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Direction of action of cornering force and lateral force

  1. Jun 29, 2013 #1
    I have attached a diagram(figure 1(b)) indicating the directions of the cornering force and lateral force when there is sufficient side-slip angle i.e the vehicle moving direction plane and wheel rotation plane are at a big enough angle a.k.a side-slip angle.

    From my understanding, the cornering force is a result of the tire deformation and lateral force is a result of the 'CENTRIFUGAL force'(am i right?). Then, in the figure 1(b), shouldn't the lateral force act in the opposite direction(outward direction of the turn) than what is being indicated? i.e the lateral force and cornering force should be in opposite directions but not essentially 180 degree.
    In the figure, it looks like the vehicle is traveling straight and the wheel is turned right, so my diagram would have the centrifugal force acting to the left and the cornering force to the right. But,this is not the case as given in the book 'vehicle dynamics handling' by Masato Abe.Why?

    I'm assuming: both the lateral force and cornering forces are reaction forces.

    Q2.Is it the cornering force of the lateral force that produces the self-aligning torque as a result of the asymmetrical nature of the tire deformation in the contact patch?

    Wiki:"The diagram is misleading because the reaction force would appear to be acting in the wrong direction. It is simply a matter of convention to quote positive cornering force as acting in the opposite direction to positive tire slip so that calculations are simplified, since a vehicle cornering under the influence of a cornering force to the left will generate a tire slip to the right."
     

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  3. Jun 29, 2013 #2

    Simon Bridge

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    We would usually run the problem as having the lateral static friction of the tires with the road providing centripetal force. Centrifugal force is normally used in the reference frame of the car.
     
  4. Jun 29, 2013 #3
    That just explains the occurrence of the 'lateral force'.Q1. How does the 'cornering force' arise then? Is this a frictional force also? Q2. Why do you classify the lateral force as a static frictional force?Is it because you are assuming the maximum use of traction and that the tire is just at the point of slipping?

    Q3.Why is the lateral force perpendicular to the wheel axis and NOT the 'vehicle movement direction' axis? because in fact it is a frictional force with reference to the movement direction of the car? i.e centripetal force with respect to the turning radius centre gives rise to the lateral force.
     
  5. Jun 29, 2013 #4

    Simon Bridge

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    The centripital force is the "cornering force" in the sense that it's what pushes the vehicle around the corner.

    All the forces that accelerate a vehicle (per your example) arise from friction between the tires and the ground. This should be clear from examining what pushes on what in an inertial frame.

    The rest is fine details ... btw: I am unable to resolve your attachment.
     
  6. Jun 30, 2013 #5
    Rephrased the question with a new diagram

    I understand that cornering force and lateral force(aka disturbing side force in below fig) are different things, but why does the 'disturbing side force' in the below diagram arise? Is it because of the centripetal force in this case i.e the car is moving to the right.?? If yes, why does this centripetal force act at the centre of the tire and not eccentrically like the cornering force? clip_image0063.jpg

    By analogy, a flexible rubber ball on the ground attached to a string which applies a centripetal force on it would act at the point where the contact patch has maximum deflection? or otherwise?
     
  7. Jun 30, 2013 #6

    Simon Bridge

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    Is the diagram supposed to be showing a tire (from below) while skidding?
    (The tire does not appear to be moving in the direction it is pointing.)
    I suspect the diagram will need to be understood in context of the passage in which it appears (or is fig-referenced from).

    Anyway - when a tire is rolling, the pressure across the contact area is not generally going to be uniform.

    If you are cornering in your car, and you let go of the steering wheel - what happens?
    When you turn the wheel, what happens to the lateral forces on the surface of the tire where it touches the road?

    Fs acts through the center because otherwise it would produce an additional torque turning the wheel - that torque is already accounted for elsewhere. The diagram does not show each individual contributing force, but the important results of them. One of the results is a component that acts through the center of mass.

