# Building a personal helicopter, I have a 5kw electric hub motor, how big will the rotors need to be? [closed]

I am building a personal electric helicopter, it consists of a lightweight steel frame that attaches to the axle of the hub motor. If the motor outputs 155nm of torque @ 5kw, and a max rpm of 700. How much area will the rotors need to have to reach at least 200 pounds of thrust, I weigh 150 pounds plus 50 for the motor frame and battery. Something like this but with a single 4 blade rotor attached to the hub motor.

• I'd start looking at some similar design, like the Hiller ROE Rotorcycle, Gyrodyne RON Rotorcycle, Ka-56 and kit-helicopters Commented Apr 10, 2023 at 19:37
• Wait, a single 4-blade rotor? How are you going to compensate the torque?
– Zeus
Commented Apr 13, 2023 at 8:42
• I was thinking with a rather large tail rudder, it might be a bit twisty at the beginning but hopefully once it gets up to speed there will be enough side thrust to compensate for the torque Commented Apr 14, 2023 at 3:51
• "Hopefully"? With such attitude I wouldn't advise to test it out yourself ;) On this device, once you're off the ground, it will be much easier to spin you than the rotor.
– Zeus
Commented Apr 14, 2023 at 7:58
• It can be as large as you want, still in hoover it won't produce any force... Commented Apr 14, 2023 at 8:34

If you are asking this question here, like this, then I humbly suggest that you probably shouldn't be attempting this.

This is life or death stuff, and without years of professional aviation knowledge and experience the chances of something going horribly wrong are very real.

A helicopter needs the highest power in hover. As a first approximation we can use the momentum theory which estimates the thrust as:

$$T = 0.5\sqrt[3]{2 \rho A P^2}$$

Where:

• $$A$$ is the rotor area;
• $$\rho$$ is the density;
• $$P$$ is the power needed to generate that thrust;
• and the $$0.5$$ is just to take into account the simplifications introduced by the momentum theory.

Obviously in hover thrust equals weight and therefore:

$$A = \frac{(2T)³}{2 \rho P^2}$$

If you are going to use a contra-rotating configuration then you have obviously to split the thrust between the two rotors but taking into account that the lower rotor will need some 15% more power the thrust being equal due to the fact that it works in the wake of the upper rotor.

• If the pilot and machine weigh 100kg, that's 1000N of thrust required, which gives an area of 320 square meters, or a blade length of 10meters. 5kW is pretty low power for a helicopter. For comparison, the Mosquito single-person helicopter has 60-95hp engine options. Commented Apr 17, 2023 at 8:03