# Moving bike stabilizes but standing bike falls!

by hellboy4444
Tags: stability
 P: 23 A bike "B1" is on the go. Another one, "B2" is standing upright and still. Which one is more stable, in the sense that, if I push one, which one shall fall down more easily?
 Homework Sci Advisor HW Helper Thanks P: 12,854 The moving bike has rotating wheels - so it is gyro-stabilized. I remember letting a 2-wheeler roll down a hill as a kid, I expected it to fall over almost right away but it went all the way down the hill first (about 20m).
 HW Helper P: 7,117 The moving bike is stable because of steering geometry. You could replace the wheels and tires with circular skate blades that don't rotate, give it a shove on an ice rink and it would still be stable. Most of the stability is due to trail (weight distribution on the front tire also has an effect). Trail is the distance from the extention of the steering axis to the ground, back to the contact point between the tire (or skate blade) and ground. Generally, the longer the trail, the slower the bike can move and still be stable. What the trail does is cause the front wheel to steer inwards in response to a lean of the bicycle. This results in self-correction that returns the bike to an upright position (although it's velocity direction will have changed somewhat).
 Homework Sci Advisor HW Helper Thanks P: 12,854 Moving bike stabilizes but standing bike falls! Lowel and McKell 1982 (which actually puts it after I last looked - so I'm out of date!) http://fisica.ciencias.uchile.cl/~go...ability_82.pdf ...conclude that the "castor" (trail) of the front wheel is the most significant contribution to overall self-stability. If gyroscopic effects are ignored, the bicycle is almost self-stable - a small perturbation that may otherwise push it over just leased to a circular path - except that oscillations grow in magnitude (it gets more and more wobbly) till it does fall over. The gyroscopic effects improve stability but increase the tendency to wobble. Though the graphs show a smaller amplitude of wobble at a higher frequency. The biggest contribution to a ridden bike's stability is the rider reacting to smooth out wobbles and correct lean. The rider is also the biggest contribution to the bike instability by raising the center of mass. Motorcycles have a significant gyroscopic effect, being heavier and faster, - the trail contributes strongly to stability while maneuvering. You'll feel it through the handlebars. The less-stable configurations seem to be more "cool" though(?) The practical research has mostly been done with a rider though. This has turned out to be more interesting - and a good example why answering these questions is a good idea.
 HW Helper P: 7,117 Non-stable and stable modes demonstrated in videos of a riderless bicycle at tudelft. The bike is instrumented with some sensors and a lap top attached behind the seat to record the readings. treadmill measurments .htm The test results at 30 kph (8.33 m/s) conflicted with the mathematical model they made for that same bike that calculated the bike would be in capsize mode (slowly fall inwards) at 8 m/s and faster speeds, but the video showed the bike to be very stable at 8.33 m/s. http://home.tudelft.nl/index.php?id=13322&L=1
 P: 4,512 Bikes are not mainly stable due to gyroscopic stabilization. When the bike tilts, the front wheel tends to precesses, turning the steering column and front wheel into the turn, tending to right the bike. That all there is to the basics of stability of a standard bicyle.
P: 4,033
 Quote by rcgldr Non-stable and stable modes demonstrated in videos of a riderless bicycle at tudelft. The bike is instrumented with some sensors and a lap top attached behind the seat to record the readings. treadmill measurments .htm The test results at 30 kph (8.33 m/s) conflicted with the mathematical model they made for that same bike that calculated the bike would be in capsize mode (slowly fall inwards) at 8 m/s and faster speeds, but the video showed the bike to be very stable at 8.33 m/s. http://home.tudelft.nl/index.php?id=13322&L=1
I wanted to follow up on this, because I exchanged e-mails with some of the involved:

The capsize mode predicted by theory above certain speed is not very prominent and remains around zero. It basically means that the bike fails to straighten up and recover from the turn. You cannot show this on a treadmill, because the bike cannot do a long continuous turn there. My own idea: The perturbation they introduce on the treadmill above capsize speed is more like the weave mode for low speeds, which might explain why it doesn't grow at high speeds, and the bike appears stable.

Others did experiments with releasing a bike above capsize speed bike in free terrain, and the bike remained in a circular turn until it slowed down to stable speeds. Then it straightened up again. I suggested to use electric powered bikes to see what happens if the bike doesn't slow down.

