# Questions on how helicopters fly

• B
• Whitestar
In summary: The lift force that each blade can induce is proportional to the square of the average (root to tip) speed with which it moves within the surrounding air.

#### Whitestar

In a previous thread, I brought up the topic of Airwolf and the wonderful people here have explained why it is impossible for Airwolf (or any helicopter) can't fly at supersonic speeds.

Interestingly enough, it was Airwolf that got me fascinated with helicopters in the first place! LOL! Anyway, I did some further research and was surprised to discover that when a helicopter flies forward, the right side of the main rotor decreases the angle of attack in order to maintain equal lift across the rotor disc, while on the left side of the main rotor the angle of attack of the retreating blade increases to compensate for the lower blade tip speed. Once the critical angle of attack is reached, the blade stalls at the blade tip in the 9 o'clock position and moves inwards.

Please see the following diagram below:

The thing that I find puzzling is why does the advancing blade (the right side) decreases the angle of attack while the retreating blade (the left side) increases the angle of attack? Does it have to do with the fact that it is rotating from the right to the left?

Whitestar said:
Does it have to do with the fact that it is rotating from the right to the left?
No, it has to do with the aerodynamics. If you reversed it then you would have the advancing blade increasing its angle of attack and the air would be pushing very hard on the bottom of the high-speed blade, creating a lot of lift and on the other side the retreating blade would have less speed (relative to the air) AND have a lowered angle of attack and therefor very little lift. SO ... the helicopter would quickly flip over to the left.

There's a Youtube video that contains a short cut of a the view from a perspective stationary with respect to the rotors. It starts at 0:33s. You can see the pitch variation.

(Never mind the rest of the video; I am 99.9% certain it's mostly fake. There are lots of knockoffs of it online with various less believable mishaps).

Whitestar said:
The thing that I find puzzling is why does the advancing blade (the right side) decreases the angle of attack while the retreating blade (the left side) increases the angle of attack? Does it have to do with the fact that it is rotating from the right to the left?
The lift generated by a blade depends on its angle of attack (all else being the same more AoA mean more lift until the thing stalls) and its speed through the air around it (all else being the same, more speed means more lift) .

Because of the forward motion of the helicopter the advancing blade is moving faster through the air than the retreating blade. Thus the retreating blade needs more agle of attack to keep the lift the same on both sides.

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Bystander
Whitestar said:
The thing that I find puzzling is why does the advancing blade (the right side) decreases the angle of attack while the retreating blade (the left side) increases the angle of attack? Does it have to do with the fact that it is rotating from the right to the left?
In that diagram, do you clearly understand the reason behind the difference of the blade tip tangential velocity between 3 o'clock and 9 o'clock?

Lnewqban said:
In that diagram, do you clearly understand the reason behind the difference of the blade tip tangential velocity between 3 o'clock and 9 o'clock?
Well, from what I understand is the blade tip at the 3 o'clock reference equals forward speed and the blade tip at the 9 o'clock reference equals negative speed, correct?

Whitestar said:
Well, from what I understand is the blade tip at the 3 o'clock reference equals forward speed and the blade tip at the 9 o'clock reference equals negative speed, correct?
Yes. So did you understand post #2?

Whitestar said:
Well, from what I understand is the blade tip at the 3 o'clock reference equals forward speed and the blade tip at the 9 o'clock reference equals negative speed, correct?
Let's assume that the helicopter is not moving forward at first.
In that situation, the tangetial velocity of the tip of any blade is 10 m/s respect to ground and non-moving surrounding air.
Because of that, each blades keeps the same angle of incidence all around the rotation.

The lift force that each blade can induce is proportional to the square of the average (root to tip) speed with which it moves within the surrounding air.
https://www.grc.nasa.gov/www/k-12/airplane/lifteq.html

The lift force that each blade can induce is also proportional to the average angle of attack (AOA).
https://www.grc.nasa.gov/www/k-12/airplane/incline.html

Please, note that I have used the word average in both cases, because in reality, the tangential velocity of each section of the blade increases from the root toward the tip.
For that reason, each blade must have a built twist, which decreases the angle of incidence of each section from the root toward the tip.
By having that shape, we can achieve a more or less constant value of lift for the whole span of the blade.

But let's go back to focus on the rotating tip of one blade.
Respect to ground and still air, that tip was moving at 10 m/s.

Lnewqban said:
For that reason, each blade must have a built twist, which decreases the angle of incidence of each section from the root toward the tip.
That is probably true for all propellers, but not for all helicopters.
Many helicopter blades are free of twist or have reduced twist so that lift is generated over a zone somewhere along the blade that can move over a full rotation. That is safer because there will always be somewhere on the blade providing lift.

hutchphd and phinds
Normally air flows downwards through the helicopter rotor, while the blade lift supports the aircraft.

