A 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?
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
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.
F*Δt = ρ*(πd^2/4)*v^2*Δt
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.