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Thread: Extrapolating Load sensitivity characteristics

  1. #11
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    Example of this processing on TTC Forum

    Engineering, not wrenching.

    New School.

  2. #12
    Quote Originally Posted by Zac C View Post
    There's a stupid easy way for you to learn about what's going on first hand. Go get yourself a mountain bike, take the front pressure down to about 5 psi, and throw it into a corner (helmet and mouth guard are recommended for this experiment). You'll get a nice physical sensation and view of the tire rolling over under cornering force, before (probably) hitting the deck.
    I have clear this problem, but it is related, as you say, to pressure.
    With more load it is possible to increase the tyre pressure (and perhaps camber) in order to keep it solid and in a correct position relatively to the ground.

    However, after correcting any stability and contact patch difference with this kind of settings, the more loaded tyre will still have a lower friction coefficient than the lighter loaded tyre. Why is this?

  3. #13
    Ehm, load sensitivity? Or you mean the physics behind load sensitivity?

  4. #14
    Quote Originally Posted by mech5496 View Post
    Ehm, load sensitivity? Or you mean the physics behind load sensitivity?
    load sensitivity is just a name, it's not a reason:

    Quote Originally Posted by Fra881 View Post
    Can you please elaborate more on the tyre overload study with Mx, and on the physical reasons of load sensitivity?

  5. #15
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    Where the Rubbers Meet the Road.

    The Chemistry and Physics of micro and macro adhesion between two surfaces.

  6. #16
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    Smile Another possibility

    If you don't care for that explanation, think of it as 'ROUNDOFF ERROR".

  7. #17
    Quote Originally Posted by Fra881 View Post
    I have clear this problem, but it is related, as you say, to pressure.
    With more load it is possible to increase the tyre pressure (and perhaps camber) in order to keep it solid and in a correct position relatively to the ground.

    However, after correcting any stability and contact patch difference with this kind of settings, the more loaded tyre will still have a lower friction coefficient than the lighter loaded tyre. Why is this?
    As a good insulator, rubber doesn't have any coulombs.

  8. #18
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    admission

    That's why you have two tires: One from coulomb A and one from coulomb B.

  9. #19
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    Jason, Fra881,

    Here is my dumbed-down, big-picture, explanation of Tyre-Load-Sensitivity.

    Firstly, let's define:
    Coefficient-of-Friction = Force-Tangential/Force-Normal,
    or Mu = Ft/Fn, with (Ft = Fx,y, Fn = Fz) in the usual car terminology.

    This Mu = Ft/Fn equation is so beloved by Physicists that many of them consider it to be a fundamental, unbreakable, Law of Nature. Truth is, it is only a rough and ready "Rule of Thumb". It applies reasonably well to contact between HARD and SMOOTH surfaces, but only over a VERY SMALL RANGE OF FORCES.

    At very low levels of Fn and Ft the equation fails. This is typically because of the "Van der Waal's forces" (ie. the weak electrostatic forces between molecules of the two surfaces) that allow flies and geckos to walk on walls and ceilings. Similarly, a small piece of sticky tyre rubber will easily cling to a vertical wall (ie. Fn = 0, so Mu = Ft/0 = INFINITY!), or even to a ceiling (ie. Fn = -ve, so Mu = -ve!).

    At the opposite extreme of forces the problem becomes one of "Strength of Materials". For a given cross-sectional contact-area, say, the area of your tyreprint, as the Fn and Ft forces increase the various compressive and shear stresses also increase, and eventually something breaks. In the case of your tyre rubber it is its shear strength that gives up first, so Ft necessarily has a maximum level. (Technically, shear is compression + tension, both at 45 degrees to the Ft loads, and ultimately the rubber fails in tension, though after much stretching.)

    Anyway, consider the typical FSAE tyreprint area, about as big as your hand. In the miniscule range of shear forces from about Ft = 10 N to say 1,000 N (1 -100 kg) the "Friction Equation" holds reasonably well. But down around 0.001 N (tiny), or up around 10 MN (1,000 tons!) you should not put any trust in "Mu".

    Looking at it a slightly different way, for a given tyreprint area you can do multiple "Mu tests" and then plot Ft on a vertical axis against Fn on horizontal axis. The slope of the line from origin to a point on the curve is Mu. But from the above "Strength of Materials" considerations it should be obvious that Ft has a definite upper limit, or ceiling, above which the curve cannot go. This limit is the "ultimate shear strength" of the whole tyreprint. In practice, this limit is approached gradually as little bits of the tyreprint fail one after the other.

    Interestingly, it is not always the "tyre" that does the failing. Off-road tyres, such as on farm tractors, have hard "knobs" that embed themselves into the softer soil. Bulldozers have steel "cleats" that do the same thing. As the Ft loads increase, it is the soil that first starts to fail in shear. As all farmers know, more Fn (ie. Fz) helps a little, but NOT much. The best way to get more Ft (ie. drawbar pull) is to fit wider tyres, because these provide more tyre/soil AREA to carry the shear loads.

    Similar applies in FS/FSAE.

    Z

  10. #20
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    Scrub a Dub-Dub

    From some older (Round 3? Data). This is at a constant, moderate vertical load, which is not the way a tire sees a cornering scenario. The offset at zero slip angle is typical of some classes of tires (as in 'race' tires). What happens under traction and braking is another thing to contemplate over a few pints.
    So, if you are using scrub radius values in any of your hand-waving design reviews, this result changes everything. I suppose I should have reversed the X axis direction. Rights turns are to the left. Oops...

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