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Frequency of a cars tire?

  1. Jun 15, 2010 #1
    So I'm trying to determine the frequency of a vehicles tire which is 3 ft in diameter, traveling 65 mph.

    This is what I have determined so far.
    r = 1.5 ft v = 65mph = 95.34 ft/s

    circumference = 2*pi*r = 9.42 ft

    T = distance/speed = 9.42/95.34 = .0988 so f should be the inverse of that right

    so does f = 10.12 hz or rev/s

    i found omega to be 607.26 RPM or 63.6 rad/s Am I missing anything. Or just making this too complicated.

    suggestions would be appreciated.
     
  2. jcsd
  3. Jun 15, 2010 #2
    My final goal is to try and graph a wave demonstrating what a given point on that tire goes through under the above circumstances.

    Where the wave is deformed by the pressure which acts on that point.

    So whenever the tire is in contact with the ground it will have a peak, and when it is farthest from that point it will trough.
     
  4. Jun 15, 2010 #3

    sophiecentaur

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    The peripheral speed of the deformation due to the road will be quite a bit less than the natural wavespeed of a wave travelling around the periphery under its own steam, I think. It will definitely be lower than the speed of any wave set up in the air which is in the tyre (speed of sound in air, I reckon).
    As for the actual shape of the wave, I think it is more likely to be a narrow pulse, just a bit wider than the footprint, rather than a sine-like shape. i.e. the majority of the deformation will be very local to the footprint. This will be a result of the high modulus and low mass of the tyre material.
     
  5. Jun 15, 2010 #4
    Trying to understand what you said in the first sentence.
    why woud the deformation be less, when the force of the car is acting upon the tire, rather than just rotating in the air.

    Would a narrow pulse be like that of a heart beat? Thats what I would expect the wave to be most similar to.
     
  6. Jun 15, 2010 #5

    sophiecentaur

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    More or less, yes. I mean that the shape of the tyre will very soon recover to its undisturbed (circular) shape after that portion leaves the road. So it's not so much a "peak and a trough" as a small region of displacement.
    When you use a term like 'wave', it implies, to me, energy travelling through a medium as opposed to, in this case, an impressed force causing an 'unsustained' wave. I guess what I am saying is that a wave, as such, only exists in the region of the footprint- and attenuates very soon in front and behind the point of contact.

    One time I did come across the idea of waves and tyres was in a discussion about maximum allowable speeds of heavy trucks on roads. The idea was that the truck sets up a slight surface wave on the tarmac and extra damage soon occurs when the rear wheel are going fast enough to catch up the wave launched by the front wheels, causing extra net deformation of the road surface.

    I think that a good approach to use for (at least a part of) your investigation could just use the deformation of each point on the tyre and exploring the energy loss due to the hysteresis of the tyre material. The 'area under the curve' would represent the lost energy and the rate of wheel rotation would give you the power dissipated. There's friction with the road, too, where the tyre surface changes shape when it hits the road, of course. . . .
     
  7. Jun 15, 2010 #6
    I have found the area under the curve in in^2, and also the rate of the wheel rotation. But what units should that be in?

    And how do you correlate the two to find the power dissipated.

    I also don't even now where to begin with the hysteresis, I'm sure I would need to find some coefficient for the material. Lost on that aspect.
     
  8. Jun 15, 2010 #7

    sophiecentaur

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    I'm afraid that Imperial Units bring me out in spots but if you change everything to feet and pounds (poundals? perhaps) you should get the foot pounds and Horsepower (OMG).
    You would need to measure the actual load / displacement curve for the tyre being loaded and unloaded and that will give you an idea of the energy involved. It would be hard to do by calculation because the structure of the tyre is so complex and the loss part of the modulus is anyone's guess. I bet the info wouldn't be readily available from tyre manufacturers, either - sounds like commercially valuable knowledge.
     
  9. Jun 15, 2010 #8

    dlgoff

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    This probably won't make too much difference but if you look at just one point on the tire, it will follow a http://en.wikipedia.org/wiki/Cycloid" [Broken].
     
    Last edited by a moderator: May 4, 2017
  10. Jun 16, 2010 #9
    It does follow the cycloid. but I'm looking at that point on the tire as it defelects under load to generate a pulse/ or wave. so like i said a little earlier, you would essentially see a heart beat like pulse.
     
  11. Jun 16, 2010 #10

    sophiecentaur

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    I'm not sure where you are going with this study. If you are dealing with the stress and energy losses / wear / heating then the actual path of the tyre through space isn't really relevant but the change of shape as time goes on is what counts.
     
  12. Jun 16, 2010 #11
    Yes I am trying to follow the shape essentially as the tire rolls. and possibly put it on a graph.
     
  13. Jun 16, 2010 #12

    sophiecentaur

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    OK. That's definitely not a cycloid, then!
    I could imagine several practical ways of plotting the actual shape. Using a position sensor bolted to the hub, rolling the car forward over a few turns of the wheel (few enough turns and you could get away with a simple twisted wire connection). OR how about a photograph? That would show the shape.
     
  14. Jun 16, 2010 #13
    yes. I was hoping to pinpoint the location on the outer part of the tire, in the tread area. To observe the forces which it encounters as it rolls around. I would imagine a compression when in contact with the ground, and an expansion when it is at the opposite point from the centrifical force.
     
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