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

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.

• welcome to aviation.se this is a homework question, we do not do homework, but we help you understand. in order to receive a good answer can you specify what have you tried? what do you not understand?
– Federico
May 10, 2017 at 20:20
• sir i want to know that if a full sized rotor is lifting lets say 635kg weight at given rpm and power at a given altitude at ft/min than how much half sized rotor will lift at same rpm and power at same altitude at ft/min will it be half the weight like 635/2=318kg ? May 10, 2017 at 20:38
• Your question is very hard to read because you have not used spacing and paragraphs. It is also hard to answer because an R22 cannot lift 635kg at 14,000 feet. May 11, 2017 at 6:23

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 $635 \cdot 9,8 \cdot 6,09 m/s = 37,9 kW...$

• Nice math... Though I am not convinced. OP says same power, which would mean a much higher RPM and all the extra losses associated with that. I guess it depends on your definition of $P$. May 11, 2017 at 13:15
• @Trevor From the Froude-Rankine 'actuator disk' approach, a similar formula (but with the constant...) can be derived, expressing the thrust of a propeller as a function of the power applied to the prop, its diameter and the density of the fluid... May 11, 2017 at 13:32
• Yes, but I think it worth mentioning, switching rotors wont actually get you "the same power applied to the rotor". What the appropriate derate factor /RPM is, is of course is indeterminate. May 11, 2017 at 13:43
• @Trevor Of course, the 'actuator disk theory' is not an exact solution, but just a useful, approximate approach... May 11, 2017 at 13:54
• For the smaller rotor: higher RPM or higher blade pitch? May 11, 2017 at 15:34

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.

• "I would look for a helicopter design rule-of-thumb to find the answer.": It's what the OP is asking: How to calculate the lift vs the rotor diameter, all other parameter unchanged?
– mins
May 10, 2017 at 23:22
• Exactly that is what i want to know May 11, 2017 at 10:09
• The loading per blade of the rotor is roughly the same for a wide variety of helicopters. If the number of blades is kept constant, the possible lift of the half-sized rotor will be a quarter of the original rotor. Of course, it will also absorb less power. However, in this case the engine is the same, so @xxavier's conclusion gives a better answer. Ideally, the smaller rotor would have deeper and more blades. May 12, 2017 at 10:41
• @PeterKämpf But that wasn't part of the problem statement ... which leaves the answers having to guess at the nature of "all else being equal." (Your points are well taken). It might be useful to engage with the OP and get a tad more definition for the question. May 12, 2017 at 12:29
• @KorvinStarmast: Yes, you're right. That's why I started the sentence with "Ideally". Anyway, this smells a lot like some homework question, so my level of engagement is low. May 12, 2017 at 13:32

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