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Determining Tire Pressure

  1. May 12, 2010 #1
    Good day to everyone,
    I hope it is appropriate to post my practical statement of a problem here, to seek physical knowledge. My main background is Computer Science.

    The Problem:

    I want to algorithmically determine the tire pressure of a car, using informations of the car's footprint and it's tire-load (half axle load).

    Example:

    The car's tire has a load of
    [tex]F = 420 \cdot 9.81 N [/tex]

    The tire's footprint length (in driving direction) is
    [tex]L = 0.13 m[/tex]

    The tire's footprint width (orthogonal to driving direction) is
    [tex]W = 0.175 m[/tex]

    Now i estimate the footprint area as
    [tex]A = L \cdot W[/tex]

    Then, the tire pressure in bar could be estimated by
    [tex]p = \frac{F}{A\cdot 10^5} \approx 1.811 bar[/tex]

    But the true pressure is much higher:
    [tex]p_{true} = 2.30 bar[/tex]

    I think footprint area is overestimated (too big), and F needs to be reduced by a value according to tire type and F, as the tire already carries some load due to its material.

    What things are missing? Is there any function (e.g. f(tireType, N)) i could use as correction factor of this value?

    Thanks to all of You in advance!
     
  2. jcsd
  3. May 12, 2010 #2
    Hi there!

    Look, I admit to being entirely out of my league here in that I can't give you a definitive answer to your problem, but I think you're going wrong in calculating pressure as force per area. For this particular case, I'd recommend looking at the Ideal Gas Law or perhaps even at equations for Internal Pressure.

    Anyhow, that's where I would've started :smile:

    Hopefully someone more knowledgeable on the subject will come along soon with a more detailed strategy!
     
  4. May 12, 2010 #3
    Thanks a lot, every bit of information is highly appreciated!
     
  5. May 12, 2010 #4

    rcgldr

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    Another factor to consider is tire wall stiffness. In the case of run flat tires, the sidewalls are stiff enough that even at 1 atm pressure (zero reading with a gauge), the tire will support the car.
     
  6. May 12, 2010 #5
    Tyre pressure has bugger all to do with the amount of tread or anything like that. (In a very broad sense, tread and stiffness will have an effect on how the tyre reacts and heats up, but as all you are looking for is the pressure, we can conveniently ignore it).

    Boiled down with assumptions it's acutally a very easy problem. The tyre pressure involves, base pressure (ie reading when tyres haven't been used) and the effect of tyre temeprature. You can simply treat it as a control volume with heat applied. You can also treat it as an ideal gas depending on what it's filled with (im assuming it's air or N2 or something like that).

    That means you have to get operating tyre temp, and the transfer to the internal air.

    EDIT: So phyzmatrix have more confidence as you were pretty much bang on.
     
    Last edited: May 12, 2010
  7. May 12, 2010 #6
    *feeling chuffed*

    :biggrin:
     
  8. May 12, 2010 #7
    There's absolutely nothing wrong with calculating pressure as force per area. It's true
    that you can also calculate the pressure with the Ideal gas law, but you have to know
    exactly how much air you put in, the volume of the tire and the temperature.

    The problem with finding out the pressure by measuring the size of the contact patch
    is that the contact patch is not a rectangle, so the estimate given for the area is certainly
    too high. Another problem is the thickness of the tire walls, You can only be certain that
    the pressure on the ground is equal to the pressure inside the tire in the middle where
    the tire is pressed flat against the ground.
     
  9. May 12, 2010 #8
    What. Have you actually read this somewhere or seen it done? If so post a link, or a book that you got it from. Or have you just thought about that and made it up.

    As it certainly doesn't make sense as you've written it.
     
  10. May 12, 2010 #9

    jack action

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    I don't quite understand your problem. Why do you want to find the tire pressure based on the footprint area? The tire pressure is pretty much constant (given a temperature) and the footprint area varies with tire load.

    Furthermore, how did you get those numbers:

    Did you actually measure it? How? Was it a slick tire or one with threads? Because you will have to subtract the area of the grooves that don't touch the ground to get the actual area.
     
  11. May 12, 2010 #10
    The contact patch is determined by the air pressure supporting the load on the tire. The unit pressure of the rubber on the road would be affected by the percentage of void in the contact patch area but a tread pattern won't affect the contact patch size.
     
  12. May 12, 2010 #11
    Tire construction varies widely but the object is to achieve an even load distribution at the design load and pressure. For most purposes, that can be assumed.
     
  13. May 13, 2010 #12
    I know that, but I don't see why the argument is made that you can find the tyre pressure from the size of the contact patch and the load on the tyre, in the way he said it.

    The way he wrote it was that :

    He's making it sound like it's a totally trivial thing to calcualte a tyre pressure purely from the load on it and the contact patch. I'd argure that it's an utterly stupid way to calculate it, as it;s far more effort and less reliable than using a pressure and temperature relation.

