42 replies to this topic

[desmo]

We've all read how aero loads at high speeds can more than double the downward force fed into the tires compared to the static weight of the car. My questions are: Does this increase in force manifest itself as a proportional increase in the area of the contact patch of the tires? How does the aero downforce affect the Coefficient of Friction of the contact patch? And finally: If one was to have access to dynamic telemetry from a wheel tire pressure sensor, how would this change as a car accelerated from a standstill to 350kph as enormous aero loads are fed through the tire into the track? [p][Edited by desmo on 06-23-2000]

[gunner]

1. No
2. Increases
3. Increases

Gunner

[John Oliver]

Other than heat, I don't think there's any real reason to think that the tire pressures would change signficantly. So I think we could assume that the contact patch would increase in size with downforce by itself. There may be some interaction with the increase in rolling diameter of the tire due to centrifugal growth at speed, but I don't know how significant that might be.

As tire loads INCREASE, contact patch coefficient of friction tends to DECREASE. The effective cornering stiffness of the tire thus decreases with load. Combined with the front/rear aero balance of the car, this will obviously have significant implications for understeer and handling balance.

As far as the dynamic pressure trace, I think pressures would increase due to the tractive load causing a temperature increase. At high speeds, the coefficient of friction would be less, due to the aero load, so the temperature increase might slow down, but as long as you're applying tractive force (even at constant Vmax), it seems like you'd continue to get some temperature and pressure increase.

[gunner]

Doesn't Dry Nitrogen eliminate the pressure increase due to the increase in tire temperature? In fact in turns you would have a variation in air pressure and patch size due to inertia and body lean shifting weight to the tires on the outside of the turn.

Gunner[p][Edited by gunner on 06-23-2000]

[John Oliver]

I believe using dry nitrogen reduces the change in pressure with temperature, but you can't eliminate it. If temperature increases, volume and/or pressure must increase in some combination. That gets at my comment about centrifugal tire growth -- the volume does increase with speed, but I doubt if it's enough to allow pressure to remain constant.

[gunner]

John.

I guess the answers are really only known by the engineers who design and test them in the lab. They have so many ways of changing materials and laying out ply patterns that it can change from race to race.

Gunner

[PDA]

The relationship of pressure to temperature of gasses is simple and straight forward.

The reason for using dry nitrogen is to ensure that there is no water in the gas used to fill the tyre. If there are droplets of water in the gas, there will be an enormous increase in pressure when it turns to steam.

[Wolf]

PDA has a point. Nytrogen is also used because of variations in air properties (contents of consituents vary- thus eliminating possibility of detemining exact behaviour of air with temperature and pressure changes). For this purpose in aviation is CO2 used.
As for friction, coefficient of rolling (am not sure if it's correct term, friction due to revolution of tyres) varies more significantly with speed than with pressure or downward force.

[Wolf]

Pertaining Desmo's last question is: it would definitely and significantly increase the pressure with speed (since the downward force, same as the drag, is proportional with cube of the speed)

[desmo]

Actually Wolf it increases as the square of speed.

A simple mathematical model predicts that the increase in contact patch would be proportional to the downward force on the tire. This also predicts that the tire pressure would remain the same as the force increased. The pressure between the tire and the ground vertically should not exceed the pressure inside the tire no matter how much force is applied. If the pressure inside the tire remains essentially the same as more load is applied the force can only be balanced by a proportional increase in the area of the contact patch.

I would like to see some empirical evidence before I trusted the results of a mathematical model that may not consider other variables that are not obvious.

John if tractive force might potentially raise the temp. and hence the pressure of the gas inside the tire would the temp./press. more likely remain constant in a coastdown?[p][Edited by desmo on 06-24-2000]

[Wolf]

Sorry about that blunder; it's power that increases with the cube. Pressure in the tyre remains the same but, as I gather, the local pressure in vicinity of the contact surface increases, and in the rest of the tyre decreases accordingly. So the contact surface doesn't have to change proportionaly to the load.

[gunner]

Desmo.

