# Wooden Blades and the Stability of Ken Wallis' gyros

#### Jean Claude

##### Junior Member
What I said is, that with a rotor the reaction always follows the applied force or moment by a phase lag of 90° since the rotor is a second order system in resonance, so maximum flapping occurs over the nose since the maximum moment from cm is applied at the point of maximum flow velocity, i.e. at psi=90°, where psi is the angular blade position and psi=0 is over the tail of the aircraft.
I agree with you on this point.
As for the position of maximum angle of torsion, if you refer to the case of the external weight (not the case of maximum moment due to cm) then I’ll stick by my guns. The maximum torsion of the blade due to the external weight takes place at the point of maximum acceleration due to rotor flapping, which is over the nose of the aircraft.
This is where our opinions differ, Juergen,
During a revolution, the blades must cyclically correct their angle of attack to compensate for their airspeed variations during forward flight, and thus prevent roll. This is usually achieved automatically by the longitudinal flapping.
Despite of this flapping, the axis of the shaft having to continue to include the V formed by the blades (coning), must impose angular oscillations to him and overcome its inertia.
This is the 2/rev vibration explained already many times by Chuck Beaty, whose torque is maximum at 3 o'clock - 9 o'clock.
The angular inertia due to the external nose weight is only added to that of the blades V.

This cyclic torque can produce a proportional torsion of the blades, forcing to push the stick further forward to keep the level (By definition, a longitudinal flapping increased)

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#### kolibri282

##### Super Member
My notion that the maximum flapping takes place over the nose (and in the negative direction over the tail) of the aircraft comes from the equation of flapping motion. The angel of flapping is a1(t) = a1max*cos(oM*t), where a1max is the maximum longitudinal flapping angle, oM is the rotor rotational frequency oM= 2*pi*n/60 (n is rotor rotational speed per minute) and t is the time. The angular flapping speed than is the derivative of that equation with respect to time t, thus vFlap(t) = d(a1(t))/dx = - a1max*oM*sin(oM*t) and the acceleration is the second derivative aFlap(t)= d(vFlap(t))/dt = -a1max*oM^2*cos(oM*t). To get the speed and acceleration at the nose weight angular values have to be multiplied by l=0.8*R, the distance from the flapping hinge to the nose weight, thus
bt2dt = l*btm*oM^2; % flapping acceleration
which is line 25 of my program. I have used beta here because the general flapping equation is usually written as beta(t) = b1*cos(oM*t) + b2*sin(oM*t). Maximum flapping occurs for oM*t = k*pi, where k is any positive integer, so the angular positions for maximum acceleration are above the nose and the tail of the aircraft. I do not quite understand why you think that rotor coning is the source of acceleration for the nose weight, but would actually find it much more intuitive if the maximum effect of the nose weight were in the 3 and 9 o'clock position.
Since you are probably much more experienced in this type of dynamic stuff I really appreciate your scrutiny of my results.

• DonBishop

#### Jean Claude

##### Junior Member
My notion that the maximum flapping takes place over the nose (and in the negative direction over the tail) of the aircraft comes from the equation of flapping motion.
Yes! But the blade twist angle has nothing to do with the angular flapping acceleration, Juergen. See this sketch
Max angular accélération of α is at 9 o'clock and 3 o'clock PS:
Some values of torsional stiffness of blades
- AVERSO Stella with extruded aluminum alloy : 6400 Nm2/rd (measured value)
- Cierva C30 with tubular spar: 7400 Nm2/rd (NACA report 1727 p.6) 8800 Nm2/rd (ARC report 1859 table 9)
- My wooden blades as below: 230 Nm2/rd (measured value) Last edited:

#### Jean Claude

##### Junior Member
The existence of a variable distance between the blade tip and any reference plane depends solely on the choice of reference plane and is not a source of real acceleration (i.e., inertial force generating).
( If you had chosen the actual plane of rotation as reference, you would have only obtained a constant distance, and therefore zero acceleration)

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#### C. Beaty

##### Gold Supporter
The terminology causes confusion and misunderstanding; birds’ wings flap, rotorblades rotate. If a rotorblade flaps while in flight, look out below!
“Flap/drag” hinges permit the rotor to rotate about an axis that differs from the rotorhead axis but that’s not flapping.

