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

  1. Mar 3, 2014 #1
    Why is it that it's much easier to balance on a moving bike than a stationary one? I've heard a few different answers:

    1. The wheels of the bike each have their own angular momentum vectors, which like to stay pointed in the same direction, and hence resist tipping from side to side (sort of like a gyroscope).

    2. Since the bike is moving, any tipping to one side or the other can easily be corrected by turning the front wheel slightly to one side and getting a mv2/r force that will tend to counteract the tipping. The greater the bike's velocity, the bigger this force is.

    3. Something to do with speed making it easier for a rider to orient his/her center of mass in line with the frame of the bike. Not sure how that would work, admittedly.

    What do you guys think?
  2. jcsd
  3. Mar 3, 2014 #2
    Being a total noob to physics (still haven't taken a college level course), I'll chip in anyways.

    The aerodynamics around the wheels are my guess. The air splits at the front of the front wheel. Imagine you tip to one side a little bit. The spinning of the wheels, simply due to air resistance/friction, make the air swirl a bit next to the wheels. The top of this 'wheel' of swirling air pushes against you if you're tipping over, so that you stay high and upright.

    If I'm wrong, could people correct me, please? :)
  4. Mar 3, 2014 #3


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    I have not heard of an aerodynamics reason for a contribution to staying upright before, but there could be some truth to it, maybe not exactly as you have mentioned. Perhaps you are the first to think of it that way.
    Right now I won't say one way or the other cuz I have to chew on that a bit, perhaps a lot. interesting!
  5. Mar 3, 2014 #4


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    Turn the pushbike upside down, spin its wheels super fast, and try rocking it side to side. You'll notice that there is little resistance to speak of, hence one can conclude that the gyroscopic effect from the spinning wheels is negligible. The wheels are just too light and spinning too slow to provide enough of a stability.

    2&3 are the same thing, and are best understood as a centrifugal effect. As the bike leans to a side, the steering wheel turns(even if you're not riding the bike), and a centrifugal force appears.

    There's actually a good deal of research on the minutiae of the subject. Have a look here:
    There's a TEDx video further down on that page.

    @ModestyKing, air resistance is even more negligible than the gyroscopic effect. Additionally, it could never provide a stabilising force, as it is proportional to the falling speed - it may never push the bike in the direction opposite to the lean, merely slowing down the fall.
  6. Mar 3, 2014 #5
    The aerodynamics reason seems like it would have some affect, but I can't imagine it being all that significant. I would imagine that biking around in an airless room would be very similar to biking around outside, although I could be totally wrong.

    If we replaced the wheels on the bike with two wheels that had much smaller radii, would it be harder to balance on? If so, perhaps the angular momentum reason is correct. Considering that L = Iw, reducing the wheel's radii would reduce I and hence L, and the wheels would have less angular momentum to conserve. Or maybe the size of the wheels has nothing to do with it.
  7. Mar 3, 2014 #6
    Great idea about turning the bike upside down! That's a great way to disprove the angular momentum idea. Thanks!
  8. Mar 3, 2014 #7


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    No, but the gyroscopic effect on the front wheel can steer it to keep balance, if you let the handles go.

    Yes, for normal bikes it is mostly steering, which is done by the rider or caster+gyroscopic effects. But surprisingly a two wheeler can be self-stable without steering or gyroscopic effects:

  9. Mar 3, 2014 #8


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    gyroscopic effects,
    A turn, or torque, on the handle bars sets up a precession of the wheel. The precession will tip the top of the wheel away from the turn, if I have my right hand rule correct. In the end, the result is the wobble seen on a rolling wheel as it 'self-corrects' along its route.

    With a mass attached off the axis of a rolling wheel, such as the rider and the bike, the dynamics become more complicated, than just a rolling wheel will follow a straight path along the ground.
  10. Mar 3, 2014 #9


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    From P.A. Cleary and P. Mohazzabi, Eur. J. Phys. v.32, p.1293 (2011):

    And from a recent news article:

    So there is a "simple" explanation for it, but if you want to be as accurate as possible, then it is no longer that simple. You'll notice that this is true for almost anything in physics, and for almost any phenomenon that you observe.

