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

[Kevin Shasteen]

I've been reading for sometime now about g-forces in respect to a car's ability to accelerate from a standing start. I've come upon a few formula (or is that formula[e]) & they talk about the flow of torque from the flywheel, thru the trans, to the driveline, thru the differential & to the ground via wheel thrust...I understand all that; I've worked those formula (or formulae?) till I grasped the idea.

 

My question is this, none of those articles explain or suggest what effects the "Width" of the wheel/tire has on traction, or the lack therof. They all talk about wheel thrust and only require 1/2 the distance of the wheel/tire but still say nothing about the width of that wheel/tire.

 

So, I guess I'm asking, "At what point is traction gained or lost in relation to the width of the tire/wheel & how is that determined?"

 

Any response is appreciated: you can keep it as simple as you like or get as complicated as you like...I would just like some direction as to how width plays a part in traction & how it is determined.

 

Kevin,

(Yea,Still an Inliner)

[Tim240z]

Kevin,

Please take this as a comment from the uneducated , but I would imagine that the aspect ratio of the tire would have a significant impact here. I would imagine that, although low profile tires are great for lateral traction, maybe not so much for accelleration traction. just food for thought.

Tim

[Kevin Shasteen]

Thanks for your response.

 

However, My question wasnt intended to be directed towards low profile tires & their aspect ratios. My question was meant more towards the section width of the tire in general...towards acceleration in general.

 

If we would assume (look out-here comes my assumption) that the few formula I've come across only refer to the rolling radius (rolling raidius includes the aspect ratio) but doesnt even make reference to the section width in regards to traction & acceleration on a drag strip; then, if section width isnt mentioned-therefore isnt important, why arent we all putting bicycle tires on the rears of our cars...obviously the width of a tire does make a difference on acceleration; but what is the formula for determining what that section width should be for a given particular set up?

 

My inquiring mind wants to know(?).

 

Kevin,

(Yea,Still an Inliner)

[Dragonfly]

I don't have a formula for you but I have some comments. I have several books on handling, suspension, tires/wheels etc. one of the things that these books have in common is they refer to the molecular structure of the rubber in the tire and how each molecule caries a certian amount of load and reinforces the molecules around it therefore the more molecules of rubber you have creating friction between the torque and the road the more traction you are going to have. The trade off (there is always a trade off) is that the more traction you have the more friction you have which means more heat and lower fuel economy as well as more power required to move from point A to point B.

 

Dragonfly

[SleeperZ]

My thinking (and that's all it is, but I did get an engineering degree) along those lines is the tire width should follow the weight of the vehicle [for a given tire compound]. Obviously the softer rubber compunds give you more grip, but shear off easier, so by making it wider you are increasing your grip. I think there is a law of diminishing returns, where wider won't give you any more traction - this should be where the reduced weight on the tire (per square inch) gives you less friction. [frictional force is the weight/square inch times the frictional coefficient of the tire compound]

[Guest greimann]

Kevin, what you are asking essentially how does one determine when a tire system reaches the limits of static friction and transitions to kinetic friction (not to be confused with rolling friction). The standard model of friction assumes that the frictional resistance is independent of surface area, given by the formula:

fr = cf X mass

 

where FR = frictional resistance, cf = coefficiant of friction and Mass = mass.

 

So from this theoretical formula, we can see that the ability of a vehicle to accelerate is independent of the surface area of contact of the tires.

 

- BUT -

 

The wild card in this problem is the coefficient of friction, which is dependent on a variety of factors that cannot be easily boiled down to a simple formula. Several factors that come into play are: Tire compound, road roughness, dirt, or water, and many other factors that affect the kinetic coefficient of friction, including tire width.

[Kevin Shasteen]

Ok, I believe we're heading in the correct direction here. I do know that tires, especially performance tires, have a coefficient of 'g' rating applied.., how is that 'g' rating determined, is it based on SleeperZ's comment:

 

Originally posted by SleeperZ:

..,[frictional force is the weight/square inch times the frictional coefficient of the tire compound]

..,

 

If so, then what I'm understanding is: The frictional force (torque overcoming inertia?) is equal to the weight (weight of the car?) divided by the square inch of the tire (square inch of the tire's foot print where it contacts the surface of the ground?) multiplied by the frictional coefficient of that tire? In frictional coefficient of the tire are you refer'g to the 'g' rating of that particular tire?

