Keep it up Lu,

I think this topic has been one of the most interesting that I have ever read on PPRUNE. Its also good to see that 99.9% of the posts are civil, polite and balanced arguments.

GV

To all of you that have requested a copy of my report, I have made some major changes to one of the three diagrams. Please log onto:

<A HREF="http://205467.homestead.com/diagrams.html" TARGET="_blank">http://205467.homestead.com/diagrams.html</A>

The changes are highlighted in yellow. The changes were percipitated by my trying to figure out what Frank Robinson had said in response to my postings in the Rotorhead Forum. From the changes, you can see that the picture originally painted didn't go deep enough in explaining the problems in countering Zero G.

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The Cat

A rotor disk IS NOT moved by gyroscopic forces. Some call it flapping to equality, call it what you like but there is an aerodynamic effect which will act at some point (90 degrees is used a rule of thumb for basic helicopter aerodynamics and is often compared to a gyroscope to get the idea across to budding helicopter pilots). This is not a point of discussion and will not be advanced by repeating (out of context) the solid disk analogies used by Prouty and others in their articles aimed at pilots.

See my previous post and all the others that mention Lock number.

Afterthought - Autogyros are rigged with no phase lag, the disk moves in phase with the control orbit and are in that sense quite different to helicopters.

[This message has been edited by Grey Area (edited 10 December 2000).]

To: Grey Area

I don't know if I posted this in the Rotorheads Forum but I do know that I posted it on Just Helicopters several months ago.

The following post deals with your philosophy that individual blades flap to equality or,whatever they flap to. The first part deals with retreating blade stall and addresses the fact that the disc moves due to gyroscopic precession as opposed to individual blades stalling out and hitting the tail cone.

First, to fly and or work on a helicopter, you must know how it flys .

Try to visualize this, or better still, pick up a pencil and paper and draw a picture of a circle. Cut the circle into four equal parts. On the circumference of the circle where one line meets, write the letter N for North. Now, label the other lines East, West and South. North is the direction the helicopter is flying. The existing way of teaching helo aerodynamics is to say that the blade is stalled over West and because the blades have a precession angle of 90 degrees the effect will take place 90 degrees later or, over South. That sounds reasonable, but can you picture a blade that has a tip speed near the speed of sound flying from an in track position at West to a stalled position over the tail cone at South. Now, try and picture that same blade flying from that low position over South to an in track position over the right side at East. Then, because the pilot has the cyclic in a forward position, that same blade, because of precession, will be flying low over the nose or, North and back up to an in track position over West and then, it starts all over again at the rate of say 250 or more times a minute. Pilots bitch when they have a small vibration. Can you imagine what the above described action would do to you and the helicopter ?

I would suggest that you stop thinking of individual blades flying in a circle. Instead, think of the blades as a solid disc just like the rotor on a gyroscope. If you apply a force to a gyro that is on gimbals, the gyro because of precession will react 90 degrees later in the direction of rotation. The same is true on a helicopter. It does not matter how many blades and if the rotor system is rigid, semi-rigid or fully articulated. The same laws or physics apply.

Cyclic input will change the pitch relationship across the disc and will result in an imbalance of forces. If the pilot pushes the stick forward, the greater force is over the West side of the disc. This upward force (due to precession) will cause the disc to raise over South and drop over North. Aerodynamics plays a minimal part in this action. The change in disc position was caused
by the change in lift forces but the actual movement was caused by the gyroscopic forces and characteristics of the spinning disc.

Regarding retreating blade stall, the greater applied force is on the right or East side and precession causes the disc to rise at North and go down at South.
Blades do not fly to a position, the are moved by gyroscopic forces. If you want to call it blade flap that’s OK. When you enter the realm of blade stall there are ways of getting out of it but once it happens you can't stop it. Once again, individual blade do not stall. They simply change the balance of forces across the disc and physics does the rest.

In forward flight the disc is tilted down over the nose. In order to do this the pilot introduces forward cyclic. This tilts the disc and the resultant thrust line is also tilted forward and the helicopter flies forward. Discounting inflow and transverse flow the disc got to that position due to the force imbalance across the disc which caused it to tilt. The individual blades did not fly to that position, the blades as a part of the spinning rotor were moved by precession.

Here is another point taught in the text books. The books state that the advancing blade flaps up and the retreating blade flaps down. This is true for a Bell or any other helicopter but the movement up and down is almost imperceptible as any movement out of track will be countered by pitch coupling which restors the blade to the in track position. If the blade flapped the way it is described in the books the helicopter would be flying backwards.