With an autogyro, air flows upwards through the rotor causing autorotation as the aircraft is propelled forwards. The blades glide and so support the weight of the autogyro. Straight blades without twist are used for autogyros. That way the section generating lift can change dynamically throughout every rotation.

Loss of power is an emergency in a helicopter. Autorotation then becomes essential. But the change to upward flowing air would require blade twist be reversed to that for normal flight. The safe compromise is straight blades.

Klystron and hutchphd
Baluncore said:
...
Many helicopter blades are free of twist or have reduced twist so that lift is generated over a zone somewhere along the blade that can move over a full rotation.
Would you mind explaining this a little further?
Isn't the whole blade moving over a full rotation?

Lnewqban said:
Isn't the whole blade moving over a full rotation?
With an untwisted straight blade, the part close to the centre of the rotor is too slow to provide lift, while the tips are often stalled and making noise. Between those extremes, the intermediate part of the blade that is able to provide optimum lift, changes dynamically in position over each rotation.

The radius to the point of maximum lift changes with pitch control, direction, and speed of flight.

Lnewqban
Baluncore said:
while the tips are often stalled and making noise.
I wonder how long this has been included (optimally) in rotor blade design. It could be one of the features that makes old copters so noisy and new ones so quiet. I remember the sounds of the helicopters in MASH, particularly and compare it with what we hear when the Police helicopter goes over, these days. Of course, there's the engine difference too but the thwock is at the rate of the rotors.

sophiecentaur said:
It could be one of the features that makes old copters so noisy and new ones so quiet.
Blades were once extruded from aircraft aluminium, so were easily made straight and cut to length for X-ray inspection. But now that computer generated composite shapes are possible, tip modifications to shed a vortex quietly, seem to dominate over blade twist modification.

I thought the thwock was the sound of the blades passing the fuselage rather than a direct radiation of blade or tip noise. Why do/did they sound like that?

phinds said:
Yes. So did you understand post #2?
Not entirely. Could you please elaborate?

Whitestar said:
Not entirely. Could you please elaborate?
No, I can't make it any simpler than that.

Whitestar said:
Not entirely. Could you please elaborate?
"They really can't fly, they just beat the air into submission," over heard in the Nam.

russ_watters and davenn
Continuation of post #8:
I hope that you could follow the above explanation, with the helicopter hovering a fixed point on the ground and with no wind.
Let's forget about the twist of the blade, following above @Baluncore explanation.
Just consider one of the blades as a straight rectangular wing moving horizontally through still air at a speed of 10 m/s respect to the ground.

Now, the velocity of the wind increased from 0 to 1 m/s and its horizontal flow is from the nose to the tail of the helicopter.
From the point of view of the wing (or blade), what two values of air flow velocity will be felt at 9:00 and at 3:00 o'clock?

Whitestar said:
The thing that I find puzzling is why does the advancing blade (the right side) decreases the angle of attack while the retreating blade (the left side) increases the angle of attack? Does it have to do with the fact that it is rotating from the right to the left?
When you ride a bicycle at a speed of 10, the rubber at the bottom of the wheel is in contact with the ground, so is doing a speed of zero. The axle is moving with the bicycle at a speed of 10, while the rubber at the top of the wheel is moving forward at a speed of 20.

When you hover a helicopter with a blade tip velocity of 300, the blades all have the same tip velocity, from zero at the centre to 300 at the tip.

Lift from an airfoil or a rotor blade is a function of airspeed squared, but lift is also related to angle of attack.

When that helicopter flies forwards at a speed of 100, the blade tips on the right are moving through the air at; 300 + 100 = 400; while the tips on the left have an airspeed of only 300 - 100 = 200.
Due to blade airspeed, the lift from the tips on each side of the helicopter rotor will differ by, 400² to 200²; a ratio of 16:4 or 4:1.

Now consider the part of the blade that does most of the work, at about 2/3 of the tip radius. When hovering the blade speed at that point will be 300 * 2/3 = 200;
When the helicopter is flying forwards at a speed of 100, the blade airspeed on the right will be; 200 + 100 = 300, while on the left it will be, 200 - 100 = 100; which is there a ratio of 9:1.
The helicopter cyclic control makes it possible to cyclically change the blade angle of attack during the rotation, so as to balance the lift from the blades on the two sides of the rotor.

berkeman and Lnewqban
Baluncore said:
When you ride a bicycle at a speed of 10, the rubber at the bottom of the wheel is in contact with the ground, so is doing a speed of zero. The axle is moving with the bicycle at a speed of 10, while the rubber at the top of the wheel is moving forward at a speed of 20.

