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Tire Slip Coefficient

  1. Apr 2, 2010 #1
    Im working on designing a caster wheel and have a mathematical model that Im following. However im stuck with how to determine the "tire slip coefficient". Its measured in Newtons. At first I thought its a coefficient so normally would be unitless but only a parameter with Newtons satisfies the mathematical model. If i dont have newtons the formulae is not dimensionly consistent.
    Could anyone help me out with what a tire slip coefficient is?
    How do we calculate it?.
    Its my final parameter for my design.
    P.S: The model is taken from a reliable source and has been patent in 2001.

  2. jcsd
  3. Apr 2, 2010 #2

    Ranger Mike

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    see my post Race car physics
    Dec29-08, 11:31 AM
  4. Apr 2, 2010 #3
    i did go through your post . So your saying this has links to longitudinal and lateral forces. hmm the formulae that your using:
    Force = acceleration * weight* ch height/ wheel base

    Question how valid is this formulae cause this isnt dimensionly consistent as left hand side has units of N and the right hand side does not reduce to si units of newtons?
    And how does this link to tire slip coefficient?

  5. Apr 2, 2010 #4
    Weight = mass

    A common colloquial switching of words.
  6. Apr 2, 2010 #5
    acceleration = m/s^2. this wont cancel out
  7. Apr 2, 2010 #6
    What is newtons second law?

    This is simply

    F=ma * a ratio of cg height to wheelbase.
  8. Apr 2, 2010 #7
    ooh ryte. but how does ths link to tire slip coefficient?
  9. Apr 2, 2010 #8

    jack action

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    The theory for vehicle tire lateral force capabilities is as follow:

    When subjected to a lateral force, the contact patch of the tire will deform itself in a way such that, view from the top, it makes an angle with the tire's vertical plane (imagine the threads twisting at the contact patch), That angle is known as the slip angle.

    It was found that the lateral force ([tex]F_{L}[/tex]) produced by a tire is proportional to its slip angle ([tex]\alpha[/tex]), hence:

    [tex]F_{L} = C_{\alpha}\alpha[/tex]

    Where [tex]C_{\alpha}[/tex] is the tire slip coefficient. Obviously, the proper SI unit is not Newtons, but Newtons per Radian.

    Only the tire manufacturer could give you such an information as it takes special equipment to measure that ... and they usually don't like to share.

    EDIT: I checked back my info and the coefficient I gave you is called the "cornering stiffness". But I can't think of anything else that would have Newtons for a unit.

    This coefficient is also proportional to the vertical force ([tex]F_v[/tex]) applied to the tire, such that:

    [tex]C_{\alpha}=C_s F_v[/tex]

    Where [tex]C_s[/tex] is the cornering stiffness coefficient and the SI unit is Radian-1. http://zzyzxmotorsports.com/library/understanding-parameters-influencing-tire-modeling.pdf" says that typical values are 0.12 deg-1 for bias-ply tires and 0.16 deg-1 for radial tires.
    Last edited by a moderator: Apr 25, 2017
  10. Apr 2, 2010 #9
    wow thats some new information man. thanks a lot. Im screwed basically cause i cannot get my design completed without that value!.
  11. Apr 2, 2010 #10
    Dunno, can't acutally remember a bean about suspension geometry and tire dynamics. I need to reread that thread myself or dig out milliken.

    Mike's the man with the answers for questions like this.
  12. Apr 2, 2010 #11
    the tire slip coefficient (TSC) is a unitless coefficient, just like the static or sliding friction coefficient. For tires used in drag racing, TSC can be greater than 1. If a vehicle has a vertical weight Wtire= m·g Newtons on a tire, then the maximum possible horizontal traction force Froad= TSC·m·g Newtons. If the tire radius is R meters, then the maximum torque is T = R·Froad Newton-meters. The maximum power before slipping is then

    P = 2 pi RPM·T/60 watts = 2 pi RPM·T/(60·746) HP

    Use metric units to ensure correct numerical results.