I don't doubt that your theory is correct but I will run a test in the morning. I have a very accurate screw on air guage. So I will jack up the front corner of the wifes Explorer. Put on the air guage read it then remove the jack and read the guage again. The guage is graduated in 1/4 lb incraments and should tell if tire pressure increases under load.

Gunner

[gunner]

Desmo.

If the pressure between the contact patch and the road doesn't exceed the pressure inside the tire no matter how much down force is applied! Put 20 lbs of air in the front tire of your car. Then jack the car up and put your foot between the contact patch and the road and slowly release the jack! Then tell us why you are off work with a mashed foot. 20 lbs should never hurt you should it? And no I'm not going to try it.

Gunner[p][Edited by gunner on 06-24-2000]

[Yelnats]

Increased load will raise both the tire preasure and contact patch area while slightly decreasing the coeficient of friction.

Anyone whos (over)loaded a trailer will note how the tires squash down under load. This means there is a reuduced volume in the tire (increased preasure) and obviously a larger contact patch. Of course this is modified by the stretching of the tire carcass and stiffness of the sidewall so the whole thing gets pretty complicated.

The reduced co-efficient of friction comes about because part of the resistance to sliding comes from the mechanical strength of the interlock between the tire and the roughness of the track. When this limit is exceeded the rubber sheers off leaving the notorious skid marks. Because the strength of the rubber remains constant with the increased load an effective reduction in the co-efficient of friction is the result. Again this is complicated because the increased tire preasure results in more interlock per unit area but the rubber shear strength is still a constant in this equation.

[gunner]

Yelnats.

When tires squash down from over load are they not under inflated for the load they are carying? If so an under inflated tire will cause the contact patch to buckle upwards in the center and cause the tire to ride on its outer edges? And in my test this morning pressure does increase with load.

Gunner[p][Edited by gunner on 06-24-2000]

[gunner]

Can the ply design on an F1 tire hold the contact patch paralell to the road surface no matter how much air pressure the tire has?

Gunner

[desmo]

The model assumes that the tire is perfectly elastic. Obviously it isn't. Mathematical models seldom do any justice to the complexity of what they are trying to describe. Physical realities cannot easily be reduced to variables and constants in an equation.

[gunner]

Desmo.

Did you try my foot and tire patch test? I know a couple of professors of physics that I met at our Astronomy club. These guys are so convinced that they are correct at everything that they would have tryed the test then walked around for six weeks with there foot in a cast. Refusing to tell the class that they were wrong. Even though several in the class told them not to try it.

Gunner [p][Edited by gunner on 06-25-2000]

[Yelnats]

Gunner, I used the squashing of an overloaded tire to illustrate what takes place to a lesses extent when a tire is loaded within it's limits. And I'm glad to see your tests confirmed my analysis about increasing tire preasures with increasing loads that indirectly confirms my comment that the tire contact patch dosn't increase in direct relationship to the load.

And thanks for doing the test! I once berated a tire technician for setting the pressures on my new tires while the car was on the hoist and he was quite surprised to see that I ended up with overinflated tires when he set it down! {;->

[gunner]

In my thread on ply design what I was getting at was. If something like a Kevlar belting on the contact surface canot hold 100% of the contact patch paralell to the road surface between the fastest and slowest corners and turns do to pressure variations. Which are not constant do to downforce increasing with speed. A compromise has to be reached in tire pressure as we have a variation in contact patch size. If the pressure gives 100% contact on a 30 mph turn it will give less on a 150 mph turn due to under inflation. It seems like the ideal design would be to have the side walls strong enough to not flex and let the increased or decrease in pressure hold the contact patch at 100% at any speed. This may be varafied by the tire engineers checking the temperature of the contact surface after a pit stop. Any comments on this theory?

Gunner[p][Edited by gunner on 06-26-2000]

[goGoGene]

By increasing the contact force you DO NOT increase the coefficiant of friction, you increase the allowed maximum force before static friction breaks and you enter kinetic friction.

