Paging All Physicists!

Maintenance/Repairs
81

Nice graphic.

As has been pointed out, there is more surface area in the “squished” part of the tire - and that accounts for the additional surface area needed to balance the force equation…right?

So what about the fact that the surface area increase is both upwards as well as downwards?

And the “flat tire” that everyone seems to think is a key piece of evidence? Red Herring!

The additional mechanical stiffness that the air pressure magically endows the sidewall acts not only in the vertical direction, but also in the horizontal direction and no one seems to be arguing that the vehicle is moving sideway because of the inflation pressure.

I sense we are very close to the “Eureka” moment, when the true nature of pnuematic tires, and the way they work, is suddenly realized. It make take some time to get there, because there is a awful lot of “unlearning” that has to take place (One is the issue about the “flat tire”), but when you get there, it will all become clear.

82

Hear him! Hear him! Eric correctly identifies the tire as the supporting structure and the pressurized gas as a tensioning element in that structure. The question of pressure change in the gas, and the micro-analysis about it, is irrelevant and tangential at best to the original question.

I’m a mechanic (among other things) and I can testify that you don’t need to be a rocket scientist to explain the slight change in pressure between loaded and unloaded tires. The change in pressure is proportional to the work being done by the tire structure, of which the pressurized gas in an integral part.

My frustration, as a mechanic, with the discourse so far is that it seems entirely focussed on that tangential phenomenon while a wonderful topic, and the genuine answer to the young lady?s imperfectly phrased question languishes. The question arises from a widely held misperception that a car rests on its tires like a wagon on its wheels. That misperception creates the apparent paradox at the heart of the question. The prize inside the puzzle (that our dynamic duo and their scientific cadre missed) is the chance to explain how a modern tire works. Isn?t that the whole point?

83

Hi Cal,
I’m an old mechanic and I’m getting a kick out of this… sort of. You are right that the question ought to have been put to engineers before physicists. I just want to point out that when you said the tire wall carries only about 10 Lb. I’m sure you wer thinking only of the part under the axle. The tire wall really carries just about the whole load, in tension, above the wheel. The pressurized air is just a tensioning or pre-stressing element. Anyway, I wouldn’t want to tangle with the throng but I think they are missing the boat, or bus perhaps. The pneumatic tire is a wonderful example of an engineered structure that was quite likely misunderstood by its own inventor. Thinking of the sidewall as spokes in a wheel will make for more accurate understanding. Humbly, etc.

84

The tire is not holding up the rim. As CapriRacer points out, the pressure against the rim is the same all the way aroung the rim. The rim is suspended in the tire. The rim does not sit on the bottom part of the sidewall. Instead, the air is pushing against the rim with exactly the same force in all directions, suspending the rim. Here is an analogy. Put a bathtub on a scale and fill it with water. Weigh the bathtub. Now get in the bathtub. You are suspended weightless in the water (nearly so). But the scale registers your additional weight. The same thing happens, I think with a wheel and tire.

85

Almost right. It is the increase in pressure eric points to that adds to the tension on the rubber and steel belts. The tire pressure must increase. Tires do not conserve volume. To a close approximation, they preserve surface area. Deforming the tire decreases the volume, but nearly conserves surface area. Sitting on a balloon provides an example. The pressure goes up when you sit on a balloon (you know this because it is harder to poke with your finger). The volume goes down (you know this because PV=nRT, and you have not changed n, R, or T). The surface area goes up (you know this because the balloon gets more transparent). But volume changes by the cube of the radius. Surface area changes by the square of the radius. The volume effect swamps the sruface area effect, and the pressure goes up.

It is impossible for the volume to stay the same or to increase. The perfectly weak balloon or soap bubble described by one writer in this thread cannot maintain a pressure difference between the inside of the bubble and the outside of the bubble. It therefore cannot suspend any weight. Any deformation tending to decrease volume in one part of a perfectly weak bubble must be compensated by deformation elsewhere icreasing the volume, otherwise there would be a pressure difference between the inside and the outside of the bubble. If the bubble has any strenth at all (all real balloons and tires do, and tires have a lot), the volume must decrease and the pressure must increase when an external force, like the weight of a car on a tire or the weight of a person sitting on a ballon, is imposed.

I could be totally wrong about all this.

