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View Full Version : Aerodynamic Precession, Not Gyroscopic

Fliterisk1
03-19-2005, 10:28 PM
I would like to correct myself concerning statements made on gyroscopic vs. aerodynamic precession. I was wrong. The gyroplane rotor does not precess, though it acts as if that is what it does. If need be, I’m sure there are those present in this group capable of correcting this explanation.

The rotor of a gyroplane does react in a manner similar to (though not the same as) a gyroscope when determining response to control inputs, so treating it as if it were one will provide essentially the correct answer in predicting its response. It has angular momentum which follows the laws of physics. But it also has aerodynamic forces from blades at relatively high airspeeds which are quite capable of shifting the axis of spin and all its momentum, and do so rapidly. Yet they do so without creating a moment which would cause the typical gyroscopic precession. (!)

If a slim bar in the shape of a rotor is mounted in a gimbal ring, then it will precess like a gyroscope when a moment is applied to the axis of rotation. An example of this is a propeller mounted on a shaft, and it will precess as expected. The rotor of a gyroplane, however, is not mounted this way. It has that teetering hinge. It is impossible for the standard teeter hinge to apply a moment to the rotor the would result in precession. This is really too obvious to be overlooked—though I was stubborn. (I drew a pic of a gyroscope today. Mr. Beaty‘s example of a twirling rock over an air jet was a pointer.)

If the spinning mass is a gyroscope, or the propeller on a gyro, then the typical moments that are applied to them will come from a direction which is fixed in space, or move relatively slowly. However, the forces which are applied to the rotor are from the blades themselves, and so they move at speed with the rotor.

The forces applied to the blades are from the blades themselves, and cause translational movement; having little to do with change in angular momentum. (So where did the angular momentum go? Absorbed by the blade feathering, I assume.) But apparently they actually do fly to their new position as has been said before (repeatedly) in another thread. They cannot react gyroscopically against a moment because no moment is generated from the rotor head—usually. However, modifying the angle of the teeter hinge (also mentioned repeatedly in another thread) should allow for the application of moments to the rotor from the rotor head. Doing that should introduce actual precession, though it would probably be mixed with other aerodynamic effects.
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Also, with regard to a discussion in that other thread about unloaded rotors. I’m wondering why helicopters which have the teetering hinge can also strike the tail under unloaded conditions—meaning the rotor is not under control. I didn’t think precession would do that. But it’s been said that the rotor will rapidly realign to its new position with the pilot’s adjustment of the cyclic. Pilot’s adjustment of the cyclic is normally rather controlled. But suppose panic sets in, and cyclic movement became rapid and extreme. Then the blades could develop a momentum from ’aerodynamic precession’ (‘flap‘) which would cause them to bend, and putting them outside normal and safe range of control movements, resulting in a tail strike. Or can casual stick movements also do this in zero-g? Why?
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Extended formation flight is worrisome business
So be your own man
Be wary of fraternities.
Bryan

C. Beaty
03-20-2005, 09:15 AM
Perception is contingent upon perspective.

Consider the Bell gyroscopic stabilizer bar.

I think most everyone agrees that a pair of flyweights on a pivoted bar is as much a gyroscope as one with a solid rim.

The Bell stabilizer supplies rate damping to the helicopter as a whole.

If the attitude of the machine is disturbed, the gyro bar lags behind a mast tilt; lag dependant on tilt rate and as the rotor is referenced to the bar, rotor thrust moment opposes airframe tilt.

The stabilizer bar is slowly precessed into alignment with the mast by the action of dampers, items 40 in the patent drawing.

Now consider the stick drawing to the right. No dampers.

If the observer is seated on and rotates with the mast, the flyweights oscillate with sinusoidal motion if an impulse has caused the gyro bar to precess relative to the mast. The only damping tending to precess the gyro bar into alignment is solely due to pivot friction. Damping is the ratio of energy stored to energy dissipated per cycle, not very much being dissipated with ball bearing pivots.

The velocity about the pivots is the first derivative of displacement.

