How much lift is created by a half-sized rotor?

Question

I wonder if lift is directly proportional to the diameter of the rotor disc.

For example, if an engine spinning a 26 ft diameter rotor can lift 635 kg, will a 13 ft diameter rotor using the same engine lift 318 kg? The power absorbed by the rotor is the same in both cases.

Answer 1

For the hovering rotor, the stationary case, it can be safely assumed that the lift $L$ is a function of the input power $P$, the diameter $D$ of the rotor and the air density $\rho$.

Thus, $L = f(P,D,\rho)$

where $f$ is a function to be determined.

From dimensional analysis, the lift $L$ can be easily derived:

The variables are Lift $L$, dimensions $MLT^{–2}$; Power $P$, dimensions $ML^2T^{–3}$; Rotor diameter $D$, dimensions $L$ and air density $\rho$, dimensions $ML^{–3}$

The variables form a non-dimensional product $k$

$k = L^a\cdot P^b\cdot D^c\cdot \rho^d$ where $a,b,c,d$ are numbers to be determined.

Let’s form now a parallel product $k^*$ with the dimensions:

$k^* = (MLT^{–2})^a (ML^2T^{–3})^b (L)^c (ML^{–3})^d$

Clearly, $k^* = M^0 L^0 T^0$... We now take the exponents for each dimension:

$a + b + d = 0 \\ a + 2b + c – 3d = 0 \\ –2a – 3b = 0$

We make $a = 1$, since $L$ is the variable we’re going to solve for.

$b = –2/3 \\ d = –1/3 \\ c = –2/3$

Then,

$k = L^a\cdot P^b\cdot D^c\cdot \rho^d \rightarrow k = L\cdot P^{–2/3}\cdot D^{–2/3}\cdot \rho^{–1/3}$

Solving for $L$

$L = k\cdot P^{2/3}\cdot D^{2/3}\cdot \rho^{1/3}$

where $k$ is a constant

Hence, for rotor diameters $D_1$ and $D_2$, and for the same power and air density, the corresponding lifts $L_1$ and $L_2$ are:

$L_1/L_2 = (D_1/D_2)^{2/3}$

For the case of $D_1 = 13 ft$ and $D_2 = 26 ft$, $L_1/L_2 = (13/26)^{2/3} = 0,63$

In other words, the smaller (13 ft) rotor gives you, for the same power and air density, just 63% of the lift attained with the larger (26 ft) rotor.

That's for the hover. For the climb, you'll need extra power. In order to move 635 kg vertically upwards at 1200 ft/min (6,09 m/s) you would need $635 \cdot 9,8 \cdot 6,09 m/s = 37,9 kW...$

Answer 2

No, it won't be half. It will be much less than that.

The area of lift is circular, so halving the diameter decreases the area of the circle to 25% of the original. Area = πr².

Besides the effective rotor area, there are also losses for the body of the aircraft being in the center, though the rotor doesn't generate lift near the center because of low blade speed and not having the blades go all the way to the center. I would look for a helicopter design rule-of-thumb to find the answer.

Answer 3

Using momentum theory and assumption of uniform inflow distribution, rotor thrust T can be expressed as

$$ T = C_T * ρ * A * (ωR)^2 $$ And power P as

$$ P = C_P * ρ * A * (ωR)^3$$

From empirical data, for rotor solidity = 0.1:

Case 1: R = 13 ft = 3.96m, A = π * $R^2$ = 49.32 $m^2$, $T_1$ = 635*9.81 = 6229.4 N, assume ρ = 1.225 kg/$m^3$ and tip speed = ωR = 200 m/s

=> $C_T$ = 6229.4/(1.225*49.32*(200)$^2$) = 0.0026; $C_P$ = 0.00025; P = 120.8 kW

Case 2: R = 6.5 ft = 1.98m, A = π * $R^2$ = 12.32 $m^2$, P = 120.8 kW, assume ρ = 1.225 kg/$m^3$ and tip speed = ωR = 200 m/s

=> $C_P$ = 120,800/(1.225*12.32*200$^3$) = 0.001, $C_T$ = 0.0105,

$T_2$ = 0.0105 * 1.225 * 12.32 * 200$^2$ = 6338.6N

According to momentum theory and uniform inflow assumption, the smaller rotor delivers 2% more thrust at equal power. The rotor runs at twice the RPM and the blade is at maximum angle of attack, may even be stalling. It is the absolute maximum thrust it can deliver.

Surprised because it is counter-intuitive? So am I.

Source: Principles of Helicopter Aerodynamics by J. Gordon Leishman section 2.5