    Bob S
  13. Apr 4, 2010 #12

    Ranger Mike

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    Hi All..Just got back from Easter visit..will assist where i can ref: tires..i need to reread some stuff and will post later..
    thanks and Happy Easter..hunt up some fun!

    Attached Files:

  14. Apr 5, 2010 #13
    hmmm. So I did go through that reference Jack.
    The cornering stiffness would be the same as tire slip coefficient but the problem is ves can we measure the lateral force to determine the tire slip coefficient. The reference gives range of values for the cornering stiffness coefficient which is not the same as tire slip coefficient.

    Are there any range of values for the cornering stiffness i can play with or how would i measure the lateral force?
  15. Apr 5, 2010 #14
    Bob your theory is interesting as well . but it does not solve the problem about how to get the tire slip coefficient. I get its definition now, but its not unitless thats what i have learned. It has units of N/ rad but since radians are unitless it just has N as units. Jack gave good insight into this.
  16. Apr 5, 2010 #15

    Ranger Mike

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    see Gen Physics post - which car will slide first - Jan 27-09 11:33 Jeff Ried helped me a lot on this slip question
    see Gen Physics post - Race Car physics - dated Dec31-08 10:44 AM
    see Gen Physics post - circular motion of car on banked track, with friction Dec28-08 11:34 PM Doc Al was a good mentor on this post..thanks Doc.
    The answer is in these posts...If still unclear, I will try to assist you in finding it..thanks
  17. Apr 5, 2010 #16

    jack action

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    What Bob has defined is the friction coefficient of the tire and, yes, it is unitless. That's why I did not suggest that parameter.

    I never heard the term "Tire Slip Coefficient". Maybe you could show us an equation where it is used, so that we can better answer your question.
  18. Apr 5, 2010 #17
    the equation used is as follows:
    Vc = [tex]\frac{Cd*T}{2*Iw}[/tex] + 0.5*[tex]\sqrt{[Cd*T/Iw]^{2} +(2B^{2}K_{s}T)/Iw}[/tex]

    Vc, Shimmy speed (m/s)
    T, caster offset distance(m)
    Iw, mass moment of inertia(kg.m[tex]^{2}[/tex]
    Cd, torsional Damping (N-s-m)
    B, Tread Distance(m)
    Ks, Tire Slip Coefficient (N)

    Reference: US Patent 4989920 - Self-damping caster wheel by James J. Kauzlarich
    http://www.patentstorm.us/patents/pdfs/patent_id/4989920.html" [Broken]
    Last edited by a moderator: May 4, 2017
  19. Apr 5, 2010 #18
    I did go through these posts Mike. but i still cannot find a solution to my problem. What would be an appropriate value for the tire slip coefficient. I dont know how to find the lateral force on the tire because i dont have enough information unfortunately.
  20. Apr 6, 2010 #19

    Ranger Mike

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    What would be an appropriate value for the tire slip coefficient. I dont know how to find the lateral force on the tire because i dont have enough information unfortunately.

    this is a two part question. the lateral force can be found at post I did in Introductory Physics forum - Calculating G Forces or the Centripical force -Dec27-08, 01:43 AM

    tire slip coefficient - Race car tires were invented to contradict Newton's Laws of Friction. Tires can and do generate forces greater than the load applied to it. It can develop acceleratetive force, decellarative force and side force. In the case of combined lateral and longitudinal force, the sum can be considerably greater than the maximum force that can be developed in any one direction.
    It is important to realize that the coefficient of friction is DIMENSIONLESS. It is an indication of the maximum force which can be developed by one tire when compared to another tire under the same conditions. Tire manufactures use this as ONE consideration ( of many) when designing tires relative to performance and handing.

    It is important to remember that the force that can be developed by any one tire is the product of the instantaneous vertical load applied to the tire and the tires maximum coefficient of friction under the existing conditions. Both these factors change constantly with variations of road speed, load transfer, track conditions, tire temperature, and a bunch more stuff. In the lateral sense, we refer to this generated fo

    Cornering Power is centrifugal acceleration capability or in a more scientific sense Friction Force (Ff). Friction force equals the Coefficient of Friction (Cf) times the vertical load or force (Fv). Ff=CfFv.
    Tire manufacturers do not give out raw data in pounds-force as the units for lateral force so Gs is a convenient unit to show important trends without specific data. rce as the tires CORNERING POWER which is just the old centrifugal acceleration capability.

