Minimum speed that a bike won't fall over [closed]

What is the minimum speed I have to go that my bike won't fall over with a given angle $$\alpha$$ of tilt, diameter $$d$$ and weight $$m$$ of a tire (let's say the bike tilts 30 degrees at maximum when driving in a straight line and the diameter of a tire is 0.60 m with a weight of 1 kg per tire).

I guess the angular momentum of the tires has to counter the torque of the gravitational force somehow. How can I calculate this?

• Maybe this resource will be of help: ruina.tam.cornell.edu/research/topics/bicycle_mechanics/… Bicycle dynamics are very complicated.
– Evan
Aug 12, 2021 at 21:17
• People can stay upright on conventional bikes while not moving for virtually unlimited time (until they are bored or need to go to the bathroom, essentially). Forward motion is not needed to balance. Aug 12, 2021 at 21:22
• Not supposed to pick an answer after two secs , Jon C it is hard to stay up on a bike not moving Aug 12, 2021 at 23:35
• @AlBrown afaiaa the SE network imposes a minimum time limit between asking and accepting, and it's longer than 2 seconds.. Aug 13, 2021 at 9:05
• won't fall over with a given angle ... when driving in a straight line - question is confused; there is only one angle (perfectly in line with the direction of action of the force of gravity) and infinite speeds regardless the weight of the assembly. You cannot ride a bike in a straight line, on an angle (i.e. such that the center of gravity is not over the contact point of the tires with the ground) so that it "won't fall over".. A bike that is leaning over must describe a direction of travel that is an arc, not a straight line.. not that a line on the surface of a sphere is straight, but.. Aug 13, 2021 at 9:28

I guess the angular momentum of the tires has to counter the torque of the gravitational force somehow. Can someone help me calculate this?

That's the peculiar thing, the angular momentum of the tires doesn't actively provide torque to bring the bike back upright. Rather, it slows down the rate at which the bike can tilt, allowing a rider enough reaction time to make the proper adjustments to keep it upright.

If you get a bike up to high speed and then just "let it go", you'll notice it falls to the ground pretty quickly, almost as quickly as if you just let it go in your garage.

Edit: There's a lot of discussion below about how a bike can be self correcting due to the ability to turn - lots of great physics in there but it's missing the point of the question, which is the correction when

moving in a straight line

The amount a bike can precess to stay upright will depend on a number of parameters about the bike. On one extreme you have a wheel by itself, which will continue to turn until it slows to a stop and then fall. On the other extreme, you can imagine an extremely long bike with locked handle bars - it will fall to the ground quite quickly. All that would be a great answer to a separate question.

Moreover you'll notice I carefully worded my answer to say:

the angular momentum of the tires doesn't actively provide torque to bring the bike back upright

As I wanted to give a simple answer that doesn't overwhelm someone asking a simple question at their level (but I get that everyone loves to "actually..." on this site). This statement is correct that the angular momentum of the tires will slow the tilt toward the ground but never restore the bike toward its center. That was the point.

• Thanks. That makes more sense than the way I thought of it. I remember my mechanics teacher saying that parents should tell their kids to go fast and not slow because it's safer lol Aug 12, 2021 at 22:54
• If you get a bike up to high speed and then just "let it go" ... it falls to the ground ... as quickly as if you just let it go [with it stationary] - i disagree; bikes can be (and are) set up so that the "falling over" causes the steering to turn more sharply than required for the given arc being described as it falls, shifting the balance of forces and bringing the bike back upright (excess torque opposing that imposed by gravity). It exchanges some of the forward momentum to do so, so starting it off at a high speed gives it ample excess momentum to perform it for an extended period Aug 13, 2021 at 9:25
• You can see in the video associated with the linked comment under the question - ruina.tam.cornell.edu/research/topics/bicycle_mechanics/… - the human gets the ridersless bike up to a good speed before knocking it sideways. This starts a series of damped oscillations; the bike "survives" being "knocked over" and carries on (having traded some of its speed). Had that bike been stationary it would have been lying on the floor a couple of seconds later Aug 13, 2021 at 9:28
• @IlmariKaronen your last paragraph is definitely not what I would expect from experience - that's because, IMHO, the last paragraph is not true Aug 13, 2021 at 9:30
• Other than the bit about angular momentum, this answer is not correct. A moving bike is a self-righting mechanism and does not require input from the driver to remain upright. A moving bike does not fall fast when let go. It only falls when slowed down enough. The principle is as follows: suppose the bike starts leaning left. Therefore, the bike turns left. But inertia wants the bike to keep going forward, which forces the bike to lean more to the right, thereby cancelling the initial left lean. This is how it continually self-rights itself as long as it moves fast enough. Aug 13, 2021 at 12:06

