Blade Sailing

Tyger

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he threw the gyro into a hard sudden left bank at high speed and the teeter stops were definitely hit.It sounded like a piece of tin being rapidly whacked with a small hammer.He quickly straightened up and it disappeared instantly and his comment was "whoa we don't want that". So i know you can hit those stops by putting big pressure on the rotor.
"Big pressure" is a rather imprecise term...
Can you (or anyone reading this) explain why a hard left bank might cause that?
Would it be less likely in a hard right bank?
 

Vance

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I feel there is value in keeping the things that precipitate a gyroplane takeoff mishap simple and using appropriate terminology has value.

In my opinion flap may not be the best term to describe what happens when semi rigid teeter rotor has an un-commanded excursion because the blades of any rotorcraft are always flapping in forward flight to manage dissymmetry of lift.

I prefer “blade sailing” because the advancing blade sails when the retreating blade stalls.

It is my observation that the term blade flap to describe what is managing dissymmetry of lift may be poorly chosen because it seems some people are confused by it.

The retreating blade will have a higher angle of attack than the advancing blade as part of managing the dissymmetry of lift.

Part of the rotor disk is always stalled.

Put in the simplest terms; too much airspeed for the rotor rpm will cause a two blade semi rigid rotor to have an un-commanded excursion from the intended path.

As the stall of the retreating blade expands the flapping hinge allows the angle of attack of the retreating blade to increase and the angle of attack of the advancing blade to decrease further stalling the retreating blade.

Prevention is simple enough. Verify that you have enough rotor rpm for the indicated air speed.

A common error is to not have enough disk angle to accelerate the rotor allowing the gyroplane to accelerate faster. At some point people decide that it is time to rotate and add a little back stick with predictable results.

The POH for most gyroplanes are written in a way to keep as far away as practical from not enough rotor rpm for the indicated air speed.

The frequency of blade sailing events suggests to me that people are not following the POH.

I had a smart, young client Thursday who was landing and taking off on his second flight. On his third takeoff in response to the nose coming up he planted the nose and the rotor stopped accelerating as The Predator’s acceleration increased because of the reduced drag.

We had briefed extensively on this very thing and in the debrief he answered all the questions correctly indicating to me he had heard and understood what I had said.

In his two hours of flying it was the only thing he had done to put the aircraft at risk.

I continue to search for better ways to describe the critical process of getting the rotor up to speed before flying a gyroplane that will stay with people.
 

XXavier

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"Big pressure" is a rather imprecise term...
Can you (or anyone reading this) explain why a hard left bank might cause that?
Would it be less likely in a hard right bank?
For a rotor turning CCW (as seen from above), a left-hand turn would mean less airspeed as 'seen' by the retreating blade at the inside of the turn and more airspeed seen by the advancing blade, at the outside of the turn. Hence, the flapping angle needed to compensate for the dissymmetry of lift would be higher and the limiting stops might be more easily reached. For a right-hand turn, the opposite is true...
 

WaspAir

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I don't understand what happened in that stop-banging turn described above.

First, the left vs. right turn analysis seems to treat the disc as a fixed wing. Even so, given the small blade span on most gyros, the airspeed difference seen by the inner tip vs. the outer tip in a steeply banked level turn (owing to the difference in circumference traveled) won't be very big unless the overall airspeed and turn radius are very small. The overbanking tendency arising from such airspeed differences is rarely noticed in moderate span relatively fast airplanes but quite noteworthy in huge span slower sailplanes. For a VERY small span (rotor diameter) gyro, I think one would have to be already on the verge of a problem in level flight in order for the small speed difference from the roll-in to create a problem.

Further, reduced translational speed on the retreating side (inner, lower tip in a left turn for CCW rotor) from that circumference difference would not increase dissymmetry, would it? It's usually greater "backwards" flow on the retreating side that makes issues, and reduced wrong-way flow is helpful to compensation. Smaller translational speed subtracts less from the rotational airspeed seen by the retreating blade, and increases net airspeed.

One would also expect the g-load in a brisk turn to increase rotor rpm and provide margin against such problems.

I'm still troubled as to what really happened.
 
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XXavier

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I don't understand what happened in that stop-banging turn described above.

First, the left vs. right turn analysis seems to treat the disc as a fixed wing. Even so, given the small blade span on most gyros, the airspeed difference seen by the inner tip vs. the outer tip in a steeply banked level turn (owing to the difference in circumference traveled) won't be very big unless the overall airspeed and turn radius are very small. The overbanking tendency arising from such airspeed differences is rarely noticed in moderate span relatively fast airplanes but quite noteworthy in huge span slower sailplanes. For a VERY small span (rotor diameter) gyro, I think one would have to be already on the verge of a problem in level flight in order for the small speed difference from the roll-in to create a problem.

