Nosewheel Flutter: can anyone help me with the physics?

[RedAndWhiteBaron]

On a couple of recent training flights I've had some nosewheel flutter on takeoff or landing. I understand, I think, what causes it, but I don't understand why.

I'm training in a Grob 115 - it has about 15° of nosewheel rudder authority, however the nosewheel can caster to 30°, give or take, on a wacky bungee-esque setup. I ask this because on a recent takeoff I had some nosewheel flutter - and my first instinct was to reduce the weight on the nose gear - i.e. rotate too soon - so it concerns me. That's a rather deadly instinct.

This has happened once on takeoff and once on landing. When it happened on takeoff, I had a normal amount of forward yoke pressure for such a crosswind, intending to rotate 5kts faster than normal. My instructor was beside me, was able to take control, and explained to me that it was most likely caused by me forgetting to apply crosswind inputs. The crosswind was about 10kts, maybe 7, to a plane that normally rotates at 55kts. When it happened on landing, I was solo, and I was already firmly on the ground landing in a 7kt crosswind, maybe 5kts. The nosewheel fluttered violently enough that some dust was shaken loose from the canopy. I did notice that I had some forward pressure on the yoke while at (I estimate) 40kts, which I then released, along with brake pressure, and the issue resolved. I stopped safely.

So in both cases I was dealing with a crosswind, and in both cases, some forward yoke pressure. I understand in both cases that either an incorrect crosswind input or forward yoke pressure (read: more weight on the nose wheel) could cause this to happen.

But what I don't understand is why - and given that my first instinct upon takeoff flutter was to rotate, I'd really like to understand it.

Can anyone help me with the actual physics of this?

[CpnCrunch]

I think you're talking about shimmy. The nosewheel vibrates violently left and right and it feels like it's going to rip itself apart. Lots of articles out there about it. It generally indicates a maintenance issue, or landing too fast, or putting too much weight on the nosewheel, or some combination. Certainly wrong crosswind inputs could cause it. I had occasional bad shimmy in my 172 after normal landings, and it resolved after my AME properly inflated my nosewheel strut. Can also be caused by a dodgy shimmy damper.

I did some training at a school in YBW a few years ago and their 172 always had really bad shimmy on landings, and the attitude was "some 172s just do that". Yes, they do, but only because they're poorly maintained.

Anyway, in your case it sounds like you might be putting too much weight on the nosewheel during a crosswind landing. You don't need to plant the nosewheel into the ground. 5kt crosswind is pretty low. Generally after landing you should have a lot of back pressure on the yoke as you slow down, so that things like this don't happen. I've never flown a Grob, but it looks like a fairly regular nosewheel plane. Maybe talk to your instructor.

[PilotDAR]

Exactly what CpnCrunch said.

Everything in the nose [tailwheel, if so equipped] must be at the correct pressure stiffness, and no freeplay. For Cessnas, yes, the shimmy damper must be working properly, but more critical is it's no freeplay attachment to the strut, and the no freeplay of the torque links. Correct oleo and nose tire pressure, and great shimmy damper will not help if the steering is mechanically loose. That's a walkaround check item: If you can steer the wheel, and the upper part of the steering is not moving, maintenance is necessary. Sadly, that maintenance is often overlooked, because to do it right is a surprising amount of work.

[photofly]

To more precisely answer your question, if you analyze the dynamics of the nosewheel setup by looking at the castering force and how it varies with speed, the rake angle of the strut, the trail of the nosewheel relative to the axle and the springiness of the rubber, and include the weight and moment of inertia of the wheel and then include the gyroscopic effects, you’ll end up with a second order linear differential equation lurking in there somewhere. When you resolve the coefficients you’ll find there’s a combination of those factors that causes the characteristic equation to have two complex roots whose real parts are positive, which as every high school physics student will tell you, is the signature of an oscillator, and that oscillation is going to grow until the linearity you assumed in your analysis breaks down, in other words the shimmy is going to get really big really fast.

The easiest way for the designer to fix it is to throw in extra damping (make the first order coefficient a lot more negative) which makes the (b^2 - 4ac) term of the characteristic a lot more positive, forcing the roots of the equation to be real, and removing the oscillatory character.

That’s why Clyde Cessna bolted a $2000 shimmy damper to the nose gear of his nosewheel airplanes.

[PilotDAR]

That’s why Clyde Cessna bolted a $2000 shimmy damper to the nose gear of his nosewheel airplanes.

When Clyde bolted them on, they didn't cost that. He, and most every other manufacturers, included a shimmy damper in the design, to prevent wheel shimmy, while allowing effective steering. Even tailwheels have them. A well maintained plane won't shimmy a wheel. So the recertification of shimmy is maintenance. The design is established and proven, in type approved planes, and does not require change.