F = uN

N = normal force, which would be increased due to the down force

u = coeff. of static friction, this would change with respect to temp, but NOT contact force, otherwise why would we physicists bother putting "N" in there in the first place.

F is the maximum force perpendictular to the Normal force before sliding (loss of static grip) occures.

ggg

[gunner]

ggg.

How does the formula compensate for the variation in contact patch area which varies due to tire pressure which varies due to down force?


Gunner

[goGoGene]

Gunner great question...

This equation is amount of frictional force per unit area, so it would be

F = u*N*A

"A" (the contact patch) in this case varies wrt N (weight of car, plus down force), and yes tire pressure, so...

F = u*N*A(N,P)

what is the function A(N,P) equal to?

I don't know. But notice that "u" would not be a function of this, but it would vary wrt temp and humidity, as would the tire press vary wrt temp...

F = u(T,%H2O)*N*A(N,P,T)

and with that, it's time for a beer, leave solving u(T,%H) up to the materials engineers, they have to earn a living too ya know.


ggg[p][Edited by goGoGene on 06-26-2000]

[mojo jojo]

right. coefficient of friction is strictly a material property. Take a 1'x2'x3' piece of, say, steel. It will take the same amount of force to drag it along the floor no matter which side it is laying on.

[gunner]

ggg.

These equasions are way beond me. But one very inportant thing you mentioned was that it takes more than one engineer to design a tire. Cord,Ply, sidewall,bead,contact and rubber compound engineers, I use to work at Goodyear in Akron when I was a kid and saw a lot of these things done.

Gunner

[gunner]

mojo jojo.

IF we knurl one side with pointed knurling and polish the other will it take the same amount of force to drag either side?

Gunner[p][Edited by gunner on 06-27-2000]

[John Oliver]

Originally posted by goGoGene
By increasing the contact force you DO NOT increase the coefficiant of friction, you increase the allowed maximum force before static friction breaks and you enter kinetic friction.

<...>

u = coeff. of static friction, this would change with respect to temp, but NOT contact force, otherwise why would we physicists bother putting "N" in there in the first place.

You are right in your description of static friction for most materials. However, pneumatic tires are strange and poorly understood devices, and the usual static friction model does not completely describe what goes on in the contact patch. For tires, the coefficient of friction actually tends to DECREASE as normal load is increased.

In the range of any of the curves I've seen, the product of normal load and coefficient of friction continues to increase as load increases (i.e. the degradation in friction is more than offset by the increase in normal load), but at an ever-decreasing rate. So it would seem that the more downforce you have, the less you gain from the next incremental increase (diminishing returns).

[Leo]

GGG,

Unfortunately things are far more complicated than the formulae you propose. The main thing to consider is that u (fric. coeff.) depends on the sliding speed in the contact patch. Since there is a sliding area and an adhesion area within the contact patch, the sliding speed and thus u(!) will vary throughout the contact patch. Varying the load on the tyre will change the contact patch area and also change the sliding/adhesion characteritics, so it is more than likely that the 'total' u will vary also.

[gunner]

I think that the answer to the original question is. The original question was miss stated as a simple question. The answer is very complex and takes many engineers from the tire manufacturer to try to come up with the ultimate racing tire.

Gunner

[goGoGene]

Leo, are you saying that portions of the contact patch are sliding and some are static (with respect to the road)?

I was assuming that u was for the static (not sliding case). I'm not disagreeing with you, I just find it hard to picture what's going one with the contact patch where the coeff. friction varies with respect to location u(x,y).

This makes some sence to me, seeing that I read that Michilin were having problems with their tires being too grippy, the problem being that u(static) transitioned to u(kinetic) too quickly, and the cars were snapping out of control.

So what you said now makes some sence, that some of the conact patch starts to slip (though I still can't picture what's going on) so that you achieve a smooth transition from us to uk (heh, that's funny).

So the equation now stands as...

