From my understanding helicopter blades work similar to a planes wings, ie the air going over is faster due to the shape. So my question is why then are the blades rectangular? ie they are the same width the whole length of the blade? I feel like it would make more sense to have it thinner towards the base since the blade is moving slower, so you could save on material cost and weight.

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    $\begingroup$ Helicopter blades are already really highly stressed to begin with and quite small compared to the size of the aircraft when compared to airplanes. Could you thin the root without failure? $\endgroup$
    – DKNguyen
    Commented Mar 19 at 13:33
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    $\begingroup$ @DKNguyen Yes but how could any of that help here? $\endgroup$ Commented Mar 19 at 23:44
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    $\begingroup$ The lift from an airplane wing has little to do with the air being faster above the wing than below due to shape. It works by pusing (and to a certain extent, "pulling") air downwards. You can fly with a purely rectangular wing profile. Many planes can fly upside-down. The reason for the elongated teardrop profile you often see in diagrams is for optimising lift-drag ratio, not a fundamental necessity for flight. $\endgroup$
    – Arthur
    Commented Mar 21 at 9:17

7 Answers 7


The lift of a wing or propeller blade section is proportional to the area of the section (viewed from above) and the square of the velocity of the section going through the air. Ref NASA lift equation. $$ L \propto A \ V^2$$

The sections nearest the base are moving the slowest so for equal width, the inner sections are providing the least lift. If anything, the blades should be wider at the inner to mid sections, because the inner sections have the shortest leverage lengths and put the least bending torque stress on the hub for a given lift contribution. The inner sections also have to support the outer sections, so greater cross sectional area on the inner sections helps to support the outer sections. There is of course a region close to the hub where the amount the width would have to increased to compensate for the reduced velocity is excessive and counterproductive and just adds weight and drag.

enter image description here

A horizontal aircraft propeller is very similar in principle to a helicopter blade and aircraft propellers generally have wider sections nearest the base, except for the sections right next to the hub that are so slow, relatively speaking, that they are generally just shaped for strength and low drag, rather than shaped for lift or thrust. The aircraft propeller starts out narrow (because the lift very near the hub is just a waste of time, gradually increases to a maximum width where the lift is useful and then tapers off again towards the tips, where excess width just causes excessive bending torque.

Generally speaking, the outer parts of a wing or propeller are the first to stall and it is good to have a solid less sensitive wing profile near the base to provide as much lift as possible in a stall condition.

Wings and helicopter blades are perhaps surprisingly not always made to an optimal aerodynamic shape and other conditions have to be taken into account. For example, the elliptical plan shape of the spitfire aircraft wing is said to be ideal, but aircraft rarely have elliptical plan shape wings, due to the excessive complication of manufacturing them and there are many other factors and compromises to be taken into account such as the aforementioned stall characteristics.

The aerofoil cross section of a helicopter blade is highly critical part of the helicopters design and often it is easier to model the predicted performance of a blade by just using a consistent cross sectional length for the entire blade. The design of a helicopter blade is further complicated because in fast forward flight parts of the blade that are advancing are going faster than the helicopter and parts of the blade that are retreating may actually experience backward airflow over some blade sections, so the helicopter blade has to be very resilient to a wide range of flow conditions and compromises have to made.

Consider a blade of radius D that is wider at the tips. A section at the tip might provide X amount of lift. This amount of lift creates a bending torque of X*D on the blade. If instead we put the wide section of the blade nearest the hub, say at 0.01 D, then the bending torque on the blade for the same amount of lift is 0.01 X D and so the blade can be made lighter as it has to resist less bending torque.

Another factor to be taken into account is drag. One form of drag is skin friction drag and this is proportional to the cross sectional width and velocity squared. Have too large a cross sectional width near the tips where the velocity is highest increases the overall drag.

From the above it might seem that short fat stubby blades would be ideal, but there is another form of drag called lift induced drag and this is inversely proportional to the aspect ratio and favours long thin wing and blade profiles for low drag. Helicopter rotors generally have a larger radius than propellers and this is because the hover efficiency is better for a low disk loading that results from having a greater radius.

enter image description here

An examination of the image above reveals another reason helicopter rotors require a larger radius than a propeller. During forward flight the part nearest the hub on the retreating side fails to produce lift and it it was not for the part of the rotor outside the shaded area, the helicopter would have no lift on one side and be unable to fly normally. A smaller diameter rotor would limit the maximum forward motion of the helicopter.

You may have noticed that the rotors of a helicopter parked on the ground look fairly flimsy and bend towards the ground under their own weight and wonder how they manage to support the weight of the entire helicopter in flight without just bending into a useless cup shape. What prevents this happening is centrifugal force that straightens the rotors out when under power and this effect is befitted by having longer blades with higher tip velocities.

Unlike any regular aircraft wing or propeller, a helicopter blade has an adjustable angle of attack and this is continually changing cyclically with every revolution and yet another layer of complexity to the design and choice of the aerofoil section and plan form of a helicopter blade. EDIT: As mentioned by @kruemi in the comments, it is worth noting that the rapid cyclical change of pitch of the blade cause an additional stress on the system and a loner thinner blade benefits from a lower pitch moment of inertia and reduces the stress on the hub mechanism and therefore the required weight of the mechanism.

