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 Motorcycle Safety
 Physics and the theoretical
 Turning, center of gravity and camber thrust
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Andrew Dressel
Male Standard Member
244 Posts


Milwaukee, WI
USA

Moto Guzzi

California Special

Posted - 07/30/2013 :  8:05 AM
quote:
Originally posted by JanK
In summary:
- Tire's turn centre is always displaced relative to the intersection of tire's axis with the road plane. Thus tire always generates some small amount of camber torque. In a steady state turn this torque is compensated by other torques around the steering head axis to produce a zero net torque.
- Coffee cup's turn centre always coincides with the intersection of cup's axis with the road plane. Thus coffee cup never generates camber torque.



Well, to give the cone (the term Foale uses) model some credit, Foale points out that the wheel does not follow as tight a turn radius as it would if it were allowed to roll without slip. Thus "the actual corner radius described is considerably greater than the cone radius."

On the other hand, Foale appears only to describe a tire generating force due to camber, and I don't see where he describes generating torque due to camber at all. If I remember correctly, Pacejka doesn't describe separate torques due to slip and camber, either. He just lumps them all into Mz.

In any case, few models are perfect, and some may be better for illustration than for prediction and vice versa.
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The Meromorph
Male Moderator
834 Posts
[Mentor]


White House, TN
USA

BMW

R1100RT

Posted - 07/30/2013 :  12:01 PM
To clarify my own post, what I was erroneously referring to as camber thrust, is, in fact, the 'so-called' coffee cup effect.

Further, I do not understand the statements to the effect that the rolling coffee cup generates no torque. What causes it to turn (roll in a circle)?

My repeated claim is that this coffee cup effect is indeed trivial compared with all the other forces at normal riding speeds, but that at very slow speeds (alleged sub-countersteering speeds) it can have a significant effect, given the proper profile, proper tire pressures and enough lean angle. This is because this maximises the effect, while many of the other forces are small in this situation.

I'm going to just watch this thread, I think, because y'all are not only over my head, but nearly out of sight over my head.
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Andrew Dressel
Male Standard Member
244 Posts


Milwaukee, WI
USA

Moto Guzzi

California Special

Posted - 07/30/2013 :  4:33 PM
quote:
Originally posted by JanK
In summary:
- Tire's turn centre is always displaced relative to the intersection of tire's axis with the road plane. Thus tire always generates some small amount of camber torque. In a steady state turn this torque is compensated by other torques around the steering head axis to produce a zero net torque.
- Coffee cup's turn centre always coincides with the intersection of cup's axis with the road plane. Thus coffee cup never generates camber torque.



Hmmm.

A tire on a bike in a steady-state turn is constrained by dynamic equilibrium (balancing the roll moment of centripetal force against the one of gravity) to follow a certain path with a certain camber angle that usually puts its actual turn center farther away than its kinematic turn center, thus generating a torque in the contact patch that tends to steer it towards the direction it is leaning.

A slowly rolling cone is constrained by kinematics to follow a path that puts its actual turn center exactly at its kinematic turn center. If it were rolled fast enough, however, so that it started to slide and no longer followed that path, then it would be in a situation similar to that of the tire except that kinetic friction rather than static friction and tire compliance will be generating the force and torque. This is my understanding of how Foale develops his model for the centripetal force generated by camber. I just don't see where he continues it to the point of generating a torque as well.

I'm not saying the cone model is better or even as good as any other model for explaining or calculating forces or torques generated by a tire as it rolls forward with a non-zero camber angle, but I do get its appeal for making the explanation for how the torque is generated simple and easy to grasp.
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Andrew Dressel
Male Standard Member
244 Posts


Milwaukee, WI
USA

Moto Guzzi

California Special

Posted - 07/30/2013 :  4:42 PM
quote:
Originally posted by The Meromorph
Further, I do not understand the statements to the effect that the rolling coffee cup generates no torque. What causes it to turn (roll in a circle)?



I believe the distinction is that the torque generated in the contact patch of a cambered bike tire under any sort of normal riding conditions continually tends to tighten the turn radius. On the other hand the torque generated in the contact region of a rolling cone only exists when the cone starts rolling, to initiate its rotation, and to maintain its kinematic turning radius, the distance from its tip to its base, should some other effect tend to alter that radius.
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JanK
Male Junior Member
76 Posts


Ljubljana, Ljubljana
Slovenia

BMW

F650CS

Posted - 07/30/2013 :  6:41 PM
quote:
Originally posted by The Meromorph

Further, I do not understand the statements to the effect that the rolling coffee cup generates no torque. What causes it to turn (roll in a circle)?



