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WaspAir

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Now I’m getting confused. If you moved the cyclic forward such that the rotor head tilted 1° forward, that would decrease the pitch on the advancing blade (Eg from nominal 2° back to 1°) and conversely increase the pitch on the retreating blade (Eg from the nominal 2° up to 3°). Now you have 1° pitch on the advancing blade and 3° on the retreating blade. Would this not correct for the “dissymmetry” of lift?

I’ve got a feeling that we’re overlapping two different rotor behavioural characteristics here.
The extent of dissymmetry is airspeed dependent. When you push the stick forward, you can change/adjust the system to a new steady state accordingly, but there will be cyclic variation in the angle that automatically happens without pilot intervention at any fore-aft stick position / airspeed combination. Teetering or flapping (choose your term depending on how many blades you have) does this.
 

WaspAir

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Xxavier, I think I was typing when your post appeared but we seem to agree.
 

Vance

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May I suggest a few corrections...? They follow:

With an anticlockwise turning rotor as viewed from above, the advancing blade starts rising at six o’clock, reaching a maximum climbing velocity at three o’clock. At that point, it starts to slow down its climb, but keeps moving upwards until, at 12 o'clock, the blade reaches a maximum height, its climbing velocity coming down to zero. Then, reversing its motion, it starts to sink, reaching maximum sinking velocity ninety degrees later, at nine o’clock. And a further ninety degrees later, at six o'clock, the blade reaches the lowest point of its orbit...
I think he (and WaspAir) are saying that max height is at 12 o'clock (zero upward velocity), while max upward velocity (NOTmax height, as you just wrote) is what's at 3 o'clock.
In my opinion there is no dissymmetry of lift at twelve and six o’clock so the blade is in its neutral position in relation to the spindle as it would be in a vertical descent with no forward airspeed.

With an anticlockwise turning rotor as viewed from above at approximately three o’clock the blade has reached its maximum distance above its neutral position.

At approximately nine o’clock the blade has reached its maximum distance below the neutral position.

I would not attempt to calculate the blades acceleration due to dissymmetry of lift.

I feel the most important takeaway for a gyroplane pilot to understand is; no pilot input is required to manage dissymmetry of lift and the retreating will likely stall first because it is at a higher angle of attack than the advancing blade.
 
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WaspAir

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In my opinion there is no dissymmetry of lift at twelve and six o’clock so the blade is in its neutral position in relation to the spindle as it would be in a vertical descent with no forward airspeed.

With an anticlockwise turning as viewed from above at approximately three o’clock the blade has reached its maximum distance above its neutral position.

At approximately nine o’clock the blade has reached its maximum distance below the neutral position.
Vance, my understanding is that up/down velocity changes the angle of attack because it changes the apparent direction of airflow. If you are moving down, the air seems to come from a bit lower position, while if you are moving upward, the air seems to come from a higher position. But all this depends on motion/velocity, NOT on position. Merely having a blade high or low does nothing; it is the velocity that makes the angle appear to change, because it adds an extra component to the relative wind. You get a compensating effect only from speed, not from height.

At the highest position, you momentarily have no up/down velocity, for the instant between climbing and descending. That means you have no angle of attack effect at that time. This is what happens at 12:00. You are at max height, neither moving up or down, and doing nothing to angle of attack, at the same time that you have no dissymmetry of lift and don't need any compensation.

Likewise, at 6:00, you are at the lowest point, with no up/down flapping velocity, for the instant between descending and climbing. With no flapping motion, you have no effect on angle of attack, and don't need any, because there is no dissymmetry.

At the 3:00 position, you have the maximum upward velocity, for maximum angle of attack effect, but you are not yet at the top of the climbing motion (you're still moving upward quickly). As rotation continues, you lose upward velocity, eventually hitting zero when you reach 12:00.
Thus , at 3:00, the blade has its maximum upward velocity, but has not yet reached maximum displacement (which happens at 12).

One addtional thought: if the blade was highest at 3:00, the disc would appear to tilt left; with max height at 12:00, it appears to tilt back.

I feel the most important takeaway for a gyroplane pilot to understand is; no pilot input is required to manage dissymmetry of lift and the retreating will likely stall first because it is at a higher angle of attack than the advancing blade.
No argument there!
 
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mceagle

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With an anticlockwise turning rotor as viewed from above at approximately three o’clock the blade has reached its maximum distance above its neutral position.
At approximately nine o’clock the blade has reached its maximum distance below the neutral position.
This is what I find hard to accept. This would mean that the rotor disc plane (and consequently the rotor lift vector) would be inclined to the left, causing the gyro to turn left, just as surely as it would if you initiated a left turn with the cyclic (Which also inclines the rotor lift vector to the left)
 

WaspAir

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This is what I find hard to accept. This would mean that the rotor disc plane (and consequently the rotor lift vector) would be inclined to the left, causing the gyro to turn left, just as surely as it would if you initiated a left turn with the cyclic (Which also inclines the rotor lift vector to the left)
Right, Tim -- it is the up-flapping speed that is greatest at 3:00, not the height, which peaks later.
 

mceagle

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At one hundred knots of indicated air speed my advancing blade at ninety degrees is seeing two hundred knots of airspeed and my retreating blade at two hundred seventy degrees is seeing fifty knots of airspeed. I feel this produces a significant difference in lift that needs to be addressed somehow.

