3 ms·

This is not a correct picture. You shouldn't have to watch a GP race for long to realize that, at speed, the bikes don't pivot around their centres of gravity,

by lutorm 11y ago

This is not a correct picture. You shouldn't have to watch a GP race for long to realize that, at speed, the bikes don't pivot around their centres of gravity, they pivot around the contact patch. This is because applying torque to the front wheel through the handlebars (essentially around the yaw axis) results in a gyroscopic rotation around the orthogonal, roll, axis.

barrkel 11y ago
The road pushes the contact patch to one side; it would be quite remarkable if that got them to pivot around the contact patch, since the contact patch has no leverage to use itself as the centre of the roll axis. By centre of gravity, to be clear, I meant the roll axis running through it. I used the word rolling precisely to refer to the roll axis. Gyroscopic effects definitely complicate things - steering takes more effort at speed - but the fundamentals of counter-steering is that it rolls the bike due to torque from a reaction of the road against the angle of the front wheel. (FWIW, I don't own a car and use motorbikes as my main form of transport; I do about 15k a year, a mix of city commuting and long distance touring. Smaller, lighter bikes are easier to apply extreme countersteering to, and I do often in the city, so I'm well aware of what actually happens when I steer harshly, which exaggerates the effect of everything described. For example, sharply countersteering upright will push the front of the bike into the air.)
- lutorm 11y ago
  But that's what I'm saying, it's not the force from the contact patch that leans the bike, it leans despite the force from the contact patch. The situation at high speed is fundamentally different from the situation at low speed. Try this experiment (this became quite long, but I wanted to make the explanation as explicit as possible to avoid misunderstandings): 1. Put a pebble in the road as a marker and ride towards it quite slowly, like jogging speed. As you approach the pebble, initiate a sudden countersteer to the left. What will happen is that the handlebars will turn slightly to the right, the contact patch will move to the right, and the bike will roll to the left as it rotates around the CM. Your front tire will likely pass to the right of the pebble. This is the situation you describe: at low speeds, the force from the contact patch makes the bike rotate around the CM, which initially moves opposite to the direction you are trying to go. (It has to, the contact patch force is to the right, and that's the only lateral force affecting the bike initially.) 2. Now try the same maneuver at quite high speed. (Do it in a wide parking lot or something so you don't run off the road, obviously...) What you will find is that unlike in the first situation, when you countersteer, the handlebars are pretty much rock solid. Even though you attempt to turn them to the right, there is no appreciable motion. Because the wheel doesn't turn, the contact patch continues basically a straight line. However, the bike still leans over and starts turning to the left. You'll now find that the wheel will run over or pass slighly to the left of the pebble. This behavior is impossible to understand without the gyroscopic effects. If the bike still rotated around the CM, the contact patch would still move to the right, actually by a larger amount if you think that it takes a larger force to overcome the stability of the spinning wheels. If that was the case, the front wheel would still pass well to the right of the pebble, but it doesn't. What is missing in this case is the effect of gyroscopic precession in a fixed gyro. With the front wheel spinning rapidly, the attempt to turn it right when countersteering will not rotate it around in the steering direction, it will cause the wheel to rotate orthogonal to the applied torque, in the roll plane, taking the whole bike with it. If the bike was suspended in the air, it would just rotate around its center of mass, but because friction prevents the contact patch from moving laterally, there is a resulting force with a magnitude that keeps the contact patch stationary. Note that unlike in the slow-speed case, where the force from the contact patch acts to the right, it now acts to the left. The resulting motion is one where the bike rotates around the contact patch, moving the center of mass in the direction of the lean. This motion looks quite "magic" if you're not aware of what the gyroscopic precession does. Without the precession, you wouldn't need to go very fast until it would be impossible to turn the bike at all. If you just imagine that the gyroscopic effect makes the bike resist leaning over, you'd need to develop a larger and larger contact patch force to the right as you went faster and faster. This would make the bike track further and further to the right as you attempted to turn to the left, which is exactly the opposite of how a bike really behaves. Hope that explanation helps.
  - barrkel 11y ago
    I don't have enough intuition for gyroscopic precession to fully follow your high speed case (how fast, I wonder?), but the requirement to overcome gyroscopic stability otherwise causing tracking off line convinces me that I'm wrong at higher speeds. I still think my picture is more explanatory at the speeds I commute at (almost always below 45mph, with most steering at sub 25mph). How do braking forces play into this, and do you know what kinds of speeds we're talking about? A soft front tyre (i.e. lower than normal pressure) has a dramatic effect on handling at street / city speeds - the extra rolling resistance makes the steering want to steer further into the turn, causing a countersteer upright. So you need to fight the steering to maintain line. But of course when putting a bike on track, you run with lower pressures as a matter of course - and I know from anecdotal experience that it is possible, on track, to brake while leaned over a bit hard enough to start lifting the rear, something that would require an immense effort to avoid straightening up at street speeds. You've given me something to think about, but unfortunately I won't be able to solve the problem without either learning a lot more mathematics, or playing with models...