Regarding autogyros, when Helidrvr asked me to compare helicopters with autogyros I didn't want to embarrass myself so, I contacted two manufacturers of autogyros along with a designer of autogyros. All three responded that autogyros change disc attitude the same way they do on helicopters. Create a lift differential across the disc and the disc moves via gyroscopic precession.



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The Cat

[This message has been edited by Lu Zuckerman (edited 10 December 2000).]

[This message has been edited by Lu Zuckerman (edited 11 December 2000).]

[This message has been edited by Lu Zuckerman (edited 11 December 2000).]

To: Grey Area

I am very familiar with rotor blade/rotor head interlock on fully articulated rotor systems but I am not familiar with the term you use (Lock number).

Please explain.

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The Cat

Lu:

All above is correct except, retreating blade stall only occurs above VNE. If the Aircraft is operated within the envelope, no blade stall will occur. What you will get in the case of a fully articulated system, is lead and lag or hunting and seeking.

What is commonly encountered in all systems is advancing or retreating blade stall through VNE. This is recognised by a violent uncommanded Pitching/banking input. The recovery should be to reduce airspeed.

This is the simple explanation for some of our less technically gifted visitors.

Cheers, OffshoreIgor http://www.pprune.org/ubb/NonCGI/eek.gif

Lu:

Are you serious? I must say that I have stayed away of this forum, but when I saw the thread getting so large, I thought I'd have a peek.

Wow, I can't believe the comments made on this thread. You claim that Bell never had a Mast Bumping problem prior to the 206? I think if you do your homework, you will find that every Bell product is prone to Mast Bumping. For those of you out here who don't understand what I'm talking about, mast bumping is when the Rotot assembly (typically an underslung system) contacts the Mast in such a manner, so as to actually cut into the mast. The term Bumping does not do this condition justice!

This condition should not be confused with 'Spike Knock' (another Bell problem) or Pylon Whirl. Pylon whirl is when the Transmission actually moves on it's mounts in a circular or unatural motion.

Lu, you seem like a knowledgeable guy, but, you are not a Pilot. You continue to give advice to pilots and all professional pilots appreciate the technical advice you give.

You are an acomplished engineer, with a fine background in theoretical engineering.

Don't forget that we are very competent pilots with the ability to think and reason out these problems all by ourselves.

Cheers, OffshoreIgor http://www.pprune.org/ubb/NonCGI/eek.gif

Lu,
what pitch coupling are you refering to? I'm not aware of any delta effect on the Bell teetering head, I thought it was more associated with articulated systems (76 has it) or tail rotors.

Grey area,
please remind me of the meaning of lock number, my text books are 3500 miles away! I recall it was related to the number of blades and also affected control power of the rotor? 2 bladed sytem; lock number =0?