F = u(T,%H2O,x,y)*N*A(N,P,T)

ggg

p.s. I'm now just being funny, this is what we used to do during my masters, to solve a problem you realy had to simplify a bunch of stuff, but sometime we would think about how hard the real problem would be if we had to solve it.

Gunner, yes, I completely agree with you, this is a very hard problem, not to be solved in this thread, I'm pretty sure I have proved that to myself. [p][Edited by goGoGene on 06-27-2000]

[gunner]

ggg. I know that I don't fit in with you guys and your educations but I have learned that at times experience has an advantage over education. For example a team of the greatest tire engineers in the world can design the ultimate racing tire! And some uneducated pit worker can prove they were all wrong with a simple tire temperature guage.;)

Gunner

[gunner]

Another example.

If any of you guys have ever seen a tire testing machine? They mount the wheel and tire on pedastels with bearings. There is a large driven drum that is movable. They spin the tire and can controle the speed and down force and make videos and computer readouts of the test. If you could see the pile of scrap tires outback you would wonder why these bosos are paid big money. A lot of it is trial and error.

Gunner[p][Edited by gunner on 06-27-2000]

[Yelnats]

The co-efficient of friction of a tire bears little relationship to a rigid material like steel that comes very close to the theoretical model we all learned about in Elementry Phyisics.

The force needed to slide a block of steel on steel will show a linier relationshipt to load as we learned in school. This is because the two surfaces bump over each other with little loss of material (unless huge loading forces are involved).

A rolling tire instead has many of the characteristics of a gear operating on a infinite rack in that the tire interlocks with the pavement to provide grip. As soon as it starts to slide it's as though the "teeth" are breaking off and the sheer strength of the "teeth" are a major determining factor in the coefficient of friction. Because the shear strength of the interlock is independant of the loading the simple rules of elementary physics don't apply here.

If these simple rules did apply then race cars would run on the narrowest tire possible to reduce air resistance. But of course this is not the case and the wider tire provides more area allowing a softer coumpond for more interlock with less loading per "tooth". We see tires going "off" when the overheat because the shear strength of the tire has been reduced below optimum and the interlock fails at a lower level.

Incidently the co-effecient of frictiion is independent of speed and the increase in traction with speed we are discussing here is due to areodynamic loading only.

[Ray Bell]

Poor old Art has always been defensive and sensitive about his lack of education, but here he explains just why it's something he shouldn't worry about. The guy in the pit lane with the tyre gauge shows it.

[Leo]

GGG, the leading edge of the contact patch is in adhesion, while the trailing edge is sliding.

This is the best I could find on the net:
http://www-tt.wbmt.t.../module1A_2.ppt

Skip to slide 22, and you see what I mean in the bottom graph. If you want details of what's happening you should search you're local technical library for the work of prof. H.B. Pacejka, he has done extensive work on tyre modelling.

[gunner]

Ray.

I do resent not getting an education. As the Professer told me when he couldn't answer my questions about the Observatory we were building. (It's a damn shame you didn't get a dagree in engineering as you could of made one fine engineer)Quote Bob Doyle Professer of Astronomy and Physics Frostburg State University.

Gunner

[gunner]

Ray.

I'l give you old you ungreatful clod. You seem to forget that when I posted on the board the first time you were happy as a sissy in the YMCA because you weren't the oldest any longer! Straighten out you old pecker before I come down there and knock you out.

Gunner[p][Edited by gunner on 06-28-2000]

[mojo jojo]

>>The co-efficient of friction of a tire bears little
>>relationship to a rigid material like steel that comes
>>very close to the theoretical model we all learned about
>>in Elementry Phyisics

Actually, it is exactly the same. The only difference is the scale of the surface irregularities and the force required to remove material. Put a steel shaft with some rough spots in a brass journal bearing, and it will remove material from the bearing quite easily. The mechanisms of contact mechanics and friction are the same no matter what the material.

>>As soon as it starts to slide it's as though the "teeth"
>>are breaking off and the sheer strength of the "teeth"
>>are a major determining factor in the coefficient of
>>friction. Because the shear strength of the interlock is
>>independant of the loading the simple rules of elementary
>>physics don't apply here.

