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DYK proposal

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Template_talk:Did_you_know#Camber_thrust

Steer and Lean?

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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)[reply]

It is correct as currently written. Even though the rider may be applying a torque on the handlebars counter to the direction of the turn, the front end must actually point, if only slightly, in the direction of the turn and so in the direction of the lean. Additionally, though often counter, the applied torque necessary may be zero or even in the direction of the turn, depending on bike geometry, forward speed, and rider position. See bicycle and motorcycle dynamics and counter steering articles for more detail and copious references. -AndrewDressel (talk) 15:54, 18 June 2009 (UTC)[reply]

Cones etc

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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)[reply]

I doubt it was a conscious decision, at least on my part. However, Cossalter nor Pacejka mention the cone analogy, and Foale concedes at the top of page 2-24 that the actual corner radius described is considerably greater than the cone radius. -AndrewDressel (talk) 00:55, 12 March 2010 (UTC)[reply]
Indeed he does. We all know bikes lean to balance and that speed, radius and lean angle need to be matched up or the bike will fall over. The cone theory links only radius and lean angle, not speed. In his dismissive sentence Foale implies there is some element that responds to centrifugal force in a way that creates a self-balancing bike. The actual trajectory taken by the bike is not due to centrifugal force; it is because it is steered. -Rolo Tamasi (talk) 15:38, 12 March 2010 (UTC)[reply]
Cossalter discusses this in detail on pages 52-53. He asserts that up to 28° of lean "for a certain type of tire", camber thrust nearly matches the required centripetal force exactly. The amount that it is slightly excessive is balanced by a slightly negative slip angle. Above 28°, camber thrust must be augmented more and more by slip angle to provide the necessary centripetal force. -AndrewDressel (talk) 19:27, 12 March 2010 (UTC)[reply]
I didn’t explain my point well. A bike is leaned to a particular angle in order to be balanced for the particular curve not in order to generate steering forces. Whatever happens the sum of all turning forces will be modified by steering inputs to achieve the desired path that matches the lean. That is why the bike does not follow the path of a cone with the same lean, it happens whether camber thrust is greater or less than the required total turning force. The cone effect path is tighter than the balanced path but to say that the bike does not follow the cone effect path in practice because of centrifugal force is meaningless. The rider could steer the claimed cone effect path but the bike would fall over high-side.Rolo Tamasi (talk) 21:28, 12 March 2010 (UTC)[reply]
Cossalter (which I have read but don’t have a copy) expresses things in a far more disciplined way.Rolo Tamasi (talk) 21:28, 12 March 2010 (UTC)[reply]