86

Well, the bathtub weightlessness phenomenon is due to buoyancy, but that is not a significant factor here. I agree with CapriRacer that the tire holds up the rim. I disagree with the notion that air isn’t supporting anything. Here’s why: Newton’s laws.

Let’s say I exert a force F on object A, but object A is resting against object B, which holds A stationary. So Newton’s second law says that object B must be exerting a force -F on object A, since A is not accelerating. Newtons’ third law then says that A must be exerting a force F on B. I push on A, and A transmits the force to B. So a statement like “A is bearing up under the force, B just holds A in position” is nonsensical and, in fact, violates the laws of physics. If B holds A in place, it is also being pushed on, and pushes back.

Now, that being said, I am also suspicious of the statement on the Bridgestone web site that the air bears the weight and the tire is “just a container.” In my analogy above A does need to be able to transmit the force. Also, there are objects that transmit forces with gain - they are called levers.

Now, back to the tire. I’ve seen web sites that claim you can find the weight of a car by measuring the total area of the contact patches between the tires and the ground and multiplying this by the air pressure in the tire. Even if there is some correlation here, it suggests that the air is a “weight-bearing” element. The air pressure generates force and the tire, by warping and stressing, transmits the force to the rim.

So I have written up an analysis (attached as pdf) that I did to see if my intuition was right. I actually drew a complete free body diagram, analyzed the forces, and solved some equations. The results are interesting, and while I think they support my viewpoint, they also support CapriRacer’s point that you have to look at the whole thing and include the horizontal forces.

87

NOTE: NEW entries have been posted to PAGE 1 (entries 1 thru 20).

88

As an Unskilled Laborer in the 80’s I serviced farm tires, which we filled with calcium chloride solution. You would jack the rear farm tractor tire up; and; with the valve at 12 o’clock and the valve core out (valve open), lower the empty tire untill the bottom deflected to where the tire had about the same volume as a loaded in-service tire; then fill with solution- but for the purpose of this thread we’ll

leave the tire slightly off the ground. If you filled the tire with solution so it was JUST to the level of the valve, (tire off ground, valve core out), and then lowered the tire, deflecting the tire bottom; solution would start pouring out the valve, proving that the tire’s volume had diminished. (I have witnessed this.) The next part is “seat of pants” physics, so straighten me out if necessary: the farm tire

seemed to deflect in the same manner that the automotive tire would as you lowered the lift; seemingly proving that the car tire’s volume would also diminish as it became loaded. However,the difference in the two scenarios is that there is no air pressure in the farm tire. Actually, there is- one atmosphere- but its pressure doesn’t increase during deflection since the valve core is out, so I haven’t created a

perfect model here. Anyway, isn’t the MISSION of this whole discussion to determine whether the volume of the car tire decreases as the vehicle weight is transfered from the lift to the tire; or is it more involved than that? (By the way, I’m honored to be part of such a rarified discussion; and also glad there’s minimal “mudslinging” here- unlike some other posts.) KS

89

In the interest of ending this discussion in this arena, I propose we move it to:

http://www.eng-tips.com/viewthread.cfm?qid=194137&page=1

bfaull,

I am prepared to address your statement:

“…A common high-school science experiment is to measure the total area of the contact patches between the tires and the ground and multiply this by the air pressure in the tire. You get a number that comes fairly close to the weight of the car. This to me suggests that the air is a weight-bearing element…”

Here’s the statement you will have to contend with:

The average footprint pressure of a tire (load divided by the area of the footprint) does not equal the inflation pressure. Therefore, the air pressure is not carrying the load of the vehicle.

90

Actually, I agree that the first part of your statement should be true!

I would revise the statement to read: The average footprint pressure of a tire (load divided by the area of the footprint) does not equal the inflation pressure. This is because the air pressure is not carrying the load of the vehicle directly, but rather indirectly through a leverage effect.

I decline your invitation to continue the discussion elsewhere. I’m not trying to win an argument here, but wanted to understand the physics. I think I have accomplished this (with your help, by the way.) I have revised my cartalk2.pdf file to make the leverage discussion a little clearer; I would encourage you to give the folks on that other discussion a link to the pdf file and the associated post. (You are even welcome to download and repost it.)

I also wanted to get Click and Clack’s attention, so that they might call me back and give me advice on their show about a problem I have been having with my car. This, alas, has not been successful. Cheers.