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Another day in paradise

I can't believe how many posts there are on this subject but I've decided to add one more.
It seems there are many pilots/mechanics out there who do not understand some basic helicopter aerodynamics. I don't think ANYONE FULLY understands the subject, and anyone who says/thinks they do; are deluding themselves.
It seems to me that most people are confused about the issue of advancing and retreaing blades and the fact that the blades act as a flying disc.
If you take two airfoils (rotor blades), at equal angles of attack (NOT pitch angle), and accelerate them to the same airspeed, they will produce the same amount of lift. Put those same blades on a helicopter in forward flight; the advancing blade will always have a higher airspeed, and therefore more lift.
Obviuosly, without some way of countering this problem, helicopters or gyroplanes could not fly. The problem is countered by allowing the blades to flap, and therefore changing their angle of attack. The advancing blade will flap UP, REDUCING the angle of attack (and lift); and the retreating blade will flap DOWN, INCREASING the angle of attack (and lift). This equalizes the lift across the disc. That's why the retreating blade will always stall first (explanation to follow).
This flapping action happens at exactly opposite sides of the disc, so any movement will be AS a disc and NOT individual blades.
If you ever get a chance, take a look at the blades during a tracking check, with a Chadwick strobe light. It makes no difference if the helicopter is in forward flight or a no-wind hover; collective up or down, or any position of cyclic. The blades will stay in the same plane of rotation. If they didn't, it would be felt as a vertical vibration and the problem would be remedied by changing trim tabs etc.
As Lu has stated many times, the result of this flapping force will occur 90 degrees after the event due to gyroscopic precession. The purpose of Delta hinges and offset pitch horns are to REDUCE THE DEGREE of that flapping. I don't think there are many helicopters that use these things in the main rotor system, they are usually found on tail rotors.
Unfortunately, this flapping action would still result in a high point at the front of the disc (Flap back, or Blow back). The PILOT counters this by adding MORE forward cyclic.
Try this experiment:
While flying at say, 70kts, push forward on cyclic to around a 100 kt attitude and then hold (cyclic) position and note aircraft attitude. you will note that the helicopter will initially pitch down and start to accelerate. It will then pitch up due to Flap back and you will have to add more forward cyclic to keep the helicopter in the selected attitude. I have demonstrated this, hundreds of times for the benefit of student pilots, it's not my imagination.
Lu, if you try this experiment at a low enough airspeed, such as the transition from hover to forward flight, the helicopter WILL pitch up enough to fly backwards; then, if you had the guts to stay with it, forwards again like a pendulum.
Retreating blade stall is caused by a number of factors, NOT just high airspeed!
The retreating blade will stall when it reaches it's critical angle of attack.
If you wanted to experience retreating blade stall, try this:
You'll probably only do it once and we'll all read about it later, so do it on your own!
Load your helicopter to max gross, then go fly at a fairly high airspeed in turbulent conditions, beep N2 to minimum and then practice some steep turns at high altitude. If that doesn't do it, you obviously don't need a helicopter to fly! All of those things contribute to retreating blade stall.
There is no way anyone could cover all the aspects of helicopter aerodynamics on a web site such as this. Just look at the number, size and complexity of all the books that have been written on the subject.
Oh and by the way; don't try the retreating blade stall experiment.

To: offshoreigor Post 1

I realize that the onset of retreating blade stall will occur at, near or beyond VNE but it could also be influenced by other factors as well. I referenced retreating blade stall in order to make my point about blades not acting individually in the process of retreating blade stall. The onset of retreating blade stall usually manifests itself with a vibration caused by the rotor system reacting to the force imbalance across the disc. If you have ever seen a child’s string pull gyroscope you will have noted that it would stay in the position you placed it in. When the gyro rotor speed starts to decay it starts to tumble. At the first instant of tumble the gyro tilts. This puts a perturbing force into the rotor and due to precession the gyro will tilt in another direction . The second instant of tilt sets up another perturbing force and the gyro tilts again and so on until the gyro falls to the floor or what ever you had placed it on.

The same thing is going on in the rotor system although the rotor speed is constant. The first tendency for the disc to tilt back will input a perturbing force and the precession sets up a left roll. If not caught in time the disc will flap back.

To: offshoreigor Post 2

If I implied that only the 206 would suffer mast bumping I am sorry. The situation on other Bells was on the Huey in Viet Nam. At my last count, which was in 1976 there, were around 53 incidents of mast bumping on Bell Helicopters. We had a mast separation in Iran where the blades and rotor head on an AB 205 came off when the engines and rotor were running down.

The Bell fix was to thicken the mast in the area of mast bumping and on the AB 205s they placed a rubber bumper around the mast to protect it from being bumped. The reason this never occurred on earlier Bells was that there was a cable that limited the rotor travel in teetering. If the pilot came up against the cable, he got a sudden jarring as his rotor mast and trannie would move forward. The engine, which was hard mounted to the trannie also, had similar cables attaching the engine to the airframe. The whole assembly was restrained by lord mounts and was quite flexible.

Regarding pylon whirl it is my opinion that this is caused by the positioning of the lift link on the transmission in relation to the center of lift forces acting through the transmission. Under heavy lifting conditions or maneuvering conditions the displacement of the lift link in relation to the lifting forces causes the transmission to shift position.

This shifting of position in my opinion was the cause of the Bell 412s drive shaft to become disconnected in flight.


To: 212 Man

The pitch coupling on the Bell Main Rotor system will occur under two conditions. Condition (1) the pitch horn must be above or below the teetering axis and, condition (2) the blades must move in such a way that the tip path plane is not parallel to the swashplate. Both conditions must be present in order to get pitch coupling.


