This is the first in what is intended to be a simple guide to the black art of electric power systems and is intended for everybody, even if you have no knowledge of electricity at all. The idea is to try to help those who are trying electrics for the first time, particularly to try and avoid costly mistakes and disappointments. It will answer the question, “How do I choose the right motor, speed controller, battery and prop for my aeroplane?”. I will try to waffle as little as possible. So here goes…

How much power do I need?

If you were using an engine, your kit or plan would probably recommend an engine such as a .40 or .46. You would probably use an engine of this size (or if you’re Pete, put a .90 on it!) with the recommend prop and go fly. And it would work.

If you looked at the small print of the instructions for your engine, you might find that it specifies a power output such as 1 hp (one horsepower) @ 15,000 rpm. Well, there is a direct equivalent in electrics. Power is measured in watts and there are 746 watts to one horsepower. So our engine can put out 746 watts at 15,000 rpm. We will get to how to work out watts soon but first let’s look at an example.

An OS .40 FX claims to put out 1.36 horsepower at 16,000 rpm. So, how many watts is that? Well, it’s 1.36 x 746 watts, which works out to be 1015 watts. Now, the recommended prop for this engine is 10 x 6 to 11 x 8 and with that prop it is probably not going to spin at anything like 16,000 rpm, which means that the real power output of the engine will be lower, much lower. The actual power output may only be a half of the maximum claimed when you put a reasonable prop on so perhaps 500 watts.

So we can say that to fly a .40 sized aeroplane in a similar way to an OS .40 FX, we need 500 watts.

The watts per pound rule of thumb

One way to work out how much power you need is to use the watts per pound rule of thumb. You take the weight of the aeroplane in pounds and multiply by the watts per pound figure from the list below, depending on what performance you want.

• 50 watts per pound weight – minimum for taking off from the ground
• 100 watts per pound weight – sport performance
• 150 watts per pound weight – strong aerobatic performance
• 200+ watts per pound weight – unlimited, ballistic performance

So, if we took our .40 sized kit from the example above, we could probably expect it to weigh 5 lbs. If we wanted “sport performance” we would use 100 watts per pound of weight so, 5 x 100 = 500 watts needed. And that matches what we think the OS .40 FX might put out with a reasonable prop (Phew!).

Some more examples:

1. A scale WW1 biplane weighing 6 lbs. We need to take off from the ground but don’t need excessive power. So we choose 50 watts per pound. Power needed is 6 x 50 = 300 watts.
   If we wanted above the minimum power level to ensure that we could take off from our grass field we could use 75 watts per pound (you don’t have to stick to just the values in the table, you can use values in between) which would give 6 x 75 = 450 watts required.
2. A high performance aerobatic model weighing 5 pounds. For strong aerobatic performance we choose 150 watts per pound from the table above. So, power required is 5 x 150 = 750 watts.
3. A hotliner weighing 4 pounds. For unlimited vertical performance we choose 200 watts per pound. Power required is 4 x 200 = 800 watts.

So, job done – not quite

So, I can work out how much power I might need for my aeroplane but how does that help me choose the right motor, speed control (ESC), battery and prop? Well, you might say that if your model needs say 250 watts that all you need to do is get a 250 watts motor, stick on a battery and prop, wire up your ESC and fly. It might work, but it probably won’t work well and at worst you could burn up your motor and ESC, ruin your battery and crash your model!

A better way is to choose your components in the following order:

First choose the battery,
then the prop,
then the motor,
and finally the speed controller.

That’s crazy! Why would I choose the battery first and, even more bizarrely, the prop second? Let’s take a look.

Batteries

The thing about batteries is that small batteries can provide small amounts of power and big batteries can provide big amounts of power. Also more batteries can provide more power than fewer batteries. So, if our aeroplane requires a certain amount of power (which we can now hopefully work out) we must make sure that our battery pack is capable of supplying at least that much power. If it can’t, it won’t matter which motor, ESC and prop we choose because it still won’t have enough power!

That’s why it’s a good idea to choose the battery first! Okay, so how do I do that? It’s all about volts and amps (stay with me, this is easy!). Let’s use good old NiCads as an example.

A Nicad pack comes in a certain capacity and voltage. The capacity is measured in mah (milli-amp-hours) and the voltage in volts. A 9.6 volt pack of 2400 mah cells would be made up of 8 cells, each of which is 1.2 volts and 2400 mah capacity. Notice that 8 x 1.2 = 9.6 volts. 2400 mah means that the battery can provide 2400 milli-amps of current for one hour. A milliamp is 1/1000th of an amp so that is the same thing as saying 2.4 amps for 1 hour.

The battery could also give 4.8 amps for ½ hour or 7.2 amps for 20 minutes (1/3 hour) or even 24 amps for 6 minutes (1/10th hour). Simple. If you use it up faster (more current) it won’t last as long.
Now for maximum power in bursts, a good Nicad will be able to provide up to 15 times its capacity, which in our case, is 2.4 x 15 = about 36 amps.

So how does that help me get the power? Well it’s simple because:

Power in watts is volts x amps.

