Traction; How/When is it Lost at the Tires?

[johnc]

No rocket science needed but you do need to be acquianted with algebra, trig, etc. The book is $64 if I recall correctly. I don't recall a reference in the book to the "Mystery Formula" you mention.

[katman]

Originally posted by Greimann:

So why don't all drag cars have 4WD? Because of dynamic weight transfer. A car that can pull its wheels on launch has effectively put all of its weight on the rear drive wheels. Slower cars have a lesser effect but the weight transfer is significant enough that the car is faster with 2WD, than having to carry the extra weight and drivetrain friction losses of 4WD.

 

And people who try to drag race front wheel drive cars are just plain bucking the laws of physics (or is that the law of Civics?).

The law of Civics, that's good. Actually there's probably other reasons why drag cars don't all have 4wd (rules, costs, weight, etc.). When it comes to tires, F=uN is not linear. As the normal force increases the coefficient of friction is generally reduced, so 4 tires carrying a specific weight usually have more grip than 2 tires carrying that same weight. With weight transfer in a drag launch, a 4wd car with the power apportioned according to the resulting weight on each pair would probably be faster than a car with the same power and all the weight on just 2 tires.

 

As for the original question, there's way too many variables involving tire construction (belts, formulation, stiffness etc.) and the pavement it sits on to be very successful arriving at a formula that relates tire width to traction. As stated earlier, while theoretically width should be independant of traction (no area variable in the F=uN equation), we see in practice this is not so. Has to do with the fact that the tire/pavement interface is not smooth so there is additional grip due to "interaction". There may be a paper out there where somebody tested a tire to get grip using width as a variable, but I haven't seen one. Wouldn't be too hard to do and plot a curve though.

[Guest Anonymous]

All of these arguments are very correct, and i have been discussing this with my physics teacher(also avid auto-x'er). In the formula for the Ff, it is the coefficient of friction times the normal force(weight). No where in the formula calls for the surface area of the tire.

 

With wider tires, you also acheive NO greater contact patch if you have the same PSI in the tire. But the thing that is changed with wider tires is the shape of the contact patch.

 

I believe that the shape of the contact patch greatly effects the traction but who knows. The true answer could be truely simple.

[Guest SpudZ]

This months Grassroots Motorsports has an article on wider

tires and traction . It is an on going series on tires they have.

ZF they state the same thing as you do about contact patch,

but the also state that it is patch shape is different. They

compare an 8" wide tire to a 5" . The 5" contact patch

runs higher on the tire then the 8" does. This has to do

with the weight you were talking about.

 

[Kinked_Chrome]

After recently taking High school physics and learning about forces and friction, I was also curious as to why wider tires help for traction. As many of you have pointed out, in the equation: friction force = coefficient of friction x normal force, surface area has no role. Lots of you are probably familiar with Fred Puhn's book "How to make your car handle". He address this in the beganning of the book when defining slip angles and limit of adhestion. He loosely defines this mystery of why wider tires help as "grip". Or the fact that the rubber on the tire interlocks with the road surface, more rubber touching, more interlock. Which is also why the tire compound makes such a large difference as scotty described, sticky tires "interlock" better with the road surface than a regular radial. Unfortunately there is no magic formula given in his book to find optimal tire size.

[awd92gsx]

Have owned and drag raced AWD vehicles, I have to chime in...

 

There is nothing like a good launch of an AWD high powered car.

 

With full interiored street tired DSM's pulling 60 foots of 1.5, it's hard to argue the advantages of AWD. The only AWD issues are breakage of driveline parts.

 

..now...with that being said...

 

It's true that top fuelers, prostockers and the likes will never run AWD or 4WD systems, but, have you seen the size of the tires on those badboys =o) Seriously, though, with as much torque as they have, having an AWD system becomes disadvantageous because the front wheels aren't on the ground anyway and the extra drivetrain would mean extra weight.

 

Now, I'm not arguing the laws of physics, but, one cannot deny the fact that there are a multitude of WWD (Wrong Wheel Drive) 8 and 9 second cars out there. It's hard enough to make any 4 cyl. run an 8, so they must be doing something right.

[Guest Anonymous]

Hey guys,

There are three basic equations that you need to know to understand how tire width/weight/and friction affect each other and when one is greater than the other

1--- Net Force = mass*(acceleration)

2--- Net Torque = I*alpha (the fish symbol)

I = the resistance to acceleration

on a car that would be

the mass times the tire radius

and you have to square the TR.

alpha = angular acceleration

on a car that would be your

acceleration

divided by the tire radius.

