How stiff are F1 tyres?

[DaveW]

Maybe it's because you need only 2% nett change in pressure to support static load of 2011 spec F1 car?

I'm not quite sure how you arrived at your nett change in pressure, Marekk, but that would translate to around a 1 psi change with a reasonable downforce applied, assuming linearity. I hope a change of that magnitude would be spotted by somebody tasked with maintaining constant tyre pressures. Incidentally, just for info, a 185 kg change in down force on one wheel caused a 13 percent increase in vertical tyre stiffness with no observed increase in tyre pressure (in one example).

[ringo]

Marekk: while you are awaiting Ciro's response, you might like to look back in this thread to find an image of Michelin's "Tweel". That has no internal pressure to change, but manages not to collapse under load. In a conventional tyre, the spokes of that design are replaced by the sidewalls stabilised by internal air pressure. The internal volume, & hence pressure, does not have to change (& doesn't to first order) when the tyre reacts an applied load. I hope this helps a little.

The tweel has spokes which are clearly not behaving like cables in a typical tyre. I call BS on that one.


the spokes are deflecting only because they are under buckling.

[DaveW]

The tweel has spokes which are clearly not behaving like cables in a typical tyre. I call BS on that one.

Are you suggesting, Ringo, that (the belt of) a pneumatic tyre would behave differently? If so, how? Would the belt deform less, or more, do you think? If it deforms in a similar way could it be, perhaps, that the side walls of a pneumatic tyre deform (buckle) in a different direction compared with the spokes of a tweel?

As a general comment on the difference between spokes & side walls, it is probably worth pointing out that side walls will react a proportion of applied load by shear deformation (strain energy, if you wish).

[ringo]

The tweel has spokes which are clearly not behaving like cables in a typical tyre. I call BS on that one.

Are you suggesting, Ringo, that (the belt of) a pneumatic tyre would behave differently? If so, how? Would the belt deform less, or more, do you think? If it deforms in a similar way could it be, perhaps, that the side walls of a pneumatic tyre deform (buckle) in a different direction compared with the spokes of a tweel?

As a general comment on the difference between spokes & side walls, it is probably worth pointing out that side walls will react a proportion of applied load by shear deformation.

It wont behave differently but to suggest that the laod carrying capacity of the tyre is not related to the tyre pressure is not honest.
That was your purpose of pointing out a tweel is not under air pressure.

The tweels spokes are under compressive loading, while a tyre side wall is under a tensile load by the air pressure.
I can kick my tyres and deform the sidewall easier if the tyre has less air. This deformation is a result of a lowered side wall stiffness.
It's pretty straightforward, i don't why you guys are trying to mislead.

I suppose an F1 tyre may have a much stiffer sidewall than a road tyre, but you still can't take the air pressure out of the equation.

that's why this works:

[DaveW]

It wont behave differently but to suggest that the laod carrying capacity of the tyre is not related to the tyre pressure is not honest.
That was your purpose of pointing out a tweel is not under air pressure.

If you actually read the post you quoted, you might decide that your accusation is unfair (I hope you do, anyway). The reason I posted was to suggest that loading a tyre does not require a change to internal pressure.

The tweels spokes are under compressive loading, while a tyre side wall is under a tensile load by the air pressure.

Apologies, Ringo, but I don't think you understand. Ignoring the undoubted fact that the spokes of a tweel can transmit a small compressive load both before & after they buckle, I think a short ponder over the fine photograph you posted might convince you that by far the majority of tweel spokes under the loading condition shown are under tension. Those under compression are buckled, & transmit very little load.

I think the tweel is very instructive in demonstrating how a pneumatic tyre actually works. Some have stated that a tyre contact patch area is exactly (& necessarily) equal to the applied load divided by the internal pressure. If that is the case, then why does the belt of a tweel not collapse onto its wheel rim? It is because its shape is maintained (more or less) by tensile loads in the spokes (they don't like to increase in length, if you like). The side walls of a pneumatic tyre work in the same way - when they are stabilised by internal pressure. Having said that, the side walls of low profile tyres tend to be able to support serious loads without being pressurised - for example touring car race drivers complained at one time that they didn't know their 19 inch tyres had punctured until they disintegrated.