    In the longintudinal sense we call it TRACTION CAPACITY. For this discussion we assume the tires traction capacity to be equal in both directions.
    Slip angle of a pneumatic tire is defined as ' the angular displacement between the plane of rotation of the wheel ( direction the rim is pointing) and the path that the rolling tire will follow on the track surface". This path is made up of successive footprints of the contact patch laid down as the tire rolls..( any questions on contact patch?)
    In order for the vehicle to change direction, regardless of speed or track banking ( curvature) each of the vehicle's tires must assume some value of SLIP ANGLE.

    How and why the tires responds to changes in chassis geometry is a whole nother story so we will skip the part about sidewall distortion, tire compond, tire grip, and all that to address your question - What would be an appropriate value for the tire slip coefficient.

    The appropriate value would be the value of the tire that provided the maximum cornering power for that particular road surface or track, with that particular vehicle weight and suspension setting at that temperature at that time of day.

    Racing is about compromise. Can you calculate the appropriate value for the tire slip angle coefficient mathematically without knowing the tires composition, content of rubber/synthetic material, side all construction (bias vs. radial) belt material. steel, nylon, Kelvar, sidewall distortion, heat fade, tread construction if any, compound hardness ( durometer).
    Also vertical load must be know and to do this you need to know total vehicle weight, percent sprung to unsprung weight, weight transfer percentage versus vehicle speed, degree of track banking ( in cornering as well) weight jacking effect, camber change under load, bump steer impact on toe - out. Caster change, spring deflection under load, control arm impact on motion rate and much more..then throw in the track itself...current temperature, effect of sunlight on track surface, track surface roughness, composition ( concrete and asphalt or all asphalt).
    My recommendation is to look up calculating Gs on a skid pad..see above first sentence I wrote. Look over the generic graph attached. Note the passenger car tire works pretty good up to 4 degrees slip angle and to .8 Gs and is generally the results you see in road tests of street cars on a skid pad. You may get to 1 g with a hot corvette and the max you should see with a full blown formula car is 1.4 g without a lot of aero added to it.

    frasool - pls. let me know if this helped..confused ya or just muddied up things and I will try again

    Attached Files:

  21. Apr 6, 2010 #20

    jack action

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    I couldn't check your link because we have to register to access it, but this has nothing to do with traditional car tire definitions.

    After studying your equation, I think the tire slip coefficient unit is really only N. Worst, if we assume that the term [tex](2B^{2}K_{s}T)/Iw[/tex] = (angular velocity X radius)²

    Then the tire slip coefficient has the unit of N.rad². And I have no idea what it means.

    I think that this last term is related to the self-aligning torque, but that is just a guess.
    Last edited by a moderator: May 4, 2017
  22. Apr 6, 2010 #21
  23. Apr 6, 2010 #22
    Yeah you have to register for it. Just so you know its a free registration :p.
    Moving on, I did check the dimensionless nature of the equation and only Newtons fits in here. Self aligning torque would be something required for the braking purposes . i dont think it would be used in this equation as the author of this paper suggests, Ks= predetermined tire slip coefficient!
  24. Apr 7, 2010 #23

    jack action

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    The self-aligning torque has nothing to do with braking, it is the force that tends to re-align the tire when subjected to a lateral force. And I think it would be related to the equation you gave us, which determines the shimmy speed of a caster wheel.

    Actually, this term:


    is easily understood as:

    [tex]Cd = \frac{torque}{angular \ velocity}[/tex]

    [tex]Iw = \frac{torque}{angular \ acceleration}[/tex]

    [tex]T = arm \ length[/tex]


    [tex]\frac{Cd*T}{Iw}=\frac{\frac{torque}{angular \ velocity}distance}{ \frac{torque}{angular \ acceleration}}=speed[/tex]
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