Why does the car come with a jack and other tire changing tools

Discussion in 'Prime Main Forum (2017-Current)' started by EyePrime, Feb 4, 2019.

  1. Lee Jay

    Lee Jay Senior Member

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    The load on the tire wall, yes. Not the load lifting the car. The portion of the pressure pushing equal and opposite on the tire wall is not lifting the car, just stretching the rubber. Think about it - only the force pushing against the ground is lifting the car.
     
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  2. PiPLosAngeles

    PiPLosAngeles Active Member

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    It doesn't work that way. If it did, the pressure on the bottom of the tire would be 10 lbs higher than the pressure at the top of the tire. This can easily be confirmed.
     
  3. fuzzy1

    fuzzy1 Senior Member

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    But with your flat tire, the 'piston' on the ground is only about 100 square inches, not 900+ square inches. The portions of the tire away from the ground are not supporting the car at all.
     
  4. fuzzy1

    fuzzy1 Senior Member

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    Huh? Please explain further.

    Do note that where the tire is pushing on the ground, the rubber is not getting stretched out as much by the interior air pressure.
     
  5. Lee Jay

    Lee Jay Senior Member

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    You aren't using your head.

    Obviously, the pressure on the inside of the tire is constant throughout the volume of the tire. But do you think the upward force on the top of the tire is lifting the car? How about the forward and backward force on the left and right side of the tire? They are just pushing against the tire itself.

    If you were right, the pressure inside of a tire hanging on the wall would always be zero (gauge pressure) because it's not holding anything up. But you aren't right and the pressure can be pushing against nothing but the hoop stresses of the tire material.
     
  6. PiPLosAngeles

    PiPLosAngeles Active Member

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    Yes, they are. That's how it works. What's holding the car up is every square inch of the entire tire. The car's weight pushes down on the tire with 1,000 lbs of force. That weight increases the pressure in the tire by 1000 lbs ÷ y square inches of tire surface area - 1 psi in the case of a tire with 1,000 square inches of surface area.

    You guys are getting confused about the difference between where the weight of the car is transferred from the rubber to the ground and where the weight is transferred from the air to the tire. Those are two very different things.
     
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  7. Lee Jay

    Lee Jay Senior Member

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    By the way, this is why the size of the tire's contact patch with the ground changes with internal pressure, and thus why you can see the pressure inside of the tire just by judging the size of the contact patch with larger (and flatter) meaning less pressure - thus the name "flat tire".
     
  8. Lee Jay

    Lee Jay Senior Member

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    You are extremely confused. The car's weight is supported by the ground. The tire transfers the weight of the car to the ground, distributing that load around the contact patch with the ground. If you assume that pressure is constant (which it isn't, for other than obvious reasons), then the pressure against the ground is the weight of the car divided by the contact area. The rest of the tire does nothing other than containing the pressure.
     
  9. pghyndman

    pghyndman Active Member

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    Newer slime designed to work with TPMS equipped wheels.

     
  10. fuzzy1

    fuzzy1 Senior Member

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    The front, back, and sides of the tire aren't holding up squat. They aren't even pointing in the proper directions to hold up the car against gravity. They are merely containing the internal air, and balancing out the force vectors on their symmetrically offsetting components on the diametrically opposite side of the tire.

    In this sense, the only part of the tire holding up the car is the top side opposite the contact patch. This part is not being flattened like its opposite-side contact patch, but instead is being stretched out by the internal pressure.

    You can simplify your tire load analysis down to just these two non-identical, non-offsetting portions of the tire. The entire vehicle load is carried within these portions. All other tire portions perfectly offset the forces on their diametrically opposite portions, for a net of zero force, thus holding up no weight.

    The tire's pressure change (P) is driven by its volume change under the Ideal Gas Law, PV=nRT. That volume change (V) is in turn driven by how much flattening happens at the contact patch, compared to the total tire volume. That flattening is very dependent on initial tire pressure, not a constant depending solely on the vehicle's weight.

    You could put a constant pressure regulator on the valve, or connect the valve stem to a very large constant pressure tank, so that very little or no pressure change happens at all when the vehicle weight is placed on the tire hub. The tire will still support the vehicle's weight despite no pressure increase.

    :rolleyes:
     
    #50 fuzzy1, Feb 5, 2019
    Last edited: Feb 5, 2019
  11. Trollbait

    Trollbait It's a D&D thing

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    First, these are parking brakes, not emergency brakes. Applying just the rear brakes with just arm or leg strength isn't going to do much to slow a speeding car going forward. The service brakes in all cars have at least two hydraulic circuits, with each one operating the opposite front and rear brakes. That is your emergency brake.

