Pressure question - is inside pressure the same as road pressure? 1

[SirJohnAMacDonald]

I am reading on another forum a discussion that the pressure inside the tire is directly related to the pressure exerted on the road.

http://forums.vwvortex.com/zerothread?id=3360620&page=2

It's a bit of a circus.

Im not sold on it, but the hump might be onto something. Rock crawlers use very low tire pressures. I still don't believe it.

Now Im not a tire engineer, but maybe some of you are, Is there any truth to this? Or are tires really strong enough to hold themselves together that it isn't true? Is there an ENGINEERING resource to look up?

??

 

[patprimmer]

Three factors are involved.

Tyre carcase stiffness from the rubber and reinforcmet fabric.

Air pressure.

Dynamic effects.

The air pressure is full adjustable, but the structure is fixed.

When the tyre is dead flat, structure controls 100% of the contact area. The higher the air pressure, the more it over rides the structure. Dynamic effect vary with speed.

Regards

eng-tips, by professional engineers for professional engineers
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[NormPeterson]

"Mechanics of Pneumatic Tires" has some interesting data that show contact pressure between the tread and the road varying on the order of ~2:1 over the area of the contact patch.


Norm

 

[cibachrome]

How about this:

Take a BMW 5 Series or a Corvette. They use "run flat" tires. Take an air conditioning vacuum pump (same connections !!) and reduce the tire's internal pressure (-5 psi). Does the World end (Doomsday Scenario) or does the tire continue to lay on the road? And is the car still driveable? Does the ride & handling suck?

 

[SirJohnAMacDonald]

I think a good way to answer that is also - does the contact area change in proportion to that pressure change (even as an average).

However is this true for the tires of a giant truck used at a strip mine?

 

[SirJohnAMacDonald]

Since I'm not in the tire business, it is a bit confusing, but I believe that the stiffness in the tire structure and tread resists uncontrolled expansion. That will mean the pressure in the tire will increase, but the pressure to the ground won't increase at the same rate, since the contact area won't increase at the same rate.

Do I have this right?

 

[CapriRacer]

That is exactly right.

 

[ubrben]

If you have a lightweight car or bike the potential for a very stiff construction with low inflation pressure to work becomes much greater. The main dynamic effect to overcome is of course heat generation.

Ben

 

[ssn596]

Lets get a little practical and pragmatic here.

ASSUME the tire has no tread.
The pressure inside the tire is 32 psi.
Pressure outside the tire where it contacts the road is 32 psi. If you measured the downward force the tire exerts on road (axle weight) and divided it by contact area you should come up with 32 psi.

 

[NormPeterson]

Nope.

Inflation pressure is not the only factor. See post #2. Nor is contact patch unit loading (your external pressure) uniform. At best, inflation pressure is one variable spring in parallel with another.

I don't think I'd want tires for which contact patch P/A was uniform and equal to the inflation pressure anyway.


Norm

 

[CapriRacer]

SSN596,

Sorry, but that is not true. The flaw in your arguement is that the pressure inside the tire is holding up the vehicle. The inflation pressure is primarily a stiffening agent for the structural elements of the tire.

Proof? - In a month there will be a Tire Society meeting in Akron, OH. Buried somewhere in one of the papers will be data of footprint measurements at some load / inflation pressure combination. This data will not be the point of the paper - it will be data that is used for some other purpose. It is probable that this data will show that the average footprint pressure will not be equal to inflation pressure.

Please note: There are some situations where the average footprint pressure is equal to inflation pressure, but that is coincidental.

 

[ssn596]

My estimate was based on a tire with no tread and a completely flat "road" surface such as a piece of plate steel. And this is static ie stationary. Once you switch to dynamic then my "theory"?? goes right out the window centifugal force would enter into the equation for instance.

OK so the sidewall and tread section of tire can support some weight no doubt about it since an empty casing would collapse becuase of its own weight.

If I bleed a tire down to 0 psi it totally collapses so the air pressure is supporting the vast majority of the downward load.

So how much of the average weight on an axle is actually supported by the force exerted by air pressure on the tire wall and how much by the tire structural components for the AVERAGE street tire. Rough value "thumbrule" value will do very well. Thanks.

Dan Bentler

 

[Compositepro]

Air pressure is by far the biggest factor supporting the weight of the car for most tires. I'm sure exceptions can be found, and there are other smaller factors involved such as sidewall and tread flexibility. But, basically, what supports the car is the force of air pressure times the contact patch area. At the edge of this area is a zone where the contact pressure goes from air pressure down to zero. If the tire is on a stone, then the conact pressure will be greater than the air pressure at that point. It will be proportional to the radius of curvature of the tread.

If you analyze how load transfers from the ground contact area to the axle you will find that the car is actually hanging from the top of the tire through the reinforcing cords and air pressure is keeping the top of the tire from collapsing toward the ground. There are a lot of complexities but that doesn't change the fundamentals.

