Choosing power system for Electric RC Planes!

This guideline is from my personal experience on how to select a power system for my electric setups. different people go by different theories but this was has worked for me in the past. All values used are estimates and empirical values; they may change based on your setups but these could give you a starting point. Experimentation is key!

We know there are 4 forces that act on a flying machine.

a. Gravity (Weight)

b. Lift

c. Thrust

d. Drag

Out of these only Lift and Thrust are completely controllable. We need to work with these as inputs.

Weight acts in the opposite direction of lift. Objective: Minimize this.

Drag is a function of lift and depends on how aerodynamic the air-frame design is. Objective: Minimize this.

Moving on, we're left with two parameters of interest, Lift and Thrust. Lift and thrust both are affected by the power system and that's what we're interested in! Thrust is what imparts velocity and can be looked ar the way it affects the velocity, for the purpose of this discussion

A. Start with the Weight and assume the Velocity constant

1. Start with an estimated weight, for a 48 inch (1.2m) wingspan plane, a good flying weight (AUW) could be say between 1.2-1.6 kg for a trainer type plane. The lighter the better. See below how we got to that figure

2. The wing loading defines an AUW you want to achieve. The weight (kg) of the plane is the wing area in square meters, multiplied by the wing loading L gives the weight W

Wt = WL x L

3. For every kg, if

0<=L<=5, you need about 80Watt/kg to fly - gliders

5<=L<=12, you need about 120W/kg to fly - trainer type

12<=L<=18, you need about 180W/kg to fly - acrobat

>18, advanced - jets, need very high speeds and high wattage for stability. typically 300W/kg

4. Multiplying the wattage per kg with weight gives us the total wattage (power) required. In our case, it would be about 192W for a 1.6 kg trainer type plane. Add about 20% for emergency pull outs during landing etc and you arrive at a value of 230W. Remember 1.6 kg is a  heavy plane for that size. It all depends on weight, lighter the better.

P = Wt x Watts

5. This is the point where we start choosing a motor. Remember Ohms law.

V = I x R

Every motor has a max current specified on it; consult the specs (We will not get into Kv at this point, that discussion comes later). So let us say you choose a 2836 type motor with a max current of 21A. Running it on a 11.1V Lithium Polymer battery will give you 233W. This is a close match so you will need

- 3s 11.1V battery (its 12.6 at full charge, but we calculate against the lowest possible voltage)

- 21A is the max current before the motor fries so you will need an esc that can throughput 21A, 30A is a good choice.

So we get about 233W of power (@ 21A & 11.1V) to manage that weight with a 2836 motor. We have a seemingly good choice of motor.

Keep in mind that electric motors will try to drive almost any sized propeller and in turn keep pulling larger currents from the battery, and they might burn out while doing so if the max current crosses what the coils can handle. This is unlike engines that will stall without much damage.

Engines are source and transformers of energy. Motors are only transformers of energy, battery being the source!

B. Lets figure out velocity, with the power constant now

Lets look at the velocity at which a plane must fly, so it does not fall out of the sky. The speed at which a plane loses its lift is the "stall speed". Landing is carried out at a notch above this stall speed. A plane normally flies at about 1.5X the stall speed and takes off at about 2X the stall speed. So we need a motor (+ prop) that can give our plane that speed. Here are some empirical values

glider - 20kmph - 30kmph

trainer - 30kmph - 50kmph - lets assume 40kmph

acrobats - 50kmph - 80kmph

jets - >80kmph

So for our kind of plane we need about 2 x 40kmph = 80kmph

Like V = IR, thus I is a ratio if V/R; Similarly, Kv is a ratio of V/RPM so V*Kv=RPM which implies, that at no load the motor will spin Kv times faster with each increase in volt. The motor in turn drives a propeller at a lower efficiency typically 70% for a mated propeller which means that if the motor can run at 100 RPM on no load, it will run at 70 RPM with a prop on. This efficiency goes down as the prop dia and pitch increase and the motor spins slower with higher props.

A propeller has two components to it, the diameter and the pitch (you might want to read up on google). The diameter defines the thrust that the prop will produce and the pitch defines the forward movement per revolution so a 9x6 prop will move 6 inch forward for every complete revolution,

Kv = 1100

V = 11.1

Prop Efficiency = 70% (0.7)

RPM = Kv * V * Eff 

plugging in the values we get 8547 rev/min

Forward movement per revolution = 6 inches = 15 cm

Forward movement per min = 15 cm * 8547 (RPM) = 128205 cm/min

= 1282.05 m/min = 1.28205 Km/min = 76.9 kmph,

which is a good figure so our choice of 2836 motor @1100 Kv can be used and the motor will stay within the prescribed limit of 21A. If you increase the prop size, the thrust may increase but trades off the speed and vice versa. From here on, you need to play with a couple of setups to find the right fit. Keeping the plane as light (low wing loading) helps significantly. Also planes do not fly at 100% throttle at all times while the calculations are done with maxed out figures. By effectively managing the throttle one can get significant savings.

Experimentation is key!