    Second, lose power, and the brake booster holds enough vacuum pressure for 2 to 3 applications of the brake pedal; plenty for an emergency stop.

    Regardless of whether you are confused or not, inflating a tire with no load on it is less strain for the low bid compressor than with the load.

    Technicians still have to clean it out to patch the tire. Depending on how their day went, you could be left buying a new tire to replace an "unrepairable" one.
     
  12. frodoz737

    frodoz737 Top Wrench

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    :rolleyes: Round-n-round this goes. Those that can and will, prefer and use a spare. Those that can't or won't, use their cell phone and loose at least half a day. All this other talk is just BS justification for one's chosen side...and neither side is going to budge.
     
    #52 frodoz737, Feb 5, 2019
    Last edited: Feb 5, 2019
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  13. Dave Lane

    Dave Lane New Member

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    There is a simple answer to this. Many have a set of winter tires on rims that need switching out twice a year. Without a jack that would be impossible.

     
  14. Lee Jay

    Lee Jay Senior Member

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    I just used a spare on my wife's car, and it cost me more than half a day. An hour to change the tire (including driving to her location and back) and five hours the next day to drive the car to the tire shop, get a ride home, get four new tires, get a ride back, and drive both cars home.

    Without a spare, the same operation would likely have taken an extra hour and would have gone something like: Drive to her location, remove her damaged tire and put it in my trunk, drive to the tire shop the next day and have that one tire replaced, drive back to the car and mount the new tire, drive to the tire shop to have the other three tires replaced, and drive home. The extra hour is for the trip back and forth to the car the next day.
     
  15. frodoz737

    frodoz737 Top Wrench

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    But for folks like me...10-12 minutes and I'm on my way. Oh...my Wife changes her flats too.
     
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  16. bisco

    bisco cookie crumbler

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    it's a chat room :cool:
     
  17. Lee Jay

    Lee Jay Senior Member

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    I don't see how you buy four new tires, get them mounted and balanced in 10-12 minutes. And my wife can change her own tires but she had our 10 year old daughter in the car and she was in the dark on ice so it was much safer for me to drive over and help out.
     
  18. CharlesH

    CharlesH CA HOV Decal #5 on former PiP

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    Uggh. Brain fart.Thanks for the reminder that what was mentioned is intended to be a parking brake, not an emergency brake. I really do know about the redundant hydraulic brake system on the service brake, but got off on the wrong direction. Maybe I needed more coffee before posting....
     
  19. PiPLosAngeles

    PiPLosAngeles Active Member

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    That's exactly what they're doing. Try knifing one and see what happens to the car. Is it still being held up? Nope. Strange, it's like the sidewall was holding something up.

    Now, someone who wasn't paying attention in physics might say, "of course the car isn't held up anymore! You let all the air out," and they might say it without realizing that they're making the point. The air is what distributes the load equally to ALL points "inside" the tire - the tread, the sidewalls, the top, the bottom, the sides. That's what holds the car up. The tire's resistance to popping is literally holding the car up.

    If this were true, it would require that the pressure at the top side of the contact patch is higher than at the the parts of the tire that aren't "holding up squat." Why? If we have point a on the tire "holding up" a weight and point b on the sidewall that isn't "holding up" anything, the pressure (weight) on point a would be a + c more than point b, where c is the amount of weight being "held up" by point a and not by point b.

    All that being said, the original issue was how much extra force would be required to inflate a tire that was bearing the weight of the car versus a tire that was not. The fact remains that the extra force can be mathematically approximated as:

    f = (x ÷ y) × z

    where:

    x = the total weight supported by the tire
    y = the total surface area exposed to the pressurized air
    z = the surface area of the pump's piston
     
  20. CharlesH

    CharlesH CA HOV Decal #5 on former PiP

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    Thank you for explaining the downside of the old technology in small GA aircraft. I would think that proper maintenance and pilot training goes a long way to offset those issues. The failure modes in those simple aircraft seem to me to be limited and more straightforward to deal with. There is something to said about a machine whose failure can kill someone that can maintained by a skilled mechanic without a degree in computer technology. My knowledge of all this is entirely second hand, from my son who is working his way up to the big aircraft. (Please note that I am not technology-adverse; I have been working in advanced computer science research and enterprise software development for over 30 years. But people don't die if the stuff I work on fails.)
     
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