91

Well, if anybody is still following this, here is how I think it works. The pressurized air inside the tire applies perpendicular to the walls of the inside of the tire and the inside surface of the wheel. The compressed air pressure holds the inside surface of the tire away from the inside surface of the opposite side of the tire. The tread on one side of the tire is pushed away from the tread on the opposite side of the tire. The sidewalls are also pushed away from each other by the compressed air. There are tens of thousands of pounds of force making the tire hold its shape, 32 pounds per square inch for a typical car. That air pressure resists the force of the ground imposed upon the outside of the tire when the car is lowered off the lift and placed on the ground.

Even without air in the tire, the sidewall of the tire has some resistance to being bent under a load. An unmounted tire will support its own weight, and a mounted tire with only atmospheric pressure will support the weight of itself and the wheel. But if you balance your weight on the bead of an unmounted tire, you will flex it beyond the amount of flex generated by lowering the car onto an inflated tire. Therefore the sidewall of the tire does not have sufficient strength to hold up the car when the car is lowered off the lift.

For a two ton SUV, there is a load of approximately 1,000 pounds imposed upon the ground under each tire (recognizing the load is not even between the front and back of the car). That means there is one thousand pounds of directed force available to deform the tire. Some of the 1,000 pounds of force imposed will be taken up by flexing the sidewall of the tire, but not much.

The remainder of the 1,000 pounds of force flattens the tire against the pavement until the surface area of the flattened part of the inside of the tire times the inflation air pressure inside of the tire equals the remainder of the 1,000 pounds of force to be opposed. That force is translated to the pavement by the tread of the tire. The local pressure exerted by the outside of the tire against the ground can and does vary from place to place because the inside of the tire is smooth and has a smaller diameter, and the outside of the tire is treaded and has a larger diameter. The tire must therefore deform to change shape from circular to flat on the part that contacts the pavement. (There would also be some resistance to bending the tread, which wound take up some of the 1,000 pounds of force, and local effects all around the edge of the tire where the force imposed on the outside of the tire changes from about 32 psi to atmospheric pressure. These effects oppose some part of the 1,000 pounds of force.)

How does the weight of the car get translated to the pavement? I think it works like this. A tire has a high sheer resistance and a high tensile strength, but little resistance to bending. The sidewall of the tire resists deformation parallel to the surface of the tire very strongly, but has only weak resistance to being bent or straightened. The weight of the car is translated to the wheel by the axle pressing against the wheel through the lugs and contact surface between the axle and the wheel. The weight is translated from the wheel to the tire by the wheel pressing on the bead of the tire. The bead of the tire is held to the wheel by the air pressure directed perpendicular to the overlap between the tire and the rim times the surface area of the overlap times the friction between the tire and the rim. As the bottom of the tire bead (the part of the bead closest to the ground) is pressed down, the side of the bead (the part at 9:00 and 3:00 from the axle) follows because the wheel holds the bead stiff. As the side of the bead presses down, the part of the sidewall horizontal from the side of the bead is pulled along with it because the shear strength of the sidewall is greater than the shear stress created by the load imposed on it by the bead pulling down. The sidewall is pulled down with 1,000 pounds of force by the bead. The sidewall pulls the rest of the top half of the tire down because of the high tensile strength and shear strength of the tire. The actual distribution of stress and strain is likely very complicated. The top of the tire is pulled down with a total of 1,000 pounds of force. The inside surface of the tread on the top half of the tire likewise carries the air contained within the tire down. The opposite side of the tire is repelled by the contained air and presses against the ground. There is more than enough upward directed air pressure contained in the tire to suspend the tire. There is almost no deformation (strain) in the top of the tire because the shear and tensile strengths of the tire are high and the 1,000 pound load is distributed over the entire downward facing surface of the inside of the tire.

When the car is lowered from the lift, the ground imposes a point load on the surface of the tire (actually a line load considering the width of the tread). The air inside the tire does not push out strongly enough at that point (on that line) to resist a 1,000 pound load. Instead, the surface area must increase until the surface area equals the load divided by the inflation pressure (ignoring the strength of the sidewall). I don?t know whether the sidewall is stiffened by the air pressure, but the strength of the sidewall and other effects can be determined by subtracting from 1,000 pounds the surface area times the inflation pressure.