 

[patprimmer]

Even if the tyre side wall and tread structure could support 90% of the weight on the wheel, it would still go flat with no air.

Regards

eng-tips, by professional engineers for professional engineers
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[CapriRacer]

Just so everyone understands: The air pressure is not carrying the load of the vehicle.

Like Compositepro said: "....the car is actually hanging from the top of the tire through the reinforcing cords and air pressure is keeping the top of the tire from collapsing toward the ground...."

This is exactly right.

I like to think of it as the air pressure is tiffeneing the sidewalls and prevent them from buckling.

But the important point is there is a lot hard data on footprint pressure that says that the average ground pressure is not always equal to the inflation pressure. For those who have a theory that the pressure inside the tire is holding up the vehicle, the hard data proves your theory is wrong.

 

[crysta1c1ear]

Pressure question - is inside pressure the same as road pressure?

Now that I'm a bit drunk, I can say what I thought of saying before but didn't dare be so flippant .

The tire doesn't know what is outside it - air, road, water when aquaplaning etc. So the question boils down to is inside pressure the same as outside pressure? Now if that were the case, there would be no need for a valve and no need to inflate the tires!

The reason the skin on a ballon can stay in place when there is a greater pressure on the inside than the outside is that there is a third force needed in the balancing of forces and that is an effect from the tension in the ballon's skin. With a rolling tire there will be centrifugal effects too so the front of the contact patch (reducing radius as the wheel turns) won't behave the same as the back of the contact patch (increasing radius), even if they are similar at rest.

You can press on a balloon and increase the pressure at some point on its surface. But the pressure inside a ballon is presumably uniform (as the air inside is free to flow around inside it and balance the pressure). So what happens? You deform the surface of the ballon. If the pressure on the outside and on the inside are equal then the surface is presumably stationary and flat at that point.

Press hard with a finger on a ballon and you can make the surface concave instead of convex. In this case, the rubber is trying to push your finger back out, whereas elsewhere on the ballon the surface is still convex and trying to contract the ballon to a smaller size.

So with uniform pressure in the ballon, you can have a greater external pressure at some points and a lower external pressure elsewhere.

This is what I think is happening with tires.

Rather than try to equate/balance internal and external pressure, I think you have to balance the net effect of [ul]
[li]internal pressure[/li]
[li]external pressure[/li]
[li]centrifugal effects, and[/li]
[li]the elasticity effects of the rubber and bands etc.[/li]
[/ul]

 

[Compositepro]

It seems that some people are interested in discussing every factor that maybe involved in how a tire works under every possible condition. That's fine. But, again, fundamentally in pneumatic tires it is the air pressure at the contact patch that supports the vehicle.

The ideal tire would have a tread and sidewall with no flexural stiffness but very high tensile stiffness (i.e., no stretch. Woven fabric has this property. But fabric will not hold air and doesn't wear well in contact with the ground. Rubber is used to seal in the air and for traction and wear resistance. The action of flexing rubber dissapates energy and increases rolling resistance which is undireable. The thickness of the rubber is determined mainly by durability issues. A bicycle tire is a good example.

Air pressure in the tire is contained by the tensile strength of the tire cords. When the tire is off the ground it is round and there is and all forces are in balance. When load is applied to the tire it flattens at the area of contact. A very flexible membrane cannot apply load to a flat surface. It is air pressure that pushes on one side of the membrane and pushes it against the ground. This is an analysis of the loads in the contact patch area only. But it is accurate to say it is the air pressure that pushes against the ground.

How does the load actually get to the axle? The sidewall at the bottom of the tire flexes more and at the top of the tire the sidewal flattens slightly. This causes the tension in the sidewall cords on the bottom to go down and at the top of the tire to increase. So as I said in a previous post
the vehicle is actually hanging from the top of the tire.

In any inflated structure the tension in the wall due to contained pressure is inversely to the radius of curvature of the wall.

 

[ssn596]

Compositepro - you said
In any inflated structure the tension in the wall due to contained pressure is inversely to the radius of curvature of the wall.

So as the radius increases the tension in the wall structure decreases??

Does this apply to steel pipe, boilers and pressure vessels? So if I double the radius I then (assume inversely proportional and linear - ???) decrease the tension in the metal wall by half? Therefore for a given pressure I can use thinner metal as I increase diameter??

Dan Bentler

 

[Compositepro]

Sorry, thanks for catching that. The tension is directly proportional to the radius of curvature, not inversely. So, if you have a cylinder the wall tension (stress) will double if you double the diameter (radius of curvature) or if you double the internal pressure.

I guess the confusion was that increased curvature means decreased radius of curvature.

 

[crysta1c1ear]

But, again, fundamentally in pneumatic tires it is the air pressure at the contact patch that supports the vehicle.

A force pressing down cannot hold a vehicle up.