• All_In and Jean Claude

#### kolibri282

##### Super Member
The speed of the air at the rotor leading edge at any station r along that edge depends on the blade angular position as:

Vres = V*sin(psi) + oM*r/R

where V is the flight speed, psi is the angular position (psi=0 is over the tail of the aircraft) and oM is the rotational frequnency oM=2*pi*n/60 where n is the rotor angular speed in revolutions per minute. The magnitude of this speed variation is independent of the choice of coordinate system and leads to a change in the aerodynamic force perpendicular to the blade. If we consider the resultant of the aerodynamic force perpendicular to the blade then it is also independent of the choice of coordinate system. The only possible way for the blade to remain in constant, unaccelerated motion would be if this variable aerodynamic force were balanced, at every angle psi, by a force of equal magnitude and opposite direction. The only other force acting perpendicular to the blade is the component of rotational acceleration that acts when the blade is flapping up or down, i.e. when the blade is not exactly perpendicular to the axis of rotation. Now the extreme values of aerodynamic force occur on the advancing and retreating side (psi=90° and psi=270°) while the maximum out of plane angle of the blade occurs near the nose and the tail of the aircraft, therefore the rotational acceleration can not balance the aerodynamic force and according to Newton's law

F(t) = m*a(t)

under the action of this time varying force the blade undergoes a periodic acceleration perpendicular to the plane of rotation.

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#### Jean Claude

##### Junior Member
Now the extreme values of aerodynamic force occur on the advancing and retreating side (psi=90° and psi=270°)
Your mistake is to assume that the variations of aerodynamic forces during the rotation are only due to airspeed variations of blades elements.
While it also depends on the angle of attack changes

In the actual plane of rotation (i.e, the tip path plane), the angle of attack of the blades also varies cyclically, which exactly cancels the effects of airspeed variations. Thus, since there are no longer cyclic variations in aerodynamic forces when the final plane of rotation is established, the blade is no longer subject to cyclic vertical acceleration. Last edited:

#### kolibri282

##### Super Member
The formulae used here are from "Helicopter Performance, Stability and Control" by Raymond Prouty.

In the general case, the axis normal to the tip path plane (TPP) is not aligned with the shaft axis of rotation. The angle is (Prouty page 169):

a1s = mu*(8/3*thtaN + 2*(mu*alfS - vi/oMR) - (1 + 3/2*mu^2)*B1)/(1 - mu^2))

All the variables are uniquely determined by the aircraft and the flight state, there is no free parameter available.

Due to a1s the blades make a different angle with the shaft depending on their position, which leads to a difference in lever arm of the centrifugal force trying to bring the blade back to the axis of rotation. This generates a fluctuation in the accelleration force component normal to the blade which follows a cosine (accB, the yellow curve in the diagram). The extreme values of this out of plane acceleration are located at psi=0 and psi=180 degrees, i.e. over the tail and the nose of the aircraft.

/Quote Your mistake is to assume that the variations of aerodynamic forces during the rotation are only due to airspeed variations of blades elements, while it also depends on the angle of attack changes /Quote

The formula for the resultant blade angle of attack (the blue curve alfa) is given on page 161, I have omitted all terms describing the influence of rotational velocities p,q of the aircraft. i.e. straight and level flight are considered. As can be seen the angle of attack is larger on the retreating side.

The angle of attack is multiplied by the square of the resultant airspeed giving the resultant force (the red curve FalV) which basically follows the angle of attack curve, only that due to the diminished resultant speed the force is small on the retreating side.

As can be seen the force and acceleration curve are out of phase by 90°, which was to be expected since the rotor is a second order system in resonance.

If the blade acceleration were to be zero all along the rotor blade path the resultant force at each blade element must be zero for every station psi around the rotor disk.

How does the rotor force in your opinion, Jean-Claude, balance the blade centrifugal acceleration when the force and acceleration maximum is 90° apart, or is there any additional force acting on the blade?

I have attached the files used to calculate the diagram.  #### Attachments

• TestClcAlfa.txt
927 bytes · Views: 1
• clcAlfa.txt
506 bytes · Views: 1
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#### Jean Claude

##### Junior Member
As with Coriolis, your accelerations relative to the bearing axis are fictitious and thus so are your forces (m.acc).
Watch this video of a bar rotating around its axis (string)
Relative to the plane perpendicular to the string, it undergoes cyclic accelerations. But this is no due from cyclic lift .

Quite simply, the real plane of rotation (tip path plane) does not change, because there is no cyclic force,

#### XXavier

##### Member
I find it very surprising that, in this very basic issue, there may be a discrepancy between Jean Claude and kolibri282... It has to be due to a misunderstanding...