by lutorm 11y ago

barrkel 11y ago

The road pushes the contact patch to one side; it would be quite remarkable if that got them to pivot around the contact patch, since the contact patch has no leverage to use itself as the centre of the roll axis. By centre of gravity, to be clear, I meant the roll axis running through it. I used the word rolling precisely to refer to the roll axis. Gyroscopic effects definitely complicate things - steering takes more effort at speed - but the fundamentals of counter-steering is that it rolls the bike due to torque from a reaction of the road against the angle of the front wheel. (FWIW, I don't own a car and use motorbikes as my main form of transport; I do about 15k a year, a mix of city commuting and long distance touring. Smaller, lighter bikes are easier to apply extreme countersteering to, and I do often in the city, so I'm well aware of what actually happens when I steer harshly, which exaggerates the effect of everything described. For example, sharply countersteering upright will push the front of the bike into the air.)

lutorm 11y ago
But that's what I'm saying, it's not the force from the contact patch that leans the bike, it leans despite the force from the contact patch. The situation at high speed is fundamentally different from the situation at low speed. Try this experiment (this became quite long, but I wanted to make the explanation as explicit as possible to avoid misunderstandings): 1. Put a pebble in the road as a marker and ride towards it quite slowly, like jogging speed. As you approach the pebble, initiate a sudden countersteer to the left. What will happen is that the handlebars will turn slightly to the right, the contact patch will move to the right, and the bike will roll to the left as it rotates around the CM. Your front tire will likely pass to the right of the pebble. This is the situation you describe: at low speeds, the force from the contact patch makes the bike rotate around the CM, which initially moves opposite to the direction you are trying to go. (It has to, the contact patch force is to the right, and that's the only lateral force affecting the bike initially.) 2. Now try the same maneuver at quite high speed. (Do it in a wide parking lot or something so you don't run off the road, obviously...) What you will find is that unlike in the first situation, when you countersteer, the handlebars are pretty much rock solid. Even though you attempt to turn them to the right, there is no appreciable motion. Because the wheel doesn't turn, the contact patch continues basically a straight line. However, the bike still leans over and starts turning to the left. You'll now find that the wheel will run over or pass slighly to the left of the pebble. This behavior is impossible to understand without the gyroscopic effects. If the bike still rotated around the CM, the contact patch would still move to the right, actually by a larger amount if you think that it takes a larger force to overcome the stability of the spinning wheels. If that was the case, the front wheel would still pass well to the right of the pebble, but it doesn't. What is missing in this case is the effect of gyroscopic precession in a fixed gyro. With the front wheel spinning rapidly, the attempt to turn it right when countersteering will not rotate it around in the steering direction, it will cause the wheel to rotate orthogonal to the applied torque, in the roll plane, taking the whole bike with it. If the bike was suspended in the air, it would just rotate around its center of mass, but because friction prevents the contact patch from moving laterally, there is a resulting force with a magnitude that keeps the contact patch stationary. Note that unlike in the slow-speed case, where the force from the contact patch acts to the right, it now acts to the left. The resulting motion is one where the bike rotates around the contact patch, moving the center of mass in the direction of the lean. This motion looks quite "magic" if you're not aware of what the gyroscopic precession does. Without the precession, you wouldn't need to go very fast until it would be impossible to turn the bike at all. If you just imagine that the gyroscopic effect makes the bike resist leaning over, you'd need to develop a larger and larger contact patch force to the right as you went faster and faster. This would make the bike track further and further to the right as you attempted to turn to the left, which is exactly the opposite of how a bike really behaves. Hope that explanation helps.
- barrkel 11y ago
  I don't have enough intuition for gyroscopic precession to fully follow your high speed case (how fast, I wonder?), but the requirement to overcome gyroscopic stability otherwise causing tracking off line convinces me that I'm wrong at higher speeds. I still think my picture is more explanatory at the speeds I commute at (almost always below 45mph, with most steering at sub 25mph). How do braking forces play into this, and do you know what kinds of speeds we're talking about? A soft front tyre (i.e. lower than normal pressure) has a dramatic effect on handling at street / city speeds - the extra rolling resistance makes the steering want to steer further into the turn, causing a countersteer upright. So you need to fight the steering to maintain line. But of course when putting a bike on track, you run with lower pressures as a matter of course - and I know from anecdotal experience that it is possible, on track, to brake while leaned over a bit hard enough to start lifting the rear, something that would require an immense effort to avoid straightening up at street speeds. You've given me something to think about, but unfortunately I won't be able to solve the problem without either learning a lot more mathematics, or playing with models...