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I am currently making a model helicopter. I wanted to know how fast must my blades spin so that it can takeoff. I guess, for that I would need to calculate lift produced right? Can someone help me find what rpm must the blades spin at, so that it can takeoff? I amn't sure about the parameters required to calculate this, so I'll just give some of the important parameters (atleast, the ones I feel might be needed) :

  • No. of Rotors = 1
  • Surface Area covered by blades = $400 cm^2$
  • Mass of helicopter = 1.5 kg

If any more parameters are needed, please let me know. Thank you in advance.

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  • $\begingroup$ At the very least, you'll need the lift coefficient for this. $\endgroup$
    – Jacob B
    Nov 30, 2021 at 6:09
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    $\begingroup$ This question implies that you can assume a lift coefficient of 1.3 for a symmetrical aerofoil (as used by most RC helicopters) aviation.stackexchange.com/questions/46518/… $\endgroup$ Nov 30, 2021 at 11:18
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    $\begingroup$ We also need to know your rotor diameter. There are equations and explanations here: aviation.stackexchange.com/questions/80622/… $\endgroup$ Nov 30, 2021 at 14:52
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    $\begingroup$ @RobinBennett that question you just linked also says that the normal lift coefficient for small helicopters is 0.4, so which would be the right coefficient: 1.3 or 0.4? $\endgroup$
    – Jacob B
    Nov 30, 2021 at 16:02
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    $\begingroup$ The lift coefficient changes depending on the angle of attack, which is controlled by the pilot via the collective stick. 1.3 might be the maximum right before the blade stalls, and 0.4 a value for normal flight that leaves enough margin for steep turns and rough air. Ishaan is asking for the minimum possible rotor speed, with no margin. $\endgroup$ Nov 30, 2021 at 16:14

2 Answers 2

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Using the equations from this question

$$L = C_l \cdot A \cdot 0.5 \cdot r \cdot V^2$$

Where:

  • $L$ is lift force
  • $C_l$ is the lift coefficient
  • $A$ is the blade area (0.04 square meters)
  • $r$ is the air density (approximately 1 kg/m^3)
  • $V$ is the speed of the blades at about 70% of the radius

Taking the speed at $0.7r$ is only an approximation, as we should really integrate the lift over the length of the blade, to account for the tips moving faster than the root.

The required lift is 15N. (to lift 1.5kg) The maximum lift coefficient is 1.3 for a symmetrical aerofoil, according to this question.

That gives $15 = 1.3 \cdot 0.04 \cdot .5 \cdot 1 \cdot V^2$, or $V = 24m/s$

You don't specify the dimensions of your blades, only the area, so I'm going to guess that you have a 100cm diameter and 4cm chord to make the numbers easier. That is similar to a '500' size RC helicopter (which is about right for the 1.5kg weight too).

0.7 of the radius is 0.35m, and a circle of that radius has a circumference of 2.2m

24m/s divided by 2.2m is 11 revolutions per second, or 660rpm.

That seems low (normal head speed for a T-Rex 500 is 2500-3000rpm) but we're calculating the absolute lowest speed to hover, with the blades right on the edge of a stall, while this sort of heli is designed for aggressive aerobatics.

This approach also ignores losses at the blade tips and the non-lifting section near the hub and ignores the effect of downdraft on the body of the heli.