Now imagine that instead of dampers, springs are connected between bar and mast.

The rotating observer wouldn’t notice anything had changed if he had no way of determining frequency and assuming stationary objects are a blur. The gyro bar would oscillate with sinusoidal motion just as previously, only a little faster.

Stationary observers would see an entirely different picture.

The spring restrained gyro bar would nutate at a rate equal to the frequency difference between 1/rev and the frequency of bar with spring. Nutate as the Earth nutates or a top nutates before falling over.

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The Hiller Rotormatic gyroscopic stabilizer works on the same principle but control mechanism is quite different.

The flyweights of the Hiller gyro bar are contained inside of small symmetrical airfoils at the tips of the bar.

The Hiller paddleblade airfoils supply precessional torque of the same order and phase as the friction dampers of the Bell bar.

The behavior of the main rotor is identical with that of the gyro bars except damping is so great, spring restraint won’t ordinarily excite a nutational mode. I have observed nutation in tail rotors with more that 45º of delta-3 coupling.

All depends upon viewing perspective. Rotorblades don’t flap at all if viewed from their tip plane axis.

Al_Hammer
03-20-2005, 10:05 AM
Chuck mentioned some good points. I'll add my 2 cents worth...

However, the forces which are applied to the rotor are from the blades themselves, and so they move at speed with the rotor.

The forces applied to the blades are from the blades themselves, and cause translational movement; having little to do with change in angular momentum. (So where did the angular momentum go? Absorbed by the blade feathering, I assume.)

It is true that no moment can be applied across the teetering hinge to the blades (or from the blades to the mast.) That is why the pilot's stick input does not cause precession to occur from control forces "directly."
However, the rotor does have a moment applied to it by the aerodynamic forces brought about by cylical angle of attack changes made by the movement of the rotorhead. These forces while "moving at the speed of the rotor" in one sense, are really varying in magnitude,cyclically. The force is always greatest at some point and minimum at the opposite point. It is fine to say that the rotor flies to a new position, but that is another way of saying that the applied moment comes from the air and not from the spindle.

modifying the angle of the teeter hinge (also mentioned repeatedly in another thread) should allow for the application of moments to the rotor from the rotor head.

If you are referring to the delta hinge, it does not allow for moments to be applied, but it does change the angle of "precession" to something less than 90 deg. The addition of an offset flapping hinge, on the other hand, does allow for moments to be transmitted from the rotor to the mast. You may be confusing the two types of hinges. Offset flapping hinges are described in any good helicopter reference book and on many web sites.

I'm wondering why helicopters which have the teetering hinge can also strike the tail under unloaded conditions—meaning the rotor is not under control. I didn't think precession would do that. But it's been said that the rotor will rapidly realign to its new position with the pilot's adjustment of the cyclic. Pilot's adjustment of the cyclic is normally rather controlled. But suppose panic sets in, and cyclic movement became rapid and extreme. Then the blades could develop a momentum from 'aerodynamic precession' ('flap') which would cause them to bend, and putting them outside normal and safe range of control movements, resulting in a tail strike. Or can casual stick movements also do this in zero-g?

The rotor can strike the tail during low g events because flapping from dysymmetry of lift becomes excessive as the rotor slows down. The rotor slows whenever the load is reduced, as in zero g. The problem actually occurs when the load is reapplied as a result of sudden back stick. A gradual application of back stick is a good thing, since it will reload the rotor. Sudden, excessive stick movements , either forward or back, always must be avoided in a teetering rotor system.

C. Beaty
03-20-2005, 11:02 AM
The rotor can strike the tail during low g events because flapping from dysymmetry of lift becomes excessive as the rotor slows down. The rotor slows whenever the load is reduced, as in zero g. The problem actually occurs when the load is reapplied as a result of sudden back stick. A gradual application of back stick is a good thing, since it will reload the rotor. Sudden, excessive stick movements , either forward or back, always must be avoided in a teetering rotor system.

I have to take some exceptions to that statement, Al.