When you hover a helicopter with a blade tip velocity of 300, the blades all have the same tip velocity, from zero at the centre to 300 at the tip.

Lift from an airfoil or a rotor blade is a function of airspeed squared, but lift is also related to angle of attack.

When that helicopter flies forwards at a speed of 100, the blade tips on the right are moving through the air at; 300 + 100 = 400; while the tips on the left have an airspeed of only 300 - 100 = 200.
Due to blade airspeed, the lift from the tips on each side of the helicopter rotor will differ by, 400² to 200²; a ratio of 16:4 or 4:1.

Now consider the part of the blade that does most of the work, at about 2/3 of the tip radius. When hovering the blade speed at that point will be 300 * 2/3 = 200;
When the helicopter is flying forwards at a speed of 100, the blade airspeed on the right will be; 200 + 100 = 300, while on the left it will be, 200 - 100 = 100; which is there a ratio of 9:1.
The helicopter cyclic control makes it possible to cyclically change the blade angle of attack during the rotation, so as to balance the lift from the blades on the two sides of the rotor.
Interesting comparison! So, what would be the case if the blade was rotating from the left to the right?

* Note: I flipped the image in the opposite direction for you to see.

Whitestar said:
So, what would be the case if the blade was rotating from the left to the right?
Seriously? Which word did you not understand?

berkeman said:
Seriously? Which word did you not understand?
I'll admit I am somewhat confused because I am attempting to understand why on the left rotor the speed is less. I get that the left side of the blade is the retreating blade because it is moving away from the helicopter, but if you were to move the retreating blade in the opposite direction, would it make any difference?

http://www.av8n.com/how/htm/yaw.html#sec-p-factor

Perhaps we could analyze a regular airplane first.
Have you noted that fixed-wing airplanes take-off and land straight against the wind when possible?
The purpose is to have a higher air speed flowing around the wing.
By doing so, less runway length has to be used; therefore, more safety margin is available if something goes wrong.

The fixed wing does not know or care about its speed related to the ground, that is a business to be attended by the landing gear.
The wing only reacts with a combined lift-drag force to the velocity of the airflow hitting its leading edge.

If the recommended take-off speed is 100 knots (51 m/s), it means that the wing needs an airflow of 51 m/s for safe take off.
No wind means that the airplane needs to roll over the ground at 51 m/s.
A desirable head wing of 6 m/s means that the airplane needs to roll over the ground at 51-6=45 m/s.
A undesirable tail wing of 6 m/s means that the airplane needs to roll over the ground at 51+6=57 m/s.

In the case of our helicopter, the 3 o'clock blade tip is moving through the air in the same direction of the fuselage, only than faster.
That tip does not know or care about the velocity of the fuselage, it only feels a great deal of airflow above and below it; therefore, it generates abundant lift.

Simultaneously, the 9 o'clock blade tip is moving through the air in the opposite direction of the fuselage, only than slower than it (and much slower than the opposite 3 o'clock blade tip).
That tip does not know or care about the velocity of the fuselage, it only feels a marginal airflow above and below it; therefore, it generates poor lift.

Because of the above, the relative-to-the air velocity of the 3 o'clock blade is greater than the relative-to-the air velocity of the 9 o'clock blade.
If we reverse the rotation, the opposite would be true.
In both cases there will be a lack of right-left balance of lift forces, and the helicopter will roll over.

Because there is no much to do about those velocities, which affect the amount of generated lift on a square ratio, as we have discussed above, the only thing to modify in order to regain the needed balance is the angle of attack of the blades as they successively approach the 3 o'clock and 9 o'clock positions.

As we have discussed above, the amount of generated lift depends more or less linearly on the angle of attack (for small angles between 0 and around 10) of a fixed wing or rotating blade.

Using that trick, we want to reduce some the AOA of the faster (respect to the surrounding air) 3 o'clock blade, and to increase some the AOA of the slower (respect to the surrounding air) 9 o'clock blade.

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Whitestar
Whitestar said:
I'll admit I am somewhat confused because I am attempting to understand why on the left rotor the speed is less. I get that the left side of the blade is the retreating blade because it is moving away from the helicopter, but if you were to move the retreating blade in the opposite direction, would it make any difference?
Hmm, maybe you're not seeing how the "apparent wind" from the forward motion of the helicopter interacts with the blades? It's been explained several times (including with numbers), but do you see how the advancing/right blade's apparent speed through the air is the addition of the blade's speed in still air *plus* the apparent wind speed of the helicopter moving forward? And the apparent speed through the air of the retreating blade on the left side is the blade speed in still air *minus* the apparent wind speed from the helicopter moving forward? The difference in those apparent speeds through the air of the blades in different rotational positions is what requires modulating the AOA to maintain the same lift by the blades in the different rotational positions.