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The Cat

To: fishboy

Regarding you comment that there is no one person that fully understands helicopter flight theory, I beg to differ with you. There are many many individuals that totally understand this subject but I am not one of them. After expounding on the fact that no one person fully understands the subject you put forth one of your own theories. You stated that the advancing blade would always have a higher degree of lift than the retreating blade. If that were true, the helicopter would roll violently to the left.

One of the things you forgot was that in order to have forward flight the disc must tip. In order to make the disc tilt, the pilot had to introduce a forward cyclic input. This creates a differential of lift across the disc and this lift differential causes the disc to tilt due to precession.

Now, the helicopter is flying forward and discounting induced flow and transverse flow effect you will note that the advancing blade has less pitch than the retreating blade. This coupled with the airflow across the disc equalizes the lift across the disc.

The flapping up or down of the blades has nothing to do with retreating blade stall. The flapping that you are addressing is the result of transverse flow effect. Once you are through this flight area (around 20-30 knots) the flapping goes away. The flapping I am addressing is due to minor outside inputs such as gusting or turbulence. This flapping is countered by the delta hinge effect on the main rotor.

You quoted me as saying that the flapping force will occur 90-degrees after the event due to gyroscopic precession. That is not what I said. I indicated that when the pilot introduces any cyclic input it would create a short-term imbalance of forces across the disc. The greatest force will be on one side of the disc and the disc will tilt up 90-degrees after the pitch input. The resultant on Bell Blade will operate around the teeter bearing. On a Robinson it will operate around the teeter bearing and to some extent around the cone hinges. On a fully articulated rotor head this action takes place around the flapping hinges. On flex beam rotor heads this action takes place on or about a theoretical hinge or bending point on the rotorhead.


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The Cat

Lu please, if you are going to reply to, and consequently destroy someones comments, take some time to read it thouroughly.
I pointed out already, that if dissymmetry of lift was not compensated for, no helicopter or gyro could fly.
Also, I did not say that flapping was anything to do with retreating blade stall. The retreating blade will always stall first because it has a higher AoA.
The rotor disc does NOT need to be tilted forwards in order to have forward flight. It can actually be tipped backwards and the helicopter can still have forward motion. It needs to be tipped forwards to INITIATE or MAINTAIN forward flight.
During forward flight a helicopter will continue to fly with considerable airspeed and the cyclic/swash plate in the neutral position. ie all the blades having the same PITCH ANGLE. During that scenario, unless something is automatically done about it, the helicopter would, as you say, violently roll to the retreating side.
That is what the flapping does. It compensates for the different airspeeds of the blades by changing ANGLE OF ATTACK, NOT PITCH ANGLE.
This was a problem that early helicopter designers struggled with. Juan de la Cierva developed the flap hinge. Without it, helicopters would not be able to fly.
Before you mention two blade systems with no flap hinge. A two bladed system will flap as a unit; as one blade is rising, the other is descending.
I am not confused with transverse flow effect, I agree that above around 20-30kts there is no transverse flow effect.
As far as Gyroscopic precession is concerned, I wholeheartedly agree with you.
If I am wrong about this, I will have to throw away all of my books, forget everything I have ever been taught and apologise to all my old students for talking crap for the last few years.

To: fishboy

In your comment about dissymetry of lift not being compensated for no helicopter or gyro would fly. You have placed the cart in front of the horse. For the helicopter to fly in any direction the disc must be tilted. In effecting this tilt the pilot must input some cyclic control. This means that the advancing blade has decreased pitch when over the right side of the helicopter (Not French design) and the retreating blade has increased pitch. This causes the disc to tilt and the helicopter to fly in the direction of cyclic movement (Not Robinson). With the helicopter flying in the commanded direction there is no need to compensate for any thing (discounting inflow and transverse flow).

In your second and third paragraphs you address transient conditions and not conditions that would result in forward flight and maintain forward flight.

What we need is a common ground in addressing flapping. We can address the flapping of the disc when in the first 20-30 knots of forward flight caused by inflow and transverse flow effect and we can talk about the flapping caused by gusting and or turbulence which is countered by pitch coupling. Flapping does change the angle of attack in relation to the relative wind and, this helps in equalizing the lift across the disc. If you want to address the tilting of the disc as flapping that’s OK with me.

Regarding retreating blade stall, the blade is not stalling, it is generating less lift and this causes a force imbalance across the disc and now the advancing blade is generating more lift. In this condition there is a disymmetry of lift and with gyroscopic precession the disc tilts back.


