These pages are designed to give the beginner an understanding of what all the numbers mean, and how to do some basic calculations. It is NOT meant to be a complex article, or delve into all the variables involved.

When you do any calculations, there are always losses involved in the electric products, or efficiency losses. For example, if you have an electric motor that is 300 watts, and 90% efficient at power use, you would actually have to supply the motor with around 333 watts for it to produce 300 watts. This is because no system is 100% efficient.

Bear in mind, then, that if you come to a conclusion on a motor/gearbox/battery/esc, that you may have to increase or decrease values to allow for losses in efficiency.

If you see any obvious mistakes (both spelling and interpretation), then email me so I can fix them

Power

Power = Volts x Amps.

So, a motor supplied with 11 volts, drawing 27 Amps, is producing 297 watts of power (11 x 27 = 297)

Most model motors will have a power rating.

However, this power rating can vary with both the input voltage, and the current draw.

The input voltage is varied by the number of cells in series we have in the battery pack. The current drawn is varied by the prop size & pitch.

Most motors will include a piece of paper that will list various power outputs, based on a number of voltage and prop ranges.

So, the same motor that pulls 32 Amps, at 11volts, with a particular prop, is drawing 352 watts. The same motor pulling 18 amps, at 7 volts, with different prop is only drawing 125 watts of power.

All motors will have a maximum current draw. This is usually determined by the thickness of wire used in the motor. Thicker wire = more current. However, the smaller the motor, the smaller the wire needs to be to get the required internal windings.

So, the formula, Power = Volts x Amps, with basic maths, can be used to find any value when 2 of the other values are known.

• Power = Volts x Amps
• Volts = Power / Amps
• Amps = Power / Volts

These formulas can then help us determine which battery back, speed controller, etc to use.

If we know our motor can produce a maximum of 540 watts, and the maximum current it can draw is 35 amps, we can find the volts

Volts = 540 (watts) / 35 (amps), which equals 15.4 volts.

So, with the right battery pack, we can supply around 15.4 volts.

But how do we get our motor to draw 35 amps? We use the propeller.

The diameter and the pitch will determine how much amps the motor will draw. If you increase the diameter or pitch, you increase the load on the motor, which will make it draw more amps (which will increase the power).  If you decrease the diameter or pitch, you decrease the load on the motor, which will make it draw less amps (which will decrease the power).

Similarly, if we are going to use an 11.1 volt LiPo pack and need 240 watts of power for the model, the motor has to be able to draw 21.6 amps (240 / 11.1 = 21.6), and also the speed controller has to handle 11.1 volts and 21.6 amp.

In reality, the voltage drops as you increase the load on the pack, the motor may only be 90% efficient, and there will be losses in the wiring and the speed controller. In the above example, our motor (and speed controller, battery, wiring, etc) may have to be able to handle 25 or 27 amps.

General input power requirements

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