3--- Frictional Force = Cf * N

cF - coefficient of friction

N - Mass*gravity (32ft/s/s).

The torque equation makes the assumptions:

that there is no air resistance, that you could actually determine the coefficient of friction.

Basically with the first equation we get the second equation. The difference is just that one deals with linear motion and the second deals with rotational motion. (I.E. the rotation of the tire.)Now to incorporate friction into the fun... you just take the Fnet - friction and set that = to mass*acceleration. You would find the acceleration by taking your quartermile speed and divide it by the ET Then you want to solve for Fnet that is needed to make you accelerate given the frictional force and the mass of your car. After you solve for Fnet you simply muliply that number by the tire radius to determine what torque is needed to make your tires rotate. If you understand everything I just said then you can see that the tire width is not incorporated anywhere in Newton's Second Law of Physics. That may not make any sense... but i think it is right tell me what yall think...

[Kevin Shasteen]

Okay, two months later, from when this issue began I now feel I have come to a few conclusions.

 

I have had a few light bulbs come on, specifically in the area of traction/friction of a car's tire's during acceleration and negative acceleration.

 

Without going into a full 1000 word report I will basically say this:

 

I "feel" (remember-it is politically correct to "Feel") that a car's drive wheels/tires will lose traction if and when the torque being applied to the drive wheels exceeds both the "Normal Force" already applied downwardly on to the tire pushing onto its "Contact Patch", in combination with the tire's load and friction ratings.

 

Normal Force being the amount of weight on the drive wheels at any given moment; such as the Weight of the Vehicle or the %of Weight of the Vehicle on the Drive Wheels. For our purposes "at any given moment" will be under "WOT Acceleration" from an initial standing stop.

 

Notwithstanding an optimum set up steering/suspensioned car, weight transfer-better termed "Load Transfer" will shift weight to the drive wheels. For our purposes-that will be the rear wheels on a RWD vehicle. The amount of weight shifted is determined by the cars Front:Rear weight Ratio, Center of Gravity (Cg) of that particular Vehicle within a latitudinal, longitudinal & Vertical manner.

 

This should beg the question what or how much torque is required, at the rear drive wheels, to match or exceed the "Normal Force". Well, each operator will have to determine their car's Cg. before that can correctly be calculated. But if you accept the usual Front:Rear 58:42 ratio for a GT car-such as the Z then you can do the math and pretty much get close in making an "Educated Guess". Guessing may not be absolute but it will get you in the ball park & give you a reference point of what I am attempting to say.

 

Nonetheless; to determine your car's torque at the drive wheels-simply work the math from the Flywheel thru your Transmission's Gear Coupling (whatever gear you are in at the time), to the Differential Gear (your rear end's gear ratio), to the Wheels (Rolling Ratio). This gives you the amount of weight, in lbs/ft torque at the wheel, and is to be divided into the "Total Amount of Weight" (Mass), which is applied downwardly on your drive wheels at the moment of our hypothetical acceleration. Dont forget to add the Load Transfer weight to the already existing Normal Force.

 

It is "THIS" moment where the rear suspension is shocked into a transistion from static to dynamic motion in its attempt to overcome inertia. If your engine has the "umph" to exceed the "Normal Force" and the tire's load/friction ratings-then your car will achieve that "Burn Out" effect and determine acceleration; or the lack of acceleration depending on how much of a burn out one prefers.

 

This moment, this blink of an eye moment is further measured in increments of Gravity units of 1; or a 1 "g"; which is 32 ft/sec^2 or the fraction thereof if your car is not pulling a full g.

 

The approach given above, if you have not noticed yet, does not include the Tire's Cf. Another approach would be to multiply:

 

Tire's Cf x %Weight on Drive Wheels x Weight of Car

 

This "Tire's Cf" would be the "Static Cf Value" of the tire (go figure).

 

This again begs the question; when and how is the tire's Cf determined...guess we will have to get this number from the tire mfg; if they even release those numbers. I have not checked with any tire mfg's so I dont even know if that figure is available to Joe Public.

 

Anyway, thought I would throw this out there hopefully to trigger someone's thought &/or responses-any responses, as usual, will be greatly appreciated.

 

Unfortunately I still have not found any info on the tire's "Width" or "Contact Patch" and how it effects the ability to keep/lose traction-DOH!

