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OT? Weigh your car by checking tire pressure?

Discussion in 'Electronic Design' started by Rich Grise, Sep 23, 2007.

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  1. Rich Grise

    Rich Grise Guest

    Saturday morning TV is mostly cartoons, yes, but there's one show that's
    not a cartoon, but it's still fun:

    It's like Stealth Educational TV - like Mr. Wizard for the 21st century
    kid. And they do Real Science.

    But today, they said that you can weigh your car by using your tire
    pressure. What you do is measure the footprint of each tire, take its
    pressure, and then the footprint of the tire times the PSI equals
    the number of pounds that the tire is supporting, and their sum is the
    weight of the car. (except for the tires themselves, I presume).

    But they did the experiment - they measured the footprints of the four
    tires on Some very well-preserved Nash Rambler, and their PSI, and did
    the arithmetic, and came up with a number that was within 10% of the car's
    "official" weight.

    Well, I'm a little uncomfortable with that. What about the pressure that's
    in a tire when it's not on the car? Where does that pressure go? Is that
    that 10% fudge factor that they admitted to in the show?

    How much does the pressure change when you take a standalone tire, mount
    it on the car, and let the car down on it?

    Or do they get around that by saying, "Well, you can ignore that, because
    the tire's not supporting any weight". Or is it entirely ( or mostly) due
    to the flattening of the bottom of the tire? Does the "bias pressure" (a
    term I just made up now, for the pressure that's there when it's not on
    the car) get lost below the noise floor?

  2. Guest

    It took me a while before I decided that they are right. When the
    tire is not on the ground you multiply the pressure ( Say 30 psi ) by
    the footprint ( 0 ) and get zero. The big problem is measuring the
    footprint and the stiffness of the tire sidewalls and tread.

  3. Dave Baker

    Dave Baker Guest

    There are two sources of error in that method. The first, which is the one I
    think you are referring to is the increase in pressure as you put weight on
    the tyre. This will be the inverse of the reduction in volume of the tyre as
    the bottom of the tread flattens and will be very small.

    The second, which is the significant one, is the amount of the load which is
    being supported by the stiffness of the tyre tread and sidewalls. If the
    tyre were a balloon, i.e. perfectly elastic, the measurement as above would
    be exactly correct.

    So the result they should have obtained is a calculated weight somewhat less
    than the actual weight of the car.
  4. In the days of cross-ply tires, it took energy to flex the sidewall as
    the tire rotated and deflected to bear the vehicle weight.

    With regular radial tires, this energy loss is minimized both when
    rotating, and as weight is applied to the tire - hence the cautions
    not to over inflate them.

    You can verify this for yourself by taking tire pressures at one end
    of your car, then jacking that end up.
    You should not see much change at all.

    Brian W
  5. Phil Allison

    Phil Allison Guest

    "Brian Whatcott"

    ** Classic example of a non-sequitur.

    ** Perfect example of a illogical conclusion drawn from a non- sequitur.

    ....... Phil

  6. There you go folks, that's an Australian contribution for you:
    pithy, to the point, incisive, witty, analytical, carefully reasoned.

    And they still cling to men-only bars there, isn't it Dai?


    Brian W
  7. Phil Allison

    Phil Allison Guest

    "Brian Whatcott"

    ( snip this utter moron's' ASD fucked drivel )

    ** Hey - Mr Autistic Shit for Brains.

    Go **** your mother !!.

    ...... luv, Phil
  8. E Z Peaces

    E Z Peaces Guest

    Well, if you measure the tire radius in a horizontal plane, then measure
    the distance from the center of the wheel to the ground, you will know
    how much the tire is flexing in mm. Then if you unscrew the valve core
    from your spare, you can stand the wheel up and see how much weight it
    takes to push the center of the wheel down to that height. With a
    radial, I wonder if the weight of the rim would flex the tire that much.

    It seems to me that except those few pounds, it's air pressure that
    holds up an inflated tire. In ordinary circumstances I wouldn't know
    how to measure the footprint accurately. The car would have to be on a
    very smooth, flat surface and I would need a thin "feeler gage."

    I've read that tires of larger diameter have less rolling resistance,
    but for some reason tires of smaller diameter hold better on snow. The
    contact patches are the same area.
  9. jk

    jk Guest

    The method also assumes (I think) that the pressure distribution
    across the footprint is uniform.
  10. Nick Mueller

    Nick Mueller Guest

    Hahaha! They pulled your leg!
    The reason why this works is, that there is an efficient tire pressure for
    each load and tire. The tire manufacturers tell what pressure for what
    load. And that number (psi per lb) is in the same ballpark. It depends on
    the construction of the tire (stiff/soft side walls, width, diameter, etc).

  11. Guest

    I'm tempted to come round and let your tyres down, and watch you try
    to catch your now weightless car as it drifts away across the sky...

  12. Here's a question on an engineering group that has an
    engineering answer!

    There is a pressure sensitive material available (for gaskets etc.)
    that color codes the pressure that it experiences.
    That would be one analytical approach.

    Brian W
  13. Joseph Gwinn

    Joseph Gwinn Guest

    It is true that one can weigh a car this way, using the principle of the
    tonometer, which is most often used in medicine to measure the pressure
    within the eyeball.

    The basic principle is that in the flat section (the footprint), tension
    in the membrane (the tire) is perpendicular to the surface, and thus
    cancels out, so the net force generated is the internal pressure times
    the area of the flat section.

    The tonometer was invented a century ago, so most web references assume
    that one already knows how it works, and tries to sell you one. But
    here is one article with a good explanation:


    (Reprinted from the Archives of Ophtalmology,
    January 1961, Vol. 65, pp. 67-74, Copyright 1961, by American Medical
    Association, "Corneal Bending and Buckling in Tonometry",
    Berkeley, Calif.)

    Joe Gwinn

  14. There are far too many variables involved in getting an accurate
    reading in this manner for it to EVER be feasible.

    What DUMB BASTARD that doesn't know a fucking thing about Usenet made
    this stupid, cross-posted SHIT?
  15. E Z Peaces

    E Z Peaces Guest

    When I posted, it occurred to me that the contact patch was probably
    larger than weight/pressure because at least at the front and back of
    the patch, the tread would press down with less pressure. Now I've
    found this article by a BMW motorcycle rider explaining that the subject
    is very complicated:

    Because the pressure of the tread approaches zero at some points, it
    seems a water film will carry some of the weight any time you roll on a
    wet road. Traction is diminished, and you hydroplane when you go fast
    enough and the fluid is deep and viscous enough that the film supports
    even the parts of the tread with the most pressure.
  16. Gunner Asch

    Gunner Asch Guest

    Evidently it was you.

  17. CBFalconer

    CBFalconer Guest

  18. mpm

    mpm Guest

    Take the car and launch it into orbit.
    Then, measure the tire pressure and determine the weight.
  19. I think the important factor is width, not diameter.

    The contact patches are the same size, but a different shape. After
    the first part of the contact patch has displaced the snow, there's
    more contact patch left in a narrow tire to contact the pavement. The
    amount of snow which needs to be displaced is proportional to the
    width of the contact patch, not the area.

    I don't know why a tire of larger width would have less rolling
    resistance, unless it's just that tires of larger width usually have a
    smaller sidewall (proportionally) when used on the same car.

  20. Responses don't count, dipshit. I have to ensure that my post gets read
    by the twit that posted across several groups, many of which he may not
    even visit or pull headers from. Get a clue.
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