  11. Mar 3, 2014 #10


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  12. Mar 3, 2014 #11


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    For a standard bicyle, the key factor in self stability is due to the steering geometry, specifically "trail". If you extend the steering axis line to where it intercepts the ground, it will intercept the ground in front of the contact patch. This will cause the front tire to steer in the direction of any lean, and within a speed range, the trail will lead to self stability where a bicycle tends to remain vertical despite any disturbances. At very high speeds, the mathematical models show that a bicycle will become slightly unstable, with a tendency to fall inwards at an extremely slow rate due to gyroscopic forces. In real life, this appears to be countered by the fact that the contact patch is on the "inside" of a leaned tire, generating an outwards torque, and in the case of racing motorcycles, a bike at high speed tends to hold the current lean angle (or the rate of inwards falling is so slow that it's imperceptible), and the counter steering effort to straighten up a bike is about the same as it is to lean the bike.

    Gyro effects are unlikely to contribute much to stability, since gyro effects would be a resistive response to a rate of change of lean angle as opposed to a response to lean angle. As mentioned at high speeds, the resistive response dominates, and the bike tends hold a lean angle (perhaps with a very slow rate of falling inwards).

    Forward inertia is relative to a frame of reference. Riding a bicycle on a large flat surfaced treadmill is the equivalent of riding on a street with a tail wind blowing at the same speed as the bicycle. One issue with rollers is mostly due to having a pair of contact points between the front tire and the rollers, which creates torque resistance to steering inputs. The other issue is visual, but the effect will differ with riders.

    A collection of videos showing ridden and riderless bicycle testing done on a large treadmill at TUDelft. One issue with a mathematical model for the riderless bicycle shows that it should tend to fall inwards at a very slow rate above 8 m/s (barely unstable capsize mode) , but the actual bicycle turns out to be very stable at 8.33 m/s (the 30 kph video).


    Wiki article contains the eigenvalue model for the same bicycle as seen the treadmill tests:


    It's also on page 4 from the orignal article from TUDelft:

    http://home.tudelft.nl/fileadmin/UD/MenC/Support/Internet/TU_Website/TU_Delft_portal/Onderzoek/Wetenschapsprojecten/Bicycle_Research/Dynamics_and_Stability/doc/Koo06.pdf [Broken]

    Instead of using trail to steer the front tire inwards due to lean, they attach a weight at the end of a rod that extends beyond the front tire which generates a yaw torque when the bicycle is leaned, which in turn causes the front tire to steer into the direction of the lean. The core principle is the same, to cause the front tire to steer inwards in response to lean. More info on this from TUDelft in the link below as well as on the ...schwab/Bicycle/index.htm ... web page linked to above.


    One a side note, since the front tire contact patch is "behind" the extended steering axis line, the contact patch can be moved somewhat relative to a bicycle. For skilled riders, it's possible to balance while not moving forward, using steering inputs to generate a lateral contact patch force to generate a corrective torque if the amount of lean to be corrected is very small. Trias events involve situations where a very light motorcycle (or mountain type bicycle) is not moved forwards. The riders swing their legs to generate torque to balance the bike (similar to a tight rope walker moving their arms) while at the same time "hopping" the bike to change direction while confined to a very small area.
    Last edited by a moderator: May 6, 2017
  13. Sep 14, 2014 #12
    Moving bike stays up for a simple reason

    Tire forces. Look at a bike as it falls over. The tires are holding the bike from sliding sideways. This camber force is the product of the angle of lean and the weight. As the bike rotates, the center of mass accelerates mostly to the side. The camber force reacts to this. Now the bike moves and the rotating wheels are no longer held by the road. The bike tips but the camber forces push the bike into the fall. Instead of rotating, the frame moves sideways and does not fall. The center of mass is now moving forward and also slightly to the side. The velocity vector is not aligned with the frame but is pointing slightly to one side. This creates slip angle forces at the tires in the opposite direction of the camber forces. The front end pivots to align with the velocity vector driven by the trail and the slip angle force. The front slip angle force goes to zero. The rear slip angle fades more slowly and pushes the back of the bike away from the falling side. There is a moment then rotating the bike about it's yaw axis and sending the bicycle into the lean. The bike goes over center. A failure of much physics on this matter has been the assumption that the bike is steered into a fall by manipulating the front wheel. Centrifugal force is then generated with the right coordinate system. The bicycle is a more passive device with the steering force coming from the rear wheel. A video at youtube titled Bicycle Stability 101 explains this.
    Last edited: Sep 14, 2014
  14. Sep 14, 2014 #13


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    Both camber force and slip angle forces are newton third law pairs of forces with the same direction, the tires exert an outwards force on the pavement, the pavement exerts an inwards force on the tires.