For me, the tweel demonstrates that it is the belt that transmits load to the road surface. Internal pressure is important in a pneumatic tyre because it stabilises both the belt & the side walls that serve to maintain the shape of the belt (as do the spokes of the tweel). Loads are transmitted from the rim to the belt mainly (though not entirely) by the spokes (or section of side wall) furthest away from the road surface - as suggested by Ciro & others. I say not entirely because the side walls do, I think, transmit some load to the rim both by compression & shear. I would concede, however, that the proportion of load transmitted in that way is normally quite small.

p.s. Apologies again, Ringo, but I can't see much similarity between the skirts of a hovercraft & a pneumatic tyre. A hovercraft "sits" directly on a layer of air under pressure. It could (at least in theory) do that with or without skirts. Skirts simply make the concept more efficient by reducing leakage.

[DaveW]

To amplify my last post, consider the tweel you posted. Assume zero degrees is at TDC & think of the spokes attached to the belt in the range +/- 180 degrees (don't mind which direction you assume to be positive).

Cut the spokes between +45 & +135 degrees, & those between -45 & -135 degrees & load the wheel vertically. The spokes between -45 & +45 degrees will cause the upper part of the belt to remain at a fixed distance from rim, but the load will cause the unsupported belt sections to bow outwards (away from the rim) until the rim contacts the belt at the contact patch. Conclusion? The cut spokes would have been under tension.

Now take another tweel sample, cut the spokes lying between -45 & +45 degrees & load the wheel vertically again. This time the upper part of the belt will deform upwards (away from the rim), & the spokes remaining attached will bend & buckle, distorting the belt until the rim yet again contacts the belt at the contact patch. Conclusion? The cut spokes would have been under tension.

It follows that those spokes of an unbutchered tweel attached between -135 & 0 & 0 & +135 degrees (at least) will be under tension & working to maintain the shape of the belt when the tweel is loaded vertically. Note that it was not necessary to pretension the spokes, nor to rely on internal pressure for the belt to support the load.....

[marekk]

Maybe it's because you need only 2% nett change in pressure to support static load of 2011 spec F1 car?

I'm not quite sure how you arrived at your nett change in pressure, Marekk, but that would translate to around a 1 psi change with a reasonable downforce applied, assuming linearity. I hope a change of that magnitude would be spotted by somebody tasked with maintaining constant tyre pressures. Incidentally, just for info, a 185 kg change in down force on one wheel caused a 13 percent increase in vertical tyre stiffness with no observed increase in tyre pressure (in one example).

Dave,
As i wrote in this thread many times, people tend to underestimate air, even looking at 560 tonnes aircraft going airborne at just 160mph.

With a very rough approximation (13" rim, 13" thread witdh and 26" diameter, square profile to make it easy to calculate), at 18 psi, assuming 14 psi atmospheric pressure, total pressure acting on inner sides of tire/rim system is about 10,000 pound. 1 additional psi is worth about 2,400 pound.
185kg change in down force will be worth in this case about 0,15 psi change in pressure and even less considering that part of this load goes to stretch tire's ctructure (no idea how much, hovewer).
I'm not an expert on tires in any way, but i can imagine that F1 tires with very tiny sidewalls carrying vertical loads are quite sensible to tension change.
And sorry, but you can't assume linearity - with additional load you are changing volume - this relationship is cubic.

[ringo]

Apologies, Ringo, but I don't think you understand. Ignoring the undoubted fact that the spokes of a tweel can transmit a small compressive load both before & after they buckle, I think a short ponder over the fine photograph you posted might convince you that by far the majority of tweel spokes under the loading condition shown are under tension. Those under compression are buckled, & transmit very little load.

They are still under compression even when buckling. They haven't passed their yeild as yet. They will return to their original shape like any other spring under a compressive load.
Yes the spokes on top are under tension because the load is being distributed by the band of the contact surface, which is very stiff tube if you will.
But again you cannot ignore the load on the lower spokes; their deflection determine how the displacement of the band and hence how much tension is reaching the upper spokes.
The air will play a similar role in the vertical direction. It has a stiffness, no different than a pneumatic valve in an F1 car.
So, the air that is missing in the tweel is being replaced by the spokes. The sidewall is what's missing here without a replacement.

I think the tweel is very instructive in demonstrating how a pneumatic tyre actually works.

The tweel is demonstrative of how contained air works.