Any point of a rotating 'flapping' blade describes a circumference, that, as such, is contained in a plane. It isn't a 'warped circumference' due to 'flapping'... Hence, the only acceleration that can be felt at any point of the blade is the constant centripetal acceleration due to the rotation. There's no periodic 'acceleration due to flapping' at all anywhere... It's true that –due to feathering– not every point at a given station describes the same orbit, but there's no acceleration anywhere other than the centripetal acceleration...

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#### kolibri282

##### Super Member
The freely rotating beam is a simple spring dash pot system. When the beam makes an angle with the axis of rotation the centrifugal force tries to bring it back into the plane perpendicular to the axis and the moment of inertia of the beam makes it continue to swing downwards when it is perpendicular to the string. This all works since we do not have aerodynamic forces.

Now for the rotor blade the plane of the swash plate is called plane of no feathering, because the blade angle of attack does not change with respect to this plane. The tip path plane on the other hand is one in which there is no cyclic flapping of the blade but the angle of attack changes and so does the resultant velocity on the leading edge, so a cyclic force is generated in the tip path plane which is the red curve in my diagram. This force is not fictous, you can measure it by e.g. placing small orifices in the blade and measuring the difference in pressure on the blade surface. If a force is applied that varies with time then there is a time dependent acceleration according to Newton's dF(t) = m*da(t). The only way to have the blade move around without acceleration would be to apply a force of equal magnitude and opposite direction at every point around the rotor path.

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#### XXavier

##### Member
If the angular velocity of the blades in their real (tip-path) motion plane is constant, it doesn't matter which cyclic external forces may act on the blades. Thus, the angular velocity being constant, there is no cyclic acceleration at any point of the blades at any time. The only acceleration present here is the constant centripetal acceleration due to the rotation...

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#### Mayfield

##### Gold Supporter
George Bernard Shaw allegedly said something akin to “England and America are two countries divided by a common language.”

Something similar happens when lay persons try to visualize the path of the rotor.

Chuck Beatty has been explaining this to the community for more than 50 years. Many of us still have trouble really understanding it.

I for one am very grateful that Chuck, Doug, Javier, Jean Claude, Juergen, and others continue to try to educate us.

I can't speak for others, but I believe, no matter how many times it's explained, using the term “flap” incorrectly denotes a low point, a high point, and movement/acceleration between those points.

Many of us, think in static terms. We divide the circular path of the rotor into 360 degrees, or 6400 mils, or some other division. In our mind we freeze the rotor at some point in it's path then have a tendency to say:

“The rotor tip is at it's lowest point at six o'clock with zero vertical velocity. At three o'clock the rotor tip is at maximum up flap velocity, and at twelve o'clock it is at it's highest point with zero vertical velocity. And finally, at nine o'clock it's at maximum down flap velocity.”

This is the way “flapping” has been explained to student pilots for a long time.

It's much harder for many to accept that “flapping” is a myth, a construct. Hopefully the folks mentioned above won't tire of explaining this to us.

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#### Jean Claude

##### Junior Member
When the beam makes an angle with the axis of rotation the centrifugal force tries to bring it back into the plane perpendicular to the axis
No, Juergen. As the name implies, centrifugal force is a force that moves away from the center of rotation. So does it not try to bring the blade into the plane perpendicular to the bearing axis.
When you give up your false center, then the cyclic flapping aerodynamic force also disappears, i.e. the cyclic effect of the A.o.A cancels the cyclic effect of airspeed The true axis of rotation is not the bearing axis

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#### EI-GYRO

##### 21st Century Crankhandler
Having watched Ken Wallis perform the hands-off hover on several occasions, pretty close-up, (say 30-40feet away), I noted he spent a bit of time
'juggling' the throttle and stick before releasing the controls, and not for very long (say about 10 seconds max), I didn't come away feeling the aircraft was particularly stable. More that it was an experienced and skillful pilot doing a tricky demonstration, rather than any demonstration of any inherent stability of the machine.

Just my impression, feel free to disregard it.

The G-AXAR report is very interesting, but IIRC Pee-wee Judge was only flying it that day due Wallis not being available. Yes, he overdid it, but whether by too much balls or inexperience on type, I can't.

Ernie Brooks (?) did similar in the Brooklands Hornet. PIO/PPO was not understood then.

