Steering related doubt

  • Thread starter sganesh88
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Consider this situation.
A non-rotating wheel is cruising through space such that its velocity vector is at some angle to the wheel plane. (Like when a car slides through ice when the driver turns the steering wheel). When its base touches a surface with friction, it takes a turn and its new velocity would point along the wheel plane. (assuming there's a mechanism that prevents it from toppling). This is because a frictional force in the direction opposite to the initial velocity acts on the wheel at the instant the wheel base touches the surface. It can be split into two components Fx and Fy(see images in the link) one along the wheel plane, one perpendicular to it at the point of contact with the surface. Fy completely destroys the momentum along the y direction while Fx rotates the wheel reducing the velocity till Vx=R*W(omega). So finally the wheel's velocity vector will be along Vx.
am i right in this explanation? Replies to this will take us forward to my doubt. :)
http://img25.imageshack.us/img25/892/dounbt.png [Broken]
 
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Answers and Replies

  • #2
tiny-tim
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Hi sganesh88! :smile:
… This is because a frictional force in the direction opposite to the initial velocity acts on the wheel at the instant the wheel base touches the surface. It can be split into two components Fx and Fy(see images in the link) one along the wheel plane, one perpendicular to it at the point of contact with the surface. Fy completely destroys the momentum along the y direction while Fx rotates the wheel reducing the velocity till Vx=R*W(omega). So finally the wheel's velocity vector will be along Vx.
Sorry, I'm not understanding any of this :confused:

once the wheel hits the surface (and if it doesn't bounce off again), it must have a velocity parallel to the tangent to the surface (this is geometry, not physics :wink:), and it may also have a rotation, both about its own axis ("rolling") and perpendicular ("leaning"). :smile:
 
  • #3
Lok
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Is the friction force at the front of the wheel or the lower middle. Assuming we are looking from above? and the plane is in front of us?

Just asking because it counts.
 
  • #4
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Hi sganesh88! :smile:


Sorry, I'm not understanding any of this :confused:

once the wheel hits the surface (and if it doesn't bounce off again), it must have a velocity parallel to the tangent to the surface (this is geometry, not physics :wink:), and it may also have a rotation, both about its own axis ("rolling") and perpendicular ("leaning"). :smile:
I knew it could confuse. :smile: I can just ask you to read again patiently because i have tried to explain it to my level best. If you have specific doubts in the qn i can clear it correctly. Responding to ur post, it wouldnt bounce because its not a collision and no impulse force would act on it. Just friction, because only the base of it would be grazing over the surface(Like placing a rotating wheel on the ground and it springs forward!)
If you did understand the initial condition, what i want to know is what exactly happens when the wheel contacts the surface.(of course including a hypothetical toppling prevention mechanism)
 
  • #5
rcgldr
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The direction of friction force is opposite of the direction of velocity of the wheel with respect to the surface the wheel is sliding on. There isn't any force perpendicular to the direction of travel, so the wheel will slide in a straight line.

If the center of mass were to the "side" of the contact patch, then the wheel would yaw and curve a bit, until the wheel yawed so that the center of mass was directly in front of the contact patch.

Steering requires some means of generating a force perpendicular to the direction of travel.
 
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  • #6
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There isn't any force perpendicular to the direction of travel, so the wheel will slide in a straight line.

Steering requires some means of generating a force perpendicular to the direction of travel.
Just because there is no force perpendicular to the velocity doesn't mean the wheel will slide in a straight line. If its just a square block, what you say is true. The block will slide with reducing velocity till friction brings it to rest. But in case of the wheel, the situation is different. The geometry of the wheel is the main reason for our ability to steer.
 
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  • #7
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I'd add this comment by one of the writers of the "Bicycle and motorcycle dynamics" wikipedia article in the talk page.
Suppose that suddenly the front wheel is turned 45° with respect to the direction of travel. Then, as with a sliding block, you would suppose that the friction force vector is exactly opposite the direction of travel. As with any vector, this force vector can be decomposed into perpendicular components: one aligned with the direction the wheel is pointing, and the other aligned with the axis of the wheel. If we suppose that the wheel is free to rotate about its axis, then the component of the friction vector aligned with the direction of the wheel must go to zero (in the idealized case where there are no disipative forces, such as bearing friction or aerodynamic drag, slowing the rotation of the wheel) because any non-zero friction force would simply cause the wheel rotation rate to change and quickly become zero. The only significant non-zero component of the friction vector remaining is the one aligned with the axis of the wheel.
I think Jeff too participated in that. :)
link: http://en.wikipedia.org/wiki/Talk:Bicycle_and_motorcycle_dynamics#Turning._Explanation.3F
 
  • #8
rcgldr
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Assuming the wheel has mass, then some component of force perpendicular to the direction of travel is required for the path to deviate from a straight line, resulting in some type of curve. I assume that a single sliding, non-rotating tire isn't going to generate any signficant "side" forces regardless of it's orientation if the center of mass is aligned with the contact patch.