Klystron and Lnewqban
Whitestar said:
I get that the left side of the blade is the retreating blade because it is moving away from the helicopter, ...
The advancing blade is moving forwards relative to the helicopter direction of flight.
The retreating blade is moving backwards relative to the helicopter direction of flight.

Klystron and russ_watters
Whitestar, if you try and get your kite to fly by running in a circle as if you are standing on the tip of the blades with the wind representing your helicopter flying in one direction, you and your kite will feel more wind and get more lift when you are running into the wind [as the blades on the right side are] than when you are running with the wind at your back getting less lift [as the blades on the left side are]. Less lift on the left side needs more lift on the left side by angling the blades when they are on the left side. Below is a typical explanation.

The diagram wording is misleading. It states that the blade on the left side has a "lower tip speed," but the speed of the blade rotation is the same on both sides. However, the air is moving slower across the blade on the left side than on the right because the blade on the left side is moving in the same direction as the air (from the front of the helicopter toward the back). The air on the right side is moving faster across the blades because the blades are moving "into the wind" as the helicopter moves forward. The air moving slower across the blades on the left creates less lift. The air moving faster across the blades on the right creates more lift.

When you start to fly a kite, you run in the opposite direction that the wind is flowing so the air is flowing faster across the kite [and you as you feel more wind], creating more lift for the kite to take off. If you were to run in the same direction that the wind is flowing, less air would be flowing across the kite [and you as you feel less wind], so the kite would have less lift. Similarly, the helicopter blades on the left side are moving in the same direction as the air when the helicopter is moving forward, so there is less air flow across the blades, thus less lift than the blades on the right side. The blades on the right are moving in the opposite direction of the air as the helicopter moves forward and the air moves toward the front of the helicopter (from front to back). More lift on the right side would cause the helicopter's right side to lift up, and the helicopter would roll over unless more lift is created on the left side to equalize the lift on both sides. More lift on the left side is created to equalize the lift on both sides by increasing the blade's angle (leading edge pointing more upward) only when the blades are on the left side.

If the helicopter was on the ground with it's blades rotating, the air would be moving at the same speed across the blades on both sides; the lift on the right and left sides would be equal; no blade angle adjustments would be necessary. When the helicopter takes off and moves forward, the air flow across the blades on the right side, which are also moving forward, are moving at the speed of the blades before take off plus the speed of the helicopter moving forward. As the helicopter flies forward, the speed of the air moving across the blades on the left side that are moving backward and "with the wind" are moving at the speed of the blades before take off minus the speed of the helicopter because the helicopter's forward movement is moving the air toward the back/tail edge of the blades, which is the wrong direction to create lift, which reduces lift.

Lnewqban

Klystron, Nugatory and Lnewqban
Worth mentioning also that there is a 'centre of pressure' of the rotor disc of a single rotor helicopter, the centre being where there is zero lift, where the retreating blade is the same reverse speed as the helicopter is forward (i.e. zero air speed). Thus, the centre of pressure is the zero pressure point and is set away from the rotor's centre, towards the retreating side.

Now consider that not only is the retreating blade pushing more slowly against the air, the rotor has to balance one side to the other. The maximum lift and speed of a single rotor helicopter is therefore dominated by the aerodynamics of the retreating blade, the one on the advancing side is easy to configure to create as much lift as needed.

The limit on the advancing blade is how fast it can go, which is close up to supersonic. As the retreating blade needs to make lift, the tip speed has to be reasonably faster than the speed of the helicopter (so it can go backwards relative to the aircraft but, from its POV, forwards into air).

These conditions put the practical limit on the speed of single rotor helicopters, being sufficiently less than a half of the speed of sound that the retreating blade can still generate lift. So helicopter maximum speeds are about 1/4 to 1/3 of the speed of sound.

Nugatory and Lnewqban
Lnewqban said:
In the case of our helicopter, the 3 o'clock blade tip is moving through the air in the same direction of the fuselage, only than faster.
That tip does not know or care about the velocity of the fuselage, it only feels a great deal of airflow above and below it; therefore, it generates abundant lift.

Simultaneously, the 9 o'clock blade tip is moving through the air in the opposite direction of the fuselage, only than slower than it (and much slower than the opposite 3 o'clock blade tip).
That tip does not know or care about the velocity of the fuselage, it only feels a marginal airflow above and below it; therefore, it generates poor lift.