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    $\begingroup$ One question, why did you substitute Area with 0.04? How did you get 0.04? $\endgroup$ Dec 1, 2021 at 10:37
  • $\begingroup$ 0.04 square meters is 400 square centimetres, as there are 100*100 square centimetres in a square meter. $\endgroup$ Dec 1, 2021 at 19:49
  • $\begingroup$ Also, when should you use 1.3 as coefficient and when should you use 0.4 as coefficient? $\endgroup$ Dec 2, 2021 at 2:12
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    $\begingroup$ The lift coefficient depends on the angle of the blades, which is controlled by the pilot via the collective stick. 1.3 is the absolute maximum when the collective stick is all the way up. When you're on the ground with the control all the way down it will be zero and it can even be negative if set up for aerobatics and inverted flight. During normal flight it will be somewhere between 0 and 1.3, depending on how you are flying and 0.4 could be a reasonable figure for 'normal flight'. $\endgroup$ Dec 2, 2021 at 10:11
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    $\begingroup$ Cardboard may not be strong enough to handle the centrifugal forces from being spun around, and it might flutter if you don't add some weight to the leading edge. You can get lift from a flat plate but it's not as good as an aerofoil. This question says that the maximum lift will only be 0.7-0.8 though. aviation.stackexchange.com/questions/21391/… Bending it to look like the top surface of a wing is a significant improvement. $\endgroup$ Dec 2, 2021 at 12:21
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THERE IS A BIT OF MISUNDERSTANDING HERE, THE LIFT EQUATION CANNOT BE DIRECTLY USED FOR HELICOPTERS

The correct equation for rotary-wing is:

$T = ½ \rho (\omega R)^2 \pi R^2 C_T$

where $\omega$ is the rotating speed of the rotor and $C_T$ is the thrust coefficient ("thrust" since rotors generate no lift) which cannot be more than some 0.03 for very aerodynamically optimised rotors.


The lift equation given in the chosen answer is used to calculate lift generated by a wing (or, more generally, an aerodynamic body) with surface $S$ when it moves in a fluid of density $\rho$ with a well defined speed $V$:

$L = ½ \rho V^2 S C_L$

Why is this equation not suitable for rotary-wing applications? The speed $V$ seen by an helicopter blade is basically given by the sum of 3 different terms:

  1. the speed due to its rotation around the rotor shaft; this speed varies linearly along the bladespan from 0 at the rotor head to $\omega R$ at the tip, where $\omega$ is the rotating speed and $R$ the bladespan;
  2. the speed at which the whole helicopter is flying;
  3. during its rotation, the blade bumps into the wake shed by the previous blade; and after one complete rotation, it bumps into its own wake; these wakes also modify the local speed seen by the blade.

So, the sum of these 3 terms gives a total speed which continuously changes in space and time and that definitely cannot be represented by one single constant value $V$.

So, can the lift equation be used for an helicopter? No, because the blade does not move within a uniform and well defined airflow with steady speed $V$.

That equation can be used for airplane, cars, cyclists on bike, bridges and so on, but not for helicopter or, more generally, not for rotary-wing aircraft.

So, is it possible to use an equation as easy as that one for rotary-wing applications? Yes it is, but in the helicopter world nomenclature is a bit different: lift, per definition, is the component of the aerodynamic force perpendicular to the airflow. As discussed, the airflow seen by a rotor is the sum of 1.+2.+3. and it is definitely not constant, it changes continuously. Therefore defining a lift for a rotor is quite difficult simply because the airflow (and it's perpendicular component) changes continuously. Therefore for a rotor not a lift rather a thrust is defined: it is the aerodynamic force parallel to the rotor shaft and positive upward.

Analogously to the lift equation, also the thrust can be represented as the product of a density, a velocity (squared) and a surface but with the following distinctions:

  • as a velocity, the velocity of the rotating blade tip is used, that is $\omega R$
  • as a surface, the area $A$ of the rotor disk is used, that is $\pi R^2$

Therefore for rotary-wing application, the following thrust equation is used:

$T = ½ \rho (\omega R)^2 \pi R^2 C_T$

(Note: in USA the $½$ is normally dropped)

$C_T$, a bit like $C_L$, depends on several geometrical factors and trade-off, but in general its maximum value is within the range 0.008 and 0.028, according to how much optimised the rotor is.


Intermission

Let's do the math for Ingenuity making the following assumptions

  • $\rho = 0.01 kg/m^3$ Note: density on Mars changes according to season from 0.01 to 0.02; we use the lowest value in order to be conservative;
  • $\omega$ = 2400$rpm$ = 250$rad/s$
  • $C_T$, we suppose that the rotors are of a very optimised design = 0.028.

