# Why is it easier to balance on a moving bike than a stationary one?

rcgldr
Homework Helper
Nice summary on self-stability (video):
One issue with the video at around 2:00 demonstrating gyroscopic effect is that the weight of the wheel is still in front of the axis so that weight is also causing the wheel to steer.

Note that gyroscopic effect is a reaction to a roll torque. Once the wheel has steered inwards enough to produce a coordinated turn, the roll torque and the related gyroscopic effect become zero. As the wheel continues to steer inwards enough to start reducing (correcting) the lean angle to return to vertical, the roll torque becomes outwards, and the gyroscopic reaction opposes (dampens) the inwards steering needed for correction back to vertical. So the net gyroscopic effect is to dampen (oppose) the steering correction related to bike geometry.

It would have helped if just enough weight was placed behind the front tire to eliminate the weight related steering effect then note the reaction to leaning the bike with a spinning front wheel in mid-air, or to orient the bicycle nose down, but then the weight would tend to oppose any steering movement related to gyroscopic reaction.

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A.T.
So the net gyroscopic effect is to dampen (oppose) the steering correction related to bike geometry.
The gyroscopic effect reduces both, the roll that would happen due to gravity and then the over correction that the steering geometry would introduce. If you want to see the stabilizing effect of gyroscopic steering in an isolated manner, look at a single wheel rolling.

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rcgldr
Homework Helper
The gyroscopic effect reduces both, the roll that would happen due to gravity and then the over correction that the steering geometry would introduce.
True, but I was pointing out the fact that the same gyroscopic reaction that would tend to steer inwards while a bike experiences an inward roll torque while going into a lean, would tend to steer outwards while a bike experiences an outwards roll torque while recovering from a lean, so the fact that a bike recovers from a lean is because the steering geometry overcomes the opposing (dampening) gyroscopic reaction, except for high speeds, when the gyroscopic reaction dominates, resulting in "capsize" mode, where a bike falls inwards at an almost imperceptible rate. I don't know if the fact that the contact patch is on the side of the tires of a leaned bike, resulting in an outwards roll torque, is taken into account with these mathematical models.

"Science doesn't currently know what it is about the combination of variables...." This is a disturbing thought. What are we supposed to understand from this statement? There's this guy, Hugo Science, and he isn't omniscient in the correct selection of variables to create a stable bicycle. And we want his opinion, why? Clearly his opinion does not matter as there are excellent, capable, safe riding two wheelers being popped out by guys with different names. If you were to ask those who express their understanding of these variables in the execution of each kinetic sculpture, they might say:

The yaw inertia of the frame should be a minimum as should the rotational inertia of the steering frame. The center of mass will be dictated by the rider and will be determined by the wheelbase and rider position. The center of mass of the front axis should be near the axis and as low as possible. It is not important which side of the axis it is except that too forward a position tends to induce shimmy instability. The moment of inertia increases as the square of the distance between the axis and the mass center.

Notice in the film how he added a brace to hold a weight behind the front wheel thus moving the center of mass to the rear while boosting the rotational impedance significantly. Yet, he could still ride the bike. Even with the slower steering front end the bike was ridden. For sure, it was like the time I shifted paving bricks on my front carrier but we can clearly rule out the mass in front of the steering axis as any meaningful player in this game of pick your favorite variable.

We can also toss away the idea that the normal force turns the wheel when the bike leans. The camber force creates an.equal and opposite moment. See Steering in bicycles and motorcycles by Fajans, 1999.

That leaves the gyroscopic moment created by rotation about another axis. Magic, as he points out. In Koojiman, you get this litany of what we don't have to have and prominent on the list is that gyroscopic moment. Interestingly, when the discussion turns to stability the gyroscopic seems to swoop in and save the day any time the Moments about the steering axis become zero. You can see why they preferred the mass in front of the steering axis.

The gyroscopic action does provide a moment in the right direction to couple lean and steer. The bugger is that it is velocity dependent, negative damping. It gets in the way at times. The amount of trail needs to be increased to compensate for it and even then it can promote front end shimmy at resonant speeds.Bike's would be better off without it and some are. Ski bikes are amazingly easy to handle thanks to this omission.