Again, it's exactly the same. Two metals in contact also exhibit exactly the same behavior as you describe. The difference is that the shear strength of metals is much higher and the surface irregularities are much smaller.

Frictional forces vary over the contact patch as parts of the tire approach and receed from the pavement, but because the applied normal force is varying, not because of a change in coefficient of friction.

>>If these simple rules did apply then race cars would run
>>on the narrowest tire possible to reduce air resistance.
>>But of course this is not the case and the wider tire
>>provides more area allowing a softer coumpond for more
>>interlock with less loading per "tooth". We see tires
>>going "off" when the overheat because the shear strength
>>of the tire has been reduced below optimum and the
>>interlock fails at a lower level.

Well, no, because what we are talking about here is the maximum allowable shear stress, or force per area. Every material has a maximum shear stress that can be applied before bits start to tear away. Decreasing the area of the contact patch increases the stress while force remains constant(since area is decreasing). This is why a wider tire allows faster cornering speeds.

Wether or not the coefficient of friction varies with load depends on the reaction of the material on a molecular level. It is at least possible that with a big enough load, the structure of the cross-linked polymers that make up tire rubber will change, giving a different reaction to shearing forces. Temperature definitely has an effect. Heating the tire sufficiently will degrade the molecular structure (by breaking the links between polymer chains), resulting in a decreased coefficient of friction.

I'm just trying to get across that coefficient of friction is strictly a material property. It has no relation to applied load, unless that load can somehow alter the molecular structure of the material.

[goGoGene]

A thoughtful reply mojo jojo...hey didn't I see three little girls kicking the crap out of you the other day?

ggg

[mojo jojo]

Originally posted by goGoGene
...hey didn't I see three little girls kicking the crap out of you the other day?

Yeah, but I'll get 'em next time.

[Yelnats]

posted by mojo jojo - "I'm just trying to get across that coefficient of friction is strictly a material property. It has no relation to applied load, unless that load can somehow alter the molecular structure of the material."

And I'm just trying to get across that when the load and lateral forces exceed the shear strength of one of the materials at the interface, then the coeffecient of friction is no longer linear. The material properties of rubber are not linear because the rubber starts to shear before it slides. So the shear strength of rubber is an important part of determining it's coefficient of friction and because it remains constant and as the load increases the coeeficient of friction varies as the load increases.

Just to complicate things a bit I should mention that we are talking about the coeeficient of friction of asphalt also but that does not vary with the small loads applied by a tire. The shear strength of the asphaltic cement bonding the aggregates in the pavement exceeds that of the rubber by such a large amount (assuming temperatures below 90c)that it's cooefficient of friction is essentially linear.

It wasn't long ago that race car designers proved physicists wrong who had theorised that a car cannot exceed 180 mph from a standing start in the quarter mile. I hope we're not in a similar situation here! (and bumble bees can't fly) {;->
[p][Edited by Yelnats on 07-03-2000]

[mojo jojo]

Originally posted by Yelnats
And I'm just trying to get across that when the load and lateral forces exceed the shear strength of one of the materials at the interface, then the coeffecient of friction is no longer linear.

Well, you'll get no argument from me when you say that the situation changes once the tire starts spinning faster or slower than it should, or starts to slide perpendicluar to the direction of motion. When the tire slides the rubber melts, and the bits and pieces of rubber act sort of as a lubricant.

Originally posted by Yelnats
The material properties of rubber are not linear because the rubber starts to shear before it slides. So the shear strength of rubber is an important part of determining it's coefficient of friction and because it remains constant and as the load increases the coeeficient of friction varies as the load increases.

Not quite sure what you're getting at here when you say "starts to shear before it slides". Are you saying shear as in deflection due to a shear stress, or shear as in tearing away material? Rubber is classified as a non-linear, elastic material because of its response under load (shear or otherwise). Rubber's displacement under load is not linear WRT load. Elastic means it returns to its original dimensions when unloaded, as long as it wasn't stretched to failure.

