Questions on how helicopters fly

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  • #1
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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.

See link:
https://www.physicsforums.com/threads/a-plausible-jet-helicopter.968992/

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:

600px-Retreatingbladestall.png



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?
 

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  • #2
phinds
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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.
 
  • #3
DaveC426913
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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).
 
  • #4
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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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  • #5
Lnewqban
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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?
 
  • #6
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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?
 
  • #7
phinds
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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?
 
  • #8
Lnewqban
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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.
Please, see:
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).
Please, see:
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.
 
  • #9
Baluncore
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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.
 
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  • #10
Baluncore
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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.

To see many straight helicopter blades; Google Images “helicopter blades”
 
  • #11
Lnewqban
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...
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?
 
  • #12
Baluncore
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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.
 
  • #13
sophiecentaur
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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.
 
  • #14
Baluncore
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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?
 
  • #15
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Yes. So did you understand post #2?
Not entirely. Could you please elaborate?
 
  • #17
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Not entirely. Could you please elaborate?
"They really can't fly, they just beat the air into submission," over heard in the Nam.
 
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  • #18
Lnewqban
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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?
 
  • #19
Baluncore
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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.
 
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  • #20
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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.


helicopter retreating blade stall (mirrored).png
 
  • #21
berkeman
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So, what would be the case if the blade was rotating from the left to the right?
Seriously? Which word did you not understand?
 
  • #22
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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?
 
  • #23
Lnewqban
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Please read about P-factor here:
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.
 
  • #24
Lnewqban
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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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  • #25
berkeman
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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.
 

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