oldfrt Posted February 24, 2007 Report Share Posted February 24, 2007 Definition of going in a straight line: The front wheel is in a plane with the rest of the bike, the steering head is pointing straight ahead, and the back wheel is exactly tracking the front wheel. Isn't this exactly what is happening when you're leaned over in a constant radius turn and maintaining a constant speed? Quote Link to comment Share on other sites More sharing options...
racer Posted February 24, 2007 Report Share Posted February 24, 2007 No. The front wheel is turned slightly into the turn. Quote Link to comment Share on other sites More sharing options...
oldfrt Posted February 25, 2007 Author Report Share Posted February 25, 2007 Are you sure? Quote Link to comment Share on other sites More sharing options...
oldfrt Posted February 25, 2007 Author Report Share Posted February 25, 2007 I think you're right that the wheel would or could be pointed slightly into the turn in some or maybe most turns, but here's what I'm thinking. If the bike were going around a circular track that was banked such that the lean was perpindicular to the track then there would be no turn in the bike geometry. Taking that concept to the extreme, think about a bike in a spherical cage. It can actually do vertical loops. Certainly it's not turning at that point. Then again, back to the banked circular track. Take away the banking and the track is flat. The bike is still leaned but now the contact patch is on the side of the wheel instead of centered. As far as the bike is concerned it might as well just be riding on a banked straight road maintaining a contact patch off-center of the tires. The difference is just more g's in the turn. I believe the amount of turn in the bike geometry is more due to the lean than to the turn radius in the road or track. On the other hand, if you stand holding the bike and keeping it vertical push it around in a circle with your body as the center it's obvious that the wheel has to be turned to the locks to go throught the turn. Leaning the bike to the inside will allow it to turn tighter. If the bike could be leaned all the way as if you were standing in a spherical cage, then it could do the turn with no turn in the bike geometry. I just talked myself into agreement with you. The bike is always turned slightly in the turn direction but the bike geometry turn decreases as the lean increases and would go to zero if the bike were leaned 90 degrees. This is just something I have been thinking about for some time and have wanted to understand a little better. Hope the subject is not boring or annoying to others. Or even useless. Quote Link to comment Share on other sites More sharing options...
Woody Posted February 26, 2007 Report Share Posted February 26, 2007 Whoa, that set the old nogging spinning. For a given speed, the more a bike leans, the more the front wheel points into the turn. Also, the bike lean should be understood in relation to the road. If the road is already banked, the natural line for a bike that is 'upright' (90 degrees in relation to the road) is to follow the banking and at this point it is safe to say that the bike is not 'turning'. To put it another way, if a turn is banked you need less lean to get around it than if the turn was flat. Your example of the 'wall of death' is a perfect example of taking this to the extreme. A bike going straight, yet also changing direction. Quote Link to comment Share on other sites More sharing options...
oldfrt Posted February 26, 2007 Author Report Share Posted February 26, 2007 I'm thinking of "lean" relative to a horizontal plane instead of the road surface. All the tilt of the road surface does is to move the contact patch. Here's another way to see it. Travelling in a circular path, the lean of the bike would generate a cone. For a given circle, as the lean increases the apex of the cone flattens until if the lean went all the way to 90 degrees the cone would flatten and become a disc. At that point the steering head is not turning. That's also why you can negotiate slow speed turns by letting the bike lean into the turn that you can't accomplish by just pushing the bike through upright. Quote Link to comment Share on other sites More sharing options...
Woody Posted February 26, 2007 Report Share Posted February 26, 2007 I'm thinking of "lean" relative to a horizontal plane instead of the road surface. We're heading toward some very rough waters here. Lean is a necessary byproduct of cornering forces. It is what stops the bike from flipping over whilst cornering. It is what gives the bike balance through a corner. The amount of lean is relative to the speed of the bike and the rate of turn. Lean more and you'll either take a tighter line or you need to go faster to retain the same rate of turn. The front wheel points into the turn. The amount it points into the turn is just the right amount to achieve the right arc for the given lean angle and speed (we set the arc through steering input and the lean follows). Turn the front wheel more or less and the bike's amount of lean changes. (Change your speed, by the way, and you change the arc.) Without cornering forces, the bike needs no lean and to generate cornering forces, the front wheel needs traction. What sort of traction would be present if the bike was at 90 degrees from upright? This will decide the state of the front wheel. The only scenarios I can think of where traction would be present are: 1. A banked corner 2. A wall of death (which is effectively an extreme banked corner) For 1. the bike would be cornering just as normal, with the front wheel pointing into the corner. For 2. the front wheel would be straight. For a flat turn, the bike would be on its side, no traction would be presented to the tyre and the direction of the front wheel is immaterial. Quote Link to comment Share on other sites More sharing options...
oldfrt Posted February 27, 2007 Author Report Share Posted February 27, 2007 What sort of traction would be present if the bike was at 90 degrees from upright?This will decide the state of the front wheel. The only scenarios I can think of where traction would be present are: 1. A banked corner 2. A wall of death (which is effectively an extreme banked corner) Actually, 90 degrees is not possible because the g force would go to infinity. It's just a theoretical way to view the mechanics and geometry of turning. Quote Link to comment Share on other sites More sharing options...
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