Some have stated that a tyre contact patch area is exactly (& necessarily) equal to the applied load divided by the internal pressure. If that is the case, then why does the belt of a tweel not collapse onto its wheel rim? It is because its shape is maintained (more or less) by tensile loads in the spokes (they don't like to increase in length, if you like). The side walls of a pneumatic tyre work in the same way - when they are stabilised by internal pressure. Having said that, the side walls of low profile tyres tend to be able to support serious loads without being pressurised - for example touring car race drivers complained at one time that they didn't know their 19 inch tyres had punctured until they disintegrated.

I don't think the total contact patch area equal to the applied load / air pressure.
The side wall takes some of the load, but the air pressure is largely responsible for pre laoding the sidewall radially.

For me, the tweel demonstrates that it is the belt that transmits load to the road surface. Internal pressure is important in a pneumatic tyre because it stabilises both the belt & the side walls that serve to maintain the shape of the belt (as do the spokes of the tweel). Loads are transmitted from the rim to the belt mainly (though not entirely) by the spokes (or section of side wall) furthest away from the road surface - as suggested by Ciro & others. I say not entirely because the side walls do, I think, transmit some load to the rim both by compression & shear. I would concede, however, that the proportion of load transmitted in that way is normally quite small.

The belt transmits the load of course, but for the tweel, the spokes would not be deflecting from a zero force. You can't ignore that spokes of the same strenght as the ones on top are buckling.
When you then go on to say the air "stabilizes" a normal tyre, it even shows that you admit that the air is contributing to the stiffness. I can't see what else stabilizes is suppose to mean.

p.s. Apologies again, Ringo, but I can't see much similarity between the skirts of a hovercraft & a pneumatic tyre. A hovercraft "sits" directly on a layer of air under pressure. It could (at least in theory) do that with or without skirts. Skirts simply make the concept more efficient by reducing leakage.

Ahhh you can't see it.
But these very same cables of the suspension bridge analogy have to be sitting on something. They cannot be sitting on the sidewall, for that would mean they would be sitting on themselves and under no loading; which is not logical.
Using the same bridge analogy, what is it that represents the towers that resist the cables?

I see it as the air pressure transmitting uniform load to it's container providing hoop stress to the walls and belt and in effect the cables which is what's supporting the upper side of the tyre.
Deflating the tyre would affect it's stiffness. So a tyre's stiffness is dependent
on it's air pressure.

A run flat tyre is an example of increasing the stiffness of the tyre wall to compensate for the missing contribution of stiffness from the air.

[DaveW]

With a very rough approximation (13" rim, 13" thread witdh and 26" diameter, square profile to make it easy to calculate), at 18 psi, assuming 14 psi atmospheric pressure, total pressure acting on inner sides of tire/rim system is about 10,000 pound. 1 additional psi is worth about 2,400 pound.

I'd be grateful if you could set out how, from the dimensions & pressures you quoted, you arrive at the total load acting on the tyre/rim system...

[DaveW]

Ringo,

Interestingly, I don't disagree with much of your response. Of course the pressure of an inflated pneumatic tyre usually affects its stiffness. I don't think I have suggested otherwise, not intentionally, anyway. My point was, to repeat myself, that loading a tyre does not require a change in volume (& hence pressure) for it to react an applied load. I also stated that the internal air pressure does not change to first order when a tyre is loaded.

I do have a couple of issues, however.

I think you might need to think again about the effect of buckling on the stiffness of a tweel spoke. Wikipedia might help.

I believe there is a difference between, say, putting a plate on top of a balloon & applying a load to the plate (as in your hovercraft analogy), & wrapping the balloon around a wheel & applying a load to the wheel axle.

[ringo]

You posted before i could post this response.
I understand your position on stiffness and it's right, becuase this would have been a knock out punch.
But i'll post it anyway since it's on topic.

RIDE PROPERTIES OF TIRES
Supporting the weight of the vehicle and cushioning it over surface irregularities are two of the basic functions of a pneumatic tire. When a normal
load is applied to an inflated tire, the tire progressively deflects as the load
increases. Figure 1.53 shows the static load–deflection relationship for a 5.60
13 bias-ply tire at various inflation pressures [1.35]. The type of diagram
shown in Fig. 1.53 is usually referred to as a lattice plot, in which the origin
of each load–deflection curve is displaced along the deflection axis by an
amount proportional to the inflation pressure. The relationship between the
load and the inflation pressure for constant deflection can also be shown in
the lattice plot. Figure 1.54 shows the interrelationship among the static load,
inflation pressure, and deflections for a 165 13 radial-ply passenger car
tire. The lattice plots of the load–deflection data at various inflation pressures
for tractor tires 11–36 and 7.50–16 are shown in Figs. 1.55 and 1.56, respectively [1.36]. The load–deflection curves at various inflation pressures for
a terra tire 26 12.00–12 are shown in Fig. 1.57. The vertical load–deflection
curves are useful in estimating the static vertical stiffness of tires

It follows that the slope of the load vs deflection is a stiffness. Therefore for differing slopes across differing inflation pressures, we have an influence on stiffness from the pressure.

