Human-Powered Helicopter

HPHA student builds a human-powered helicopter (HPH). It is a helicopter powered by one person carried on board and the power is generated by pedaling.
This human-powered helicopter has the following parameters: diameter of helicopter rotor blades is 15 meters and total mass is 100 kg. The air density 1.3 kg/m^3.

Can the HPH take off?


Solution.

The helicopter flies upward against the force of gravity by using their rotors to throw air down beneath them.
A cross-section of an air flow is

πd^2/4 ,

where d is the diameter of helicopter rotor blades.

An amount of air that flows during a time Δt is given by the formula

Δm = ρ*(πd^2/4)*v*Δt

where ρ is an air density and v is a velocity of the air flow.

A momentum is a product of the mass and the velocity.

Hence, a change of the momentum
Δp = ρ*(πd^2/4)*v^2*Δt

The change in momentum equals the impulse applied to it.
Thereby,

F*Δt = ρ*(πd^2/4)*v^2*Δt

or

F = ρ*(πd^2/4)*v^2

As I mentioned before, the helicopter flies upward against the force of gravity, then

F = m*g , where m is total mass of HPH.

From previous equation

m*g = ρ*(πd^2/4)*v^2

Now, we could rewrite the equation for v

v = (2/d) * √(m*g/(ρ*π))

and calculate v

v = (2/15) * √(100*9.8/(1.3*π))≈ 2.065 m/s

Now, let’s speak about the kinetic energy of the air flow. An equation is used to represent the kinetic energy

Ek = 0.5*m*v^2

then a change of the kinetic energy of the air flow

ΔEk = ρ*(πd^2/8)*v^3*Δt

Power P is the rate at which work is done

P = ΔW/Δt

If we neglect energy losses then all work done equals the change in kinetic energy. Hence,

P = ρ*(πd^2/8)*v^3

Let’s calculate the power

P = 1.3*(π*15^2/8)*2.065^3 = 1011.45 watt

The professional cyclist could produce this power, but for very short period of time.
Image Source: Atlas Human-Powered Helicopter – AHS Sikorsky Prize Flight.

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