#### kolibri282

##### Super Member
First I would like to thank everyone who has been bearing with me and so patiently contributed to this thread! Last night I came about a line in the book by Padfield "Helicopter Flight Dynamics" that finally gave me the impression that here was an explanation which I completely understood, see link below. Let me first start with a tiny bit of math. I still have difficulties to understand Jean Claude's idea of a false and true rotation center. For me the rotation center always is the shaft, which I find easy to understand because if you follow math rigorously it leads you to this point of view. I found it very helpful to rigorously stick to math in cases of highly complex, three dimensional motions because you only have to handle the math correctly to arrive at the correct result. The centrifugal acceleration is described by the term:

ac = oM x (oM x r)

where oM is the angular velocity of the shaft about the direction of the shaft (oM actually is a vector, as is r), r is the location vector of the point you consider and x is the vector cross product. Now oM x r is the velocity V of the particle. Let us say that this, in our case, is a tip weight, and by virtue of the cross product is perpendicular to both the axis of rotation and r. So one may write the formula as

ac = oM x V

where V is the tip velocity, as mentioned before. But again oM x V is perpendicular to both oM and V, so following the math the rotation is about the axis oM and ac is perpendicular to oM. Consequently Jean-Claude's "False Rotation Center" is my choice and correctly describes the physics, which is important in what is to follow.

On the page I found, Padfield offers a formula for the out of plane blade acceleration (sic!) azb. Omitting the fuselage pitch/roll contributions the formula is:

azb = rb*( -oM^2*beta - beta2dot)

where beta is the total flapping angle with respect to the shaft and beta2dot the flapping acceleration of the blade. Padfield writes:
Quote: The term azb is the normal acceleration out of plane of rotation, and there are three important components. The first is the acceleration due to blade flapping (note by JH: this is beta2dot), the final term in the expression. When the blade flaps up and down once every revolution this term is equal and opposite to the oM^2*beta term in azb, the component normal to the blade of the centripetal acceleration. /Quote.

I am very glad Padfield rewrote this part of the book, because in the old version, I have in print, this is not stated as clearly. So finally now every piece of the puzzle is in place:
a) there is an acceleration of the blade due to the aerodynamic forces which leads to an out of plane blade acceleration with a 90° phase lag and a change in blade flap angle with respect to the shaft. So rotor flapping does exist in a physical sense!
b) by the odd physics of this very complex system, that is a rotor, this blade acceleration is exactly canceled at every point of the path by the component of centripetal acceleration which is perpendicular to the blade (see sketch below)

If we write the longitudinal flapping angle as:

beta = bN + b1c*cos(oM*t)

and differentiate twice we get

beta2dot = -b1c*oM^2*cos(oM*t)

If we plug this into Padfield's formula for blade acceleration we get:

azb = rb*{ (- oM^2*[bN + b1c*cos(oM*t)]) - (-b1c*oM^2*cos(oM*t) ) }

expanding the square brackets gives

azb = rb*{ - oM^2*bN + ( - oM^2* b1c*cos(oM*t)) - (-b1c*oM^2*cos(oM*t) ) }

or

azb = rb*{ - oM^2*bN - oM^2* b1c*cos(oM*t) + b1c*oM^2*cos(oM*t) ) }

and the two cyclic oM^2*b1c*cos(oM*t) terms cancel. The remaining constant coning angle (bN) term is balanced by the constant part of blade thrust (please note that in the sketch only the cyclic part of centripetal acceleration is shown, you would have to add the constant part that balances the averaged blade thrust, responsible for the blade coning angle). This is identical to the result that Jean-Claude presented, in that there is no resultant acceleration out of the plane of the blade, but for me it is much more rigorous and thus much better to understand. The result actually means that my initial idea was wrong, since the offset blade weight also moves with zero acceleration perpendicular to the blade and there will be no blade torsion. The remaining terms in the azb formula are, in my opinion, too small to give the effect I was considering in this thread.

Of course a direct observation, like the one related by Fergus (thanks goes out for that!), is a strong argument, although I seem to remember that the aircraft with which Ken made his hands off landings is not the one he flew for his hands off photo demonstrations, where his hands off flying was much longer than a few seconds. The one remaining question for me is, why Ken Wallis stuck so stubbornly to his offset blade balancing weights, (something that very obviously deteriorates blade performance) when it would have been much easier to do all the balancing inside the blade.

