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Thread: Calculating tire forces under combined braking and lateral acceleration for 1 wheel

  1. #11
    Edward,

    thanks for your valuable input. It is great to have this forum to get your questions answered.
    It has been a great resource for the thesis for a first year team.
    Somebody suggested in a former thread, when I was still determining what to write about in the thesis, to
    write a sort of "handbook" for a team at the old university to follow along and to get good inputs. I followed that route
    with a concentration on the suspension.

    It is indeed the final chapter I had to finish. All the others have been finished before!

    For the case of pure braking:

    If we take 1,5 G and assume a mu of 0,9 in the longitudinal direction (which I recon is a little bit on the low side for a FS tire) the front axle
    would experience a vertical loading of 239,4 kg = 2348,5 N. With a coefficent of 0,9
    that would give me a braking force of 2113,6 N on the front axle. Making it 1056,7 N at one wheel.

    As for pure cornering at 1,5 G:
    Taking Wl=W/2+(W*Ay*h)/t

    Wl= 148,84 kg/2 + (148,8 kg * 1,5 G * 0,3) = 141,38 kg = 1386,94 N vertical load. Higher than
    just braking!
    In the lateral direction we have to look at the centrifugal force. The centrifugal force acts on
    the CG towards the outside and both tires have to counteract that force.
    The outer tire takes more load because of the weight transfer. The higher the CG, the higher the weight transfer.
    1,5 G * 148,8 kg * 9,81 m/sē = 2189,59 N

    Lateral force is not linearly increasing with increased wheel load. Fy= Fz*mu is not entirely
    correct for tires, but I will use it here:
    Right tire:
    1291,29 N*0,9 = 1162,16 N lateral force

    Left tire:
    168,43 N*0,9 = 151,587 N lateral force

    combining these two give 1313,747 N lateral force that the axle can produce. Countering the 2189,59 N under the 1,5G lat. acc
    the tire with that mu won't even go there and sliding before.
    As I have said the mu in reality is probably higher than that.
    I've seen diagrams with 2,47 mu lateral friction coefficient at 1000 N Fz.
    So this would become more a question of not tipping the car

    So concluding the findings out of this single investigation, my value I have now found for Fy are now higher
    than in the last post. The vertical force is higher in the combined case.

    What I really need, as you have said Edward, are some realistic numbers for the coefficient of friction for the tire.
    But soon this will clear up
    Last edited by MP4/4; 03-23-2014 at 05:41 PM.
    Suspension Engineer
    University Of Stuttgart

  2. #12
    MP4/4

    Think the car is just a big mass on one tire.

    F = M*A (whether it is Fy - M * Ay or Fx = M * Ax) and at the same time Fy - Fz * Muy or Fx = Fz Mux.

    Unless you have aerodynamic downforce (which you did not stated) how can you have 1.5 G with a coefficient of friction of 0.9?

    If you think 4 tires and if you want to make sure your calculations are correct you will have to make a "sanity" check to be sure that the SUM of all Fy on the 4 tires = M * Ay and that the SUM of all Fx on the 4 tires = m * AX

    FYI 0.8 to 1.1 is the coefficient of friction of a good passenger car tire on dry, clean asphalt. 1.4 to 1.8 is the coefficient of friction of a warm slick tire on dry, clean asphalt
    Claude Rouelle
    OptimumG president
    Vehicle Dynamics & Race Car Engineering
    Training / Consulting / Simulation Software
    FS & FSAE design judge USA / Canada / UK / Germany / Spain / Italy / China / Brazil / Australia
    [url]www.optimumg.com[/u

  3. #13
    MP4/4,

    Going back to my original reply, if you assume 0.9 for mu on your most heavily loaded tire, what mu will be required from your other tires to achieve 1.5g braking and cornering? I think you'll get some unrealistically large mu values to counter your unrealistically small 0.9. For example, your original straightline braking calculations require the rear tire mu to be about 3.0. Having done that calculation you can ask yourself if that makes sense (the answer is "no". Not if you're assuming mu = 0.9 for the front tires--they're not *that* load sensitive!)

    And as Claude said, if all the tires were at mu=0.9 you couldn't do 1.5g.

    Claude's comments are perhaps a clearer version of my encouragement to make sure you aren't "losing" any load and to think about all four tires as a system, even though you are doing detail analysis only one. In the end this is the level I want you to get to. Once you see what's happening at all four tires you'll have a much easier time deciding whether or not you have a reasonable condition for your analysis.

    FSAE tires are at the grippy end of the spectrum. Doubling your mu assumption would not be a bad place to start. Talk with your Stuttgart teammates, review their data, poke at the TTC data and then you'll make an informed decision about your test case.
    Dr. Edward M. Kasprzak
    President: EMK Vehicle Dynamics, LLC
    Associate: Milliken Research Associates, Inc.
    Co-Director: FSAE Tire Test Consortium
    Lecturer: SAE Industrial Lecture Program
    FSAE Design Judge

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