Remember the tire is in quasi-static contact with the pavement. Friction can be considered a force which resists motion, given (simply), as stated before, by the formula F=u*N. As an example, consider a single wheel with mounted tire. Just a wheel, no car. Balanced on this wheel we find one elephant. The wheel is fixed (can not rotate). Attatched to the wheel/elephant system is a length of sturdy nylon rope. Spliced in the rope is a load sensor with radio telemetry. Waiting to pull on the rope is Dr. Bruce Banner from the Biology department who graciously volunteered to help us with our study. The position of the wheel is measured by a laser rangefinder. We (from a safe distance) monitor the telemetry and signal Dr. Banner to begin pulling. We see on the computer screen that the load increases while the wheel/elephant system remains stationary. We then order our underpaid research assitant to rush out and slap Dr. Banner silly and if necessary, to question his manhood. He is hesitant at first, but we offer to include his name when the paper is published and he rushes off to complete his task. This action has the desired effect. Dr. Banner's skin pigmentation begins to turn green and he begins to grow in size, his clothing ripped to shreds by the enormous increase in muscle mass (all except for his pants, as his waist size appears to be unaffected by this transformation). The load increases rapidly now untill reaching a maximum value, after which the tire slides along the pavement under the influence of a load somewhat less than the peak value.

Maximum allowable shear stress (or shear strength) only determines at what point the tire begins to slide, and the amount of force (coefficient of friction) required to keep it in motion. In normal rolling, we are concerned with what happens before the tire begins to slide.

(Yelnats)
"So the shear strength of rubber is an important part of determining it's coefficient of friction and because it remains constant and as the load increases the coeeficient of friction varies as the load increases."

Yes, once the tire is sliding the shear strength will determine how much force is required. Are you saying here that in F=u*N that F remains constant, so if N increases u must decrease? I would suggest that F is not constant as N increases. The irregularities of the road surface are pushed deeper into the tire by the greater force, and have to remove more material to continue sliding. F would increase with increasing N. Wether u is constant in this case or not I can't say. I would have to do the experiment, or call Goodyear. My guess would be that it does not vary much.

Originally posted by Yelnats
Just to complicate things a bit I should mention that we are talking about the coeeficient of friction of asphalt also but that does not vary with the small loads applied by a tire. The shear strength of the asphaltic cement bonding the aggregates in the pavement exceeds that of the rubber by such a large amount (assuming temperatures below 90c)that it's cooefficient of friction is essentially linear.

Yeah, 'coefficient of friction' only has meaning when defined in terms of two materials in contact. Otherwise you've got one hand clapping, and we don't want to open that can of worms.

Originally posted by Yelnats
It wasn't long ago that race car designers proved physicists wrong who had theorised that a car cannot exceed 180 mph from a standing start in the quarter mile. I hope we're not in a similar situation here! (and bumble bees can't fly) {;->

well geee willeekers. dem dar's sum pertty intelleegent-like physeesits ain't they!

[Bennett]

Just my two cents on the proceedings, but in Carol Smith's fine book Tuning to Win, there are several diagrams of tire slip angles vs. cornering load, etc. Slip angle is not news to anyone but I believe that it states that even without any lateral loading there is still slip occuring at the interface between the tire contact patch and the road.

This slip means that there is always local deformation and movement of the contact patch. I am not sure exactly how that would affect the coefficient of friction, i.e. how much is dynamic, and how much is static?

This implies that the different layers of the tire are deforming at different rates at all times due to both vertical and lateral loads. This would also make sense as tires heat up even without the "scrubbing" caused by cornering.

Also in Smith's book I believe it also states that dry nitrogen is used not because of its coefficient of thermal expansion, but because of its predictable expansion vs. air which is a mixture of gasses, and also because of its chemical inertness - i.e. non-flammable!

This discussion of all of the factors in tire construction and design really make me appreciate the great lengths that companies like Bridgestone and now Michelin are going through to compete in F1.

Does anyone miss Goodyear?