I had a lot of fun in this thread and have finally understood one part of rotary wing physics I had been struggling with, ever since I started out on my quest to get to grips with rotary wing stability and control. Last edited:
• Vance

#### Jean Claude

##### Junior Member
This is identical to the result that Jean-Claude presented, in that there is no resultant acceleration out of the plane of the blade, but for me it is much more rigorous and thus much better to understand.
For me, if a blade is articulated to the root, then it cannot transmit any flapping moment. The plane of the trajectory of the tips is therefore not linked to the inclination of the shaft: it is just the one that passes through the root joint. And if this inertial plane is stable, then it is enough to consider that the aerodynamic flapping forces cancel each other out.
Whereas with the "academic" vision linked to the shaft, it is necessary to suppose that the real trajectory is decomposed in two simultaneous movements (flapping + circular around the axis), whose mathematical complexity does not give more rigor, but just leaves a greater risk of confusion

It is likely that Ken Wallis sought to reduce the twisting of his blades by placing their Center of Gravity on their Center of Pressure.
With a profile at Cm>0, the center of pressure is in front of the aerodynamic center, thus less than 0.25 chord.

He managed to obtain GC at 0.25c, but it is difficult to still advance without external ballast.

The steel strip starting from the leading edge to advance GC, is then located rather close to the center of inertia, and participates only slightly in the torsional stiffness. Hence, perhaps, the choice of the additional external ballast to reduce the twisting (?)

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#### kolibri282

##### Super Member
Quote: ... it is enough to consider that the aerodynamic flapping forces cancel each other out. .... Whereas with the "academic" vision linked to the shaft, it is necessary to suppose that the real trajectory is decomposed in two simultaneous movements (flapping + circular around the axis), whose mathematical complexity does not give more rigor, but just leaves a greater risk of confusion /Quote

I find it much more confusing if you have to make assumptions for your model which are squarely at odds with the measurable physical reality.

Quote: In the actual plane of rotation (i.e, the tip path plane), the angle of attack of the blades also varies cyclically, which exactly cancels the effects of airspeed variations. /Quote

As my diagram in post #28 shows the speed variation does not compensate for the cyclic change in angle of attack and the resultant rotor force is by no means constant around the rotor disk, this assumption is physically simply wrong.

If you start out with the observable and measurable facts, namely that the angle between shaft and blade varies cyclically around the disk, and then apply first principles to explain the resultant motion and build a correct mathematical/physical model of your system you can step by step build on that. It really is necessary to decompose the actions on the blade in two components because that is the physical reality. In my engineering career, which includes six years as a teacher, I found it the most simple way for students to understand a phenomenon when you just step by step develop a correct mathematical/physical model, because then you can, for our example, in a next step e.g. very easily explain the need for lead/lag hinges in a multiblade rotor. You just have to apply the law of the conservation of momentum and, voila, the need becomes obvious . I wonder what kind of mental contorsionism is needed to incorporate lead/lag hinges in your model.

#### Jean Claude

##### Junior Member
If you start out with the observable and measurable facts, namely that the angle between shaft and blade varies cyclically around the disk, and then apply first principles to explain the resultant motion and build a correct mathematical/physical model of your system you can step by step build on that. It really is necessary to decompose the actions on the blade in two components because that is the physical reality. In my engineering career, which includes six years as a teacher, I found it the most simple way for students to understand a phenomenon when you just step by step develop a correct mathematical/physical model, because then you can, for our example, in a next step e.g. very easily explain the need for lead/lag hinges in a multiblade rotor. You just have to apply the law of the conservation of momentum and, voila, the need becomes obvious . I wonder what kind of mental contorsionism is needed to incorporate lead/lag hinges in your model.
Juergen, it is not necessary to decompose. The physical reality is simply a regular rotation in the Tip Path Plane, without any reason to take into account its inclination to the shaft except for its cyclic pitch control action in our case.
Moreover, in the case of the helicopters pitch control by the usual swashplate, you can even totally ignore the shaft position and conclude very simply that, in the TPP, the pitch cyclic variations speed/air cancels the cyclic pich variations To incorporate the drag hinges, no mental contortion is necessary: It is enough to observe that this constant rotation in the TPP is no longer constant in the drive plane, for a simple reason of geometric projection on this oblique plane in which the perfect circle becomes elliptical. So, even the invocation of the law of conservation of momentum becomes a complication useless in my model.

It is unfortunate that due to an inappropriate reference plane, many engineers still do not understand why Doman or Doblhoff rotors can do without any joints and generate no Coriolis stress.

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