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Force exerted by air in an inflated unloaded tire

  1. Oct 6, 2016 #1
    Hello,
    Im asking myself how to calculate the force exerted by air in a inflated unloaded tire. We now that a tire can be modeled as a spring and a damper, with vertical deflection used as spring compression and deflection velocity as damper velocity.
    But how about a tire with no deflection (unloaded)? the force exerted by air is not zero. As far as I know it should be pressure x area, but how to calc the area where air pressure exerts its force?
     
  2. jcsd
  3. Oct 6, 2016 #2

    phinds

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    Did you even bother to do ANY research on this? A 3 second Google search gives the answer:

    http://www.had2know.com/academics/torus-volume-surface-area-calculator.html
     
  4. Oct 6, 2016 #3
    Hello phinds,
    thank you for your reply, even if its a bit rough :)
    Of course I searched online before posting here, but i just found a link that explain that force is equal to pressure x area (wrong search).
    Anyway, if I get it right, on all points of the tire (both internal and external) air pressure will exert a force given by pressure x torus area. Is this correct?
     
  5. Oct 6, 2016 #4

    phinds

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    As indicated by the shading in quote of your question, I was clearly answering your question "how to calc the area", which is why it appeared to me (and still does) that you had done no research on that part of the question.

    I think so.
     
  6. Oct 6, 2016 #5

    Nidum

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    I don't understand the question . The air pressure is acting on all the internal surface area of the tyre .
     
  7. Oct 6, 2016 #6
    phinds,
    I didnt make any research on that part of the question cause i didnt understand that (as for you reply) it was so simple.
    Anyway, I searched the exact meaning of pressure and, quoting wikipedia, "is the force applied perpendicular to the surface of an object per unit area over which that force is distributed". So, for example, 200 kpa means that for every square meters of surface, the air will exert a force of 200 kN.
    In my case, since the tire external radius is 0.33 meters and tire internal radius is 0.2 meters, the surface area should be, from the formula PI*PI*(externalRadius*externalRadius - internalRadius*internalRadius), 0.68 meters.
    This means that, if the tire is inflated at 200 kpa, air pressure will exert, on every point of the tire, a force of 200000*0.68 N = 136000 N.
    This seems to me a bit too much, cause in my simulation the force exerted by the tire (inflated at 200 kpa) under a load of 375 kg (1500/4) is around 4700 N, with a vertical deflection of 0.034 meters.
    What is is wrong here?
     
  8. Oct 6, 2016 #7

    Nidum

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    The tyre is resisting the internal air pressure in essentially the same way as a boiler or a balloon . The internal pressure is balanced by stresses in the tyre material .

    The presence of the hub complicates the actual analysis but not the principles involved .
     
  9. Oct 7, 2016 #8

    Nidum

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    What is the real problem that you are trying to solve ?
     
  10. Oct 7, 2016 #9
    So this means that my calcs are correct? It seems a bit strange that force exerted only by air is much higher than force exerted by air under a load.
     
  11. Oct 7, 2016 #10

    russ_watters

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    It isn't. You are doing a totally different....and I suspect meaningless calculation.

    When a tire is supporting a car, the supported weight is pressure times contact area. But what about the rest of the tire? It is still under pressure, right? So why would you include it in the unloaded case but not the loaded case? And what does the calculation tell you? (Not a trick question)
     
  12. Oct 7, 2016 #11
    Russ,
    what Im trying to implement is a "physical" simulation of an inflated tire using a 3D engine.
    I mean, till now I modeled the tire as a spring and a damper. Spring force is proportional to tire deflection and damper force is proportional to deflection velocity.
    So when there is no deflection, the force exerted by the tire is zero (or better, the force exerted by the air is zero).
    Now I added a sort of "physical" layer to the simulation using several rigidbodies connected radially to the hub with a spring and a damper. When there is a deflection the calculation of the force exerted is the same, but what about the case with no deflection? how to calc the force exerted on every single rigidbody?
     
  13. Oct 7, 2016 #12

    russ_watters

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    How about this: model it as a collection of springs with wires limiting their maximum elongation, making them compressed even when the tire isn't touching the ground?
     
  14. Oct 7, 2016 #13
    If the tire is not accelerating (in the unloaded case), then the net pressure force on the tire is zero. The compressed air inside exerts a pressure force per unit area perpendicular to the internal surface at each location on the surface. You need to add up these forces vectorially to get the total net force that the internal pressure exerts on the contour surface. Since the inflated tire is axi-symmetric, this will yield a net force of zero. The rim also exerts a force per unit length on the tire contour at the bead, but this is axi-symmetric also, and its net force is zero. You need to pay attention to what Nidum said in post #7.
     
  15. Oct 8, 2016 #14
    Yes but how to calc this force? Should I use the area of the rigidbody instead of the total area of the torus?

    This is interesting, cause im adding the force to each rigidbody (in radial direction), and the rigidbodies are places radially around the rim, so theoretically the sum of the forces should be zero (when the tire is not loaded). Strangely enough,this doesn't seem the case, cause the tire start oscillating (slowly) like a pendulum especially with an high pressure setting. Probably there is something related to gravity force (if I disable gravity, the problem doesnt seem to be present anymore), or some numerical error (the problem is proportional to physic timestep).
    Also, Im applying the same force (with opposite direction) to the rim too, in order to apply the force to the vehicle too. Is this correct?
     
  16. Oct 8, 2016 #15
    I think you are tremendously underestimating the complexity of the analysis that is required. Here is a reference to a paper I wrote several years ago on the mechanics of a radial tire under pure inflation loading:

    Miller, C., Popper, P., Gilmour, P.W., and Schaffers, W.J., Textile Mechanics Model of a Pneumatic Tire, Tire Science and Technology, 13, 4, 187-226 (1985)

    See the Appendix for the detailed tire mechanics analysis.

    Chet
     
  17. Oct 9, 2016 #16

    Nidum

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    I've read Chestermiller's paper before while researching corded reinforcement for an entirely different application .

    You can find a copy here : http://tiresciencetechnology.org/doi/pdf/10.2346/1.2150994

    You'll have to pay $25 to read it but that's really a very small fee for a lot of useful information .
     
  18. Oct 9, 2016 #17

    Nidum

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    There is incidentally a vast amount of published research data on vehicle dynamics generally . Many books and papers about detail topics and many more about whole vehicle response .

    There is also a fair amount of vehicle dynamics modelling software .

    As well as the application specific software there is useful general purpose finite element and dynamic modelling software .

    A period of intensive research into what existing resources are available could make completing your project much easier for you .
     
    Last edited: Oct 9, 2016
  19. Oct 9, 2016 #18
    Chestermiller,
    thank you for your reply.
    Im simplify this complex problem cause my goal is to make a "simple" model suitable for real time games and simulation. No claiming to make an accurate and complex model.
    Anyway, what do you think about this approach (using Casing Tension)?

    http://flocycling.blogspot.it/2014/11/flo-cycling-why-do-you-use-less-tire.html
     
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