    Note: that last sentence appears to be incomplete - does what "act"? The rubder ball? What would it mean for a rubber ball to "act at the point where the contact patch has maximum deflection" or anywhere else for that matter?
     
  8. Jun 30, 2013 #7

    Ranger Mike

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    slip angle

    Whoa up there fellows – please read post on Tire Slip Coefficient dated Apr.-10, 07:57 AM in the Mechanical Engineering forum posted here.
    Tells you about Slip Angle. Your sketch 1(b) needs a little update. Once we have the concept of slip angle and cornering force understood we can discuss in detail. Should you opt not to look it up then the most important factor regarding this subject will not be addressed. And it has not been addressed to date...Hint- quit thinking in 2D and think 3D...ifin you really want to understand what is going on read Race Car Suspension Class posts...
     

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  9. Jun 30, 2013 #8
    The paragraph from the book:The scanned copy of the page in the book is attached,figure 1(b)
    "Generally,when a vehicle is traveling in a straight line, the heading direction of the wheel coincides with the travelling direction.In other words,the wheel traveling direction is inline with the wheel rotational plane.However,when the vehicle has lateral motion and/or yaw motion,the traveling direction can be out of line with the wheel rotational plane.

    The figure corresponds to the wheel looked at from above.
    The wheel is also acted upon by a traction force if the wheel is moving the vehicle in the traveling direction,or by braking force if braking is applied. Also, a rolling resistance force is at work.If the wheel has side slip,as the figure above,a force that is perpendicular to its rotational plane is generated.This force could be regarded as a reaction force that prevents sideslip when the wheel produces a side-slip angle. This is an important force that the vehicle depends on for its independent motion.Normally,this force is called lateral force,while the component that is perpendicular to the wheel rotational plane is called cornering force.When the side-slip angle is small,the two are treated as the same."
     

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  10. Jun 30, 2013 #9

    Simon Bridge

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    Not coming up on simple searches - please provide a link.
     
  11. Jun 30, 2013 #10

    jack action

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    Referring to the drawing above:

    The disturbing side force is a force applied at the wheel axle, parallel to it. It can be caused by lateral acceleration of the vehicle, but it could also be caused by a side wind or any other force you can think of.

    If the vehicle is under a side force and the wheel is not aligned with the vehicle, then the vehicle' side force will be split into 2 forces at the tire (vector wise); One which will be the lateral component of the wheel (or disturbing side force) and the other which will be a longitudinal component of the wheel (perpendicular to the wheel axis), thus promoting acceleration or deceleration of the wheel's rotation.

    The cornering force is the lateral friction force that is the tire's reaction to that disturbing side force. Due to the wheel rotation, tire deformation is involved and rolling resistance is introduced. This causes the displacement of the cornering force with respect to the axis of rotation, creating the pneumatic trail. If there is no rotation, there is no longitudinal tire deformation, so the cornering force is directly opposed to the disturbing side force (hence, no self-aligning torque).
     
  12. Jun 30, 2013 #11
    Is there any reason why the cornering force [itex]F_c[/itex] would be perpendicular to the vehicle direction of motion and not the wheel rotation plane? Again, I would ask you to see the attached image in my post T,03:40pm. Notice that the cornering force and the disturbing side force happen to be in the same direction and each perpendicular to their respective planes!
    Q1.Can I presume the lateral acceleration is caused by the centrifugal force at the turn? In that case, I could arrive at an equation for the disturbing side force:
    $$F_s=mr\omega ^2$$
    OR
    $$F_s=side wind force in Newtons$$

    Also, the cornering force can be given by considering the tire as a spring and hence this force would be divided into 2 separate regions [itex]F_{c1}[/itex] and [itex]F_{c2}[/itex], with the rear contact patch being greater than the front contact patch for most cases. The forces in the regions are found out experimentally using the pressure distribution, which is ofcourse not symmetric.
    Cornering force per unit side slip angle is given by
    [itex]F_c=K\alpha [/itex]
    where K is called the cornering stiffness.

