Below is a very lengthy explanation of gyroscopic precession. I tried to make it so that everyone on this thread could fully comprehend it, but I did gloss over some points. I tried explaining this to other groups at different times, and it has only worked after much dialogue afterwards. Feel free to ask for clarification either privately or in this thread.
Keep in mind, I'm trying to address only gyroscopic precession in helicopter rotors. I do understand that the truth is much more complex and that there are factors that affect gamma other than what I will be discussing.
There are responses to posts made by Dave, ShyTorque, Lu, HeloTeacher, and Nick following this lengthy explanation
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I'm not an expert in helicopter theory, but I do have a solid understanding of rotational dynamics. It seems consensus in this group that gyroscopic precession does not cause a 90 degree phase lag in helicopter rotors. I agree. It also seems evident from your posts that most of you don't fully understand what gyroscopic precession is. I'll try help with that and with other terms. I think I can illustrate why experts in this industry don't agree whether or not gyroscopic precession actually applies.
First of all, if you wanted to describe how hard a non-rotating hockey puck was about to hit me, you could give me it's velocity and it's kinetic energy for me to fully understand what kind of hit to expect. Or you could give it's mass and velocity, or it's momentum and kinetic energy, or it's mass and momentum, etc. The point is many of the words that we use to describe physical processes are functionally redundant. Functionally because in some instances alternate descriptions are easier to understand or apply. I believe that this is one of the reasons why gyroscopic precession is being dropped from basic helicopter theory: other quantities give a more understandable description and, in fact, more accurate.
A review of the concepts we need to understand gyroscopic precession. Conservation of momentum is a law of physics that deal with bodies in motion. Simply put, conservation of momentum is Newton's First Law of Motion, "
Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon." (Motte's 1729 translation of
Philosophiae Naturalis Principia Mathematica, Sir Isaac Newton).
Conservation of angular momentum is an identical concept except that it applies for objects rotating rather than objects in linear motion.
We understand easily how to conserve linear momentum. If a hockey puck is travelling north and we leave it alone, it keeps on travelling north. If a hockey stick imparts a force on it towards the east, we know that the puck will then travel in approximately a northeasterly direction. Since we can all do wind triangles, we can add vectors to come up with a resultant vector for this hockey puck.
If we hang that puck on a string connected at it's center and begin to spin it counter clockwise when viewed from above, it will keep on spinning if we leave it alone (ignore the torsion in the string, air friction, etc.) If we stop the puck spinning about the vertical axis, it will continue to not spin. If we impart a downwards force on the southernmost part of the non-spinning puck, that puck will want to rotate about the east-west axis counter-clockwise when viewed from the east. This should all seem intuitive. The next bit is harder to comprehend.
We now impart a force on the southernmost point of the SPINNING puck. That force wants to cause the puck to spin about the east-west axis, but the puck already has a spin. In the same way we add vectors to figure out what the hockey stick does to the moving puck, we must add these spins to get the resultant spin.
Consider the point of your right index finger to be the southernmost part of the hockey puck. Hold that finger behind your head, level with your nose, and practice the motion that would result in the case of a non-spinning puck, a rotation around the ear-ear axis going from behind your head, to your neck, your nose, your scalp, etc. You probably just did a universal hand signal to indicate madness. Haha. Now move your right index finger around your head slowly in the ear-ear-nose plane to simulate the revolution of the spinning puck prior to the force being impressed upon it, go from back of head, to right ear, to nose, left ear, etc.
Now you must add those two spins. Starting at the back of your head, the original spin starts moving your finger towards your right ear, but the rotation due to the force wants to move your finger down towards your neck, so the initial information we get on the resultant spin is that from the back of your head your finger wants to move down and right. What we don't yet know is how much to spin around the ear-to-ear axis and how much to spin about the vertical axis.
An easy way to relate to this motion is a small mass on a string, spinning around the vertical axis. If you were sitting on that mass and unaware of the spinning, you would feel a force away from where the string is attached, a centrifugal force. In fact, you would feel like you are on a pendulum with the restoring force being the centrifugal force (for a clock pendulum gravity is normally the restoring force).
When you perturb a pendulum, it swings one way, slows down, swings the other way, etc. It's initial velocity depends on the strength of the force that perturbs it, as does the amplitude of the swing. However, the time required for a full cycle, called the pendulum's period, is always the same for a given pendulum, independent of the perturbation. For our mass on a string it is given as the square root of the length of the pendulum divided by the acceleration due to the restoring force [P=(l/a)^.5 , if you like formulas]. It’s important to note that the period of these oscillations is independent of the mass.