The bike will not go straight if it is left alone and tipping. Secondly, it will tip slightly faster if not moving, but it will still tip over from any non-vertical angle whether moving or not.

All assumes no rider:

The right-had rule can tell you which way it’s path will curve as it tips, but your intuition will too: if leaning right the bike will gently turn right. And the handlebars/front-wheel as a unit has a lower moment of inertia around a vertical axis than the body of the bike, and hence they will turn faster.

Also, the gyroscopic “force” is just our way of trying to understand and communicate the dynamics of a spinning object when an external force is applied to its axis. The bike will tip over by itself any angle from vertical, however small, it may just take longer if moving.

Here’s one rigorous way to see the situation: the wheels are spinning and have an axis of rotation which is the same as each wheel’s axis. The direction of the vector representing the angular velocity is pointed along the axis to the left, determined from the right-had rule with your fingers going the spinning wheel’s direction, and its length determined by how fast the wheel spins.

If the bike is tipped a little, let’s say to the right, then gravity is pushing downward, and it is displaced slightly from the center of the wheel’s axis (displaced horizontally). That creates a torque around the very center of the wheel. We know a torque angularly accelerates something by $$T = I \alpha , \alpha = I / T$$ and the direction of this angular acceleration vector is, by the right-hand rule, pointed in the direction of the bike’s travel.

How will this change the angular velocity vector of the wheel. Just add vectors. It will transform the angular velocity vector (which points left) by moving it forward, and this means the wheel is spinning about a new axis and will move perpendicular to that new axis, ie bend to the right. The front wheels can turn more easily as they only have to move the wheel/handle-bar assembly.

If not moving, the torque will go entirely into rotating the bike itself about the point where the tires touch the pavement and not have to slowly bend the existing axis of angular momentum. If moving, the torque modifies the current dynamic by moving axis of rotation.

As an aside, the right-hand rule is an arbitrary convention; there is nothing special about right vs left. In fact, handedness is a fundamental physical parameter, which Immanuel Kant did not realize. He thought a hand in space away from any object or orientation would be neither right- nor left-handed https://philosophy.stackexchange.com/a/84452/53366

• That creates a torque around the very center of the wheel - i do not agree that the center of the wheel is the point around which the bicycle assembly rotates, if that's what you're saying Aug 13, 2021 at 8:41
• I think we probably agree: “the torque will go entirely into rotating the bike itself about the point where the tires touch the pavement and not have to slowly bend the existing axis of angular momentum” Aug 13, 2021 at 8:46
• So what's the "center of the wheel" got to do with anything? The center of rotation is the contact point of tires with ground, and the rotational torque is derived from the assembly conceptually acting as if it were a point mass at the center of gravity, which isn't the center of a wheel either.. Aug 13, 2021 at 8:51
• (I'm also slightly confused as to what the significance of the lack of a rider has; are you considering a rider as an active control mechanism keeping the bike upright, or as simply a weight fixed to the bicycle?) Aug 13, 2021 at 8:55
• Theres two things: one is the rotation of a bike tipping over. We agree thats around the point on the ground. But the wheel is spinning about its center (as seen from the bike). If we assume the wheel is instantaneously rotating around the point where it hits the ground (which technically is true) then it is angularly and linearly accelerating. We cant add the two rotation vectors if one is already moving. That was my thinking Aug 13, 2021 at 8:59