 

Newton never had to deal w/Cf's of a tire-if he had to I'm sure we would not be having this conversational thread. Still, any and all hot rodders know that a wider contact patch at the drive wheels aids traction much better than a skinny tire with less of a contact patch.

 

Still got a few questions regarding a "Tire's Contact Patch" to answer; baby steps I guess.

 

Kevin,

(Yea,Still an Inliner)

[pparaska]

I can't remember if it was Puhn's book (How to make your car handle) or some other book, but you can't use the Friction-force=Normal-force TIMES coefficient of friction equation as you might expect. You COULD measure the friction and normal forces and back out a normal force dependent coeficient of friction, but now all you've done is create an empirical solution.

 

The tire acting on the road is NOT your typical surface friction problem that only depends on normal force and a fixed coefficient of friction. As the tire is loaded, it grabs onto all the minute jaggies of the road surface. The creates a coefficient of friction that is a function of normal force and to some extent the traction force pushing the rubber that is in the little depressions underneath little cliffs and jaggies in the walls of the little depression.

 

Some things just can't be represented with a simple equation.

 

Pete Paraska

MSME

[Kevin Shasteen]

Originally posted by pparaska:

[QB].., empirical solution..,

 

Is this where the "Research & Development" takes over? Once all theory has done its deed-then give it over to the R&D Dept?

 

.., The tire acting on the road is NOT your typical surface friction problem that only depends on normal force and a fixed coefficient of friction..,

 

I know I have seen Cf's in books regarding surface frictions; would not one be able to use the the appropriate Cf in regard to the tire and the road......"AND" and this is a big and if you have noticed, also factor in the "Static Cf" rating of the tire applied to it from the mfg. Are these two Cf's not different or are they one in the same?

 

.., Some things just can't be represented with a simple equation..,

 

Dont tell me that; that is not the answer I wanted to hear Perhaps getting absolutes from an equation isnt or should not always be the appropriate approach. Instead, use the equation to get the user in the ballpark of his intended objective....then allow the R&D to take over from there as this is where the real fun begins; yes/no?

 

Kevin,

(Yea,Still an Inliner)

[johnc]

There is also a chemical bonding of the tread rubber to the road surface. Race Car Engineering had a good discussion of this topic a couple issues ago in regards to the traction control systems on F1 cars. Basically, the top F1 teams in conjunction with Bridgestone and Michelin are still not able to develop accurate algorithms that model tire CF (both physical and chemical) for traction control software. There' still some guesswork involved (temperature plays havoc with the models) and lots of trial and error testing during practice and qualifying.

[pparaska]

Yep, build it and test it. Change something, build and test again. Well, they're a little more intellegent than that on the matter, but ultimately, they build and test!

[Kevin Shasteen]

Perhaps, for now-up until someone comes up with the answer, we should be content in only knowing what we do know.

 

I still think that the approach to the Power:Weight Ratio in comparison to a skinny tire -vs- a wider tire cant be dismissed; even if we dont fully understand all elements involved.

 

Perhaps we should be content in knowing that a skinny performance tire will lose its ability to "adhere" quicker than a wider performance tire.

 

I guess what I'm getting at is this; if Race Car Engineering, the F1 Teams, Bridgestone and Michilin cant put a finger on the "final elements" involved in tire Cf's as JohnC commented on, then perhaps we lowly Hot Rodders should not worry about it either-let the experts toil in that respect.

 

My purpose for attempting to understand the traction/friction aspect was an attempt to alleviate the often seen problem of our V8z's inability to launch w/o excessive tire spin.

 

It is my suggestion that if an individual were to begin with an engine with a known power output, work the math from the flywheel to the rolling radius and finally to the tire, and if one wanted to start with the tire and work the math backwards. Anyway,is still my suggestion that one could get somewhat close in knowing "How Much Power Is Too Much Power" for that particular set up....much less for any vehicle and not just our Z's. In that attempt I quickly learned that the limit of the tire's ability to adhere under that WOT Acceleration was the limiting factor: the 'True' unknown variable if you will.

 

Let's face it, if we were to put a 7"BiasPly 1970's tire on our V8z's (myself not included as I'm still an Inliner) would not the V8z experience more wheel spin? This excessive wheel spin (any wheel spin for that matter) is evidence that traction has been lost and/or the tire's ability to adhere has been exceeded (Overloaded).