In my opinion the teeter hinge manages this without significant pilot input and the angle of attack of the rotor blade is changed as described.
Just a small correction here - that you are probably aware of anyway. (Remembering that I am not a mathematician or a scientist) Its not the blade but the rotor tip speed that is most significant. At 100 kts forward speed, given that the rotor tip speed is approx 300 kts then the advancing blade is seeing 400 kts and the retreating blade is seeing 200 kts. - just the same, it’s still a considerable difference.
i agree with your second sentence here. Pilot inputs on a Gyroplane are normally only minimal at best and most pilots would barely notice inputs required for normal flight (unless you chuck the gyroplane about)
 

Tyger

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At 100 kts forward speed, given that the rotor tip speed is approx 300 kts then the advancing blade is seeing 400 kts and the retreating blade is seeing 200 kts. - just the same, it’s still a considerable difference.
Don't forget that the speeds you just quoted are only at the instant the blade is orthogonal to the direction of flight.
 

mceagle

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I will concede that “teetering” plays an important part in rotor behaviour, especially in :-
a( bringing rotors up to speed, where the advancing blade starts to lift and autorotate before the retreating blade,
b( the initial stick movement in a change of rotor orientation, (until the rotor spin axis realigns itself once again with the rotor head spin axis),
c( hitting a sudden “wind gust”, but this is normally a very temporary misalignment.
d( a misalignment of axises in flight due to other influences, like engine torque, uncoordinated flight, and possibly even teeter friction.

However, the question relates to “in level flight at normal cruise speed” so these few considerations do not come into it.
If there were any teetering in flight it would be minimal considering rotor tip speeds, and easily compensated for by the pilot without even noticing it.
Teetering is probably a misleading description because it suggests a flapping or see-sawing motion, which is not the case. The rotor disc as a whole simply flys at an inclined angle, as directed by the rotor head spin axis. (Which is directly controlled by the pilot).

Consider this exert from the following :-

PRECESSION and GYROPLANE CONTROL
by Don McCoy PhD. FGAA, A 894
Senior Lecturer in Physics, University of Adelaide

It is well known that the action of the teeter bolt is to equalise the lift on the advancing and retarding rotors. What must also be understood is that the teeter bolt has a totally independent and equally important role of making the gyro controllable in flight. The teeter bolt causes the plane of rotation of the rotors to tilt in the correct direction for turn and bank, climb and descend, again by gyroscopic precession. It was explained earlier that, because of the teeter bolt, the rotor head cannot produce a direct torque on the plane of rotation of the rotors so the question then arises, what does? The answer is the aerodynamic effect of the air acting on the rotors. One can think of the rotors as ‘flying’ to align the rotor axis and the rotor head axis through gyroscopic precession.

References
Wheatley J.B. An aerodynamic analysis of the autogyro rotor with comparison between calculated and experimental results. NACA TR 4871.1934
Schad. J.L. Small Autogyro Performance, J American Helicopter Society, 10, 1965.
McKillip. R.M. and Chlh. M.H. Instrumented blade experiments using a light autogyro. Proceedings of the 16th European Rotorcraft Forum. Glasgow. Scotland. September 1990.
 

XXavier

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I will concede that “teetering” plays an important part in rotor behaviour, especially in :-
a( bringing rotors up to speed, where the advancing blade starts to lift and autorotate before the retreating blade,
b( the initial stick movement in a change of rotor orientation, (until the rotor spin axis realigns itself once again with the rotor head spin axis),
c( hitting a sudden “wind gust”, but this is normally a very temporary misalignment.
d( a misalignment of axises in flight due to other influences, like engine torque, uncoordinated flight, and possibly even teeter friction.

However, the question relates to “in level flight at normal cruise speed” so these few considerations do not come into it.
If there were any teetering in flight it would be minimal considering rotor tip speeds, and easily compensated for by the pilot without even noticing it.
Teetering is probably a misleading description because it suggests a flapping or see-sawing motion, which is not the case. The rotor disc as a whole simply flys at an inclined angle, as directed by the rotor head spin axis. (Which is directly controlled by the pilot).
(...)
(...)
In the see-saw rotors used by most gyros, and in normal flight, the blades have two periodic motions with a phase lag of 90º, the revolving motion of the blades and the flapping oscillatory motion, the 'teetering'. The sum of those two motions result in the 'blownback' rotor disk.
Flapping is zero only when the forward flight of the gyro is zero too. (No blowback). But in normal flight, the flapping motion is always present, and its amplitude is directly proportional to the gyro's airspeed.
 

mceagle

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Thanks for that Xxavier. The original question/answer was :-

Lift of the advancing blade and the retreating rotor blades is equalised by ...the advancing blade teetering up and the retreating blade teetering down.
Given your good explanation, is this correct?