The acceleration due to the centrifugal force is given by the radius of the rotation times the square of the angular velocity (how fast it's spinning), [a=rw^2 for the formula inclined]. Since the length of the pendulum is the same as the radius of the rotation then the period of the perturbed mass oscillations equals the inverse of the angular velocity.
This means the pendulum does one full cycle for every rotation about the vertical axis.
You now know enough to complete the motion of your right finger. You know that the downwards force on the south point now is just a perturbation of a pendulum, that the pendulum wants to drop down some, return to middle, go up some, and return to middle while your finger goes once around your head. The amount that it goes down (pendulum's amplitude) is dependant on the strength of the perturbing force, so lets keep that force small enough to cause a maximum two inch displacement. Your finger now starts at the back of your head, swings towards the right and down to a point two inches below your right ear, continues to the front of your head and to your nose, around to the left side and two inches above your left ear, and towards the back and down to the starting position.
The work above is just for a point mass on a string. The hockey puck can be split into many point masses. Since the puck is solid, the perturbing force is spread out to the whole puck and each point mass gets a perturbation whose direction and magnitude is based on its location in the puck. This can be shown easily but exhaustively. To convince yourself, first turn the point mass into a barbell by adding an opposing point mass, then keep adding more point masses.
Phew. We just added two spins. Physicists just add vectors. We draw a vector perpendicular to the disk of revolution to represent spin. It's direction is such that it points the same way as your right thumb if your right fingers were curled around it in the direction of rotation. The vectors length represents the speed of the spin (ie rpm).
For our hockey puck it originally spins ccw from above so it gets a big up vector. Our force imparted on the south point causes a slow rotation about the east-west axis so that gets a small vector pointing east. Do a vector sum and you get a big vector pointing up but leaning to the east, which is consistent with a disk spinning fast in a plane that is tipped from the horizontal down towards the east. (Note that the magnitude of the vector has slightly increased. This is very small for these types of perturbations.) Isn’t that much easier than all that pendulum and finger around the head stuff?
Back to the original purpose: gyroscopic precession...what is it? It is just a manifestation of conservation of angular momentum with a few stipulations. We talk about a 90 degree phase lag of movement from an applied force. In fact, the force creates a moment about the center of rotation, which causes a rotation. That rotation is immediate...there is no lag. We say that there is a lag because our inability to easily perceive rotational dynamics causes us to fall back on linear dynamics.
"I push here, disk should go down here. Disk goes down over there instead. Therefore that disk lags." nope. "I push here, disc rotates. Disk is already rotating, so it IMMEDIATELY changes it's plane and magnitude of rotation." yup.
What are these stipulations? One of them is that the perturbations aren’t that large. As you get larger perturbations, other factors cause the resulting motion to be even more complex, but still based on conservation of angular momentum. Since the magnitude of the perturbations depends on the relative size of the perturbations to the spin, gyroscopic precession isn’t seen when the angular velocities are slowed considerably and perturbations are the same size.
Another stipulation is that the radius in the centrifugal acceleration equation is the same as the length of the pendulum. This automatically holds true for solid objects that are free to rotate on all axes through one point, but more complex objects may induce restraints. An example would be a helicopter rotor that flaps about a hinge that is far from the mast. Except for rigid rotors and teetering rotors, this is the case. What happens, going back to my equations is l<r so the period of the pendulum (blade flapping) is less than the period of the spin (rotor rpm). This would reduce the apparent lag, but since cyclic forces have the same period as the rotor, this effect is only noticed by a change in gamma.
There are other stipulations. A general solution for a solid spinning object subjected to external forces would be much too complicated to explain on this forum, or anywhere for that matter. Because it is so complex, assumptions that are valid for the problem at hand are made. The assumptions in the case of gyroscopic precession are these stipulations.
That’s gyroscopic precession. Question that remains is this. Does it apply to helicopter theory?
Yes, if you accept that there are stipulations. But No, to be completely accurate. Truth is we aren’t completely accurate anyhow.