    Camber force is the force related to lateral deformation at the contact patch, while slip angle force is related to angular deformation (twisted "outwards") at the contact patch. The ratio of camber versus slip angle force depends on the design of a tire. Cars tend to have relatively more slip angle force than motorcycles.

    Lateral force is a function of steering input and speed, not lean angle. If a bike is steered left, it turns left. It may lean outwards (right) and crash if the bike is not leaned inwards (left via countersteering) before turning, but the turning force is a function of steering input. For example, a rider can steer in the "wrong direction" to quickly move the tires away from a small hazard like a pothole and then recover to continue going straight. The geometrical radius can be described as the point on the pavement above (or at if not leaning) the intersection of the extension of the front and rear tire axis. Due to contact patch lateral and angular deformation, the actual cornering radius will be somewhat larger.
    Last edited: Sep 15, 2014
  15. Sep 15, 2014 #14
    Camber force for a bicycle is the side force generated for a lean angle and does not depend on tire design. As a bike falls a force is generated equal to the product of the weight and angle of lean. This is easily verified by calculations of a bike as an inverted pendulum and during cornering. When perturbed, a moving bike reacts not by tipping but by moving sideways, pushed by these "camber" forces. The slip angle force is a function of tire properties and angle of steer. They can occur two ways, steer a wheel at some angle to forward momentum or change the direction of the forward momentum in relation to the wheel. The latter is occurring when a bicycle starts to tip. The weight is no longer over center leading to a lateral acceleration of the frame. A small lateral velocity results and the bikes velocity will change from a straight ahead position to one slightly to the side. Now, which way is the slip angle force? The velocity vector is at a small angle to the original which was aligned with the frame centerline. If the bike should fall to the right, the vector points a little that way. The wheels are still aligned with the frame centerline but the momentum vector now reflects a motion to the side due to the tip. The slip angle force will be the same as if the momentum were aligned with the frame and you steered left. Without an input torque to keep it there, the wheel would be pushed back into line by a force pointing left thanks to the trail in the front end. So, when the bike tips right, it moves right. This causes a change in the momentum direction and creates slip angle forces at both wheels. The front wheel force tends to zero as the tire swings into alignment with the velocity. The rear wheel is slower to react and a slip angle persists creating a force pushing to the left and a moment about the yaw axis turning the bike into the fall. Both camber and slip forces disappear when the bike is moving straight ahead and upright. Without the steering, the camber and slip angle forces would be in opposition and the bike would crash. At low speeds you may be more active in steering but very little input is needed when cruising. In this case, the front wheel is not steering but only complying with the bikes direction of travel. The trail could be zero because there is a self aligning moment that would also cause the wheel to align. The steering moment results from the rear slip angle. This disappears when the momentum and frame centerline are in alignment.
  16. Sep 16, 2014 #15


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    You can lean a bike while still going straight, once off balance, it will continue lean futher inwards until the rider steers inwards to correct the lean (counter steering). Motorcycle racers will sometimes hang off a bike, shifting the center of mass off to the inside of a bike while still going straight to setup the entry for a turn.

    When perturbed, a moving bike will lean in the direction of applied force, and then recover due to vertical orientation (but in a different path) if within the stable range of steering geometry. Scroll down this web page to the treadmill videos where the riderless bicycle is disturbed:


    I've noticed a similar effect when hit by a gusting cross wind while riding a motorcycle. Up to a point, the sensation is that the wind blows the tires out from underneath you and little correction is required to keep going straight, but if the gust is strong enough, you have to input additional countersteering input (left torque on the handlebars to lean right agains a cross wind coming from the right), to keep the bike from being blown to the left.

