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![]() | A fact from Camber thrust appeared on Wikipedia's
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Template_talk:Did_you_know#Camber_thrust
I'm no engineer (for sure) but I've put in a few mile on various motorcycles. Unless I am misreading I believe the statement "When a bike is steered and leaned in the same direction..." is somewhat inaccurate. Unless you are going very slow, you steer away from the lean. If you are making a right hand turn, you would lean right while pushing the right handlebar to the left. That may be what is meant here, but perhaps it is not stated clearly?
- Pagosa Joe —Preceding unsigned comment added by Pagosajoe ( talk • contribs) 14:51, 18 June 2009 (UTC)
There are published articles that put a lot of faith in the cone principle of camber thrust. (e.g. http://www.tonyfoale.com/Articles/Tyres/TYRES.htm ) on the other hand others adopt the argumentation used here. The two appear to be fundamentally in conflict as the theory expounded here is dependent upon tyre deformation, with no deformation there is no camber thrust.
I am interested to discuss why we decide that the cone theory has so little merit as not to be worth even mentioning here. I must stress that I am not pressing for its inclusion, just seeking to discuss the issue. Rolo Tamasi ( talk) 19:41, 11 March 2010 (UTC)
I derived a motorcycle model from lagrangian mechanics (11 dof, 6 bodies). Since the whole dynamics depend on the externally applied forces, I wanted to refine the tyre model. In the first version, tyre forces are linear in slip. For my new version, I started with deriving the equations of motion of a cone (6 dof, 1 body). I believe that the force between the surface of the cone and the plane can only depend on local variables (variables measured in the origin of the force). I have very satisfying results and would like to share that.
First, the normal force is calculated on the lowest point of the cone:
Fn = (a*x +b*v)*(x<0)
Fn is the normal force, a and b are constant parameters. x is the position in the direction perpendicular to the plane of the origin of the force, and v the velocity (time derivative of x)
now the tangential force (Ft) I calculate with lagrange multipliers that arise from the holonomic constraint that there should be no slip between the surfaces (every point that is the intersection of the body with the plane). If this force is larger than
Ft = mu * Fn * sign(y),
where mu is a constant, and y is the time derivative of the position vector of the origin of the force (local velocity vector), the force is replaced by this one before continuing the differential equation solver.
Camber thrust is a consequence of this approach, since the no slip condition (which turns into a ‘minimum slip condition’ once the maximum static force has been reached) forces the inner part of the contact patch area to slide as fast as the outer part, while the radius is smaller, thus forcing the cone or torus to yaw.
I did simulations and made animations to convince myself.
Thanks for reading my message —Preceding unsigned comment added by 134.221.172.194 ( talk) 13:01, 3 January 2011 (UTC)
I'm confused by the graphic. If the thrust direction shown is the force developed by the tire I think it's in the wrong direction. I also think the new contact patch is curved on the camber side resulting in a leading edge pointed in the direction of the camber adding to slip angle if present. Comments please. Thanks, Paul Haney, paul@tvmotorsports.com — Preceding unsigned comment added by 76.186.20.191 ( talk) 21:54, 31 August 2012 (UTC)
![]() | This article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||||||||||||||||||||||||||||||||||||||
|
![]() | A fact from Camber thrust appeared on Wikipedia's
Main Page in the
Did you know column on 18 June 2009 (
check views). The text of the entry was as follows:
| ![]() |
Template_talk:Did_you_know#Camber_thrust
I'm no engineer (for sure) but I've put in a few mile on various motorcycles. Unless I am misreading I believe the statement "When a bike is steered and leaned in the same direction..." is somewhat inaccurate. Unless you are going very slow, you steer away from the lean. If you are making a right hand turn, you would lean right while pushing the right handlebar to the left. That may be what is meant here, but perhaps it is not stated clearly?
- Pagosa Joe —Preceding unsigned comment added by Pagosajoe ( talk • contribs) 14:51, 18 June 2009 (UTC)
There are published articles that put a lot of faith in the cone principle of camber thrust. (e.g. http://www.tonyfoale.com/Articles/Tyres/TYRES.htm ) on the other hand others adopt the argumentation used here. The two appear to be fundamentally in conflict as the theory expounded here is dependent upon tyre deformation, with no deformation there is no camber thrust.
I am interested to discuss why we decide that the cone theory has so little merit as not to be worth even mentioning here. I must stress that I am not pressing for its inclusion, just seeking to discuss the issue. Rolo Tamasi ( talk) 19:41, 11 March 2010 (UTC)
I derived a motorcycle model from lagrangian mechanics (11 dof, 6 bodies). Since the whole dynamics depend on the externally applied forces, I wanted to refine the tyre model. In the first version, tyre forces are linear in slip. For my new version, I started with deriving the equations of motion of a cone (6 dof, 1 body). I believe that the force between the surface of the cone and the plane can only depend on local variables (variables measured in the origin of the force). I have very satisfying results and would like to share that.
First, the normal force is calculated on the lowest point of the cone:
Fn = (a*x +b*v)*(x<0)
Fn is the normal force, a and b are constant parameters. x is the position in the direction perpendicular to the plane of the origin of the force, and v the velocity (time derivative of x)
now the tangential force (Ft) I calculate with lagrange multipliers that arise from the holonomic constraint that there should be no slip between the surfaces (every point that is the intersection of the body with the plane). If this force is larger than
Ft = mu * Fn * sign(y),
where mu is a constant, and y is the time derivative of the position vector of the origin of the force (local velocity vector), the force is replaced by this one before continuing the differential equation solver.
Camber thrust is a consequence of this approach, since the no slip condition (which turns into a ‘minimum slip condition’ once the maximum static force has been reached) forces the inner part of the contact patch area to slide as fast as the outer part, while the radius is smaller, thus forcing the cone or torus to yaw.
I did simulations and made animations to convince myself.
Thanks for reading my message —Preceding unsigned comment added by 134.221.172.194 ( talk) 13:01, 3 January 2011 (UTC)
I'm confused by the graphic. If the thrust direction shown is the force developed by the tire I think it's in the wrong direction. I also think the new contact patch is curved on the camber side resulting in a leading edge pointed in the direction of the camber adding to slip angle if present. Comments please. Thanks, Paul Haney, paul@tvmotorsports.com — Preceding unsigned comment added by 76.186.20.191 ( talk) 21:54, 31 August 2012 (UTC)