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The Cat

I apologise for the delay, length and pedantic nature of this post but for the benefit of those who insist that rotor disk is controlled as a result of gyroscopic precession I have attempted to merge and précis the relevant areas of aerodynamic blade element theory and certain simplifications used in training circles whilst avoiding mathematics as much as possible. I have avoided the use of metaphor, as the metaphorical usage of precession is what has caused so much misunderstanding. For further reading on the subject try “Aerodynamics of the Helicopter” Alfred Gessow/ Garry C. Myers Jr (ISBN 0-8044-4275-4); seen by many as a definitive work on the subject, I would recommend it as without it I would not have been selected for Test Pilot training.

For the purposes of this piece I have made the following assumptions; a two bladed teetering head rotating anti-clockwise with no physical damping or stops; symmetrical blade profile; ideal twist giving constant lift along the span; no tip vortices and no higher order harmonics. (For the mathematicians out there, please replace the latter with the assumption of infinite blade mass when in a steady state unless you want to calculate the Fourier series yourself!)

First, one needs a reference; there is a choice of three standard references.

The control axis; which is defined as the axis about which no cyclic pitch changes occur, i.e. with reference to the control axis the pitch angle is constant around the disk.

The tip path plane axis; this is perpendicular to the tip path plane, the plane through which all blade tips can be imagined to pass.

The shaft axis. The axis of the rotor shaft.

Suppose, in a still air hover, the shaft axis is held vertically with the control axis coincident. With reference to the control axis the tip path plane will be displaced upward (flapping upward) because of the coning angle (a0) but will be parallel to the control plane. If the cyclic is moved rapidly aft, the control axis will rotate aft. A blade in the 3 o’clock position will experience an instant increase in angle of attack and therefore lift. Because of this increased lift the blade moves of “flaps” upward, likewise in the 9 o’clock position the blade will experience a decrease in lift and will move down; the tip path plane will move, the disk will tilt freely so as to maintain a balance between the moments on opposite sides of the disk.

But, how will it tilt? If one applies the basic rules of Simple Harmonic Motion (SHM), it is an undisputed fact that a rotating or oscillating harmonic system will experience maximum acceleration and minimum velocity at a point of maximum displacement. This does not give us the solution but it does provide the tools. We have already accepted that the disk will tilt to balance forces so it will tilt in a manner that will equalise the lift. On the advancing side to tilt the disk aft the blade will be moving upwards at maximum velocity at the 3 o’clock position, therefore maximum vertical acceleration downward will occur at 12 o’clock. In order to equalise lift the tip path plane must move to equalise the ANGLE OF ATTACK. As this disk is experiencing maximum upward velocity at the 3 o’clock position it will encounter an airflow from above which would tend to decrease the ANGLE OF ATTACK, therefore to equalise lift the PITCH ANGLE must be increased hence increasing the ANGLE OF ATTACK to the disks’ average level; therefore the PITCH ANGLE is proportional to the vertical velocity. Using the mathematics of SHM the vertical velocity is the product of differentiating the displacement, thus the PITCH ANGLE is proportional the differential of the disk displacement.
In the simplest case, for a disk tilted aft the displacement (ß) at any point on the disk (ψ measured anticlockwise from 12 o’clock) with a coning angle coefficient (a0) and a maximum displacement in the longitudinal axis (a1) [proportional to cyclic longitudinal input] will be defined as follows:

ß = a0 + a1cosψ

Differentiating we get the velocity which is proportional to PITCH ANGLE

dß / dt = -a1sinψ

i.e. at 3 o’clock (270 deg) displacement = 0 and pitch = a1

With reference to the control axis the feathering angles are constant (feathering is defined as the difference in pitch from the mean disk pitch), with reference to the tip path plane the flapping is constant i.e.

0 deg = Flapping full down = Feathering Zero
90 deg = No flapping = Feathering max down
180 deg = Flapping full up = Feathering zero
270 deg = No flapping = Feathering max up

The result of these aerodynamic effects, in this case the control input must be applied 90° prior to the desired displacement.

