Pressure, surface area and force in tire

[Kevin Shen]

Assuming that the temperature in the tire suddenly rises.
My question is that would that cause the contact surface area between the tires and the floor increase, decrease, or stay the same?
(given that the volume of air in the tires is a constant and cannot be changed)
Also, why?

 

[BvU]

Hello Kevin,

You familiar with the ideal gas law pV = nRT ?
According to that, the pressure would go up with T.
If the load doesn't change, and the load is approximately p*A (A the contact area), then that means A decreases.

 

[Kevin Shen]

Hello Kevin,

You familiar with the ideal gas law pV = nRT ?
According to that, the pressure would go up with T.
If the load doesn't change, and the load is approximately p*A (A the contact area), then that means A decreases.

No..I'm not very familiar with this equation, but I thought that A stays the same due to P=F/A, and so due to higher kinetic energy causing an increased force and pressure, P and F will rise proportionally so F1/P1=F2/P2 and the area is left unchanged...
Is this not right?

 

[BvU]

What is F ?

 

[Kevin Shen]

What is F ?

F is force

 

[BvU]

Why would that change ? e.g. for a car or bicycle tyre ?

 

[Kevin Shen]

Why would that change ? e.g. for a car or bicycle tyre ?

What I mean is that in a given volume like in a tire, when there is higher temperature due to P1/T1 = P2/T2, there will be higher pressure. And the higher pressure is caused by the larger momentum change when particles interact with the wall so that they create a higher force over the area. And hence if the pressure and force of the particles inside the same volume increases, I thought that wouldn't the surface area stay the same?

 

[BvU]

The surface area of the tyre stays more or less the same. But you were talking about the contact surface area. The load on the tyre is approximately Pressure times contact area.

 

[Kevin Shen]

The surface area of the tyre stays more or less the same. But you were talking about the contact surface area. The load on the tyre is approximately Pressure times contact area.

The surface area of the tyre stays more or less the same. But you were talking about the contact surface area. The load on the tyre is approximately Pressure times contact area.

Thanks again, but how do you calculate the contact surface area and know why it decreases?

 

[BvU]

If the tyre exerts a pressure on the ground that is equal to the internal pressure, then the force it exercises on the ground is that pressure times the contact surface area. The contact area normally doesn't accelerate in the vertical direction, so the ground pushes back with that same force (called the normal force).
There's a nice lady explaining this here

If p increases and the weight of the car stays the same, then the footprint decreases.

 

[jbriggs444]

If you've ever put air into a flat tire, you can see this in action. You start with a huge squashed out contact area. Then, as air is pumped in the tire and wheel rise up and rest on a much smaller contact patch.

 

[cosmik debris]

Car and motorbike enthusiasts argue endlessly about contact patches of tyres. You hear race guys on the TV say wider tyres give more grip, and fast street cars always have wide tyres etc.

There are two relevant formulae. The one above A = F/P and the friction formula StaticFriction = kF where F is the normal force and in this case the weight of the vehicle. Simplistically the contact patch area depends only on the pressure in the tyre as the weight stays the same. For friction it is only the weight times some friction coefficient k. k for good tyres can be 1.4 or so, but this probably goes against intuition hence the arguments. The answers probably lay in the tyre construction. Thin tyres heat quickly and are mechanically weak, wide tyres can be made stronger so you can make them stickier, hence more grip.

I'm sure it is not that simple but I do think there is a lot of myth associated with tyres. Incidentally tyre data is extremely difficult to obtain probably for commercial reasons. One study I found had data varying considerably for different pressures.

Cheers

 

[BvU]

True, but I think we want to keep it as simple as possible...

 

[rcgldr]

The stiffness of a tire is also a factor in contact patch area. If a tire is stiff enough, it can run without any pressure (internal pressure equals ambient pressure); such tires are referred to as "run flat" tires.

You hear race guys on the TV say wider tyres give more grip

A wider tire results in a contact patch better suited to handling lateral loads (less percentage deformation of the contact patch due to lateral loads), and the total surface area of the tread is a factor in dissipation of heat in a tire. Tires with weak sidewalls (called "wrinkle wall" tires), combined with lower pressure provide more contact area. Due to a load sensitivity factor with rubber, a larger contact patch results in a somewhat higher coefficient of friction, due to a decreased load per unit area.

 

[cosmik debris]

Due to a load sensitivity factor with rubber, a larger contact patch results in a somewhat higher coefficient of friction, due to a decreased load per unit area.

I hoped you would reply as this is something I think about a bit. Can you elaborate on this last sentence, I don't quite get it?

 

[rcgldr]

Due to a load sensitivity factor with rubber, a larger contact patch results in a somewhat higher coefficient of friction, due to a decreased load per unit area.

I hoped you would reply as this is something I think about a bit. Can you elaborate on this last sentence, I don't quite get it?

Wiki has an article about tire load sensitivity.

https://en.wikipedia.org/wiki/Tire_load_sensitivity

The wiki article mentions the slip angle for peak lateral force at around 6 to 10 degrees for street tires, 3 degrees for Formula 1 tires. Indy Racing League tires, at least the ones used for high speed ovals, are the stiffest where the slip angle for peak lateral force is around 2 degrees.

Not mentioned is that a larger (wider) tire can be used to compensate for the load.