We know our Nicad cell can provide up to 36 amps but what is the voltage? Well a Nicad cell is actually about 1.35 volts when it is fully charged, about 1.2 volts when it is empty and when it is running at fairly high power, the voltage drops to around 1 volt (I’ll explain why in a future article). So, now we know that each cell can provide 36 amps at 1 volt, which means that it can provide a power of 36 x 1 = 36 watts!

So, let’s go back to my WW1 biplane from above that we said needed 300 watts to fly. How many cells do I need in my battery pack?

We need 300 / 36 = 8.33 or a least 9 cells.

So, a 9 or 10 cell pack would give us the power we need. 9 cells is a 10.8 volt pack (9 x 1.2) and 10 cells is a 12 volt pack (10 x 1.2). We also now know that we need to draw 36 amps to get the power we need so that helps with the choice of speed controller. We need one that can cope with at least 36 amps.

What about the prop?

The prop is the right thing to tie down next. Once we have that, we can find a motor to suit it. To get the best performance, we need to choose the best prop for the job. A pylon racer will want a small, high pitch, fast prop whereas a 3D aeroplane will want a nice big, low pitch prop.

Here are a few useful facts about props:

• Bigger, slower turning props are more efficient than small fast ones.
• Bigger props produce more thrust.
• Higher pitch produces more speed but needs more power.

So, generally speaking, for sport/aerobatic aeroplanes, it is best to use the largest prop that you can fit. You will probably be limited by the ground clearance. For all out speed, choose a smaller, higher pitched prop. You may also want to choose a particular prop for scale appearance.

How do props affect motors?

If we hooked a motor up to an ESC and battery but did not fit a prop, the motor would spin very fast. If we used a meter to measure the voltage across the motor terminals and the current being drawn from the battery we could work out the power being used because Power = Volts x Amps. In this case, the current would be low, perhaps just 1 or 2 amps. If we measured 10 volts from the battery we could see that the power was 10 x 2 = 20 watts. This is the power when the motor is unloaded.

If we then put a small prop on the motor and measured again we would find that the motor had slowed down a bit and the current had increased, let’s say to 10 amps. Now the power being used is 10 x 10 = 100 watts. So with this prop the motor gives 100 watts.

Now, let’s put a bigger prop on and measure again. The motor is found to have slowed down some more and the current has increased again, let’s say to 20 amps. So now the power is 10 x 20 = 200 watts.

What you can see is that, as you increase the size of the prop on a particular motor and battery, the current increases and the power increases. In fact, if we put too large a prop on, the current may increase to the point where the motor overheats and is damaged. This is why motors have a maximum current rating on them. Put too large a prop on and you risk frying the motor.

So can I choose a motor yet?

Yes! Let’s use our WW1 biplane example and say that we want to use a 13 x 8 prop. We know we want our motor to spin a 13 x 8 prop on a 9 or 10 cell Nicad pack and we also know that the motor must be able to handle at least 36 amps and 300 watts.

I am going to use Model Motors AXI as an example series of motors. Looking at the Model Motors web site for AXI motors, here is the spec sheet for the AXI 2826/10:

Specification
No. of cells 10 - 163 - 5 Li-Poly
RPM/V 920 RMP/V
Max. efficiency 84%
Max. efficiency current 20 - 30 A (>78%)
No load current / 10 V 1.7 A
Current capacity 42 A/60 s
Internal Resistance 42 mohm
Dimensions (diameter. x lenght) 35x48 mm
Shaft diameter 5 mm
Weight with cables 181 g
Recomended model weight 1500 - 2800 g
Recomended prop without gearbox 10 cells 14"x7"12 cells 12"x7"16 cells 11"x5,5"4s Li-Poly 11"x8"

• First of all, it says that we can use 10 – 16 cells. We need 9 or 10 so that’s OK.
• Current capacity is 42 amps (for a maximum of 60 seconds). We need at least 36 amps so that’s OK.
• On 10 cells it suggests a 14 x 7 prop. Our 13 x 8 is smaller diameter (draws less current) but bigger pitch (draws more current). Close enough so that’s OK.
• From the table of example setups there is the following:

Motor Prop. w/o gearbox Battery I/A RPM U (V) P- OUT (W) P- IN (W) Efficiency (%)
        
2826/10 13x7 Gr. CAM fold- prop 10x Sanyo RC2400 34,7 7720 10,6 289 368 79

This says that the 2826/10 motor with a 13 x 7 prop on 10 Sanyo RC2400 cells will draw 34.7 amps at 10.6 volts for a power (P-IN) of 368 watts. Props of similar size and pitch such as 12 x 8, 13 x 6.5 and 13 x 8 will draw slightly different currents and consequently provide slightly different powers but will all be fairly close.

So, we have found our motor! Let’s use 10 cells and a 13 x 7 prop for 368 watts of power.

Finally the Speed Controller

This part’s easy. We can choose any speed controller that is rated for at least the current we know we are going to draw. In this example we know that we will be around 35 amps. It is better to have a little headroom so I would choose a 45 amp or bigger speed control. We also need to make sure that it can handle the number of cells that we need, 10 in our case.

Stick it all together and go fly! More next time…