So that leaves us with nothing is the one thing that steers the front wheel to make the bike come back over balance. Logically, then, the front wheel does not steer the bike. Which gets us back to," How does a bike steer?" Now that is the question and not "What steers the front wheel?" If you want to move a vehicle laterally you have to either thrust it sideways or rotate it about the yaw axis and change heading. The latter is the steering strategy while the former is the camber philosophy. To steer we need a moment about the yaw axis. We don't need a steering axis at all, until we actually try to develop such a moment and then we find it quite handy. Still, you want to know what is the moment turning the wheel toward where the bike should go? Mathematically it is the front slip angle multiplied by the trail times the cosine of the rake from vertical on a Wednesday afternoon when no one has visited the pub at lunch. Notice that it goes to zero as the slip angle goes away. This means that when the wheel is heading where the bike is going, there will be no steering force generated, and in fact, the slip angle force is causing the wheel to turn. It is opposite of a car where the driver cranks the wheel and the tires point at some angle to forward and the slip angle created pushes the car toward that direction. Single track vehicles are more clever than those brutish four wheelers. When they tip, the camber force pushes them in the direction of the tip but it pushes in accordance to the compliance of the bicycle. The front complies and allows the camber to push the front of the bike sideways. The rear is not compliant and so the bike yaws, steering in the direction that will bring it back over balance.

A.T.
True, but I was pointing out the fact that the same gyroscopic reaction that would tend to steer inwards while a bike experiences an inward roll torque while going into a lean, would tend to steer outwards while a bike experiences an outwards roll torque while recovering from a lean,...
What it would tend to do in a different situation is irrelevant. The situation when the bike is recovering from a lean, is that the handlebar is turned inwards. And the handlebar needs to be straightened before the bike is straightened, otherwise it will overshot and lean to the other side.

so the fact that a bike recovers from a lean is because the steering geometry overcomes the opposing (dampening) gyroscopic reaction,
Then explain the stability of a single rolling wheel, where there is no steering geometry, just the gyroscopic steering, which in your opinion just opposes recovery from lean.

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rcgldr
Homework Helper
The situation when the bike is recovering from a lean, is that the handlebar is turned inwards. And the handlebar needs to be straightened before the bike is straightened, otherwise it will overshoot and lean to the other side.
My issue is with the demonstration that shows the wheel turning inwards as the lean increases, but not showing that gyroscopic reaction alone would steer the wheel outwards while the lean decreases, but still leaning, which would prevent recovery (and is what happens at high speeds in capsize mode). The demonstrations imply that gyroscopic reaction adds to the corrective steering response related to steering geometry, when instead it acts as a damper, opposing the corrective steering response.

explain the stability of a single rolling wheel, where there is no steering geometry, just the gyroscopic steering, which in your opinion just opposes recovery from lean.
A single rolling wheel experiences a yaw torque when turning that translates into a corrective roll reaction. If a single rolling wheel is released at a leaned angle, it's rate of recovery (if recovery even occurs) is slow compared to that of a bicycle released at the same speed and lean angle, and if the initial lean angle is large enough and/or the speed low enough, the single rolling wheel travels in a circle (a spiral as it slows down) and doesn't recover to a vertical orientation. There's also the small effect of the contact patch being on the side of the tire when leaned, which creates a small outwards roll torque, but the gyroscopic reaction would be to steer outwards opposing the correction.

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A.T.
opposing the corrective steering response.
The gyroscopic reaction doesn't oppose what the steering geometry would do on it's own. Without the gyroscopic effect the steering geometry would also straighten the handlebar during recovery from lean.

rcgldr
Homework Helper
The gyroscopic reaction doesn't oppose what the steering geometry would do on it's own. Without the gyroscopic effect the steering geometry would also straighten the handlebar during recovery from lean.
During lean recovery, as the lean decreases, the inwards steering also decreases, but it's still inwards, while the gyroscopic reaction would tend to steer outwards (opposing / dampening steering geometry reaction), but the steering geometry dominates so the steering remains inwards (unless in high speed capsize mode). Absent the steering geometry, there would be no lean recovery, as gyroscopic reaction is zero when the roll torque is zero, which occurs once the steering results in a coordinated and leaned turn. If while leaned, a disturbance causes the lean to decrease, an outwards roll torque and an outwards gyroscopic steering reaction would occur, until the roll torque returns to zero, and the bike would remain leaned.

A.T.
During lean recovery, as the lean decreases, the inwards steering also decreases,
Due to both: steering geometry and gyrosopic reaction.

but it's still inwards, while the gyroscopic reaction would tend to steer outwards
If by "steer outwards" you mean "straighten the inwards steer", then the steering geometry is doing that too.

(opposing / dampening steering geometry reaction),
I see no opposition, as both are straightening the inwards steer.

rcgldr
Homework Helper
I see no opposition, as both are straightening the inwards steer.
During lean recovery, geometry steering reaction generates an outwards roll torque, while gyroscopic steering reaction tries to oppose outwards roll torque (absent other steering inputs, gyroscopic steering steers to eliminate any roll torque). At moderate speeds, geometry dominates and a bike is self stable. At high speed, gyroscopic reaction dominates and a bike is unstable, falling inwards at a very slow rate, called capsize mode. On a racing motorcycle at high speed, a bike tends to hold a lean angle, so if the bike is falling inwards, the rate is so slow that it's imperceptible.

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A.T.