[photofly]

A well maintained plane will absolutely shimmy if you operate it outside the approved operating procedure in the Flight Manual.

[RedAndWhiteBaron]

To more precisely answer your question, if you analyze the dynamics of the nosewheel setup by looking at the castering force and how it varies with speed, the rake angle of the strut, the trail of the nosewheel relative to the axle and the springiness of the rubber, and include the weight and moment of inertia of the wheel and then include the gyroscopic effects, you’ll end up with a second order linear differential equation lurking in there somewhere. When you resolve the coefficients you’ll find there’s a combination of those factors that causes the characteristic equation to have two complex roots whose real parts are positive, which as every high school physics student will tell you, is the signature of an oscillator, and that oscillation is going to grow until the linearity you assumed in your analysis breaks down, in other words the shimmy is going to get really big really fast.

That actually helps, thanks.

A well maintained plane won't shimmy a wheel.

You are assuming a well maintained pilot, methinks.

[lownslow]

my first instinct was to reduce the weight on the nose gear - i.e. rotate too soon - so it concerns me. That's a rather deadly instinct.

Nosewheel dynamics and maintenance aside, there’s world of difference between unloading the wheel a bit and horsing the thing off the ground below flying speed.

[photofly]

however the nosewheel can caster to 30°, give or take, on a wacky bungee-esque setup.

Every nosewheel aircraft has to allow the nosegear to caster (that is, move independently of the pedal position) otherwise a crosswind takeoff or landing would be impossible: as soon as the nosegear touched down with the rudder displaced the plane would drive itself off the runway. That doesn't happen, because the nosewheel is able to caster in the direction of motion as soon as the rubber touches the tarmac.

To that extent, steering an airplane with the nose-gear is different from steering a car (with the hands) or a child's cart (with foot steering) because in both of those cases the angle of the wheels is firmly fixed to the steering control (steering wheel or foot bar). That's not true in a nosewheel airplane.

Lots of small airplanes have free-castering nosewheels with no connection to the rudder at all. But if you have a connection, it has to be springy, and that springy force will contribute to shimmy if the factors I mentioned above align wrongly.

Additionally If you have free play in the steering linkage then you're more likely to get shimmy at least to the limits of that free play because you've got no damping at all over that range.

This wikipedia page is on point:
https://en.wikipedia.org/wiki/Speed_wobble

If I were going to investigate further I would follow up on its comment about a Hopf bifurcation - why does a shimmy occur suddenly?

[PilotDAR]

A well maintained plane will absolutely shimmy if you operate it outside the approved operating procedure in the Flight Manual.

'Never done it, but okay... I've guessing you'd have to be way outside the approved operating procedure, but anything is possible. That said, for every time I have experienced wheel shimmy, it's been a maintenance issue. I do my best though, to always keep the third wheel as light as practical.

There are a few types which do not have castoring nosewheels, Twin Otter, and I believe BN Islander come to mind - what you steer is what you get, nose tire and runway centerline be dammed. The Twin Otter has a mark which you must check for being centered before landing, the Islander has a pilot sitting beside you telling you so! But yes, Most castor, and many without steering. I have never seen a steering nose wheel which did not have some form of shimmy damper, or stiff castoring.

For interest sake, the two outside rib nose and tailwheel tires, often common to UK aeroplanes (Mosquito tailwheel, Dove nosewheel, Beaver tailwheel, for example) are specially designed to minimize shinny with the two ribs. I'm not quite sure of the physics of it, but I guess it worked, DHC certified it that way.

You don't have to over rotate, nor pre rotate a tricycle plane, you can just lighten the nose wheel to reduce the effect. But, once a shimmy starts on landing, your only way to stop it, is to stop the airplane. Tailwheels have some different factors, with old, sagging tailwheel leaf springs (so wrong castor angle) being a big one, again, maintenance.

[RedAndWhiteBaron]

however the nosewheel can caster to 30°, give or take, on a wacky bungee-esque setup.

Every nosewheel aircraft has to allow the nosegear to caster (that is, move independently of the pedal position) otherwise a crosswind takeoff or landing would be impossible: as soon as the nosegear touched down with the rudder displaced the plane would drive itself off the runway. That doesn't happen, because the nosewheel is able to caster in the direction of motion as soon as the rubber touches the tarmac.

I'm aware that a castering nosewheel is normal, but I'm not sure how common it is for it to caster beyond the limits of rudder authority where you have it, so I'm not sure if that extra 15° or so of free caster is a factor. Looks like it isn't.