In a prior post I wrote something about Buoyancy. I agree I was all wet. The point I was trying to make, badly, is that the weight of the car can be mechanically uncoupled from the bottom of the tire, but that does not mean the bottom of the tire will not feel the weight of the car. In this case, the weight of the car is uncoupled from the bottom of the tire because the sidewall below the rim is weak and cannot bear the weight of the car. The sidewall retains its tensile and shear strength, though. Otherwise the car would not move forward when the transmission turns the axle.

Fire away.

92

Thankfully nobody is reading this any more or they would have pointed this out: Never mind the stuff in my fifth paragraph above about the sidewall holding up the tire with shear strength. That’s all wrong. The sidewall to the side of the tire doesn’t have anything to do with it. The force of the wheel pushing down is held up by the bead pushing up against the bottom of the rim. The bead pushes up against the rim because there is tension created between the rim of one side of the wheel pulling against the inflation pressure of the tire pushing in the opposite direction on the opposite side of the wheel. That tension is in the sidewall of the tire equals the force created by the inflation pressure times the area of the tire facing radially inward.

When the car is lowered off of the lift, the bottom part of the sidewall does not hold the car up. The bead holds the car up because it is pulled from above, not because it is pushed from below. The sidewall at the bottom of the tire is actually pulling down against the tension of the bead pulling up. When the car is on the lift, the tension in the sidewall is the same all the way around the rim, and is equal to the inflation pressure times the surface area of the tire on the opposite side of the wheel. When the car is lowered off of the lift, the tension in the sidewall above the wheel stays the same. The tension in the sidewall below the wheel is reduced by an amount equal to the force pushing down due to the weight of the car. Since the tension is decreased on that side of the wheel, there is no longer enough tension to oppose the air pushing outward against the tread on the bottom of the tire. The force to oppose the air pushing downward comes from the ground pushing up against the tire. The force of the ground pushing up is equal to the reduced tension in the sidewall of the tire.

This can be modeled as a toilet paper holder (like a piston or shock absorber) that is sprung with compressed air instead of a spring. The open ends of each half of the toilet paper holder will need a flange to overlap with the open end of the other half to hold the toilet paper holder together in one piece under the air pressure within. If the air pressure is 100 PSI and the surface area of each end of the holder is 1 square inch, then the sides of the toilet paper holder are under 100 pounds of tension. If the holder is set on end and a ten pound weight is put on one end, then the tension in the sides of the holder is reduced to 90 pounds, and the remaining 100 pounds of air pushing down is resisted by the ground pushing up. There is no deformation in the holder until the weight imposed is greater than 100 pounds. At 100 pounds, there is no tension in the sides, and further weight will begin to shorten the length of the holder.

To make it like a tire, compress the toilet paper holder and stick a block between the flanges holding the toilet paper holder together. That represents the wheel. The flange on the bottom end of the top half of the toilet paper holder (representing the tire bead on the bottom of the wheel) presses up against the bottom of the block. The flange on the top end of the bottom half of the toilet paper holder (representing the tire bead on the top of the wheel) pushes down on the top of the block. The block is under 100 pounds of compressive force, the sides of the holder are still under 100 pounds of tension. Push down on the block (not the top of the toilet paper holder, but the block between the two halves) with 10 pounds of force. The tension on the top half of the holder stays at 100 pounds. The tension on the bottom half decreases to 90 pounds of tension. The flange on the bottom of the block pushes up with 100 pounds of pressure, the flange on the top of the block pushes down with 90 pounds of pressure, and the block pushes down with ten pounds of pressure. The air at the bottom of the holder pushes down with 100 pounds of pressure. The bottom of the holder pushes up with 90 pounds of pressure, and the ground pushes up with 10 pounds of pressure.

Anyway, I think it works like that.

What I don’t think I got wrong is that the force of the ground pushing up flattens the bottom surface of the tire, and the contact area equals the force of the car pushing down times the inflation pressure, less the inherent strength of the uninflated tire, resistance to bending, etc. If the tire were lowered onto a U-shaped cradle that matched the shape of the tread surface of the tire instead of onto a flat surface, the tire would not deform at all.

If I’m still wrong about this, I quit. Constructive criticism and pleasantries are always welcome.