P: 4,033
 Quote by Simon Bridge ...conclude that the "castor" (trail) of the front wheel is the most significant contribution to overall self-stability....The gyroscopic effects improve stability but increase the tendency to wobble.
You can have a self stable "bike" without castor and gyroscopic effects:

HW Helper
P: 7,117
 Quote by rcgldr The test results at 30 kph (8.33 m/s) conflicted with the mathematical model ... capsize mode.
 Quote by A.T. I wanted to follow up on this, because I exchanged e-mails with some of the involved: The capsize mode predicted by theory above certain speed is not very prominent and remains around zero. It basically means that the bike fails to straighten up and recover from the turn. You cannot show this on a treadmill, because the bike cannot do a long continuous turn there. My own idea: The perturbation they introduce on the treadmill above capsize speed is more like the weave mode for low speeds, which might explain why it doesn't grow at high speeds, and the bike appears stable.
One possible issue I've wondered about; did the mathematical model take into account that the tires are not infinitely thin? When the bike is leaned over, the contact patch moves off center, providing a small amount of torque that would oppose lean, and if capsize mode was close to zero, then that small amount of corrective torque could be enough to prevent capsize and perhaps enough for some self-correction, but it seems unlikely it would result in the rapid correction shown in the video at 8.33 m/s (30 kph). I'm also thinking that if there is a capsize speed for that bicycle, it's significantly higher than the 8 m/s that the model predicted.

I'm thinking the reason for the transition into "lean" stablity (tendency to hold a lean angle) at high speed is due to the gyroscopic forces (at front and rear tire) resisting any change in lean angle, and dominating the self-correction tendencies. Also at high speed, the rate of precession and reaction to leaning (roll axis to pivot axis coupling) of the front wheel will be reduced, because of the relatively large amount of angular velocity in the tires forward rotation compared to the angular velocity along the roll (lean) axis. The bike may actually be falling inwards or outwards, but the rate of change in lean angle is so small that it's imperceptible by the rider.

 Quote by A.T. Others did experiments with releasing a bike above capsize speed bike in free terrain, and the bike remained in a circular turn until it slowed down to stable speeds. Then it straightened up again. I suggested to use electric powered bikes to see what happens if the bike doesn't slow down.
In the case of sport type or racing motorcycles, at high speeds (100+ mph 161+kph), the motorcycles tend to hold a lean angle as opposed to straightening up. This would be similar to what you described with the bike holding a turn at high speed, but I wonder what the required speed was before this behavior appeared. My guess is that it was well above the 8 m/s speed originally predicted by the model. Also, capsize mode predicts the bike will slowly fall inwards, not hold a lean angle and remain in a near circular turn. Again, I wonder if the width of the tires and the offset contact patch is playing a role here.

 Quote by A.T. You can have a self stable "bike" without castor and gyroscopic effects.
As I mentioned in post #3, weight distribution can also be used. In the video, the specialized "bicycle" has the center of mass (weight) in "front" of the pivot axis, so the front tire will steer in the direction of lean, which the author states that the key aspect of self-stability (designing a bike so the front tire turns into the direction of lean). Not mentioned in the video, but noticable in the video is that the special test bike has very thin "tires", which I assume was done to minimize any effects related to an offset contact patch when the test bike was leaned.
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 Quote by rcgldr The moving bike is stable because of steering geometry. You could replace the wheels and tires with circular skate blades that don't rotate, give it a shove on an ice rink and it would still be stable. Most of the stability is due to trail (weight distribution on the front tire also has an effect). Trail is the distance from the extention of the steering axis to the ground, back to the contact point between the tire (or skate blade) and ground. Generally, the longer the trail, the slower the bike can move and still be stable. What the trail does is cause the front wheel to steer inwards in response to a lean of the bicycle. This results in self-correction that returns the bike to an upright position (although it's velocity direction will have changed somewhat).
1. "replace ...with circular skate blades...it would still be stable"
Are you sure? I don't think so. What do you think?

2. "....This results in self correction that returns the bike to an upright position"
Do you mean that the "self-correction returns the bike to an upright position"? Really?
I don't get it. What exactly returns the bike to an upright position?
Honestly, I don't believe the bike shall return to an upright position ever! If at all, it may tilt more instead. What do you think?
HW Helper
P: 7,117
 Quote by hellboy4444 1. "replace ...with circular skate blades...it would still be stable"
This has been approximated with the "bicycle" in the link, below, but I'd like to see an actual test done with two curved skates with radius similar to wheel radius on an otherwise conventional bicycle (to avoid altering steering geometry) and the bike pushed around on an ice rink or frozen lake.

http://en.wikipedia.org/wiki/Two-mass-skate_bicycle

There are kits to replace the wheels on a bicycle with skis (downhill usage), and while these are ridable, they aren't self stable (the skis will often slide instead of turning).

The reason for self-stability is that the front tires turns into the direction of lean, and this results in literally steering the tires back under the bicycle. The tires apply an outwards force to the ground, with the ground applying an equal and opposing inwards force that steers the wheels back under the bike (since the inwards force from the ground is applied below the center of mass of the bike) and straightens the bike back upwards (although the direction will have changed, depending on the amount of disturbance that caused the lean).

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