Further, reduced translational speed on the retreating side (inner, lower tip in a left turn for CCW rotor) from that circumference difference would not increase dissymmetry, would it? It's usually greater "backwards" flow on the retreating side that makes issues, and reduced wrong-way flow is helpful to compensation. Smaller translatioal speed subtracts less from the rotational airspeed seen by the retreating blade, and increases net airspeed.

One would also expect the g-load in a brisk turn to increase rotor rpm and provide margin against such problems.

I'm still troubled as to what really happened.

If we imagine the gyro turning in a circle, with no translational velocity of the center of the circle, the difference between the airspeeds 'seen' by the blades will be the same in the case of a left- or a right-hand turn, so the flapping angle will stay constant during the full circle. If we now add a translational velocity to the center of the circle, there will be another flapping angle due to that velocity, but the total flapping angle will also stay constant during the 360º turn. However, the lift of a blade is more or less directly proportional to the AoA only within limits. If –for the gyro at, say, 9 o'clock on the left-hand turn– the critical, stall angle is reached by the retreating blade for a CCW rotor, lift will drop almost entirely for that blade, and the dissymmetry of lift would be much higher than –all other things being equal– for the rotor of the gyro when turning right at 3 o'clock, with the retreating blade not reaching the stall...

Concerning the g-load during a sharp turn, it could take some time until the rotor reacts to a higher g-load by increasing the RRPMs, and –for a few critical fractions of second– the higher upwards flow will increase the mean AoA of the blades, and the danger of a retreating-blade stall...
 
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Aerofoam

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I am just wondering if the bank was so abrupt (with only roll input) that the rotor banked at a rate faster than the airframe could keep up
with and the disk, which was nowhere near any kind of stall, simply ran into the stops perpendicular to the line of flight because of the sudden
hard banking of the disk.
It seems if this is what happened, it could happen in either direction because it is not a retreating blade issue at all.
Is that possible? Could the stops be set too high?
 

WaspAir

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If we imagine the gyro turning in a circle, with no translational velocity of the center of the circle, the difference between the airspeeds 'seen' by the blades will be the same in the case of a left- or a right-hand turn, so the flapping angle will stay constant during the full circle. If we now add a translational velocity to the center of the circle...
... that translation of the center of the circle sounds like groundspeed to me, not airspeed. You can get that kind of motion by drifting with the wind while flying a steady circle through the airmass. The wind will not change airspeed.

What maneuver are you suggesting the pilot is attempting?
 
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XXavier

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... that translation of the center of the circle sounds like groundspeed to me, not airspeed. You can get that kind of motion by drifting with the wind while flying a steady circle. The wind will not change airspeed.

What maneuver are you suggesting the pilot is attempting?

Translation of the center of the circle with relation to the mass of air. Or may be contemplated as the motion of the mass of air with relation to the center of the circle. In the example that I posted, the gyro first turned in a 'stationary' circle w.r.t the mass of air, and then I added a translational airspeed for the center of that circle. Another way to see it: imagine you are circling an airship with its engines stopped. Then, the engines are started and the airship (the center of the circle) moves w.r.t the mass of air...

The pilot is obviously turning left (or right), and not (usually) flying a 360º circle, either right- or left-hand. If I mentioned a full turn, was only in relation to the flapping angle originated by it...
 
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WaspAir

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Sorry, can't picture that.

Rolling into a turn with an initial tangential velocity can establish the circumferential airspeed, but won't make the center travel through the airmass. Are you describing a skidding turn? I can't see what you're proposing for a coordinated turn.
 

XXavier

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Sorry, can't picture that.

Rolling into a turn with an initial tangential velocity can establish the circumferential airspeed, but won't make the center travel through the airmass. Are you describing a skidding turn? I can't see what you're proposing for a coordinated turn.

It doesn't matter much whether it's a coordinated turn or not. The gyro of the example follows a circle, and the important fact is that there's a difference in retreating blade stall between a left-hand and a right-hand turn. I doubt you can fly the 360º moving circle in a coordinated turn, but coordinated or not, you can keep turning till you reach 180º and the retreating blade gets close to (or reaches) the stalling point.
 
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WaspAir

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We disagree on the physics for a "moving circle". You can decrease a turn radius by increasing bank (and g-load) or by slowing airspeed, or increase radius by losing coordination, but that's not the same as a center with an airspeed.

P.S. I don't understand how a 180 gets you close to retreating blade stall. Sounds too much like the old downwind turn argument.
 