In my opinion, in basic helicopter theory, we should explain conservation of angular momentum well enough to understand what I’ve written, use gyroscopic precession as an example of a very specific manifestation of conservation of angular momentum, and then state where helicopter rotors differ. I don’t think we should say helicopter rotors have a phase lag because of gyroscopic precession, but I do think we should say that the reasons that gyroscopic precession occurs are also reasons that explain why helicopter rotors behave they way they do, but it’s not that simple.
As I mentioned, I’m not an expert in helicopter theory. I don’t know fully how they fly. I do know some of the physics behind it and that except for very specific cases gyroscopic precession does not really apply. Why it doesn’t apply is because the rotor system does not meet the stipulations that must define a gyroscope. It’s got nothing to do with whether aerodynamic forces are stronger or if the proper explanation is called aerodynamic precession. I recognize that the truth is much more encompassing than what I've written here, but I believe many have misled themselves by discounting gyroscopic precession and not knowing how to conserve angular momentum.
Matthew.
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Responses to individual posts follow (for all to read):
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Dave said, “
I believe that a simple teetering rotor, with no delta3, exhibits a 90-degree phase lag, which is totally caused by the blade flying to position. Perhaps aerodynamic precession, but not gyroscopic precession.”
You’re right in a way, because you can demonstrate this type of movement with pure kinematics. However, conservation of angular momentum is buried inside those kinematics, same as you can determine mass and velocity of an object if given it’s kinetic energy and momentum. You’ll find that the work leads you to the simplified explanation of the spinning pendulum that I gave.
Also, “
In [ShyTorque’s]
first instance, because the rotor is rigidly attached to the helicopter, the mast and helicopter will be pried (rolled) to the right”
Actually, gyroscopic precession is a very good way of describing the rigid rotor. The mast and the helicopter would roll right only if the rotor wasn’t turning. With the rotor turning, everything pitches nose down.
Finally, “
I do not believe that the activities of an individual blade can be related to gyroscopic precession.”
As a point mass can show gyroscopic precession, so can a single blade. If you have a bunch of blades connected to one hub, they don’t have to react to each other because they each generate their own perturbing forces. In effect you have a number of gyroscopes spinning in close formation. (Of course, I say a single blade can do this, but the blade must be built within the mentioned stipulations. Most blades do not exhibit true gyroscopic precession.)
ShyTorque, I enjoyed your analogy. It does make things clear. However, you seem to be considering that gyroscopic precession somehow needs the force to be exerted through the mast. The force only has to result in a moment about the center of revolution. You can do this by torquing the mast, blowing on the tubes, using a magnet, etc.
You mentioned that your rigid system, “
…will exhibit the two properties of a gyroscope, namely rigidity (it wants to keep spinning in the same plane) and precession (any force applied to the rotor mast in an attempt to tilt the system will appear to act 90 degrees further round to the direction of application until the plane of rotation of the system aligns with the force and it can no longer act).”
It is true that rigidity and precession are qualities of a gyroscope but they are also both direct results of conservation of angular momentum.
Lu, I hope this isn’t a private thread…I wasn’t invited. Please comment on what I’ve written.
HeloTeacher, as the blades slow down, the perturbations have to be even smaller to stop other conservation of angular momentum effects from dominating. The obvious test of this is dangerous and destructive…comparing the effects of large cyclic inputs at different rotor RPMs. I think leaving this to models and engineers is a better plan than any of us trying it ourselves. But I suspect that is part of the reason why cautions have arisen for cyclic inputs at low Nr.
Nick Sezzed, “
The old bugaboo about Gyroscopic Precession is quite mythological, but almost impossible to squelch, because it seems so plausible, and the real issues are so difficult to describe intuitively.”
Correct, but it’s not really mythological and conveniently plausible, it’s just close and based on the exact same concepts as the truth.
And, “
It turns out (in math that gets pretty stinky) that the blade resonates at 1 per rev, because the centrifugal spring changes its force with rpm, so it always allows the blade to resonate at its whirling rpm frequency.”
Actually, for small amplitudes, the natural frequency doesn’t depend on the centrifugal force or the rpm. The resonant frequency is dependant only on mass distribution in the blade, and the center of rotation. For larger amplitudes the one per rev doesn't hold. I doubt if large enough blade movements occur.
The math you’re talking about that describes the truth about helicopter rotors is simplified in this post, but here it describes gyroscopic precession. That’s not just coincidence. Gyroscopic precession really is quite close.
Matthew Parsons
[email protected]
[ 06 August 2001: Message edited by: heedm ]