A quick google image search of helicopter blades seems to show the majority either have the widest part of the blade nearest to the rotor hub

enter image description here

or the edges are parallel. They almost never have the widest part of the blade near the tip. To be fair the blades with the widest part nearest the hub are generally for model RC helicopters, where simplicity of manufacture is not such a big concern as the blade can just be injection molded, which is probably not usually the case for full scale manufacture. Models also operate at lower Reynold's numbers and so may not be directly comparable with full scale.

Lastly, lets consider the total planform area of the blade required to provide the required lift. A blade with the widest section at the tip would require the least total area of blade to provide the required lift (due to the higher velocities near the tip) and this should correspond to a lower total material cost, but the increased bending torque stress on the hub negates this advantage due having to build a stronger blade to withstand the bending. A blade with the widest part near the hub requires a larger total planform area, but does not have to endure high bending stress. In the end, a parallel blade is the best compromise between the two extremes (and easier to manufacture).

  • $\begingroup$ There might be structural concerns too not thinning the root. $\endgroup$
    – DKNguyen
    Commented Mar 19 at 13:32
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    $\begingroup$ I have mentioned that the root sections have to support all the remaining outer sections and that the design of the root sections of a propeller are mainly about strength rather than lift or thrust. $\endgroup$
    – KDP
    Commented Mar 19 at 13:37
  • $\begingroup$ Since the pitch of the blade changes twice per rotation: could it be that a wider blade would require more force (more inertia around that axis with a wider blade) and make it infeasible? Also the force for the pitch change has to be the same at the tip and at the root so the root of the blad has to be able to withstand that force too... so it might be just easier for engineering again to go with a more or less square shape than a maybe more optimized shape. $\endgroup$
    – kruemi
    Commented Mar 22 at 11:14

From my understanding helicopter blades work similarly to a plane's wings, ie the air going over is faster due to the shape.

This is a widespread misconception. Lift is produced due to air mass getting displaced by the blades due to the angle of attack. The idea that flow at the bottom and top converges at the same time and hence one is faster than the other is a fallacy. For more information, please read this s.e post.

So you don't need the aerofoil shape to make the helicopter fly. Even a flat rectangular blade cross-section can provide lift if you have the right amount of angle of attack.

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    $\begingroup$ Calling it a fallacy is too strong. The higher speed flow above the wing is inextricably connected to the downward deflection by pressure and continuity. It happens even if the wing is flat. You can't really say which is cause and which is effect. $\endgroup$
    – John Doty
    Commented Mar 19 at 17:15
  • $\begingroup$ even airplane can fly with flat wings. You only need more power, in the limit you could have no wing at all if you have vector thrust $\endgroup$
    – basics
    Commented Mar 19 at 17:35
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    $\begingroup$ @basics "You only need more power" which I think we can conclude means that the curvature contributes to lift. So, with regard to the common 'well actually, the curve in wings isn't creating lift', well actually, that's not quite right either, to John Doty's point. $\endgroup$
    – JimmyJames
    Commented Mar 19 at 20:52
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    $\begingroup$ @JohnDoty The same-time argument is the fallacy, not the faster speed itself. The incorrect explanation for the faster speed is the problem. I have then problems explaining to people that it is actually OK to use the Bernoulli equation for the lift, e.g. with reasonable results with Euler equations for low angle of attacks, because they heard somewhere that it is wrong. The wrong thing is the equal time argument. $\endgroup$ Commented Mar 20 at 12:38
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    $\begingroup$ @VladimirFГероямслава Indeed. It is also wrong to call "the air going over is faster due to the shape" a misconception, since the air really does flow faster over the top of an unsymmetrical airfoil at "zero angle of attack". And, if you like, that's really the misconception, since angle of attack for an unsymmetrical airfoil is somewhat arbitrary. $\endgroup$
    – John Doty
    Commented Mar 20 at 14:05

Since this is a physics site, and not an engineering site, I am going to get the napkin out and solve the Spherical Cow in a vacuum version:

Coordinates: Polar, Zero at the center (duh?). Rotating, ofc.

Blade: from $r=(0, R)$

Blade width: $ w(r) = w_0 + ar $

Spherical Cow version: $w_0 = 0$

Blade Cross section : $A(r) \propto w(r) = ar$

Mass supported at $r$:

$$ m(r) \propto \int_r^R A(r)dr \propto \int_r^R rdr$$

$$ m(r) \propto (R^2 - r^2) $$

But weight (pun!) there's more: Let's respect general relativity (centrifugal force is gravity) so, the weight supported at $r$:

$$ W(r) \propto \int_r^R m(r)g_c(r)dr $$ $$ W(r) \propto \int_r^R (R^2-r^2)r^2dr$$ $$ W(r) \propto R^2r^3 - r^5 $$

Though I'm living in a material world, I am not a material guy. Nevertheless: let's estimate internal stress:

$$ S(r) \propto \frac{W(r)}{A(r)} \propto \frac{R^5-r^5}{r} = \frac{R^5}{r} - r^4 $$

That's going to break at $r=0$, but we knew that. You can make $w_0>0$ (elliptical cow), or set $A(r) \propto w^n(r) $..but that's getting into engineering.