    Ofcourse, the maximum values of cornering force is expressed in the same way as the traction maximum:
    [itex]F_{cmax}=\mu W[/itex]

    (the author has confused the notation of the cornering force and disturbing side force).
     
  13. Jul 1, 2013 #12

    Ranger Mike

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    down force add traction during cornering

    Tires can act as springs and you just hit on the key point I was trying to make. If all we have been talking about was for a child's wagon or go kart that was simply a wagon with fixed axels, then all that has been stated about tire contact patch deformation would be true. We would be dealing with the ability of the pneumatic tire to adhere to the pavement during a turn to the point the tire could no longer “ stick” and would go into tire shear.

    When we deal with automobiles with suspension we have a new dimension. We have a lot of weight transferred during the cornering event. In a left hand turn we see sprung weight transferring to the right side, we see weight transferring to the front from the rear and we have diagonal weight coming from the left rear to the right front. This weight can be used to add down force to the proper tire and thus INCREASE the tires ability to corner. In other words, the tires tractive ability is improved by adding down force up to the point it is overcome by too much weight transferred and starts to slip.


    Go to advanced search,,type in ranger mike on User name and look up race car suspension class...it is all there.

    https://www.physicsforums.com/showthread.php?t=391693&highlight=Tire+Slip+Coefficient
     
    Last edited: Jul 1, 2013
  14. Jul 1, 2013 #13

    jack action

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    Just a precision: Weight transfer happens whether you have a suspension or not. The difference a suspension makes is that you can design your suspension such that the front-rear weight transfer can be sent to the right or left wheel and the side-to-side weight transfer can be sent to the front or to rear wheel.
     
  15. Jul 1, 2013 #14

    Ranger Mike

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    you are totally correct..in typing my reply,,i did not lay best foundation..you do in fact have weight transfer..if you could race a go kart fast enuff the two inside tires would lift..in most cases the front end washes out or the rear end comes around depending upon speed and tire grip. Good one Jack!!
    thank you for the correction..!
     
  16. Jul 2, 2013 #15
    But RangerMike,JackAction: The posts you have suggested I read don't make a distinction between 'lateral force' and 'cornering force', which is what got me all confused-up in the first place. If you could go through the image I attached in the post( Jun30-13, 03:40 PM ), you will see how the author has vectored the forces and that they are infact at an angle to each other. So,in a gist,my question was dealing with the distinction between the 'lateral force' and 'cornering force' and why they are at such angles.
    P.S: Automotive engineers have just muddied up the terms dealing with lateral dynamics. Very confusing for a student to read-first.
     
  17. Jul 2, 2013 #16

    Ranger Mike

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    The Tire traction circle

    I do not worry about vectors and terminology too much because the tires don’t. Tires don’t care in which direction they generate force- lateral, longitudinal or a combination of both.
    Let us look at the Traction Circle. Traction is non directional. Tires provide a given amount of grip (traction) at the limit of adhesion. And tires do not care which direction traction is being applied as long as this limit is not exceeded. If this fact is represented by the traction circle, the circumference of the circle is this limit of traction. It relates to a race cars path through a turn. This limit can be shown as a radius of the circle drawn in any direction. Using the clock method, 12 o'clock is 100% acceleration and 6 o'clock is 100% braking. The 3 and 9 o'clock vectors are 100 percent lateral acceleration during cornering to the right or left. Radius at 7 o'clock is combination of the two where 1/3 of the force is for cornering and 2/3 of the force is for braking.

    From point A to E, both braking and cornering forces are taking place. At F, all traction is being used for cornering. From G to K the car is accelerating with less and less traction used for cornering. At L, all traction is used to accelerate. After point L, the graph drops to zero until the car begins its cornering again. Any point on the plot not falling on the vertical or horizontal axis is a resultant force of both vertical and horizontal components ( lateral and straight line acceleration/deceleration).

