Modelling the tyre as a spring in a quarter car model

[NickTheFill]

TL;DR

Is there a better way of modelling the tyre within a two degree of freedom quarter car model?

Hi everyone
I've just completed a mechanical engineering degree, and one aspect of the classic 2DoF quarter car model that still bugs me is the representation of the tyre as a linear spring attached to the ground.

Does anyone have any experience of modelling the system with the tyre spring unattached to the ground? In my industrial placement year I was regularly involved with pave road testing and such was the ground input, that the tyre was often in mid air!

Is there a more representative model out already there?

Thanks for your help.

Nick

 

[jack action]

I was regularly involved with pave road testing and such was the ground input, that the tyre was often in mid air!

It depends on the road input, but having the tire leaving the ground is a possibility.

I'm not sure what particular problem you are trying to solve.

 

[NickTheFill]

It depends on the road input, but having the tire leaving the ground is a possibility.

I'm not sure what particular problem you are trying to solve.

It depends on the road input, but having the tire leaving the ground is a possibility.

I'm not sure what particular problem you are trying to solve.

Thanks Jack,
How can the tyre leave the road in a traditional 2DoF model?

 

[jack action]

I think I understand what you are getting at.

You have to keep track of the distance between the unsprung mass and the road. At one point, the distance will be larger then what the tire can extend to. I did see this kind of solution for determining spring length by keeping track of the distance between the unsprung and sprung masses.

From the example on this page, you can find the distance between the masses as $Z_1(s) -Z_2(s)$ and the distance between the road the unsprung mass as $Z_2(s) - U(s)$

 

[NickTheFill]

I think I understand what you are getting at.

You have to keep track of the distance between the unsprung mass and the road. At one point, the distance will be larger then what the tire can extend to. I did see this kind of solution for determining spring length by keeping track of the distance between the unsprung and sprung masses.

From the example on this page, you can find the distance between the masses as $Z_1(s) -Z_2(s)$ and the distance between the road the unsprung mass as $Z_2(s) - U(s)$

Hi Jack, not quite. My query is that in the 2DoF model, the tyre spring has a restoring force towards the ground. In reality however it will be always be zero as the tyre is never attached to the ground. I'm asking if anyone has any experience of overcoming this limitation.

 

[jack action]

Is there a restoring force? When the tire is on the ground and the ground suddenly drop, does it pull the tire down?

But worst case scenario, this is where the evaluation of the distance between the road and the tire will help you. You can set a new limit (maximum deflection of the tire) where at this point you know that $dz_2 = du$, which will give another set of equations to switch to (i.e. representing free fall). When the tire reach the ground again, you switch back to the original set of equations.

 

[FEAnalyst]

Check the "Tyre and Vehicle Dynamics" book by Pacejka. It should be useful for you.

 

[NickTheFill]

Is there a restoring force? When the tire is on the ground and the ground suddenly drop, does it pull the tire down?

But worst case scenario, this is where the evaluation of the distance between the road and the tire will help you. You can set a new limit (maximum deflection of the tire) where at this point you know that $dz_2 = du$, which will give another set of equations to switch to (i.e. representing free fall). When the tire reach the ground again, you switch back to the original set of equations.

Unless very sticky, a tyre never produces an appreciable restoring force towards the ground. Hence my initial query.

 

[jack action]

My query is that in the 2DoF model, the tyre spring has a restoring force towards the ground.

Where is the restoring force in the model? The only forces are gravity, spring and damping. The input is a displacement, not a force.

 

[NickTheFill]

Where is the restoring force in the model? The only forces are gravity, spring and damping. The input is a displacement, not a force.

Whenever wheel displacement is greater than ground displacement In the equations of motion it is represented by...kt(x_us-r). Where kt is tyre spring stiffness, x_us is displacement of the wheel and r the displacement of the road.

 

[jack action]

But there should be an initial wheel displacement at rest, preloading the [tire] spring. This is where the apparent 'pulling' happens. It is only the tire releasing its initially stored energy. Therefore, it never goes 'negative'. The same kind of preload is expected from the spring between both masses.

 

[NickTheFill]

So why does the model work equally as well horizontally?

 

[jack action]

Constant forces - independent of displacement - only add an offset to the response. If you add gravity, the system finds a stable position and stays there (no motion). Once you introduce the disturbance, the oscillation revolves around that position instead of the initial 'zero'.

The quarter-car model usually care only about the amplitude and frequency response, which is the same, no matter the constant force applied to the system.

Horizontally, it is true that if there was not a force 'pulling' the spring, the mass would break free. Put putting a constant force 'pushing' on the mass instead, would have the same effect without having the need for holding the spring to the wall. As long as the preloading of the spring would be larger than the amplitude of the response, that is.

 

[NickTheFill]

Constant forces - independent of displacement - only add an offset to the response. If you add gravity, the system finds a stable position and stays there (no motion). Once you introduce the disturbance, the oscillation revolves around that position instead of the initial 'zero'.

The quarter-car model usually care only about the amplitude and frequency response, which is the same, no matter the constant force applied to the system.

Horizontally, it is true that if there was not a force 'pulling' the spring, the mass would break free. Put putting a constant force 'pushing' on the mass instead, would have the same effect without having the need for holding the spring to the wall. As long as the preloading of the spring would be larger than the amplitude of the response, that is.

Applying a constant force to the wheel mass only in the horizontal model is not the same as tethering the tyre spring.
Spring preload is not the same as a static deflection due to gravity.