Helicopter are complex systems

Helicopter are complex systems. As an example:

  • several regimes of motion:
    • blades of a rotor should be able to work well with compressibility at the tips, and reverse flow at roots in fast advancing flight
  • rotating mechanical elements:
    • complex mechanics
    • vibrations
    • wearing and fatigue, and thus maintenance, with the associated times and costs
  • as a lot of aeronautical systems, that must be light enough to be able to flight in an efficient way, helicopter usually is an aero-servo-elastic system, where aerodynamics, structure, controls and pilot interacts in a coupled system

Helicopter blades: centrifugal load

Helicopter blades are subject to high tension load due to centrifugal force.

To keep things as simple as possible, and modelling a blade as a straight 1D beam element with uniform linear mass density $m$, rotating with angular velocity $\Omega$.

Axial force at root must equilibrate centrifugal force. Approximating $R_{root} = 0$,

$$N_{root} = \int_{R_{root}}^{R_{rotor}} m r \Omega^2 dr = \frac{1}{2} m \Omega^2 R_{rotor}^2 \ .$$

Axial load along the blade span is $N(r)$

$$N(r) = N_{root} - \int_{R_{root}}^{r} m r \Omega^2 dr = \frac{1}{2} m \Omega^2 R_{rotor}^2 - \frac{1}{2} m \Omega^2 r^2 = \frac{1}{2} m \Omega^2 \left( R_{rotor}^2 - r^2 \right)$$

As you can see, the axial load in this simple model increases with $r^2$ moving from the tip to the root of the blade. This could be one of the reasons helicopters don't have highly tapered blades close to the hub.


  • If you want some approximate number. A medium-size helicopter like Augusta Westland 139 has a rotor radius $R \sim 6.5 \, m$, and the angular velocity is approximately $300 \, \text{rpm}$ (you can't increase it much more than that to avoid undesirable compressibility effects, namely shocks, at blade tips);
  • if you take a look at blade, you could easily realize that it has a cut-out or a region close to the hub that has bluff sections: this region needs to hold high loads, and faces reverse flow in forward flight. You'd better have bluff body there, since lift is mainly created in sections close to blade tips;
  • if you're interested in beam structures, you should realize that they're anisotropic structures, in general with all the forces and displacement coupled (to deal with aeroelastic effects), with structure mainly meant to hold axial loads due to centrifugal force and torsion (it usually has the typical section of aeronautical beams, with closed thin-walled bays to hold torsion)

There actually is an ideal taper shape for helicopter blades. Rather than narrow at the tip and thin at the root, it actually follows the opposite shape. It is not a linear relationship, it follows something more like 1/x (I would have to look up the exact functional shape). This shape would provide the 'minimum induced loss'.

In general, it is not worth going after a MIL design using blade chord alone. You can also use airfoil changes or blade twist to work towards a MIL design.

However, helicopter blades are very flexible and the benefits of a carefully twisted blade start to go away when the blade is constantly twisting as it goes around.

If you look at some of the rotor designs for new eVTOL aircraft being designed (say Joby or Wisk), you will see that there is a lot more going on than with a typical helicopter rotor.

In these cases, the size and rigidity of the blade make it make sense to put more nuance into the design.


First, helicopter blades have high aspect ratio, so already pretty good aerodynamically. There is no real need to pursue the small gains available ftom non-parallel forms unless performance is very much more important than cost.

Secondly, parallel is very much easier to manufacture, so utility helicopters at the lower end of the price range - almost all helicopters - are unlikely to depart from this design.


Unlike propellers, helicopter blades are not rigid. Their design is significantly constrained by dealing with the tension forces they experience.

The line of force on the rotor is ideally within the rotor itself, and it is desirable to keep it that way to reduce stress on it. The rotors structural components will be either composite, with lays aligned with these forces or extruded aluminium.

So rather than reducing near the root, where rotor blades are tapered they are tapered towards the tip, so that the lift is more linearly distributed along the blade, meaning the line of force is straighter and the blade less stressed.

The rotor is dished in flight, behaving a bit like horizontally spinning a rock tied to a string - the string forms a cone to align with the line of force.

If the weight of the copter is 1000 kgf and the rotor is dished to a 1 in 10 angle, then there is 10,000 kgf acting along the rotor - one unit pulling the aircraft up, and the rest centripetal forces horizontally. This force develops along the rotor, so the whole force acts at the root and falls to zero at the tip.

As the tip has zero tension on it and the root has full tension, it is again desirable to taper the blade towards the tip not toward the root.

But many blades get their strength by being extruded, and it is not practical to create tapered extrusions without compromising their strength or increasing cost.

There is one thing that blades do have which fits your question though, the root cut out - where the part of the blade nearest the root is thinned or replaced by a spar - as that part has least lift and complicated angle of attack.

http://www.helistart.com/RotorBladeDesign.aspx has more information


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