Because of the above, the relative-to-the air velocity of the 3 o'clock blade is greater than the relative-to-the air velocity of the 9 o'clock blade.
If we reverse the rotation, the opposite would be true.
In both cases there will be a lack of right-left balance of lift forces, and the helicopter will roll over.
Ah, so that's what I wanted to know. As you said, if you were to reverse the rotation, there would still be an imbalance, in this case, if the rotor rotated from the left to the right, at 9 o'clock it will create more lift, but at 3 o'clock it would reduce the lift, correct?

Lnewqban
FWIW, to solve the issues of stall on the retreating rotor blades, of course the other solution is to use contra-rotating blades.

By having two rotors, the retreating blades can be allowed to stall out (retreating tip speeds drop to zero relative velocity) because the two balance each other and the net centre of pressure from both rotors will then always remain along the midline of the airframe.

Thus, whereas a single rotor with sub-sonic rotor tips can achieve a maximum of 1/3 speed of sound or so (because there still has to be a finite rearward tip velocity on the retreating blade), a contra-rotating helicopter could theoretically be run close up to speed of sound tip velocity with the retreating blade generating no lift at zero air speed, therefore can get up to the maximum possible for a rotary wing aircraft of 1/2 speed of sound.

e.g. Sikorsky X2.

Lnewqban
Whitestar said:
Ah, so that's what I wanted to know. As you said, if you were to reverse the rotation, there would still be an imbalance, in this case, if the rotor rotated from the left to the right, at 9 o'clock it will create more lift, but at 3 o'clock it would reduce the lift, correct?
Exactly!

For fixed wing airplanes, we can always keep the same lifting force by playing with the AOA of the wing and the velocity forward respect to the mass of air.
In other words, a Cessna 150, for example, can keep 1,600 lbf of weight at level flight in a range of speeds between 125 mph and around 50 mph (which corresponds to a stall AOA with deployed flaps of about 14 degrees).
In order to achieve that, the pilot must modify the engine power and the AOA via elevator input in order to keep level flight.

Note that the above speeds are relative to the incoming air hitting the leading edge of the wing, not relative to ground.
In case of a strong tailwind, for example, an observer on the ground will see a faster moving Cessna than when no strong wind is present.
The opposite applies in case of the airplane facing a strong headwind.

Klystron and berkeman
Gosh, I just learned all kinds of neat stuff.

Lnewqban and berkeman
Baluncore said:
Normally air flows downwards through the helicopter rotor...
With an autogyro, air flows upwards through the rotor causing autorotation as the aircraft is propelled forwards. The blades glide and so support the weight of the autogyro.

This is almost certainly in error.
Newton will not be denied. If you want an object to resist a force, you must supply a force of equal magnitude vectored in the opposite direction. A boat floats because the force pulling it downward is offset by an equal weight of water lifted upward. An object hovers because a mass of air is given a downward acceleration that matches the mass of the object and its acceleration due to gravity. The equal and opposite reaction is the lift given to the objects (aerofoils) imparting that acceleration. There's no magic. The forces involved must balance.
Instead of gravity pulling the object down, air is driven down instead. Instead of the ground supplying the normal force to hold the vehicle up from falling to the center of the earth, the force of lift holds it in position.

Autogyros rotate because the drag on the advancing aerofoil is less than the drag on the retreating one. When the forward groundspeed and rotational airspeed become great enough, the autogryo flies. You are trading the energy expended to drive the vehicle forward for the lift needed to raise it through the mechanism of drag. But the air is moving <downward>

You can test this for yourself. Build a taketombo. Spin it with your hands or a string, and it will move upward. Take that same taketombo, stick it in a tube and hold it out the window of a car as you drive along. Once you move along fast enough the taketombo will begin spinning and will lift out of the tube.

When a helicopter loses power, the blades are still rotating and air is still being forced downward, but the blades slow down. Lift is reduced accordingly. The helicopter begins to accelerate downward. The air accelerated downward by the still-spinning-but-slowing blades meets the air still there from moments ago. There is viscosity. It has to go somewhere and that takes time and crucially creates a whack of drag as it works its way out from under the blades. If you are lucky, the drag created and ground effect add up to enough force to keep the downward velocity of your vehicle down to a survivable prang speed. If the blades stop spinning, almost all the lift force is lost and you prang in at something between zero and terminal velocity depending upon how high above the deck you were.

But in all cases, the air is moving <downward>

256bits and berkeman
: abruptly stops in the middle of Googling "taketombo" to more urgently Google "prang" :

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russ_watters and berkeman