Nonrolling Dynamic Stiffness The dynamic stiffness of a nonrolling tire
may be obtaining using various methods. One of the simplest is the so-called
drop test. In this test, the tire with a certain load is allowed to fall freely from
a height at which the tire is just in contact with the ground. Consequently,
the tire remains in contact with the ground throughout the test. The transient

the inflation pressure seems to be affecting the stiffness from analyzing these graphs.

maybe for F1 tyres and their range of pressures, the pressure does not significantly influence the stiffness. But that is a special case.

again i'm no expert, just putting forward some thoughts.
I'll respond to the change in volume argument later. As i said i don't have any firm standing on any position, i don't know a thing about tyres. if something looks logically sound then it's simpler to accept. I may need a diagram which could agree or disagree with your opinion based on what realization it comes to.

[strad]

Damn...I can't believe you're still pickin the fly --- outa the pepper on this.
Fact is,,you can talk all the fancy crap ya want,,When I put more air in the tire the sidewall deflects less and no matter what the real reason, it's harder and stiffer to ride on.
And that's what matters.

[Jersey Tom]

After 18 pages, people still aren't getting it.

I'm convinced people love to hear the sound of their own silly debate. Not even debating the same thing. Like having one person say, "Platoon was the best Vietnam movie," and the other respond with, "I disagree. Stolichnaya is the best vodka."

Ridiculous.

[ringo]

Ringo,

I believe there is a difference between, say, putting a plate on top of a balloon & applying a load to the plate (as in your hovercraft analogy), & wrapping the balloon around a wheel & applying a load to the wheel axle.

Ok let me cast out the hovercraft anaolgy since that is not exact the same loading case.

The belt of the tyre, which is of a considerable stiffness and has a defined unstressed shape, is what supports the radial cables that further support the bead then the rim.
The effective stiffness of this circumferential belt is dependent on the air pressure of the tyre. The belt will collapse without the pressure providing the tension.

This is the stability that you mention. I think the tweel spokes provide this same stability to the tweel belt. The tweel spokes are more like the air in the tyre to me.This is why it can take a compressive load or an impact load.
It acts as the side walls when subjected to lateral loading.

The same picture i posted, if you remove the buckled spokes at the bottom.
I would see the band moving in more towards the lower side of the rim and moving away from the upper side of the rim. subsequently the upper spokes would see more tension. This added tension is the compressive load that was taken up by the lower spokes.
In short the upper side supports the bead and rim, but the lower side balances some of that load by controlling the band shape. Especially under impacts such as kerbs.
This is all my opinion anyway.

[marekk]

With a very rough approximation (13" rim, 13" thread witdh and 26" diameter, square profile to make it easy to calculate), at 18 psi, assuming 14 psi atmospheric pressure, total pressure acting on inner sides of tire/rim system is about 10,000 pound. 1 additional psi is worth about 2,400 pound.

I'd be grateful if you could set out how, from the dimensions & pressures you quoted, you arrive at the total load acting on the tyre/rim system...

OK. Let's try
Rim area = 2*pi*r_rim*rim_witdh = 2*3,14*6,5*13 = 530
Tread area = 2*pi*r_tire*tire_witdh = 2*3,14*13*13 = 1060
Tire wall area = pi*(r_tire^2 - r_rim^2) = 400
Total area = rim area + tread area + 2 tire wall area = 2400 inch square
Pressure = 18 psi (pound/square inch)
Atmospheric pressure = 14 psi
Pressure difference = 4 psi
Total pressure acting on tire/rim system = 4 psi * 2400 inch^2 = 9600 pound.
Additional 1 psi acting on this surface = 1 psi * 2400 inch^2 = 2400 pound.

I know, we are used to it from our very begining, but total atmospheric pressure acting on human body is in fact about 17,000 kg at sea level. I have to withstand more then 20 tonnes every day. We are happy to be mostly incompressible