Then we get, remembering that Ingenuity has 2 rotors:

$T =2 \cdot ½ \cdot0.01(250\cdot0.6)^2 \pi 0.6^2 \cdot 0.028 = 7.125N$

Since $g_{mars} = 3.72m/s^2$ that corresponds to 2.17kg thrust, which is enough to hover and manoeuvre Ingenuity with its 1.8kg weight.


Bonus material: we do use (in the correct way) the lift equation

Is, in the helicopter world, the lift equation as given in the chosen answer really useless? If the hypothesis is done that the blade is composed by a continuous sequence of adjacent airfoils, then that equation can be locally used for each of these airfoils in the sequence.

So, for each slice of blade that equation can be used to get the local lift (and the local drag); afterward the local lift (and drag) is summed up (integrated) along the bladespan, from the rotor head till the tip, to get the total lift and drag; and finally the total lift and drag is decomposed in 1) a force parallel to the rotor shaft, which is the thrust generated by the rotor; and 2) a force perpendicular to it, which gives the torque needed to make the rotor spin.

Obviously that equation can be used if it is know the local speed $V$ on each slice of blade as given by 1.+2.+3. As seen, 1. and 2. are know (since $\omega$, $R$ and the helicopter speed are known); the speed 3. due to the wake depends itself on the lift and gives rise to a vicious circle: the lift depends on the wake and the wake depends on the lift! This vicious circle is broken by either using simplified models of the wake and/or wind tunnel measurements and/or CFD simulations.

Let's do an example that we will later test again on Ingenuity. If the simplification is done that:

  • the helicopter is hovering, i.e. term 2. is null;
  • the wake in/upon the rotor is constant everywhere, i.e. term 3. has a constant value;
  • the blades have a hyperbolic twist (this is normally true for a propeller blade, helicopter blades have a more linear twist but with a hyperbolic one the math is easier);
  • and each airfoil is operating at its $\alpha$ of maximum efficiency i.e. maximum $C_l/C_d$;

then we get that:

$C_T=¼ \frac{N_b c_{tip}}{\pi R} C_{l_{\alpha}} \alpha_{maxC_l/C_d}$

where:

  • $N_b$ is the number of blades;
  • $c_{tip}$ is the chord at the tip of the blade;
  • $C_{l_{\alpha}}$ is the slope of the airfoil's lift coefficient;
  • and $\alpha_{maxC_l/C_d}$ is the airfoil's $\alpha$ for maximum efficiency.

For Ingenuity we have:

  • $N_b=4$;
  • $c_{tip}=0.06m$ (ballpark estimate)
  • $C_{l_{\alpha}}=2\pi$ (typical thin-airfoils value);
  • and $\alpha_{maxC_l/C_d}=8°$ (this value really depends on the airfoil; I have chosen a standard NACA 2412 at a very low Reynold number). This gives:

$C_T=¼ \frac{4\cdot0.06}{\pi\cdot0.6} 2\pi \cdot0.14=0.028$ Yeah


Yet another bonus material: momentum theory

A first approximation of the thrust can be obtained also using the simple momentum theory which gives:

$ T = \sqrt[3]{2 \rho A P^2} $

Where:

  • $A$ is the rotor area;
  • $\rho$ is the density;
  • $P$ is the power needed to generate the thrust.

To see the simple momentum theory in action for Ingenuity, please have a look at this answer.

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  • $\begingroup$ Very helpful responce. I've tried looking these equations up and can't find the thrust equation anywhere but here. I am curious what these units are measured in. Such as R. Is .6 measured in meters when talking about bladespan? $\endgroup$ Oct 15, 2022 at 15:08
  • $\begingroup$ @OliviaRitter: thanks for you kind comment. Unfortunately both here and on aviation.se the wrong equation is the most voted and winner one... Anyhow, regarding the units of measurement you simply have to be consistent. I'll suggest you anyway to use SI: length in meter, density in kg/m³, velocity in m/s, rotational speed in rad/s and forces in Newton $\endgroup$
    – sophit
    Oct 16, 2022 at 5:50

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