    Some articles note that a leaned bike needs less steering angle than a car or tricycle. I mentioned the reason for this in a previous post. The geometrical center of a turn is the point on the pavement directly above where the extended front and rear wheel axis cross. When leaned θ °, then the geometrical radius is reduced compared to zero lean, multiplied by cos(θ). The actual radius is somewhat larger due to deformation at the contact patches.

    Some Links:

    archived bike1.htm

    google books motorcycle handling

    The wiki image of camber thrust shows a deformed contact patch due to the lateral force. It shows the inner edge of the contact patch as a straight line, but depending on the load, the contact patch is just deformed laterally.

    wiki camber thrust
    Last edited: Sep 16, 2014
  17. Sep 16, 2014 #16


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    Somehow lost an edit. This should be:

    When perturbed, a moving bike will lean in the direction of applied force, and then recover to a vertical orientation (but in a different path) if the speed of the bicycle is within the self stable range which depends on the design of the bike, such as steering geometry (designed to cause the front tire to steer towards the direction of lean. Trail is the most common steering geometry method used to implement this.

    - - -

    However, I've noticed an opposite effect when hit by a gusting cross wind while riding a motorcycle (the side load causes the front tire to steer downwind very quickly). Up to a point, the sensation is that the wind blows the tires out from underneath you, with only a bit of correction needed to keep going straight, but if the gust is strong enough, you have to apply signficant countersteering input (left torque on the handlebars to lean right against a cross wind coming from the right), to keep the bike from being blown to the left. This is on a Suzuki Hayabusa, and its fairing blocks most of a crosswind.
    Last edited: Sep 17, 2014
  18. Sep 17, 2014 #17


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    Assuming the disturbance is above the pavement, in the case of a bike with zero steering angle, you have a lateral force above the contact patches, and an equal but opposing camber force at the contact patches, which results in a torque that would cause tipping.

    In the TU Delft treadmill videos, the point of application of a lateral impulse is at the rear seat, well above the point of contact with the ground and the bike has little angular inertia along the roll axis, so the bike is tipped by the application of a lateral impulse at the rear seat, from which the bicycle recovers back to a vertical orientation (but in a new direction) if the bike is in "stable" mode.

    In the case of a motorcycle with a fairing in a gusting crosswind, the effective application of force is much lower, so less torque versus angular inertia than the bicycle situation, and the lateral force combined with caster effect (trail) at the front tire causes the front tire to steer in the downwind direction, resulting in the tires being steered downwind out from under the center of mass of the motorcycle, overcoming the torque from the lateral load and causing the motorcycle to lean into the crosswind. When the crosswind fades, the motorcycle then recovers back to a vertical orientation. Additional steering inputs will be needed to keep the motorcycle going straight and within it's current lane.
    Last edited: Sep 17, 2014
  19. Sep 30, 2014 #18

    I suppose you should understand the flaws inherent in math analysis. I have made many math models over my career and though mostly simple systems, I can assure you that the best math dude cannot overcome a wrong assumption. Now It has been an assumption in every discourse I have dragged my feeble intellect through that a bicycle rights itself through steering into the fall. I agree with this so far. The analyst accommodates that idea by assuming the front wheel steers the bike. It doesn't. Not, at least, during the time you are travelling straight ahead. In this case the back wheel is providing the slip angle force that causes the rear frame to track the front. What do they assume at the contact points? No side motion sure simplifies a lot. This is why, when you read Wikipedia on the subject of Bicycle dynamics, you get a very confusing notion of bicycle behavior. I have never seen anything as simple as a bicycle that operates by a spreadsheet constructed by a panel.
    The bicycle is a simple device. The trail at the front and the slip angle there work to make the front wheel a direction vector indicator. The front wheel aligns with the direction of travel. The front slip angle swings to zero as fast as it arises and keeps the force at the front zero. The front wheel is not steering; it is steered. Can you dig that? It is a passive, compliant, follower of momentum. What does your result mean if you assume that the bike recovers when the front wheel steers you into an arc? The rear wheel steers. There is a great video on you tube where two students create a device to dampen bike wobble. To create the wobble they tether the bike at the back. The bike rocks back and forth about the rear. It is highly unstable because its means of steering has been stifled. Could our friends with their silly little skate have done something similar?
    Last edited: Sep 30, 2014
  20. Oct 1, 2014 #19


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    This seems to be getting off topic from the original question. Perhaps the recent posts could be moved into another thread by one of the PF moderators?

