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 How do I match an electric power system to a given airframe? Or, how do I convert a gas powered plane to electric?

Q. How do I match an electric power system to a given airframe? Or, how do I convert a gas powered plane to electric?
A. Jim Bourke

Matching an electric power system to a plane is not as straightforward as many would expect. The problem is that electric power is very versatile so it is necessary to do a bit of juggling to find out which system will work best. For example, pretend that I have a 4 lb (64 oz or 1814g) sport plane with a 600 sq. in. (4.2 ft^2 or 39 dm^2) wing. I desire mild aerobatics (50 watts/pound) and a low wing loading (< 20 oz/ft^2 or 60 g/dm^2). I want at least 5 minutes of full power. Lets see how close I can get.

We'll try three different power systems. For simplicity we'll assume that a cell delivers 1V so the total voltage is the same as the # of cells. The formulas we need are:

• watts = # of cells x amps (amps = watts / # of cells)
• duration = 60 * (capacity/1000) / amps
• wing loading = aircraft weight (ounces) / wing area (ft^2)

#1 Astro 15 with a 10 cell, 1700 mAH pack: the weight of this power plant is roughly 2 lbs (900g). We now have enough information to figure out the aircraft weight and wing loading.

Aircraft weight = 6lbs (96 oz)
wing loading= 96 oz / 4.2 ft^2 = 22.9 oz/ft^2

In order to figure out the duration, we need to know how many amps of current the motor will draw. We can find out how many amps we need the motor to draw using the watts/pound ratio we chose:

Total watts=watts/pound ratio * aircraft weight (in pounds) therefore: watts=(50 watts/pound * 6 pounds) = 300

amps= total watts / # of cells therefore:
amps= 300 / 10 = 30A

At this point you may have to stop and try a different motor if the amps you need aren't realistic. In this case we can continue because an Astro 15 is still at least 70% efficient at 30 amps. Now that we have the amps we can figure out the duration.

Duration=60 * (capacity/1000) / amps
= 60 * (1700/1000) / 30
= 102/30 = 3.4 minutes

We have just found that an Astro 15 motor with a 10 cell 1700 mAH pack will fly this 4 lb plane for 3.4 minutes at 50 watts/pound with a wing loading of 22.9 oz/ ft^2 If we use this combination, we will have a short flying time and no reserve power. Lets try a larger motor.

#2 Astro 25 motor with a 16 cell, 1700 mAH pack this power system weighs roughly 2 3/4 pounds (44 oz or 1247g).

Aircraft weight=6.75 lbs (108 oz or 3061g)
wing loading=108 oz / 4.2 ft^2=25.7 oz/ sq. ft (77.1 g/dm^2).

amps needed=(watts/pound * aircraft weight) / 16 volts
= (50 watts/pound * 6.75 pounds) / 16 volts
= 337.5 watts / 16 volts = 21 amps

duration=60 * (capacity / 1000) / amps
= 60 * (1700 / 1000) / 21
= 102 / 21 = 4.9 minutes

This motor raises the wing loading a little bit, but it gives us the duration we want and quite a bit of reserve power. We can choose a prop that draws 25 amps at full throttle, but throttle back most of the time for 4.5 minutes of powered flight. Lets see what happens when we decrease the cell capacity. It will lower the wing loading, but it will also decrease duration:

#3 Astro 25 motor with a 16 cell, 1400 mAH pack. This power system weighs roughly 2 lbs (32 oz or 900g).

Aircraft weight=6 lbs (96 oz or 2720g).
wing loading=96 oz / 4.2 ft^2=22.9 or 68.7 g/dm^2
amps needed=(50 * 6) / 16 = 19
duration=(60 * 1.4) / 19 = 4.4 minutes

Now that you have the general idea, you know enough to run your own numbers. One thing I didn't examine was the effect of gearing in the above example. When I learn of a way to incorporate the effects of gearing in the above calculations, I'll present an example. Unfortunately, the watts/pound rule does not take gearing into account.