However his use of words often has to be given hugely generous interpretation. For example he states that the cone effect allows bikes to turn without a slip angle. Clearly this must be untrue as any transverse force on a tyre will result in a slip angle and all turns need a radial acceleration. Ergo, cones must have a slip angle. Being a generous person I decide he means to say that bikes don’t need as great a steering angle to provide the necessary slip angle as cars. But it makes me wary of his theories. -Rolo Tamasi (talk) 15:38, 12 March 2010 (UTC)[reply]
Well, as I mention above, Cossalter suggests that the necessary slip angle is small and even negative for lean angles from 0 - 28° "for a certain type of tire", which suggests that Foale isn't that far off. -AndrewDressel (talk) 19:27, 12 March 2010 (UTC)[reply]
I’m sure that negative steering angles can be required; I think I have made that point myself elsewhere. I suspect that semantics are getting in the way here. There is a tendency to relate slip angle only to steering force and use the two principles interchangeably. My thought is that the cone effect must also have a slip angle associated with it. So I’m building a conceptual model that has numerous elements (Cone effect camber; Slip angle associated with cone effect; Deformation camber effect; Steering force; Slip angle associated with steering angle) but this is rapidly dissolving into original research, there must be someone who has spent more time than I did considering this over my cornflakes this morning and has published something.Rolo Tamasi (talk) 21:28, 12 March 2010 (UTC)[reply]
That does not mean that the cone theory does not have merit. Foale did not invent it and the literature is full of references to it. What is more, it has attractions from first principles. It is clear to me that truncated cones roll on a curved path and I am quite willing to believe that, as their mass tends to decrease and the friction to increases, so the centre of the turn radius tends towards the apex of the cone. -Rolo Tamasi (talk) 15:38, 12 March 2010 (UTC)[reply]
Peer-reviewed literature? My reading is certainly not exhaustive, and I don't believe I've ever looked for this detail, but I don't recall coming across it. -AndrewDressel (talk) 19:27, 12 March 2010 (UTC)[reply]
Well I have only had a quick glance, but a one out of one sample – Jo Yung Wong, Page 43 fig 1.33 shows the “theoretical centre of rotation for a cambered wheel” to be the point at which the axle’s extended centre line cuts the road surface.Rolo Tamasi (talk) 21:28, 12 March 2010 (UTC)[reply]
It is also apparent that road forces are only generated at the tyre/road interface and that, when the wheel has camber, that interface is part of the surface of a truncated cone. -Rolo Tamasi (talk) 15:38, 12 March 2010 (UTC)[reply]
But not only/exactly: a truncated cone would contact along just a radial a line. Real tires have a much (infinitely?) larger 2D contact patch. -AndrewDressel (talk) 19:27, 12 March 2010 (UTC)[reply]
Hmmm, the same thought went through my mind and I decided not to mention it because it was an unhelpful complexity – you make me face up to my lack of integrity! To make amends I will restate the issue – the inner part of the contact area is rotating about a smaller radius than the outer part. This must generate a turning force that is not reflected in the article.Rolo Tamasi (talk) 21:28, 12 March 2010 (UTC)[reply]
What is more, the cone radius is so very much tighter than the balance radius for anything but the very lowest speeds so there is plenty of room for the majority of the turning forces to be lost in tyre deformation or whatever and still to be the primary turning force for a bike (that is a conceptual possibility, it could be a mathematical absurdity). -Rolo Tamasi (talk) 15:38, 12 March 2010 (UTC)[reply]
Perhaps the issue is, and the reason Cossalter and Pacejka don't mention it is, the cone model is not predictive. No one is able to predict actual camber thrust from it. Perhaps its utility is merely illustrative. That doesn't make it bad, per se, but suggests that it should be introduced with caveats. -AndrewDressel (talk) 19:27, 12 March 2010 (UTC)[reply]
That could be the reason; I have not yet worked out how a rolling cone exerts the radial force on the road surface. Maybe when I do it will be clear that small truncations make the effect minimal. However this is something I know little about. I must shake myself out of being an engineer and get back into being an editor of others work!Rolo Tamasi (talk) 21:28, 12 March 2010 (UTC)[reply]
It would be tempting to resolve this by saying that the cone forces are inevitably included within the process described in the article because the ellipse of each part of the width of the tyre is different. Unfortunately that does not appear to work, as I can see no turning forced from that difference. -Rolo Tamasi (talk) 15:38, 12 March 2010 (UTC)[reply]
The mechanism described in the article depends entirely upon deformation but a non-deforming conical shaped tyre will turn and thus, as it is inevitable that there is a conical element to all inclined bike tyres, the described mechanism is incomplete. -Rolo Tamasi (talk) 15:38, 12 March 2010 (UTC)[reply]
I raise this not (only) out of idle curiosity, it is frequently touched on in contemporary publications and we have a duty to put the great unwashed out of their misery on this point. -Rolo Tamasi (talk) 15:38, 12 March 2010 (UTC)[reply]
I agree. I'm just not yet sure what is the correct thing to say about it. -AndrewDressel (talk) 19:27, 12 March 2010 (UTC)[reply]
In editor mode I would say that we have credible citations that refer to the cone effect and others that refer to the deformation mechanism and some doubt that they are different ways of referring to the same phenomena. One approach is to refer to them both ignoring how they may interact or compare (not the ideal) or we find some other published discussion of them both.Rolo Tamasi (talk) 21:28, 12 March 2010 (UTC)[reply]


Lagrangian mechanics

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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)[reply]

Camber thrust diagram

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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)[reply]

The arrow in the diagram follows the convention used by Cossalter and shows the force of the ground acting on the tire, which is in the direction of the lean. The diagram does not show the contact patch. The dotted oval is the path of a point on the outer surface of the tire. -AndrewDressel (talk) 14:51, 3 September 2012 (UTC)[reply]
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