[PilotDAR]

but I'm not sure how common it is for it to caster beyond the limits of rudder authority

Different airplanes have differing nosewheel steering arrangements, so there's no one answer. Cessnas have springs in the nosewheel steering (hence the apparent lag in steering), so the wheel is mechanically able to castor further than the rudder deflection. Also note that the apparent (nose wheel to ground) angle will change a little based upon airplane pitch attitude. If the airplane is forced nose low by control input at speed the effective castor angle is changed, and shimmy resistance can be effected. Note that Cessna nose struts (steering axis) are not perpendicular to the ground, that is an element of the castor angle. Most, though not all Cessna nose forks place the wheel axle on the steering axis, and the strut angle gives you the castor angle. Other types (like Lake amphibian, or to a greater extent, the Grumman Tiger/Cheetah) have a noticeable trailing arm, placing the axle well behind the steering axis (like a tailwheel).

[photofly]

but I'm not sure how common it is for it to caster beyond the limits of rudder authority

Quick summary from various POHs and AFMs:

Cessna 172M: rudder steering ±10°, castering ±30°
Cessna 182P: ±11° / ± 29°
Grumman AA-5B Tiger: not connected / ± 90°
DA-20: not connected / unlimited
DA-42: not specified / ± 40*
PA-28-181 (Archer II): not specified / 40°
Piper Aztec: unspecified / 30°

So, pretty common, if not universal. One suspects that rudder steering out to 30° would be too sensitive on takeoff/landing, and castering limited to 10° would result in such a wide turning circle the plane would be effectively un-manoeuvrable on the ramp.

Looking at the dimensioned diagram, the fore-aft wheelbase of the C172M appears to be about 7 feet. A 30° caster gives a radius of turn of about 4', which means that with the horizontal wheelbase at 8'4" the inside wheel will be more or less stationary (which matches experience) - and in fact the only way to get the nosegear to caster to 30° from inside the aircraft is to lock the inside wheel with the brake.

At 10° of nosegear angle the radius of turn is about 40', which (again from experience) matches what you get if you use full pedal deflection on the ground but no differential braking. The centre of the turn is about 22' past the wing tip in that case.

For tailwheels, my old Luscombe had a free-castering tailwheel with a (useless) centre detent; I recall being informed that it was vital for the tailpin (the swivel axis) to lean aft at the top, to provide an anti-castering force, otherwise it would shimmy.

[Bede]

When you resolve the coefficients you’ll find there’s a combination of those factors that causes the characteristic equation to have two complex roots whose real parts are positive, which as every high school physics student will tell you, is the signature of an oscillator, and that oscillation is going to grow until the linearity you assumed in your analysis breaks down, in other words the shimmy is going to get really big really fast.

lol. Because every high school physics student can determine the bounds of a characteristic equation....

[photofly]

Depends which high school, I guess!

I did 2nd order linear DE's at school. Forcing functions had to wait until 1st year uni.

[Bede]

Depends which high school, I guess!

I did 2nd order linear DE's at school. Forcing functions had to wait until 1st year uni.

Quite the high school...
I think I barely learned integral calculus in high school. If I recall, characteristic equations are covered in 3rd year engineering, maybe second. I had to review in my process control textbook...

[photofly]

It was a long time ago, in a school far far away...

[Bede]

The easiest way for the designer to fix it is to throw in extra damping (make the first order coefficient a lot more negative) which makes the (b^2 - 4ac) term of the characteristic a lot more positive, forcing the roots of the equation to be real, and removing the oscillatory character..

Sorry, borderline out of my expertise, but wouldn't a positive discriminant be in the RHP (ie positive real roots) resulting in an unbounded output (greater oscillation)?

[photofly]

For an undriven oscillator you have to have complex roots to get the e^iωt and e^-iωt terms in the solutions, which are periodic. If the roots are both real then you have either exponential growth or decay, but no periodicity. That would be either an overdamped solution (which is ok) or the nosewheel will go into hard lock on one side (which you don’t want).

You can still have decaying oscillation even if the roots are complex; it depends on the sign of the the real part.

If the b^2 - 4ac term evaluates to zero, you have critical damping.

Maybe it’s better to analyze it as a driven oscillator (driven by an unbalanced tire for example) in which case the amplitude of the driven oscillation depends on how close to the resonant frequency (if there is one) the driving input is; more damping makes the oscillation smaller.

And then again maybe it isn’t!

I studied this from the point of view of mathematical physics and not engineering, so my terminology might not match yours.

[RedAndWhiteBaron]

You have to have complex roots to get the e^iωt and e^-iωt terms in the solutions, which are periodic. If the roots are both real then you have either exponential growth or decay, but no periodicity. That would be either a damped solution (which is what you want) or the nosewheel will go into hard lock on one side (which you don’t want).

Out of my ken as well, but I'm following and understanding just enough to find this to be a very worthwhile analysis.

Thanks, photofly.