In this example I have not made reference to the momentum of the blade (or higher order harmonics) in the vertical plane. As momentum must be a factor then it becomes obvious that without compensation the blade will overshoot its desired position in space. Thus the control inputs must be put in early, instead of 90° in advance, 85° or whatever depending on the mass of the blades, the number of blades, the density altitude, damping systems, rigid design and so on. One point that has great significance is the distance between the flapping hinge and the hub. In the case of a teetering head (d=0) the blades are freely hinged and control is achieved by purely tilting the thrust vector; as the distance increases so an increasing moment is created about the hub, which will be transmitted to the airframe. The lock number can be seen as mathematical ratio of aerodynamic to inertial forces on the rotor blade [this is a bit of simplification]. Note that as density altitude changes so will lock number and so therefore will the performance of the disk. This means that a designer will design his helicopter to have no cross couples for a specified height. If the density altitude is different so the phase angle will change but the rigging angle of the pitch change rods cannot, therefore the aircraft will have an undemanded roll component in all but the design conditions [except where lock = 0 ie undamped teetering head]. Note also that the total rotor thrust will affect the whole affair also.

On top of all that if any delta three component is present it will increase the effective rigidity of the system and consequently will alter the behaviour of the blades. Be aware also that at low AoA or with unusual aerofoils the effects of a delta three hinge can change as cyclic or collective pitch is altered; it is only where the change of coefficient of lift is proportional to pitch angle that the system works exactly as advertised.

With respect to low g I do concede that where no lift is being produced by the blades then gyroscopic precession might be noticeable but in normal flight regimes the forces involved are well below those flying the blades. I have found this to be a very misunderstood area among my students, often because the emphasis from the pilots’ point of view is the Low “G” and not what is really happening to the rotor head. A sound knowledge of one’s own helicopter is vital to understand what will happen. The key point to take on board is that in a low “G” environment the blades will be producing little or no lift unless a cyclic input is present; it is how the helicopter will respond to disk movement that is important to the pilot. For example, in a pure teetering head the airframe is acted upon by total rotor thrust, tail rotor thrust, drag and gravity; the disk is affected by centrifugal effects, lift and control inputs; there is no direct link between the disk and airframe. In a low rotor thrust environment the disk is virtually undamped and is free to move irrespective of its relative position to the airframe, the pilot, effectively, has no control over aircraft attitude. On the other hand a “rigid” head is able to apply manoeuvring forces to the airframe and conversely airframe movement is transmitted to the disk, the chance of blade strike is seriously reduced though not impossible.

To summarise, rotors do not suffer phase lag as a result of precession they are subject to the basic laws of physics that tell us that the position in space of an object can be derived through integration of its velocity. If you cannot accept these aerodynamic facts I would humbly suggest that criticism of the aerodynamic properties of any rotor system is unwise.

Grey Area.


[This message has been edited by Grey Area (edited 15 December 2000).]

Grey Area,

thanks for taking the time to cover the above. I'm intrigued by the SHM bit; surely there must be some form of damping to prevent the disc oscillating about the positions mentioned? Where is the restoring force constant derived from?

Anyway, my money's with you!

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Another day in paradise

To: Grey Area

First let me say this. Mr. Gessow has authored and co-authored many books on helicopter aerodynamics and as one book succeeds the previous book his theories change. This in itself is not bad as it indicates his acceptance of new fact.

However if this is how he explains how rotor blades get from point A to point B he is using mathematical gibberish. If he really believes that his explanation of blade dynamics resulting from pitch input then he is standing in the middle of a circle of all of his peers that believe in the laws of physics as applied to gyroscopic precession.

Relative to the example of the helicopter being in a still air hover: Mr. Gessow is addressing a two-blade helicopter (Bell). He addresses lift and does not relate the subsequent change of the tip path to gyroscopic precession.

Lets now take his example and apply it to a multi blade fully articulated rotor system. In his example the blade on the right side is at the maximum pitch point but unless there are four blades the blade on the left side has not yet reached its’ minimum pitch point. According to Mr. Gessow the blade on the right will depart the stable hover tip path and flap upwards over the nose. Mean while the blade on the left has yet to reach its’ minimum pitch point so it can’t fully flap down. While all of this is going on the other blades would still be in the hover tip path or, getting ready to depart the tip path. The blades over the nose and tail are still in the hover tip path. Can you imagine the dynamic loads that would be generated as the blades went out of track at 250 RPM or, more?

Now, I will say it again not only for your edification but also for the others that are confused by your posting.