    Lets get in our Go Kart. The one with solid axels, West Bend 2 cycle engine and twin tilotson carbs. If we race around the traffic cones we placed at the Wal-Mart parking lot we feel force trying to lift the inside (left) tires and plant the outside tires. If we can run fast enough and the tires are sticky enough to keep contact, eventually we would lift both left side tires for 100 percent weight transfer of UNSPRUNG Weight.

    We are dealing with two kinds of weight regarding weight transfer. One is lateral ( sideways) weight do to cornering force and the other is weight transferred due to body roll. In this case it’s the 200 pound fat kid from next door, driving the cart . Put down the Big Mac , Lumpy! Keep in mind this total weight will be transferred regardless of the cars suspension design.

    The tire is in contact with the track and pushes back on both of these transferred weights. The tire sidewall acts like a spring to a small degree to soften the load being dumped on the tire. Weight is transferred to the pavement until such time as the tire contact patch can not longer maintain adhesion. When this limit is exceeded we have the tire sliding on the pavement. The tire pyrometer results will show this tire temperature exceeds the other tires temperatures by a large degree.

    This weight transfer depends on the radius and speed your running, and your track width and center of gravity.

    Now lets modify the go kart frame to have a center pivot for the front axle ( the tube connecting both front spindles). If we modify the rear of the go kart frame to have a center pivot for the rear tube and mount the drive axle so it will work properly and if both pivot points are the same height, we have the essential of our modern race car chassis. One problem. The chassis will flop to one side or the other due to gravity. By placing coil springs on each side of the pivot points front and rear we can
    “suspend” this arrangement so the chassis is centered.

    If we repeat the cone course test we note that we can attain a bit more speed before both left side tires lift. The springs dampen the fat kid weight transfer to a slight degree by increasing the time it takes for the transferred weight to finally cause tire shear. The suspension effects the percent of load the front takes versus the rear. All Weight transferred is the exact same amount regardless of presence of the suspension or not. Suspensions dictate where a certain percentage of the fat kid weight is going ( front or rear).

    Lets really mess things up and put a heavier spring on the right side of the front of our go kart and try the cone killer course.
    We find that we now have an under steer condition as the right front tire wants to wash out or snow plow when we get going fast. This is because the right front spring is resisting the fat kid transfer of weight MORE THAN the right rear which has had more time ( as miniscule as it is ) to accept the weight transfer due to the springs coil compression. The right front ( compared to the right rear) is acting like a solid link and only the tires side walls are dampening the impact on the weight loading the tire contact patch.

    We could go on with more examples like moving the rear pivot point higher ( ROLL CENTER if you have not figured this out by now).

    By moving the Roll Center “ higher” (relative to the track surface) and assuming the Center of Gravity (CG) is above both ft. and rear RC, we have shortened the lever arm that is the distance between the CG and RC. This means you now have a shorter moment arm to counter fat kid weight transfer on that end of the car. Simply put, less load transfer will be going thru the coil spring and more fat kid weight be going straight to the tire ( same as original unspring cart example). The higher RC will wash out first since the rear tire is overloaded more than the front tire which had a longer moment arm and could better control the fat kid weight.

    This is a simplified version of weight transfer and we did not even get into diagonal weight transfer, sprung vs. unstrung weight, all the tech..was just my way of trying to explain why you lower spring rate when you have excessive tire temp on one end of the car...

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  18. Jul 2, 2013 #17

    jack action

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    attachment.php?attachmentid=59983&d=1372606543.jpg

    If you read carefully the text with the image you're talking about, it says that the cornering force is a component of the lateral force taken in the direction perpendicular to the wheel moving direction, i.e. considering the side-slip angle (I know you already noticed the mistake of the author confusing moving direction and rotation plane).

    So the «real» force is the lateral force. «Lateral» means with respect to the tire rotation plane. It can be divided into components that you can call whatever you like. Since the moving direction is determined by the side-slip angle, the component of the lateral force perpendicular to the moving direction is then dedicated purely to «cornering», thus the name cornering force. IMHO, it is of little value in evaluating performance. Maybe tire engineers have more use for it when trying to built better tires.
     
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