For this thought experiment, imagine a hockey puck and let's assume you have magical ice with no friction.

If you just place the puck on the ice, it will not rotate by itself, obviously.

To start a rotation of the puck around its centre, you need to supply torque by grabbing the edges and twisting it. Once you let go, the puck will keep rotating at exactly the same angular velocity forever, as there is no torque acting. Remember the ice is frictionless, so there is nothing that slows the puck down.

So a constant angular velocity is an indication that there is no torque on the object, or that any torques cancel out exactly.

Now imagine sticking pin in the middle of the puck. You cannot change the spin of the puck by pushing on the pin, because you cannot exert torque (assume the pin is so thin you cannot twirl it between the fingers). By applying horizontal force you can only cause the puck to move in different directions on the ice.

The crucial realisation is that these two motions are independent of each other. You can spin the puck by "torqueing" it at the edge. You can then push on the pin and move the puck around on the ice, but during this movement the puck will keep spinning at the exactly the same rate, regardless of the horizontal speed and acceleration.

This separation of motions is still valid even if the horizontal motion happens to be circular. Imagine sticking another pin in the ice and tying a string between the pin on the puck and the one in the ice. Tighten the string. Now give the pin on the puck a shove in a direction perpendicular to the string.

The puck will start to circle around the pin in the ice, but it will not spin around its own axis - you have not imparted any torque to start it spinning. The only force acting on the puck is the centripetal force, which causes it to travel in a circle.

You could also spin it before shoving it. In this case the puck will keep spinning at constant angular velocity while it's travelling in a circle. Again the two motions are independent. You could easily spin the puck in the opposite direction of the circular motion and the puck will happily keep spinning in one direction while circling in the other direction. You could also cut the string at any time and the puck would fly off on a tangent to the circular path while still spinning at exactly the same rate.

The harder you "torque" the puck, the faster it will spin and the harder you shove it, the faster it will circle. If you were very careful, you could "balance" the torque and the shove and the net effect would be that the puck would spin once in exactly the time it would take to travel one circle.

The point at which the string crosses the edge of the puck would not move. You could stick another pin into the puck at that point, tie the string to this pin and remove the section of string between the puck centre and its edge. The motion would not change, the puck would still spin in the same time it takes to make one circle.

Now this would appear as though the pin on the edge is somehow "turning" the puck and providing the torque for it to turn. But the fact that it spins at a constant rate implies that there is no torque on the puck, otherwise it would start spinning faster or slower rate.

The same logic applies to the cup (or cone). The peak of the cone (or the intersection between the axis of the cup and the ground plane) is the centre of rotation, around which the centre of gravity of the cone revolves. The cone spins in the But the cone also spins around its centre of gravity around a vertical axis, much as the puck does.

And since the cone does not increase or decrease its spin rate, when in a steady turn, there cannot be any torque on the cone.

This is all valid in the so-called steady state, when all the values (speeds, rotational velocities, lean angles,...) have settled and remain where they are. How the system gets there is another question. One, whose answers include torque...

This is as far as I got in the previous post. Thanks to this thread and James' and Andrew's posts I now know (hopefully) understand more.

If you roll a cup forcefully, it is unlikely that the roll's spin and speed will match exactly, so the cup will slip on the table surface. But after a short while it will settle into a steady state, where it will roll in a circle without slipping. During the settling down period there is torque generated.

If you roll the cup perpendicularly to its axis, as if it were a cylinder, the smaller radius will rotate too slowly and will try to brake the movement, while the larger radius will rotate too quickly and try to accelerate the movement. These forces generate a torque around the centre of the cup and this torque tends to turn the cup into the circle.

I just did an experiment that you can easily reproduce. Take a cup or a glass (I used a simple conical Ikea glass) and gently push it along the larger and then the smaller side. In the former case it will turn (into the corner) easily. In the latter case the larger side will advance (turning into the corner) but, if you get the push just right, will then slip back and keep slipping at some fixed angle. You will feel it trying to turn into the turn - this is camber torque.

If I understood Cossalter and Foale this is similar to what happens with a motorcycle tire and how the camber torque is generated. As Andrew has noted, the above mentioned situation, where the smaller radius travels more slowly than the larger radius, where the turn centre is outside the geometric centre, is always true for motorcycle tires. Thus a tire always generates torque that tends to turn the handlebars into the turn.

I must admit that I didn't follow the cup analogy far enough to make the connection before this debate. Perhaps I was misled by the image of a cup in a stationary state - rolling freely across a table without slipping - and could not imagine how this could be similar to a tire.
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