INDIVIDUAL BLADES DO NOT FLAP. THE BLADES ARE A PART OF A ROTATING DISC AND AS SUCH THE DISC HAS THE PROPERTIES OF A GYROSCOPE. RIGIDITY IN SPACE AND PRECESSION. AN EXTERNAL PERTURBING FORCE CAN INDUCE PRECESSION. IN THE CASE OF THE ROTOR SYSTEM, THE PERTURBING FORCE IS CYCLIC INPUT THAT CREATES A FORCE IMBALANCE, WHICH THROUGH THE LAWS OF PHYSICS CAUSES THE ROTATING DISC TO PRECESS. THIS PRECESSION INDUCED MOVEMENT IS 90-DEGREES AHEAD IN ROTATION FROM THE INPUT OF THE FORCE.

I could type in the section of the FAA Rotorcraft Flying Handbook that addresses Gyroscopic Precession but it says exactly the same as what I have typed above.

I am in total agreement with your comments about zero G.

Regarding your last comment, I can ask you the same question. As I stated previously, Mr. Gessow over the passage of time updated his theories. How about you?


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The Cat

[This message has been edited by Lu Zuckerman (edited 14 December 2000).]

[This message has been edited by Lu Zuckerman (edited 14 December 2000).]

Lu,

I am quite happy for one to imagine the rotor system as a disk, however one must analyse discreet elements to understand how it flies. Please note as I said all of the blades are independant of each other and therefore the position in space of one will not have an influence on the other (except wake turblence).

I am willing to learn so please explain in your precessive terms the following which I can explain using discrete element theory.

1. Why do the following phase angles differ at ICAO standard
a. Gazelle - 87 deg
b. Lynx - 72 deg
c. Squirrel - 78 deg

2. Why do helicopters not experience the rigidity in space that is one of the key properties of a gyro - ie why are they so unstable?

3. Why was a gyro stabilising bar fitted to a Bell 47 set 90deg from the rotor blades rather than just using the "precessive" qualities of the disk?

4. You have been quite critical of the delta-three hinge used in the R22. How can this be a factor if disk control is affected through precession? If this were so then the amplified restorative forces provided by the D3 hinge would not affect the stability of the rotor.

5. Imagine a rigid head, such as fitted to the Squirrel. The head is subject to aerodynamic loads from the blades and also subject to moments from fuselage loads. In forward flight the horizontal stabiliser on the squirrel provides a downforce to reduce the airframe attitude and so reduce drag whilst also enhancing longitudinal stability. If precession were the prime mover then the nose up moment provided by the stabiliser and passed to the head due to its rigid nature would induce a precessive roll to the left (Squirrel rotor turning clockwise), yet this is not noticeable, only the improved longitudinal stability is detectable by the pilot, lateral stability is not affected. Additionally as personnel move around the cabin, a movement from left to right would induce pitch and movement fore and aft would induce roll, again this is not so.

Grey Area


[This message has been edited by Grey Area (edited 15 December 2000).]

To: Grey Area

In response to your questions, here are my answers.

1) Regarding the Gazelle, the Lynx and the Squirrel (A Star) I can’t answer unless I can become familiar with the setup of the flight control systems and the rotorheads. I have a picture of an A Star and it “appears” that the pitch horns lead the blade by “what seems” to be 45-degrees. Other than that, I have no answer. A long time ago I heard that on the Lynx the entire gearbox was tilted to obtain forward flight but I don’t know if it was true.

2) Helicopters do not have rigidity in space the rotor system has rigidity in space. The helicopter hangs beneath the rotor disc like a pendulum and as such can be influenced by external forces such as gusting and different forms of drag. This is not to say that external forces can’t influence the rotor disc. If the rotor disc is exposed to gusting the blades will tend to rise out of track and the delta hinge effect will restore the blade to the in track position. Some degree of rigidity of the fuselage can be effected by the interlock of the blades and the rotorhead on a fully articulated rotor head or, a flex rotor such as that used on French helicopters and the newer MD Helicopters. It goes without saying that rigid rotors such as those used on the Bo/BK Helicopters have the strongest level of interlock and as such have a higher level of rigidity. In using the term rigidity as applied to the fuselage it really should be stability.

3) On the early Bell rotorheads the designer (Arthur Young) figured it out. If he didn’t incorporate the stabilizer bar he would not be able to introduce cyclic control and besides, without the stabilizer bar there would be no Bell Helicopter. The purpose of the stabilizer bar is to provide a stable device through which the cyclic inputs could be made to the blades. The cyclic inputs do not go directly to the blades. They are inputted into a small lever that is supported inside the stabilizer bar. The input to the lever is then transmitted to the pitch horn. The reason that the stabilizer bar at the 90-degree position is that the cyclic input to the blades had to be made 90-degrees ahead of the desired results. This is due to the phase angle of the blades, which is 90-degrees. If the blades are gusted out of position the stabilizer bar acts as a point of reaction to introduce the delta hinge effect. The stabilizer bar has a teeter point and as such will align itself with the blades. Since the stabilizer bar can move in relation to the tip path it incorporates dampers to maintain a constant rate of angular change.