Something called precession stall was not uncommon in tail rotors of older helicopters. Pirouette the machine too fast and run the risk of chopping the tail boom. A yaw rate too high could demand higher blade lift than could be supplied.

I’ve witnessed several tumble out of the sky accidents and have inspected the remains several more.

The Bensen tin tails were a good guide to the tumble rate.

There was always a sharp crease near the base of the vertical tail and another about 6” from the top. This provides the angle through which the airframe would have to rotate in ½ turn of the rotor. Whatever one assumes the rotor rpm to be; and it has to be fast enough to crease the tail, the airframe rotation has a rate that exceeds the rotor’s ability to follow.

And a Bensen wasn’t nearly as bad as the early edition AirCommands in respect to propeller thrustline/CG offset. Bensens would typically go through several cycles of PIO before the rotor began chopping at the tail.

The old AirCommands would be flying at what looked to be normal flight from the ground and suddenly tumble forward.

Frank Robinson, while still an engineer at Bell, co-authored an article published in the AHS Journal wherein tail rotor precession stall was discussed.

If I can locate your mailing address, I’ll send you a CD of it.

mceagle
03-20-2005, 04:00 PM
they actually do fly to their new position as has been said before (repeatedly) in another thread.
If blades actually "flew" to their new position in the circle, why is that position is always at 90 degrees to the applied force? Wouldn't the position in the circle be dependant upon other factors such as rotor speed, weight, airfoil, and forward speed? Surely it is not coincidental that maximum displacment is in line with precession laws, no matter what the other factors.

It is impossible for the standard teeter hinge to apply a moment to the rotor the would result in precession.
The teetering hinge does not supply a moment to the rotor but rather it alters the cyclic pitch which in turn supplies the moment to the rotor that results in precession.

If the same moment/force was applied to a spinning disc or push bike rim by an air jet then it would react the same way, and it has no aerodynamic considerations.

Udi
03-20-2005, 06:58 PM
Tim is correct. Both inertial and aerodynamic factors are at play. Any rotating disc is a gyroscope. A gyro rotor is a gyroscope, hence the name gyroplane. The fact that the rotor is hinged does not make the rotor any less gyroscopically stable than if it had no hinge. The purpose of the hinge is to allow teetering and aerodynamic control. Even if the rotor blades are "flying into their new position" they are still doing so with the help, or against the resistance, as the case may be, of gyroscopic stability.

Think again of what is happening. Say you are flying straight and level and suddenly you pull back on the stick. When the rotor blades are in the 0-180 position, they don't see any change. When they are in the 90-270 position, the advancing blade is seeing a higher AOA than before and the retreating blade is seeing a lower AOA than before. At that moment, the advancing blade is creating more lift than the retreating blade. It's like an airplane wing with the right aileron deflected down and the left aileron deflected up. You would expect the rotor disc to tilt to the left (in the US, right in Oz). But it doesn't! Instead the rotor disc is tilting back.

The question that you have to ask yourself is, how much would the rotor tilt back for a given stick input? In a gyroscope, the reaction - precession, is a function of the mass of the disc and the upsetting force. In the rotor of a gyro the magnitude of the reaction is actually determined by aerodynamics. The blades follow the position of the rotor head. They must follow the rotor head or there will be dissymmetry of lift between the two blades.

So, do the rotors FLY to their new position, or do they gyroscopically precess? They do BOTH! But the aerodynamics are the controls, and the "power steering". The proof? Heavy rotors are slower to respond to aerodynamic input and, in extreme cases they can experience precession stall, which happen when the aerodynamic controls can't keep up with (or overcome) gyroscopic stability.

Udi

Victor Duarte
03-20-2005, 10:12 PM
look at that http://www.alexisparkinn.com/photogallery/Videos/Heli_refueling_accident.mpeg

Fliterisk1
03-21-2005, 08:41 AM
Beaty said: I think most everyone agrees that a pair of flyweights on a pivoted bar is as much a gyroscope as one with a solid rim.
I'm one of the not most of.
The stabilizer bar is slowly precessed into alignment with the mast by the action of dampers, items 40 in the patent drawing.
Oh, come on. Precession implies force applied. There is no precession for there is no fixed moment applied. The dampers prevent the bar from being pulled immediately back into into position by centrifugal force, and then oscillate back and forth in what would be an otherwise undamped system. Of course you know that, and even explain that reaction in the following example.

