In the ongoing thread on countersteering, I'm disappointed that no one has addressed the following assertion directly, as far as I can tell from a quick read of the thread:
I'm looking for criticism of the following thought experiment, which has me nearly convinced that tire shape has little or nothing to do with causing a turning path once the bike is leaned over and stable, with no pressure on the bars.
For simplicity, imagine that instead of air in the tire, the tire is solid rubber and the treaded surface is flat and beveled at 10 degrees. (This is essentially the shape of the contact patch when a tire is leaned over 10 degrees.) If we roll this tire, obviously it will turn because one side travels farther than the other as the tire rolls.
Now imagine that we put two identical solid rubber tires on a motorcycle and fix the steering straight ahead, so no steering input is possible. Now lean the bike over 10 degrees so the tires are flat on the road (imagine a frictionless training wheel keeps it up, or a tractor beam from a hovering starship - whatever). Push the bike forward. Say the bevels on the tires have it leaning to the right. Clearly the front wheel wants to turn right. However, the rear wheel also wants to turn right, by exactly the same amount. But, because it's at the back, a right turning rear wheel will try to turn the motorcycle to the left. I think the two effects will cancel, and the bike will travel in a straight line. The line will be to the right of the direction the bike is pointing, and the two tires will not follow the same track, but there will be no curving path.
Now imagine that both the support holding the bike up and the steering are released. The bike will start to fall over, and the rider will have to steer into the turn (to the right) immediately to force a curving path to the right that generates the centrifugal force which balances gravity.
This pretty well convinces me that a motorcycle with identical front and rear tires follows a curving path only because the front tire is turned into the turn. The shape of the tires has nothing to do with it. (Bicycles often have identical tires and they feel remarkably like a motorcycle when you get them going really fast on downhill twisties - clearly they have no trouble turning.) Now, with different front and rear tires I imagine it's possible that tire shape can make a small contribution to maintaining a curving path, but the main cause is still likely to be that the front wheel is turned into the corner. (For completeness, the bike can also turn because the rear wheel is spinning under power and slipping sideways fast enough to the outside of the turn that the front wheel has to be turned outside as well to keep the bike from turning too quickly. But that's not relevant to turning when there's no wheelspin.)
Or, maybe I'm wrong . . .
Having read this statement in many magazine articles, I assumed that it was part of the working knowledge of magazine writers, who are supposed to be experts. This posting got me thinking about it for the first time, and I'm not convinced it's correct. I suppose that a proper analysis could be found (maybe on the site aschendel quoted - the descriptions of countersteering there are the best I've seen anywhere), but it's more fun to try to figure it out myself.Next topic: How a motorcycle actually turns when leaned over. (Hint: it's because of the decreasing radius of your tires from center to edge)
I'm looking for criticism of the following thought experiment, which has me nearly convinced that tire shape has little or nothing to do with causing a turning path once the bike is leaned over and stable, with no pressure on the bars.
For simplicity, imagine that instead of air in the tire, the tire is solid rubber and the treaded surface is flat and beveled at 10 degrees. (This is essentially the shape of the contact patch when a tire is leaned over 10 degrees.) If we roll this tire, obviously it will turn because one side travels farther than the other as the tire rolls.
Now imagine that we put two identical solid rubber tires on a motorcycle and fix the steering straight ahead, so no steering input is possible. Now lean the bike over 10 degrees so the tires are flat on the road (imagine a frictionless training wheel keeps it up, or a tractor beam from a hovering starship - whatever). Push the bike forward. Say the bevels on the tires have it leaning to the right. Clearly the front wheel wants to turn right. However, the rear wheel also wants to turn right, by exactly the same amount. But, because it's at the back, a right turning rear wheel will try to turn the motorcycle to the left. I think the two effects will cancel, and the bike will travel in a straight line. The line will be to the right of the direction the bike is pointing, and the two tires will not follow the same track, but there will be no curving path.
Now imagine that both the support holding the bike up and the steering are released. The bike will start to fall over, and the rider will have to steer into the turn (to the right) immediately to force a curving path to the right that generates the centrifugal force which balances gravity.
This pretty well convinces me that a motorcycle with identical front and rear tires follows a curving path only because the front tire is turned into the turn. The shape of the tires has nothing to do with it. (Bicycles often have identical tires and they feel remarkably like a motorcycle when you get them going really fast on downhill twisties - clearly they have no trouble turning.) Now, with different front and rear tires I imagine it's possible that tire shape can make a small contribution to maintaining a curving path, but the main cause is still likely to be that the front wheel is turned into the corner. (For completeness, the bike can also turn because the rear wheel is spinning under power and slipping sideways fast enough to the outside of the turn that the front wheel has to be turned outside as well to keep the bike from turning too quickly. But that's not relevant to turning when there's no wheelspin.)
Or, maybe I'm wrong . . .