Steering a motorcycle

Question

From my experience riding, at low speeds (between 0 and 10 mph) you mostly steer the bike with the handlebars. What I mean by this is if you want to turn left you rotate the handlebars counterclockwise, and the bike turns as desired. There is little to no lean caused by this. At medium speeds (between 10 and 20 mph) turning a bike seems to be a combination of steering with the handlebars and leaning. What I typically notice is that if I want to turn left I will push forward on the left handlebar ever-so-slightly, presumable turning the wheel clockwise just a bit, which will initiate a lean to the left, and I will then turn the handlebars counterclockwise to continue the turn. At speeds higher than 20 mph if I want to turn left I push the left handlebar slightly forward which initiates a lean to the left and I hold it like this to keep turning, never really turning the handlebars counterclockwise as in the previous two situations.

I don't really understand why turning a bike is different at different velocities. Why is lean more important in turning at higher speeds, whereas handlebar steering is more important at lower speeds. If I push the left handlebar forward at low speeds why don't I lean to the left like I do at higher speeds. At medium velocities, why does a combination of both seem to be the most efficient way to turn, and after initiating the lean left why doesn't the counterclockwise rotation of the handlebars put my bike back upright? In fact, since I rotate the handlebars counterclockwise to turn, how do I get back upright at all? Do I simply turn the handlebars counterclockwise more? If I was to try and steer the bike by simply shifting my weight, what would happen at these different speeds? For instance, if I am at a high speed and I shift my weight to the left to make the bike lean to the left will the handlebars automatically turn ever-so-slightly clockwise to initiate the lean even without any physical input to the handlebars?

Finally, how would one go about analyzing this quantitatively? If I am to set up the Lagrangian I presume I have nine degrees of freedom: the linear and angular coordinates of the bike and the angular coordinates of the front wheel. So does the make the Lagrangian

$$L=K_{bike} + K_{wheel} - mgh_{cm}?$$

It's not particularly clear to me what the constraints would be in a problem like this.

Answer 1

This is a classic example of torque as a vector cross product. When you push forward on the left handlebar, the torque $\textbf{r}\times\textbf{F}$ is down. The angular momentum vector of the wheel points to the left. A downward torque deflects the angular momentum vector down, which tilts the bike to the left, and causes you to start turning to the left.

Answer 2

What I typically notice is that if I want to turn left I will push forward on the left handlebar ever-so-slightly, presumable turning the wheel clockwise just a bit, which will initiate a lean to the left, and I will then turn the handlebars counterclockwise to continue the turn. At speeds higher than 20 mph if I want to turn left I push the left handlebar slightly forward which initiates a lean to the left and I hold it like this to keep turning, never really turning the handlebars counterclockwise as in the previous two situations.

Forget the leaning part for a moment.

Balancing a motorcycle is like balancing a bicycle, or even like walking. You are never in balance. You are always off balance, falling (even if only a little) in one direction or the other, and you are always in the process of moving your point of support so as to stop that fall.

So how do you turn? There's an easy way to think about it - you turn to stop yourself from falling. This is how I used to teach kids to ride a bike. Just get moving, like down a grassy slope, so you don't have to think about pedaling. Then if you find yourself falling to the right, turn to the right. That stops you from falling. If you find yourself falling to the left, turn to the left. In no time at all your cerebellum (the part of your brain that deals with coordination) says "Oh, I get it!", and takes over, and you stop having to think about it.

So that's how you prevent falling. Then, if you want to turn right, all you have to do is start falling to the right, and let your brainstem stop you from falling. How do you start falling to the right? You move your support point a little bit to the left, by forcing a little left turn. This is so automatic that you don't even notice, but you can consciously make it happen.

OK, so what about leaning? When you're falling, you're leaning. You can't not lean. But at higher speeds you lean more, because of all the physics - centripetal force and so on. A good way to prove you're still manipulating the handlebars is to clamp them straight ahead. (You will fall down instantly.)

Of course, a motorcycle has a nice heavy gyroscope in front whose precession does some of the adjustment for you, but there are ski-cycles with no wheels at all, and people ride them just the same.

Answer 3

A motorcycle at low speed remains upright because the slip angle forces steer the bike toward whichever way it is tipping. As speed increases, tipping produces smaller slip angles and the bike is kept upright largely by camber forces which push the bike toward any tip. So at low speed you work with the steering by steering a little more or less than what the front wheel does in following the direction vector. You are largely altering slip angle forces. At higher speeds you need to lean to create camber force to balance the high centrifugal forces of turning. Camber forces increase with lean so it is necessary to establish a lean angle to match speed and corner radius. When you lean the slip angle at the front wheel will act to push the wheel into alignment with the new direction vector so this must be overridden by holding the handlebars. The rider thus provides a negative steering torque and a positive roll moment. It is not necessary to counter steer because the direction vector is toward the lean so the bike is already steering to the outside. You can increase this steer if the bike is heavy and the need to turn is urgent. The notion of counter-steering to initiate a lean was developed to make a 2 D O F model work when four degrees are required to describe motorcycle motion adequately. This is important because the steering moment that keeps a bike upright is due to the rear slip angle producing a yawing moment.

Answer 4

This is the PROOF of NO steering AND NO counter steering, ONLY leaning achieved with fixed throttle and brake. https://www.youtube.com/watch?v=qisiO_lwWNo See REAL counter steering and HOW it can't work with a two wheeler. https://www.youtube.com/watch?v=9zilOdo2Mes And this is the complete description, with drawings, but you do have to utilize the english subtitles that do not always use the best translation. Still, it does effectively convey what is occurring. The key event is the LEAN to cause the gyroscopic precession which turns the wheel in that direction, then other forces contribute, to make a controlled turn, primarily, the centrifugal force, to establish a leaned balance (turning). https://www.youtube.com/watch?v=rJ_uJJnTCpw Counter steering does not occur on a MC, it is only used for cars. Leveraged leaning rotational torque is what must be done to turn a MC, at speed. The gyroscopic precession assures us that at the moment the leveraged rotation torque is applied, the GP TURNS the front wheel in the same direction of the LEAN. The top half of the rider/MC mass rotates in the direction of the lean, the bottom half of rider/MC rotates away from the lean due to the pivot point being center mass. The sideways movement of the tire on the ground, away from the lean direction, leads many to think the tire "steered" in that opposite direction when it DID NOT. Countless riders have been errantly brainwashed by Keith Code since 1980, into calling this MC turn action, CS, when no opposite steering occurs.