## Choosing The Correct Prop For Electric Motor

**CHOOSING THE CORRECT PROP FOR AN ELECTRIC MOTOR**

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**MOTOR LIMITS THAT ARE NOT TO BE EXCEEDED; these are given by the manufacturer and they are:**

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- Current.
- Wattage.
- Maximum RPM of the motor.

Selecting a prop is determined by considering;

- Wattage limits of the motor,
- Revs per volt of the motor,
- Current limit of the motor,
- A relationship between Watts, RPM and Prop sizes,
- What is to be expected of the aircraft, and,
- The formula
**W**(*Watts*) =**V**(*Volts*) Multiplied by**I**(*Amps*)

For example if the motor power specification is 400 watts at maximum of 40 amps and the shaft speed is 2,000 revs/volt, the voltage required to use the motor at maximum power and current simultaneously is determined using the formula transposed as **V=W/I** and dividing the 400 watts by 40 amps, giving 10 volts. Ten volts determines the RPM of the motor at 20,000 RPM Referring to a prop/power chart, a prop size in the vicinity of 7 by 4 is required to absorb 400 watts of power at 20,000 RPM. This may be OK for a small fast flying aircraft but be unsuitable for other aircraft.

Other ways of configuring the motor to handle the same power in different ways are needed. Alternatively a gearbox could be used and a 3:1 is to be used for this exercise. On the same 10 Volts the prop RPM is 20,000/3 giving a prop RPM of 6,666. From the Power/RPM/Prop chart, 400 watts at 6,666 RPM equates to a prop size of approx 13 by 8.

Another option is to use of a 2:1 gearbox which allows the prop to spin at 10,000RPM (20,000/2). From the chart, a prop size of 10 by 6 absorbs 400 watts at 10,000 RPM Using a gearbox of 4:1, the same 400 watts at 5,000 RPM (20,000/4), requires a 16 by 8 prop. If a voltage of 11 volts (3 Lipol cells) is used the new maximum current permitted is 400 watts divided by 11 volts which gives 36.3 amps. The motor shaft speed will now be 22,000 RPM.

If the voltage were to be increased to 14 volts the new motor shaft speed is 14V by 2,000 RPV, giving 28,000 RPM The 400 watt limit of the motor is maintained by reducing the current drawn on the higher voltages. The new maximum current is calculated; 400 watts/ 14 volts giving 28.6 amps. It also means that at 7,000 RPM prop speed, a gearbox 4:1 ratio is required to spin the same 13 by 8 prop as a 400 watt load.

On a 3:1 gearbox the shaft speed will be in the vicinity of 9,300 RPM which necessitates a prop size of 11 by 6. Voltage can be increased to the physical rotational limits of the motor but the load must be reduced in order to restrict current, and maintain the power parameter of the motor, in this case 400 watts. The chosen configuration to the motor/gearbox/prop determines the requirement of the speed controller and the capacity of the batteries to give desired flight times.

For example if the voltage is increased (can the controller handle the chosen number of cells), and as voltage is increased the current drawn must be reduced by fitting smaller props or a gearbox and therefore the requirement of the speed controller must be appreciated. This principle applies in reverse; if the voltage is changed from 4 to 2 Lipol cells, the current needed to maintain 400 watts increases but in this example is not permitted to rise above 40 amps. Therefore the new maximum power of the motor must be limited to 280 watts and appropriate props are to be chosen.

Have fun with math’s and electric flight, pj.