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mark biddle

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After that turn we never went into a hard right turn so could not tell you if it would do the same but for the few seconds the G load was pretty intense.
 

Tyger

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For a VERY small span (rotor diameter) gyro, I think one would have to be already on the verge of a problem in level flight in order for the small speed difference from the roll-in to create a problem.
I think I would agree with this. Especially since any such speed difference is going to be really tiny as a percentage of the overall rotor speed.
I too do not understand the phenomenon as it was described.
 

XXavier

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We disagree on the physics for a "moving circle". You can decrease a turn radius by increasing bank (and g-load) or by slowing airspeed, or increase radius by losing coordination, but that's not the same as a center with an airspeed.

P.S. I don't understand how a 180 gets you close to retreating blade stall. Sounds too much like the old downwind turn argument.

This discussion has been long, and there's no special interest from my part in having the last word. I believe that my examples and descriptions are clear enough. Having said that, I've nothing else to add...
 

Jean Claude

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It is possible to quantify the longitudinal beat angle by an easily understandable mathematical formula:
As long as we assume that the lift of a representative blade element is proportional to A.o.A (i.e. without stall), and proportional to the square of the airspeed, then the compensation of the speed difference between the advancing and retreating sides by the A.o.A difference can be translate approximately as follows:
(Vc+Vf)² / (Vc-Vf)² =a + a1) / (αa – a1) (*)
The solution to this equation is a1 = 2 αa Vc Vf / (Vc²+ Vf²)

(*) In which Vc is the circumferential velocity, Vf is the forward velocity, αa is the average angle of attack, and a1 is the longitudinal flapping angle

If, in flight, αa = 5°, Vf = 35 m/s, and Vc at ¾R = 120 m/s then a1 = 2*5*120*35/ (120²+35²) = 2.7°
So, the angle of attack of this element on the retreating side is then (αa + a1) = + 2.7° = 7.7° (i.e below the stall)

Now if, during the run, αa= 8° (due to the large disc A.o.A), Vf = 10 m/s, and Vc at ¾R = 30 m/s then a1 = 2*8*30*10/ (30²+10²) = 4.8° So, the angle of attack of this element on the retreating side is then + 4.8° = 12.8°
We see it is above the stall and the compensation is impossible. Thus, the flapping angle will increase to each round until the blades hits it stops.
 
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Tyger

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Thanks, but can you explain in simpler terms what you mean by "longitudinal beat angle"?
 

Jean Claude

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Sorry for my poor English. I meant to say "longitudinal flapping", as showed in this fig.
Sans titre.png
 
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Jean Claude

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For a rotor turning CCW (as seen from above), a left-hand turn would mean less airspeed as 'seen' by the retreating blade at the inside of the turn and more airspeed seen by the advancing blade, at the outside of the turn. Hence, the flapping angle needed to compensate for the dissymmetry of lift would be higher and the limiting stops might be more easily reached. For a right-hand turn, the opposite is true...
This is not my opinion.
By definition, the rotor of a gyrocopter is not linked in rotation to the airframe, so the rate of change of heading of a gyrocopter does not change the Rrpm/air
So, whatever the radius of the blades, the air speed of the blade tips remain mainly R* ohméga/air, even if the flight heading changes infinitely quickly
 

Jean Claude

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Well i can honestly tell you that i have been a passenger in a 160 hp dominator piloted by a very experienced friend and in demonstrating the performance of the machine he threw the gyro into a hard sudden left bank at high speed and the teeter stops were definitely hit.It sounded like a piece of tin being rapidly whacked with a small hammer.He quickly straightened up and it disappeared instantly and his comment was "whoa we don't want that".So i know you can hit those stops by putting big pressure on the rotor.
It seems possible to me, but not destructive, that it is the too fast maneuvering of the control stick that pushed the stops to meet the blades, faster than the roll or pitch rate of the rotor moves it away from them.
 

ferranrosello

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Now if, during the run, αa= 8° (due to the large disc A.o.A), Vf = 10 m/s, and Vc at ¾R = 30 m/s then a1 = 2*8*30*10/ (30²+10²) = 4.8° So, the angle of attack of this element on the retreating side is then + 4.8° = 12.8°
We see it is above the stall and the compensation is impossible. Thus, the flapping angle will increase to each round until the blades hits it stops.
First, how do you know that an AOA of 12.8º in a blade element is stalled? It depends on the airfoil and in a rotating blade the blade elements are able to fly at much bigger AOA’s than a fixed wing with no stall.

Second, you have calculated the AOA od an airfoil in the 75% of the blade. All blade elements between 75% and 100% are working at lower AOA’s.

It is possible that the flap angle is too big and then the rotor is unable to compensate lifts and hits the stops.
 
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