 

Maybe, just maybe then, we should be content in knowing 70% or 80% of the tire traction/friction problem and not concern ourselves with knowing 100% of the problem(?)!

 

Kevin,

(Yea,Still an Inliner)

[Guest Tom Scala]

I'd have to agree with those who said there is no absolute answer to this question. Have you ever gone to the strip,made 3 or 4 passes and hooked good,then on the 5th the tires break loose? Same tires,same car,same track,same burnout & launch technique,same day but there was a slight change in the track surface that caused you to loose traction.

As far as tread width goes, an 8" TWx28" tall slick will have more traction than a 9"x27" tire. The taller tire will have a longer foot print and put down a larger contact patch in the direction of travel than the wider tread tire,all else equal(whatever that means). That is why drag slicks have wrinkle down sidewalls. Aside from absorbing the shock load of the driveline the wrinkle makes the contact patch longer,not wider. For cornering you want the opposite,short stiff sidewall and widest possible tread.

Equations acn be fun to play with and will get you in the ballpark but the best combo can only be found after many controlled test-n-tune sessions at the track with a close eye on the 60' times,just like the pros do it.

[Scottie-GNZ]

Dan,

 

your post hit the nail on the head and I agree with everything you said and want to add that tire diameter also plays a big role in contact patch.

 

For the hybridZs out who have problems launching their car, you need to understand 2 things:

 

1. Street tires are not going to cut it Even the best 295 ZR street radial cannot and will never hold a candle to a lowly 235 DR with a little heat in it. Just ask RickB. You have major $$$ tied up in that beast, spend an extra $250-300 and get some DRs and I guarantee you will improve your ET by at least .5. If you want to drive them on the street, get the Nittos. People spend thousand$ to gain 1-sec in their ET.

 

2. Suspension - You have to get proper weight transfer and stop the rear suspension from unloading too quickly. "Sports car" suspension setups are a compromise at best on the drag strip. If you want both, spend the buck$ for Illuminas and have the best of both worlds.

 

No one said it was cheap

[johnc]

Unfortunately the Illuminas adjust bump and rebound together. Ideally, a set of double adjustable shocks would help considerably. Almost full soft on compression and full stiff on rebound in the rear and medium on both in the front. Use the front to control how the car handles and the rear to control the weight transfer.

[Guest AlsoRanFPrepared]

I think you already have the equation that you need. The problem is it has been using a simple linear coeffecient. This equation is far from simple and will only be able to try and model the real world. F=f(N,S,T,P,D,...)*N where N is the normal force and the F is the force resisted by friction. The difficult part is the f(N,S,T,P,D,...). This would be the function representing the frictional coefficient. I only included a few of the many factors those being N-normal force, S-surface characteristic, T-temperature, P-pressure and D-diameter. There are many others invoved. Considering we are looking at a tire and not just the rubber compound the function representing the frictional coefficient is very nonlinear. As an example the force that a given tire can transfer is not linear with normal force or weight. As the normal force is increased on the tire it deforms and changes how it is loaded and also how it transfers torque through itself to the ground. If for instance insufficient pessure is present once loading is increased the tire may actually support most of its load on the sidewalls. This will significantly reduce the area of the contact patch and will reduce the traction available. If you dont believe me you can give it a try. Next time you get a flat on both rear tires try and take a good launch. It wont be pretty. Many inexperienced drag racers try to improve their traction on regular radials by dropping the pressure too low and end up being worse off than running the tires at the manufacturers recommended pressure. Another example is the one that drag racers mention. Hot tires have better traction than the cold ones. Even if we matched the pressures inside the same tires so that the contact patch was identical in size the hot drag tire would generate more friction than the cold one. The difficulty in defining the relationship between a tire and the road in a car has led tire manufacturers to perform extensive testing of their tires. This testing gives them the emperical data where curves can be fit and equations can be generated. Unfortunately we do not completely understand this relationship and can only estimate it.

 

-Mark

[Dan Baldwin]

As has been mentioned, F=u*N doesn't begin to describe the interaction between tire and pavement, particularly a drag slick. The reason greater contact area (more a function of tire pressure than width) improves grip is that there is an adhesive effect between a hot drag slick and a grippy launch pad. There would be some tractive grip between the two even with ZERO normal force. Try and try again 'til you find which tire works best for YOUR car. No need for you to understand all the intricacies of what's going on. Probably only a handful of people in the world who really have a grasp on it anyway.