I understand the principle of “rotor blowback” due to forward speed and 90° phase lag, but I can’t get my head around 3:00 teetering up and 9:00 teetering down, without the gyro turning left.
 

XXavier

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Thanks for that Xxavier. The original question/answer was :-

Lift of the advancing blade and the retreating rotor blades is equalised by ...the advancing blade teetering up and the retreating blade teetering down.
Given your good explanation, is this correct?

I understand the principle of “rotor blowback” due to forward speed and 90° phase lag, but I can’t get my head around 3:00 teetering up and 9:00 teetering down, without the gyro turning left.
Yes, I believe you're right: the difference of velocity of the relative wind as 'seen' by the advancing and retreating blades is compensated by a decrease in the AoA of the advancing blade, and a corresponding increase of AoA for the retreating blade, those changes of AoA being caused by the change of the direction of the relative wind as seen by the blades, due to the teetering motion.

In order to visualize the disk plane with blowback, I suggest to imagine the rotor turning without flapping, the turning blades defining a plane that we term 'reference plane'. Now, we add a teetering oscillation of the same frequency but with a phase difference of 90º. The result is that the blades now turn in a different plane (the 'tip path plane'). The intersection of the two planes is a straight line passing through the 3 o'clock and 9 o'clock points. When passing through these points, the blades are in the reference plane, they are 'level' so to say...
 
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Jean Claude

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Yes, I believe you're right: the difference of velocity of the relative wind as 'seen' by the advancing and retreating blades is compensated by a decrease in the AoA of the advancing blade, and a corresponding increase of AoA for the retreating blade, those changes of AoA being caused by the change of the direction of the relative wind as seen by the blades, due to the teetering motion.
I agree. Still clearer with a squetch:
Sans titre3.png
 
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AirCommandPilot

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but I can’t get my head around 3:00 teetering up and 9:00 teetering down, without the gyro turning left.
If the rotor was not able to "teeter" you would be rolling left. The teeter allows the rotor to find it's balance without affecting the machine.
 

WaspAir

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I understand the principle of “rotor blowback” due to forward speed and 90° phase lag, but I can’t get my head around 3:00 teetering up and 9:00 teetering down, without the gyro turning left.
I'll try another cut at this, in case it helps.
The three o'clock flapping position is neutral or "level" - - higher than at the 6:00 position but lower than at the 12:00 position, midway between the two extremes. What is maximum at 3:00 is the speed at which the blade is rising. That upward speed reduces the angle of attack by adding a component to the relative wind, which compensates for the higher airspeed seen by the advancing blade. The position along the 3:00 to 9:00 line is perfectly horizontal, with no "tilt" of that line, so there is no left-turning force. Max height from all that teetering speed is reached at 12:00, which makes a blown-back disc (high in front, low in back) that is level from left to right.

The disc is left-right level, but each blade is moving upward at 3:00 and moving downward at 9:00, and that motion provides the dissymmetry compensation. At 3:00, it's headed in the up direction; it is not all the way up there yet. It is the motion (trend, heading, speed), not the position, that changes angle of attack on that side.
 
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WaspAir

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In case any of the instructors here care, the reasons I prefer describing a simple phase lag between velocity and position in repetitive motion instead of describing torque on the axis and gyroscopic precession of a virtual disc are:
1) it's much simpler, and requires no understanding of physics to comprehend, and
2) it is less misleading.
In particular, when dealing with three-blade fully articulated systems as on many helicopters and the 18A and J-2s I've spent so much time flying, the gyroscopic description is not helpful because is difficult to apply to actual separate blade motions and too often yields incorrect expectations. It often confuses students, because gyroscopes are famously stable in space, but an articulated rotor is obviously unstable, which the student quickly discovers. The rotor simply does not behave as one might expect a stable gyroscope to do, wandering all over the sky if you let it. Release the controls, and you won't be thinking you have a gyroscope on your hands, and you'll be wondering why your crazy instructor thinks it works like one.

I've pontificated this way before, suggesting that the coriolis description for lead-lag is a poor choice for pedagogy, when conservation of angular momentum is cleaner. I'm always looking for the best descriptions for flight student comprehension, not engineering student comprehension.
 

XXavier

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If the rotor was not able to "teeter" you would be rolling left. The teeter allows the rotor to find it's balance without affecting the machine.
That was exactly what happened with the first Cierva prototypes, with non-articulated blades. The tilting moment was transmitted to the aircraft's body, and the machine rolled to one side.
In our teeter-hub machines, if the teeter articulation is too tight (or dirty), thus braking or impeding the flapping, the dissymmetry of lift would roll the gyro to the left...
 

mceagle

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If the rotor was not able to "teeter" you would be rolling left.
that is correct, the aircraft would be uncontrollable.
However, the rotors are able to teeter, which allows control of the rotor orientation, not only to avert that left roll but also to allow the pilot to “tilt” the disc any other way he chooses. The teetering pivot at right angles to the span of the rotor is a ridiculously simple method of cyclic pitch control.
thanks for some of the other information - some of you more ”edicated” blokes have a more eloquent way of explaining things than me.
 
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