649
D = .917 feet or 11 inches

Based on this information I might start by checking the current with an 11x8 propeller.   After that I could try a 12x8 to see if the current is close to what I want.

For the record, our more complicated formula will predict 35 amps from an 11x11 propeller or a 12x8 propeller.  Again, the quick and dirty formula did the trick.

Please note that any time you are testing a new propeller with a motor you should be very careful.  Always use an amp-meter and always use an ESC to bring the voltage up slowly.  Stop immediately if the current is getting anywhere near the torque limit of the motor.  Do not assume that any published constants are correct and never trust your personal safety or the survival of your equipment to calculations alone.   Everything you do on paper is irrelevant if your testing results in smoke coming out of your motor.

Jim, My Calculator Won't Let Me Do That!

I've gotten some emails telling me that I need to give some more basic advice at times when it comes to these formulas.  If you are having trouble getting your calculator to understand the above, try the following trick.  It turns out that this:

D^5 = I / (V^2 * Kv^3)

Is mathematically the same as this:

D = (I / (V^2 * Kv^3)) ^ .2

Raising something to the 1/5th power is the same as taking the 5th root.

For the calculator-challenged, here are step by step instructions to the quick-and-dirty propeller guessing formula shown above:

1. Type in the Kv of the motor in thousands.
2. Push the "x^y" button.
3. Type in 3.
4. Push the multiply button.
5. Type in the Voltage.
6. Push the multiply button.
7. Type in the Voltage again.
8. Push the equal button.
9. Push the "1/x" button.
10. Push the multiply button.
11. Type in the Current.
12. Push the equal button.
13. Push the "x^y" button.
14. Type in ".2"
15. Push the equal button.
16. Push the multiply button
17. Type in "12"
18. Push the equal button and read the propeller diameter in inches.

Summary

Solving for current and propeller can require a little bit of juggling, but is fairly simple when we use the motor model we have developed over the last few months.

By now you should have a good idea of how all those electric flight simulation programs work.  Programs like ElectriCalc (which is reviewed in this month's issue) and MotoCalc (which will be reviewed in the March issue), enable modelers to plug in information about their power system (or select motors, cells, and propellers from a database which contains their constants) and immediately see the effects of the values they selected.  These programs are invaluable tools but it is important to have at least a passing familiarity with how they work.

If you routinely use these programs then take a moment to verify their calculations using the information in this column.  Experimentation and practice are the most important parts of the learning process.

Next month we will improve our understanding of batteries by adding the concept of internal resistance.

Q&A

Joe's Super Duper Thermalator 2000 R/C electric sailplane needs a new motor.  Joe wants to draw 60 amps on 10 volts.  His motor has a Kv of 3000, an Rm of .020 Ohms, and an Io of 5 amps.  What propeller should he begin his tests with?

How much current will a 12x8 propeller draw from an Astroflight 25G motor on 16 volts?   The Astroflight 25G has a Kv of 595, an Rm of .093 Ohms, and an Io of 2 amps.

Use the quick-and-dirty guessing formula to pick out a good starting place for experimentation if I want to find the right propeller for a motor with a Kv of 1200 assuming I am using 10 cells and desire a 25 amp draw.

 Dec 08, 2013, 03:54 PM Registered User Dear Mr. Bourke, I am an undergraduate student at the University of Surrey, Guildford, England. I am in my second year of study and am embarking on a group Design Make Evaluate project to manufacture a quadcopter. We have to be able to justify our propeller and motor choices so in trying to understand the relevant factors I stumbled upon your Understanding Electric Power Systems column and have followed all 6 parts of it. I have used Excel to create a model that allows me to calculate the various parameters, based on your column, so that I can get a decent, fairly realistic idea of what motor, battery and prop size we will need (with a few assumptions about our quadcopter), which I can use in my justification (as well as using websites like MotoCalc). At the end of Part 6 (February 1999), you wrote: ‘Next month we will improve our understanding of batteries by adding the concept of internal resistance.’ I was looking forward to adding this next parameter into my Excel model - and I presumed that there would be more parameters to come but alas I was unable to find Part 7 or any part after Part 6. I would really appreciate it if you could give me some guidance on the last few adjustments you wrote (or would have written) about to the Ideal Motor Model and their effects on it because this would improve the accuracy of my Excel model and thus improve my justification, which I must make within the next two weeks. Yours faithfully, FlyPhi
 Dec 10, 2013, 03:34 PM Registered User Thread OP I'm glad that my series was helpful to you. Unfortunately I haven't yet found time to finish it. I had big plans for several more topics. Perhaps some day I'll return to it. Internal resistance of a cell, in this simplified model, is a resistance value like any other, which is modeled using Ohm's law: V=IR. The voltage from the cell drops as the amperage increases. I hope that helps you with your spreadsheet. Jim Latest blog entry: Check in
 Dec 10, 2013, 05:59 PM Registered User Dear Jim, I'm sad to hear that; I've really enjoyed reading it! Just out of curiosity, what would those other topics be? I looked it up online because I couldn't remember much about internal resistance and it amounts to a voltage drop so I just factor that into the Voltage Loss Inside Motor Equation thus: Vloss = I*Rm + (EMF - Ir) ? Thanks a lot, much appreciated! If there are any other considerations you could suggest, I would be very grateful. Kind regards, FlyPhi
 Dec 10, 2013, 06:01 PM Registered User P.S. I really hope that you continue these tutorials soon because they're really good!
 Dec 10, 2013, 09:22 PM Registered User Wow, this has got to be the oldest resurrected thread.