    I'm also trying to work out what he acutally meant by it. As currently it reads less contact area for means more pressure in the tyre. Does he mean for all loads on a tyre, or a tyre only at a single working temperature and load.

    For example, in a car running at light load with a given input tyre pressure say 1.5 bar will havea contact area "x" m^2. At full load there is more contact area due to the deflections, yet as the tyre is worked more the internal pressure in the tyre acutally goes up. The large pressure is trying to reduce the contact patch size, but the increased load is increasind it. And NEITHER follow a linear relation.

    So although I know that it's possible to make a calculation of actual tyre inflation pressure from contact partch size. You need to know far more data to get an accurate result. Where as using the gas law and internal pressure gives a fairly accurate result with little data and a simple relation.


    I'd just like willem2 to clarify what he acutally meant (as I have a habit of getting the wrong end of the stick on here), as what he said at the moment I don't think makes total sense.
     
    Last edited: May 13, 2010
  14. May 13, 2010 #13
    Chris, my reply was an attempt to clarify what willem2 was referring to. I included your reply as part of the discussion. You're right, the assumptions involved need to be stated.

    Obviously, to maintain a specific contact patch area when increasing the load on the tire means increasing the tire pressure accordingly. I'm only accounting for one variable though.

    The OP is concerned with the difference between the calculated tire pressure and the actual pressure and was asking what could cause that difference. He/she then asked about the tire construction as a possible source of that variation. Without more details it's hard to pin down what the issue is but off the cuff I'd say that the tread face is quite stiff for the load. A more compliant tread face should give a result closer to what is expected.
     
    Last edited: May 13, 2010
  15. May 13, 2010 #14

    jack action

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    I have to disagree: rubber pressure = air pressure, such that the contact patch area is the area of the rubber touching the ground.

    Example:

    Two tires have the same thread pattern, i.e. with 2 ribs going around the tire (similar to http://www.lesschwab.com/tires/farm/fire_champguide4rib.asp"). For each tire the ribs are 1" wide, other measurements are:

    tire width: #1 --> 10"; #2 --> 6"
    distance between the 2 ribs: #1 --> 6"; #2 --> 2"

    If the air pressure inside the tire is 40 psi and the vertical load is 400 lb, what is the contact patch area and what is the length of the ribs in contact with the ground (assume symmetry and only the ribs are in contact with the ground)?

    For me, the area for the ribs is:

    A = (F / P) = (400 lb / 40 psi)

    A = 10 in²

    And the length of each rib in contact with the ground is:

    L = (A / 2) / Wrib = (10 in² / 2) / (1 in)

    L = 5 in

    This is valid for both tires, regardless of their widths or thread pattern differences. How would you calculate it?
     
    Last edited by a moderator: Apr 25, 2017
  16. May 13, 2010 #15
    You're not separating out the two effects.
    1) The tire carcass supports the tire load
    2) The rubber blocks of the tread support the tire carcass.

    1) The air pressure acts on the inside of the tire carcass to support the tire load.

    400 lb/40 psi = 10 in² of tire carcass area needed to support the load.

    2) The rubber ribs/blocks of the tread support the tire carcass.

    a)If the rubber of the tread fills 80% of the area available; 10in² x 0.80 = 8 in² of rubber touching the road; 40 psi/ 0.80 = 50 psi of pressure acting on the road by the rubber.

    b)If the rubber tread fills 20% of the area available (again assuming that only the rubber blocks touch the road); 10in² x 0.20 = 2 in² of rubber touching the road; 40 psi/ 0.20 = 200 psi of pressure acting on the road by the rubber.

    In both cases, the internal carcass area (contact patch) that the air acts on to support the load is the same but the unit pressure exerted by the tread on the road is quite different. Which tire would you prefer to have roll over your foot?:eek:

    This is why having more tread void results in lower dry traction; the unit pressure acting directly on the road is higher. That's why F1 put the grooves in the tires to slow down the cars.
     
    Last edited: May 13, 2010
  17. May 13, 2010 #16

    jack action

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    Well that makes sense. But it brings me back to my original question: How did BlackLotus measured the contact patch area? If it depends on the tire carcass deformation rather than the actual rubber on the ground, that would mean that you need some pretty good knowledge of the tire construction to make the relationship.

    And then, there is still my other question: Why do you want to find the tire pressure based on the footprint area?
     
  18. May 13, 2010 #17

    rcgldr

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    This ignores the fact that run flat tires can operate with zero pressure relative to ambient. The stiffness of the sidewalls and tread area also play a role in contact area versus load.
     
  19. May 13, 2010 #18
    For the pressure to be a function of the contact patch you need a very flexable tire. Much more so than what is in general use. If you want to check tire pressure you would be better off my measuring the "belly", (or radius of curvature of the sidewall) of the tire and comparing it to an expected value.
     
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