Udi, I'm not familiar with precession stall. Is it a slowing of the rotor from increased drag from too high of an angle of attack? Or is it an actual stall? My explanation, and it is ~discussed below, is that the blades are just too heavy, period. Angular momentum is not part of that equation; I'm pretty sure.

I must say that I think my posts are being getting only a cursory glance, for the replies ignore my logic, and are content to deny the 'theory' without proving it wrong, by proving my own words wrong. :( Nevertheless, I will post the following, though I doubt I'll get a serious reply. Now there's a challenge. Any takers?

mceagle said:If blades actually "flew" to their new position in the circle, why is that position is always at 90 degrees to the applied force? Wouldn't the position in the circle be dependant upon other factors such as rotor speed, weight, airfoil, and forward speed? Surely it is not coincidental that maximum displacment is in line with precession laws, no matter what the other factors.

The solution to the 90º displacement in ‘aerodynamic precession’ requires three-dimensional dynamics to be understood. I can’t simplify it much more than that right now. But rotor disk rotation is not the result of gyroscopic precession, where a rotating mass reacts, (and does so with force), at a point 90º ahead of the applied force. Instead, disk rotation is the result of the change in the blade velocity (acceleration normal to plane of rotation) as it passes through a region of applied force; (integrating acceleration over time.) The location of the point of maximum force (and AOA) is not so important in determining the point of maximum disk displacement, as in precession. What is important is the point where the applied force changes from positive to negative, (when integration ends), and this is at the boundary between the positive and the negative force regions, which is typically 90º ahead of max AOA.

The blade of a teetering rotor is free to orient its axis in any direction, unless it hits the stops. A slowly spinning rotor can be blown about by the wind, but will hold steady as it gains rrpms. Contrary to what might be thought, it is not centrifugal force which protects a teetering rotor from being disturbed by the wind. It is the reduction in time (at higher rrpms) that the gusts of wind have to shift the blade position as it passes through the region of higher velocity. If the gust lasts only a moment then the blade can only react to it for a short time before it is blown the other way; and the stick can then control general disk position. It is the blade’s moment of inertia, and lift coefficient, which the momentary puffs of wind are working against, not its angular momentum. This is true at all speed conditions, I think.

The higher the rrpms then the less time available to shift blade position in single rotation of the rotor, but it has greater airspeed available to generate greater lift in that brief time. Lift is proportional to velocity squared, so the faster the rotor speed then the faster the rotor will shift position (especially in time, but also per rev), given the same shift in cyclic position.

I believe the following technical explanation is correct, (such as it is). The blade will adjust its position as a response to forces as adjusted by the feathering hinge. The axis of the rotorhead rotation is where the AOA shift is the greatest, so this is the point where the force is greatest. We might expect this to be the point where the disk would shift the most, but since that is not the case then precession would get the credit—since we’re dealing with a 90º phase shift of movement and angular momentum.

But what is actually going on (I think) is the circular movement of a mass as a varying force pushes at it while it makes its revolution. And that force pushes at a direction which is normal to the movement of the mass at all points in rotation. This does not occur with a gyroscope. In a teetering rotor there is no spring-like force (including centrifugal force) to restrain its motion. The rotor can rotate in any plane, as defined by plane of the rotorhead, and the hinge attached to it. Only the extra blade forces actuated by the feathering hinge determine its alignment.

A change in the feathering hinge will increase forces on one side of rotation and decrease on the other, with the force maximums located at the axis of rotorhead rotation. Since force is ~proportional to AOA then I would like to think it has a graphical representation of a sine wave. Except that relative airspeed probably modifies that. But it doesn’t really matter what graphical shape it is, reasonably speaking, as long as forces are increased on one half of the rotation and lessened in the other half. The location of the peaks of the forces, I believe, are not that important, in and of themselves.