If the contact patch is offset from the center of mass, then side forces can be generated. For example, a car with all 4 wheels pointed straight, but with one or both left side wheels locked up and the other wheels free spinnning, will turn left and yaw as it slows down.

My participation in the wiki thread was in response as to how the side forces are generated, and I responded the forces were related to deformation (stress strain) at the contact patch. I used the term "slip angle" as a generic way to describe the difference between the geometric radius (zero deformation) and the slightly larger actual radius that occurs due to deformation, as the angle between the geometric and actual paths.
 
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  • #9
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So Jeff, you say that non-rigidity of the tires alone can explain the generation of the "side" forces? What about Andrew Dressel's explanation which i elaborated in my first post in this forum? What he says is that the component along the wheel plane vanishes to zero by making the velocity along that direction equal to R*W(omega). So we are left with only one force(perpendicular to the wheel plane) that has a component perpendicular to the intended direction of travel. So steering is still possible.
This also predicts that steering is impossible if the wheel plane is perpendicular to the initial velocity vector which is actually the case.
 
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  • #10
rcgldr
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So Jeff, you say that non-rigidity of the tires alone can explain the generation of the "side" forces?
My point was that no real object is infinitely rigid, everything has a stress versus strain relationship, and forces coexist with deformations. As mentioned above, my post was in response to a question in the wiki discussion; how a solid generates a force, in this case, how a tire generates a side force.

the component along the wheel plane vanishes to zero by making the velocity along that direction equal to R*W(omega).
Except that you post states that the wheel isn't rotating, (W = 0), so the force along the wheel plane wouldn't be zero. I'm not sure what you're getting at here.
 
  • #11
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Except that you post states that the wheel isn't rotating, (W = 0), so the force along the wheel plane wouldn't be zero. I'm not sure what you're getting at here.
Initial W=0. As the component of friction that is along the wheel plane exerts a torque on the wheel increasing the W, at one point of time V(along the wheel plane) becomes equal to R*W bringing that component of the friction to zero!.
 
  • #12
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This thread is not open just for me and Jeff really. Others can share their expertise and clear my doubt. I'd appreciate that.:smile:
 
  • #13
rcgldr
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I didn't realize you meant that the wheel was free to rotate, and that only it's initial condition was not-rotating.

There is some force in the direction of the wheel as it accelerates to W, but once there the only force in the direction of the wheel is due to rolling resitance, which can be ignored for this discussion.

So once at W, then there is no force in the direction of the wheel, only a force perpendicular to the direction of the wheel.

If you take deformation into account, then the actual path is slightly "outwards" of the direction of the wheel, and force is a tiny bit "aft" of perpendicular. This is only important for understanging how forces and deformations coexist with solid objects.

Now that I'm in sync, what is the "doubt" about steering?
 
  • #14
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So once at W, then there is no force in the direction of the wheel, only a force perpendicular to the direction of the wheel.
This force perpendicular to the plane will die out when it had consumed the momentum the wheel has in the direction opposite to it. So finally the COM of the wheel will have a velocity pointed along the wheel plane. So do you agree that EVEN if we assume rigidity, we will be able to explain steering? No slip angles. No deformations entering the scenario. :smile: Of course no real body is completely rigid.

Now that I'm in sync, what is the "doubt" about steering?
The basis of Andrew's explanation is the decomposition of the initial longitudinal friction to two components. The one along the wheel plane will modify W and V(along the wheel plane) such that V=R*W. (same story. But don't yawn!) But what if the wheel's initial W is set such that the component of velocity along the wheel plane is equal to R*W. The work of the component of the friction has already been done. So will the net friction become zero? What would happen then?
 
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  • #15
rcgldr
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This force perpendicular to the plane will die out when it had consumed the momentum the wheel has in the direction opposite to it. So finally the COM of the wheel will have a velocity pointed along the wheel plane.
This assumes that the wheels yaw axis is held. It wasn't explained how the yaw axis was held, but assuming it's fixed, then steering "stops" once the wheel direction is perpendicular to it's axis.

But what if the wheel's initial W is set such that the component of velocity along the wheel plane is equal to R*W0. The work of the component of the friction has already been done. So will the net friction become zero? What would happen then?
The wheel's rate of rotation would decelerate until it's component of surface speed in the direction of travel = R * W1. During this transition from W0 to W1, the wheel would accelerate forwards and inwards (steering).
 

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