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I am new! New to this forum and new to aviation (i normally deal with statics) so thanks in advance for beeing patient with me.
Recently i was trying to figure out how much weight a rotor can lift as a function of the rotor diameter. Therefore I came across this formula. $$ T_{rotor} =(0.5*P^2*η_{rotor}^2*η_{el}^2*π*d_{rotor}^2*ρ_{Air})^{1/3} $$ This brings me already quite close to my goal as I now have $ Thrust = f(P^2, d^2)$. Now I tried to relate the engine power to the rotor diameter by using :

$ P=T * {rpm}$

with $ T=I *α=(0.5* m*r^2)*v_{sound}/r $

and $ {rpm}=v_{sound}/(\pi*d)$

Which leads me to $$ P(m)=(v_{sound}^2/(4*π))*m $$


From this calculation a few questions arise:

  1. Do you think this is a good estimate (within +-15%)?
  2. Is there a better formula to estimate $I_{rotor}$ than using $I_{zylinder}$?
  3. Is there any formula to estimate $ m_{rotor} = f(d_{rotor}) $


I want to add that this estimate is for a rough feasability study and not to design a rotor or aircraft. I also know that several of the values in this formulas are a function of altitude and aircraft velocity. Please dont take this into account. At the moment I am only interested in a stationary solution.
Thanks for your help!

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  • $\begingroup$ Please don't forget the root, which makes thrust proportional to d$^⅔$ only, less than linear. $\endgroup$ – Peter Kämpf Oct 13 '17 at 7:11
  • $\begingroup$ You lost me at the P = T * rpm. Is this T for torque? What is I and which $\alpha$ are you using, angle of attack? $\endgroup$ – Koyovis Oct 13 '17 at 7:33
  • $\begingroup$ The $T = I \cdot \alpha$ is the formula for determining the torque as a function of the angular acceleration, which you can also see here. This however is not the right formula to use, as it completely neglects all air drag, and only accounts for inertia. What would happen if we have a constant rpm, and thus $\alpha = 0$? We need 0 torque to drive the rotor? OP, please check where the formula comes from, and its applicability $\endgroup$ – ROIMaison Oct 13 '17 at 12:44
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The dimensional approach is simple... 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 to be determined...

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  • $\begingroup$ How is lift proportional to $\rho^{1/3}$? In the equation $T = C_T \cdot \rho \cdot A \cdot (\Omega \cdot R)^2$ the rotor thrust is proportional to $\rho^1$ $\endgroup$ – Koyovis Oct 16 '17 at 15:36
  • $\begingroup$ @Koyovis. The dimensional derivation is correct. Hence, rho must have the exponent 1/3... $\endgroup$ – xxavier Oct 16 '17 at 18:26
  • $\begingroup$ @Koyovis In en.wikipedia.org/wiki/Disk_loading#Momentum_theory you'll find an expression for the power required in hover as a function of the thrust, disk area, and air density. Solving that expression for thrust, you'll see that rho appears with the exponent 1/3... $\endgroup$ – xxavier Oct 16 '17 at 19:02
  • $\begingroup$ For the power, yes. But we're trying to find thrust as a function of blade radius. $\endgroup$ – Koyovis Oct 16 '17 at 23:41
  • $\begingroup$ @Koyovis If you solve that expression for thrust, you'll find that rho has the exponent 1/3, exactly as when derived by the dimensional method. $\endgroup$ – xxavier Oct 17 '17 at 6:44
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No the equation in your question is not useable. However there are tabulations of existing helicopters that provide a better method.

Maximum stationary thrust as a function of the rotor diameter. What you are looking for is related to the disk loading of a helicopter: when in hover, the rotor thrust equals weight. Heavier helicopters must unfortunately use higher disk loading, a feature dictated by the Square-Cube law. From Helicopter Aerodynamics by J. Gordon Leishman:

enter image description here

This is a direct relation between blade radius and max gross weight, no need to introduce the installed power etc into it. Note that the trend line is proportional to $W^{1/3}$.

When in hover at max gross weight, there is some more lift in reserve. How much depends on the disk loading and the air density. In another question we came to an estimation of twice the max gross weight at sea level valid for the disk loading of the Prouty example helicopter = 7.1 lb/ft$^2$ / rotor radius of 30 ft / max gross weight of 20,000 lb. For heavier helicopters the disk loading is higher, and this lift reserve reduces with disk loading and gross weight.

enter image description here

The above is a digitised plot of the line in Leishmans fig 6.3 in metric units, weight in N and radius in m. Lift in the hover is proportional to the square of the rotor radius - which makes sense, proportional to the rotor disk area as in equation:

$$T = C_T \cdot \rho \cdot A \cdot {V_{tip}}^2$$

As the rotor radius gets larger it turns slower, to prevent the blade tip from exceeding the critical Mach number.

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  • $\begingroup$ Thanks a lot for your expalantions. The graphs are very helpful and exactly what i am looking for. The lower graph is from Leishmans "Principles of Helicopter Aerodynamics"? $\endgroup$ – tschifeif Oct 25 '17 at 2:30
  • $\begingroup$ The upper graph was used as a baseline. It is on a logarithmic scale, I digitised it, converted to metric and plotted on a linear scale. $\endgroup$ – Koyovis Oct 25 '17 at 2:37
  • $\begingroup$ I see :) Thanks again $\endgroup$ – tschifeif Oct 25 '17 at 3:37

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