Helicopter rotor blade angle of attack

[Oldone]

I am confused and wondering if anyone could help. At hover, the induced velocity changes the angle of attack from the pitch angle. At forward speeds, the induced velocity becomes less important, so lets neglect it for the purposes of discussion.
So the rotor is tilted forward by cyclic, say at max forward speed. With the rotor tilted in the nose down direction, the advancing blade has its angle of attack reduced by the amount of rotor tilt. The retreating blade has its angle of attack increased by the amount of rotor tilt. Both blades are moving in the direction of stall?
With flapping, the advancing blade is moved further to a reduced or even negative AOA which would stall? And the retreating blade is moving to a larger angle or stall?
Most lift is occurring around the 12 o'clock and 6 o'clock positions of the blades. Cyclic tilt and flapping would not (in a major way) change the AOA at these positions.
So the questions are:
Does the advancing blade ever stall during a portion of the rotation?
Is it necessary to reduce collective to prevent retreating blade stall (too high an angle)? Which would also move the advancing blade towards stall (too low an angle). Therefore, should collective be tied in to the cyclic to prevent to much collective pitch angle?
Is the primary purpose of delta3 to reduce blade stall, and thereby allowing greater forward speed?
Thanks in advance for any assistance!

[Jean Claude]

Originally Posted by Oldone

the advancing blade has its angle of attack reduced by the amount of rotor tilt. The retreating blade has its angle of attack increased by the amount of rotor tilt. Both blades are moving in the direction of stall?

With the blade flapping, the moments is balanced, the blades trajectory changes. Advancing blade ascends and the retreating blade descends . Now the angle of attack of the retreating blade is greater than the angle of attack of the advancing blade. Just enough to balance the lift values​​, despite differences in speed. So, the tip of the retreating blade is approaching stall.

Originally Posted by Oldone

With flapping, the advancing blade is moved further to a reduced or even negative AOA which would stall? And the retreating blade is moving to a larger angle or stall?

Since the two blades share the weight to be lifted, then both have positive angles of attack.

Originally Posted by Oldone

So the questions are:
Does the advancing blade ever stall during a portion of the rotation?

No part of the advancing blade is stalled,

[Oldone]

Thank-you very much Jean Claude. For very rough numbers, if a helicopter had a collective of 12 degree pitch angle. And the rotor is tilted forward 12 degrees by cyclic. Is angle of attack on advancing blade 0, and retreating blade 24 degrees?
Or, if collective is 5 degrees, and cyclic is at 12 degrees, is advancing blade at -7 degrees, and retreating at 17?

[WaspAir]

The angle of attack varies with location along the span.

The figures here may help.

http://www.copters.com/aero/retreating.html

[Oldone]

Thanks WaspAir. If we had a simple blade with no twist - So the entire length of the blade at the 3 clock position (CCW American rotation) would see the same angle of attack, if we consider the airstream only, and neglect downward airflow through the rotor. Would a 5 degree collective with 12 degree cyclic give a -7 degree advancing blade, and a 17 degree retreating blade?

In an airplane propeller, the v=rw varies along the length of the blade (as in a helicopter) but the airstream is normal to the blade. I am neglecting downward flow through the rotor for simplicity which is important at hover and low speeds, but for simplicity, I want to ignore at high speeds.

I think advancing blade stall is the wrong concept, but it seems the blade would be at a negative angle of attack for a portion of rotation.

[Jean Claude]

Originally Posted by Oldone

Thank-you very much Jean Claude. For very rough numbers, if a helicopter had a collective of 12 degree pitch angle. And the rotor is tilted forward 12 degrees by cyclic. Is angle of attack on advancing blade 0, and retreating blade 24 degrees?
Or, if collective is 5 degrees, and cyclic is at 12 degrees, is advancing blade at -7 degrees, and retreating at 17?

This is not true. Not only collective and cyclic angles to consider. Consider also angle a1 (blade flapping), speeds forward, and speed circumferential. Look more closely at my diagram.

[Oldone]

Originally Posted by Jean Claude

This is not true. Not only collective and cyclic angles to consider. Consider also angle a1 (blade flapping), speeds forward, and speed circumferential. Look more closely at my diagram.

Flapping angle is small relative to cyclic tilt, or the helicopter would not move forward. Looking down on the rotor at 3 o'clock, CCW rotation for the advancing blade, the speed forward is additive to the speed circumferential. Greater at the tip than the root of course.

It still appears the advancing blade goes negative AOA at some point of forward cyclic tilt? Does this establish the limit for forward cyclic? If AOA goes negative, it would cause reverse flapping and nose down?

Thank-you for the diagram.

[WaspAir]

Originally Posted by Oldone

If we had a simple blade with no twist - So the entire length of the blade at the 3 clock position (CCW American rotation) would see the same angle of attack,

No, it doesn't happen that way. You could have constant angle of incidence, but that's not equivalent to angle of attack.