    It's the difference in angle between front and rear wheels. For a conventional bike, only the front wheel can be steered. I've already explained how steering tranlates into a curved path with a radius that is a function of steering angle, lean angle, and distance between front and rear wheels (since this affects where the extended axis will cross when turning) (deformation at the contact patches will increase the radius slightly).

    Rear wheel steering is possible, but would be difficult to control. Both wheels could steer. I've ridden on the upside down "T"s used to move trampolines, which have a castered wheel at the "front" and another castered wheel at the "back". By leaning the upside down T to the left or right, the T can be "steered" so that it moves to the left or right but without any significant yaw (it continues to point "forwards", even when "steered" left or right).

    The lateral force at the front wheel is a function of the centripetal acceleration times the component of mass supported by the front wheel, and the lateral force at the rear wheel is a function of the centripetal acceleration times the component of mass supported by the rear wheel. When turning, the lateral force is non-zero at both the front and rear wheels.
  21. May 4, 2015 #20


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    Fun experiment about human bike control:

  22. May 4, 2015 #21

    I would say it is mostly if not all gyroscopic effect, although it's not very pronounced on a push bike at slow speeds but you should talk to someone who rides decent size motor bikes, (such as myself).

    Fact is, when at speed on a motor bike you steer not by 'turning' the handle bars, you turn by causing precession of the gyro that is the front wheel.
    For example, on a motor bike at speed if I wanted to turn right, I could (would) put pressure forward on the right handlebar (which at law speeds would cause the wheel to turn left) but at speed that pressures causes the bike to lean to the right and make a right sweeping corner.

    That effect only works when at speed, at very low speeds if you push the handle bards to the left you will turn to the left, at speed if you push the handle bars to the left you will lean to the right and turn to the right.
  23. Jun 29, 2015 #22
    I believe it's because of inertia when the bikes moving the gravity dosent effect it as much due to movement but when you stop there's no inertia so the gravitas full force can effect it.
  24. Jun 29, 2015 #23


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    Any references for that?
  25. Jun 30, 2015 #24


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    Nice summary on self-stability:

  26. Jul 2, 2015 #25
    The video starts out with a good demonstration of the behavior and then loses it trying to ascribe the stability mechanism to the front wheel steering the bicycle. For the bicycle to steer it in the correct direction it would have to create a slip angle at the front wheel and thus a side force. This would yield a torque about the yaw axis and that would be the steering moment. The failure of Koojiman et al is that they demonstrated that the bike steers because of the contention between the camber force and the slip angle force but they never realized it. It is easily demonstrated that when a bike tips a lateral force at the road is created that tends to push the bike toward the lean. Pushing a wheel sideways is very limited, however, so with both wheels locked in alignments the bike still falls over. This is because as it is pushed sideways a slip angle force can quickly rise to match the camber force and any lateral motion is quite small. The bike has different lateral stiffness front and rear due to the steering. At the rear, the wheel is held between a camber force and the opposing slip angle force. At the front, the slip angle force pushes the wheel into alignment with the bikes direction of motion and so the slip angle disappears as does the force. This leaves a camber force at the front to push the bike into the fall. Mathematically, this moment is equal to the rear slip angle force and the distance from the cente of mass to the rear contact.
    The subtle point is that the front wheel doesn't actively steer the bike but the bike steers the front wheel. Instead of creating a slip angle to provide a force that creates a steering moment we have steering that occurs because a slip angle disappears. While it appears that the front wheel is steering into the fall, it is actually only following the motion of the bike
    The ideas about stability are taken directly from Koojiman and are flawed as well. The lean-steer model is similar to the idea of Timoshenko who noted that for any angle of lean an angle of steer could be created to control the motion of the bicycle, The model assumes the rider must control and so the result does as well. To describe the bike behavior a math model must also include the yaw axis and the lateral dimension as a minimum. Sharpe's model of a motorcycle is much more complete. The Whipple model cannot contain the self righting steering torque because it does not include the yaw axis.
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