A hinge adjustment will increase blade force along an entire 180º of rotation, and the blade will want to respond to it the whole time—though lifting force will be a maximum at the mid-point. Velocity (normal to the plane of rotation) will continue to increase in that direction of the force until it is removed by the hinge, and that will happen at 90º ahead of maximum AOA. If the rotor has not completely adjusted to the new hinge position then feathering will reduce AOA as it passes the 90º point and moves into the other side of the circle, and accelerate it downward.

When the plane of rotation aligns with that of the hinge then the feathering forces will stop, with the rotor having new velocities as defined by the new plane of rotation. This should explain how the momentum changes. One side of rotation has increased positive Z axis velocities, (since it experienced increased lift), and the other side has increased negative Z velocities, since it experienced reduced lift (below the average). Therefore the plane of rotation tilts, and its peak should always be at the point of transition from positive to negative feathering, where acceleration has changed directions.

In gyroscopic precession, the gyroscope reacts strongest at the point 90º ahead (and also int the opposite direction at 90º behind) of the applied moment. This because angular acceleration is strongest at those points. In aerodynamic precession, that same location is where feathering (and therefore force) has the least effect. There is no odd angular acceleration response from that part of the rotor So I can’t say they are the same, though the results are the same. There is a ‘moment’ applied, in some manner, to a spinning mass. That is the same. But a study (at least my study) of individual points in each system finds different reasons for why they do what they do. Aerodynamically, the rotor is pushed (‘flies’) into a new position, simply so.
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Bryan

Fliterisk1
03-21-2005, 01:51 PM
I said...The higher the rrpms then the less time available to shift blade position in single rotation of the rotor, but it has greater airspeed available to generate greater lift in that brief time. Lift is proportional to velocity squared, so the faster the rotor speed then the faster the rotor will shift position (especially in time, but also per rev), given the same shift in cyclic position.
I believe this is in error. Deflection is proportional to time^2. If time is cut in half by doubling the rrpm, while (acceleration due to aerodynamic lift) is increase by 2^2 by the same doubling, then the rotor should adjust to the same position per rev, regardless of rrpm. And rotor disk alignment velocity would be proportional to rrpm.
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Chuck, the stabilizer bar is pushed back into position by the damping. I'm surprised you would call that precession. But I'm pretty sure I was wrong in saying the the bar would be pulled back into position by centrifugal force! I didn't get any sleep last night, but my error is still worth a return "Oh, come on!" Without damping the bar would remain aligned as it was from the start. And perspective could make it look as if it were oscillating up and down, when it would actually remain rotating in a fixed plane after sytem was disturbed. Add damping (there's always at least a little of that) and it will realign with the mechanical spin axis.
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Bryan

Fliterisk1
03-23-2005, 09:41 AM
Okay, I feel used.

Fliterisk1
03-24-2005, 10:17 PM
Note: I’ve been calling the flapping hinge the ‘feathering’ hinge, because that is the way I think it should be considered, and because I thought someone else had referred to it likewise. But I’ve just found out that the actual feathering hinge operates the same way with the certain types of gyroplane’s rotors as it does with the airplane’s propeller (feathering away all angle of attack), and is used for jump takeoffs. Apologies for any misunderstandings.
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This I believe is the relationship between rotor rpms, force applied & rotor deflection.

Rotor alignment to a specific angle is related to rotational speed, given the same amount of flapping force. I said it wasn't, believing blade shift was only a function of time. But the model actually indicates it is a function of rrpms (of course!), though I didn't see it.

The faster the rotor is spinning then the less change in the angle of disk rotation (‘precession’) for the same amount of force applied. This because the vector component of velocity will always be the same after the same application of force. So if the rotation speed is doubled then the added velocity component must also be doubled to result in the same change in angle of disk plane. Therefore, the integral of force over time (resulting in a specific velocity) must also be doubled in order to result in the same change in rotor disk position.