It's not just a matter of twist. It's a matter of different magnitude vectors being added at different span-wise stations.

The component of airflow from rotational motion must be added to the component from translational motion to get the true relative wind. If rotation and translation are exactly co-planar, you could talk about one angle of attack for the blade, with varying airspeed along the span. But if there is any difference in angle between those components, the direction of the vector sum will be different at each station. The farther out toward the tip you look, the greater the rotational component will be with respect to the constant translational component. The sum will vary not only in magnitude, but in direction as well.

[Oldone]

I conclude that, considering transition of the Osprey, the advancing blade does see negative angle of attack at high forward cyclic and high forward speeds, but the lift on the blades at 3 oclock and 9 oclock is so low that it does not matter.

[StanFoster]

Oldone- instead of the rotor tilting 12 degrees, imagine it tilting 90 degrees and its tip path plane is now vertical. The blades still will have positive angle of attack, but of course no advancing or retreating blade now. You mentioned a delta hinge earlier. My tail rotor has a delta hinge on it. Its purpose is to depitch the advancing blade and add pitch to the retreating blade, and it does this with less flapping angle required. This allows the tail rotor to be built closer to the frame . This allows a lighter designed tail rotor gearbox. Stan

[Oldone]

Thanks for the input Stan!

[birdy]

Your retreating blade will stall long before your advancing blade gets to 0 AOA, so no, youd never get to a situation where your advancing blade is producing no lift.

[Jean Claude]

Unless rigid rotor, the two blades share the weight to be lifted to equal (or almost). We can not escape from it.
A.o.A always positive for both blades. Small for the advancing blade, great for the retreating blade. This inversely with the square of their speed / air

[Oldone]

Originally Posted by Jean Claude

Unless rigid rotor, the two blades share the weight to be lifted to equal (or almost). We can not escape from it.
A.o.A always positive for both blades. Small for the advancing blade, great for the retreating blade. This inversely with the square of their speed / air

CCW looking down on rotor - aircraft flying level at max forward speed. Two bladed helicopter - The retreating blade at 9 o'clock is stalled on the inner portion of the blade. The lift at 9 o'clock is reduced by low velocity.
A blade at tip speed of 500 fps at 6 and 12 o'clock is reduced by an air velocity of 150 ft/sec to 350 fps at 9 o'clock. The advancing blade at 3 o'clock is at 650 fps. Lift is proportional to square of the velocity. So just based on velocity, not yet accounting for flapping or tilting of the rotor, we would have the 350/500 squared, or the 9 o'clock position would provide half the lift of the blade when it is in the 6 or 12 o'clock position. The blade in the 3 o'clock position would be 650/500 squared or 1.7 times the lift of the blade in the 6 or 12 o'clock position. The advancing blade must be reduced to a near zero, or slightly minus, AOA. It has to match the lift of the much slower speed retreating blade.

Unlike an autogyro, the helicopter rotor tilted forward achieves the same objective as flapping - reducing the blade angle of attack on the advancing side, increasing it on the retreating side. So what we have is a constantly varying lift on the blades, this cyclic loading has to be designed for. The integration of the lift over time of the advancing side, has to equal the integration of the lift over time of the retreating side. The peak lift of the two bladed rotor is at a max at 6/12 o'clock. It is reduced at 3/6 o'clock. So there will be a two/rev vibration. The three bladed rotor will smooth out the peaks and valleys of lift. The optimum coning angle on a two bladed rotor will constantly be varying, so a hinge or teetering is highly desirable to reduce blade stress.

My calculations say the tip of the advancing blade has to be negative AOA for a portion of advancing rotation. The overall integral of lift vs time for the advancing side is positive. This puts reverse bending on the blade which has so be accounted for.

Regards,


The autogyro normally only tilts the rotor for pitch control. It goes forward by a pusher prop. The helicopter tilts the rotor forward to move forward. So the helicopter has an automatic tendency to reduce the lift on the advancing side to match the lift on the retreating side.

[StanFoster]

Oldone- I am not an aerodynamist....but I feel that AOA of the tips of the advancing blade must always be positive. If the tip was negative...the AOA of the blade as you go inboard towards the mast would all be negative and greater than the tip. This cant be.

Instead the advancing blade is flapping up and reducing its lift several degees while this same amount is being added to the retreating blade. The forces will balance out. The faster the helicopter goes...the more the AOA of the advancing blade is reduced...but never to 0....and the more the AOA of the retreating blade is increased. There is a hole on the retreating blade that starts tangent to the rotorshaft and steadily grows in diameter outward away from the rotorshaft on the retreating side the faster the helicopter flies. This puts more and more work on the outer tips of the retreating blade as it demands equal lift and has to have more AOA to balance out the decreasing AOA taken from the advancing blade.

The AOA of the advancing blade is greatest at its tip and becomes smaller as you go inboard to the rotorshaft.


Stan