4) The delta hinge effect acts on individual blades to restore the original pitch setting when a blade is displaced by gusting or other external forces. On the Robinson the blades act opposite to each other as when one blade flaps up pitch is removed from that blade and the opposite is true for the other blade just like on a tail rotor. On multi blade rotor systems the blades are acted on individually. If there is any blade displacement the restorative effect of the delta hinge reacts before any change in the disc path is made by gyroscopic forces. If precession of the disc is caused by external forces there will be an angular displacement between the disc and the swashplate and the delta hinge effect will restore the disc to the originally commanded position

5) As forward speed increases and the horizontal stabilizer becomes aerodynamic it forces the tail down. It would seem to me that the pilot would be aware of the nose up result of this aerodynamic effect and as such would compensate for it by adjusting his cyclic position. In doing this he reintroduces the correct angular relationship between the swashplate and the rotor disc. Or, in the mere act of introducing forward cyclic the aerodynamic forces of the stabilizer are building as the speed increases so that the downward force and the tip path are arrived at and the pilot has no reference to the downward moment. On some Bell Helicopters, the angular position of the horizontal stabilizer is linked to fore and aft cyclic position. The more forward cyclic the higher the angular deflection. In doing this, the pilot does not have to compensate for the angular difference between the swashplate and the disc as the angle will be zero. Regarding the movement of personnel in the cabin I’m afraid I am going to let this part of my response held in abeyance.

I tried to diagram the system and what I found is that if the disc is displaced in relation to the swashplate it would be the same as introducing fore and aft cyclic or left or right cyclic. You indicated that there is no pitch and no roll and I will have to accept that until I can gain some knowledge on the rotor system and the input of control


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The Cat

[This message has been edited by Lu Zuckerman (edited 15 December 2000).]

Grey Area, I'm curious, as an instructor how do you explain cyclic action and disk movement to a beginning student? The above explanation seems a little weighty, especially for someone with little or no background knowledge. Lu also has a point about the FAA handbook using the gyroscopic precession example. Could this be a source of your student's confusion?

To: Grey Area

In reference to the A Star rotor head I contacted American Eurocopter and they sent me a fax of the swasplate rotorhead interface.

Although the illustration was a bit sketchy I came up with the following: In order to follow the text you will have to make a diagram. I will eventually make a detailed sketch and have Helidrvr put it on the Internet.

First draw a circle. Draw a line from the top of the circle to the bottom of the circle passing through the center of the circle. Now draw a line from left to right intersecting the first line at the center of the circle with a disposition of 90-degrees between the lines. With that, you have a pie diagram with four equal segments.

The vertical centerline is the direction of flight. Place an arrow at the top of the circle to indicate the DOF. Now it gets tricky. Moving from the top of the circle in a counterclockwise direction measure off 30-degrees. At that point, place the letter B. This is the location of the Fore & Aft servo. From that point going in a clockwise direction measure off 90-degrees. At that point mark the letter A. From point A, draw a line that intersects the center of the circle and connect the line to the opposite side of the circle and mark this point as C. Points A and C are the locations of the lateral servos. Draw a line from point B intersecting the center point of the circle and connect the line to the edge of the circle. That is where the stationary scissors attaches to the swash plate.

Now, for some aerodynamics. If the pilot pushes forward cyclic, point B will go down and the scissors will extend. The swashplate will rotate about points A and C. From what I understand, the pitch horn leads the blade by 60 degrees. That means when the advancing blade is over the left side of the helicopter it will have the pitch horn over point B. Since point B has moved down then the advancing blade will be at its’ lowest point of pitch and according to the laws of gyroscopic precession it will be down over the nose and will immediately be followed by the next two blades in the line. It really doesn’t matter where the pitch horn is in relation to the blade as long as the control input and the pitch horn displacement adds up to 90-degrees.

I will continue to work the problem about people moving in the cabin and the fact that you don't get a gyroscopically induced pitch or roll moment.


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The Cat