But if the rotor speed is doubled then aerodynamic forces are quadrupled, while the time to integrate them is cut in half per rotation. So the integral of force over time actually is doubled for the same change in angle of attack of the blade. Therefore the amount of disk adjustment (‘precession’) per blade rotation for the same amount of stick movement should remain the same; (I think, for I‘m not necessarily sure; though confident enough to post it). Is this true?
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Bryan

Fliterisk1
03-27-2005, 09:55 PM
I said...There is a ‘moment’ applied, in some manner, to a spinning mass. That is the same.
Not quite the same. Gyroscopic precession is a result of a moment applied from within the spinning object, through the axis. Aerodynamic 'precession' is as if the moment were applied from the fluid (air) in which it spins. When the hinge is rotated to change the blade AOA then it is as if the spinning 'rod' was passing through a vortex of rotating fluid. The rotor creates its own vortex and that vortex imparts its own angular momentum to the rotor, thereby shifting its plane of rotation.

Theoretical stuff... (as far as I know)

I wrote that there is no returning force on the rotor which would return it back to the plane from which it precessed. Theoretically, there could be. Typically, the rotor spin remains planar, as in one blade follows the other. The other possible spin configuration would be double cone, in which one blade goes high while the other remains low. Imagine a rod rotating in a cylinder which has less diameter than itself. As with aerodynamic precession, the forces required to do this would also be normal to the direction of travel. The axis of rotation, however, remains the same.

In this case, centrifugal force becomes part of the model, so it becomes a study of potential energy, and therefore oscillation. If the forces retaining this spin-in-a-cylinder configuration are removed then centrifugal force would cause the rotor would return back to the original plane of spin. And it should oscillate back and forth, as if within a cylinder of varying radius, while never changing its axis of spin. It should do this because nothing was done to change the angular momentum of the rod. Therefore, it cannot immediately transition to a planar spin, which must have a different axis of rotation and therefore different vector of angular momentum.

We know the rotor spin is ~always planar. I wanted to prove why it was as so, and wondered if there were portions of the blade’s spin where some potential energy was involved.

Since the rotor extracts angular momentum from the air in which it passes then this must change its plane of rotation. It is the design of the rotor hinge, simple as it is, which forces the plane of the rotor to change, and not noticeably be converted to potential (and useless oscillating) energy. For the rotor to spin in the double-cone configuration the hinge would have to be an unusual (if not impossible) design. And it should maintain a high angle of attack, or sufficiently high enough attack (and sufficiently low AOA on the other blade) as the blade makes a complete rotation.

I wrote: But it doesn’t really matter what graphical shape it is, reasonably speaking, as long as forces are increased on one half of the rotation and lessened in the other half. The location of the peaks of the forces, I believe, are not that important, in and of themselves.

I disagree with what I wrote. It is my belief that a force applied to the blade in the form of a sine wave is the perfect force for shifting the axis of spin of the rotor, with no residual potential energy effects. It is also my belief that any shape outside this sine wave will result in potential energy (instead of axis shift); and be, at least, not constructive to the purpose of the flapping hinge. I believe the rotor would vibrate up and down instead of shifting the direction of the rotor thrust vector, which is what ‘precession‘ is all about.

I’m open to corrections, or any other explanation, or gap fills.
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Bryan

Fliterisk1
03-27-2005, 11:07 PM
OOPS! The above is probably wrong.

A long bar that is forced to spin as a 'double cone' shape inside a cylinder would, when released, probably have a planar spin. I doubt it's possible for it to oscillate as described. (Ha!) And the planar spin of the released bar would have the axis of momentum it had at the point of release; though it wouldn't be parallel to the cylinder. The forces imparted by the cylinder are a moment, and they are constantly changing the axis of angular momentum.

All portions of the bar would, after release, want to continue in a straight line. All those straight lines forms the plane of the new spin. So there would be no centrifugal force and potential energy effects into a vibration as described. (ouch!)

I probably should restrain myself from further unsolicited posts to this thread, for I'm trying to do too much of this is my head; probably.

Apologies,
Bryan