 

Kevin,

(Yea,Still an Inliner)

[Kevin Shasteen]

Doh, Greiman beat me to the punch-he answered before I could reply to SleeperZ.

 

Originally posted by Greimann:

[QB]..,where FR = frictional resistance, cf = coefficiant of friction and Mass = mass.

 

So from this theoretical formula, we can see that the ability of a vehicle to accelerate is independent of the surface area of contact of the tires...,

 

So, Greimann, are you telling me that the tire is basically just a crapshoot thing? Remember, I posed the question regarding the dragstrip; there's got to be a common approach to what tire is right & what tire is wrong for a particular setting. It seems hard for me to grasp that the pro's simply use a tire because someone asked them to try it. There had to be reasons for them wanting to try it other than, "Well, its got to be better than what we were using?"

 

So Griemann, your FR=Frictional Resistance...are you speaking about torque overcoming inertia(?), and your Cf=Coeff.Friction...is this the Cf rating of the tire(?) and Mass=being mass of the vehichle or mass of the tire?

 

Thanks for all your responses guys...I'm learning something new here-so bear w/me.

 

Kevin,

(Yea,Still an Inliner)

[Guest greimann]

I don't think tire width was just chance, it was based on experimental data that showed that all other factors being equal, a larger contact patch has a higher coefficient of kinetic friction, for what ever reason.

 

The frictional resistance is the ability of the tire to transfer force to the ground, which in-turn usually translates into forward motion, or maximum braking. With high powered vehicles, the drivetrain has the ability to provide more force to the tires, than the amount of frictional resistance the tires can provide to translate into useful work. That is the threshold of motion. At that point, the system transitions to using the kinetic coefficient of friction. In standard mechanical systems, the coefficient of static friction is greater than the coefficient of kinetic friction. That is why it is so important, for either braking or accelerating, to keep the tires from sliding or spinning.

 

The mass is essentially the weight on the driving tires. If you had a two wheel drive car that weighed 3000 pounds and weight distribution was 50:50, then 1500 pounds are on the driving tires. If the coefficient of static friction was .6 then 1500 X .6 = 900 pounds of force the tires can put to the ground before breaking the tires loose. If the vehicle was 4 wheel drive, then all 3000 pounds of the car is used to provide drive traction: 3000 x .6 = 1800 pounds of force can be sent to the ground. Once the tires break loose, the coffecient of kinetic friction is factored into the amount of power to the ground. In this case assume .5 cf. 1500 x .5 = 750 pounds to the ground while the tires are spinning.

 

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?).

[Kevin Shasteen]

Well, I appreciate your responses; I still believe the rocket scientists out there have some form of formula for determing traction required in relation to power applied, specifically-regarding the Sectional Width of the Tire...and to further that formula w/in relation to an auto at the dragstrip...I may be in denial; but I'm sure its out there.

 

Looks like I've got my work cut out for me. Hopefully others can chime in....I believe, I believe-I really believe (am I crazy or what) that a physics formula does exists. Its just a matter of finding it or running into the right person.

 

Kevin,

(Yea,Still an Inliner)

[Kevin Shasteen]

I hate it when it does this...its mocking me...someone delete this thing.

[Kevin Shasteen]

someone delete this; its repeating me again/or I clicked on the wrong icon.

[Guest greimann]

I don't think it is that simple. If you really want to make your brain hurt, do a search on

 

Pacejka tire formula

 

This is supposedly a body of research that has developed a friction model for automobile tires.

[Kevin Shasteen]

Thanks Griemann, I will check that out. For anyone else reading...I've been searching the net for answers & feel (feeling is good-politically correct now-a-days) that I can clearify my question.

 

Griemann got me close; I feel my question is this: "What determines the Breakout Force of a tire, and how is it determined when that breakout force turns maximizes the slip ratio?

 

I'm gonna go read Griemanns topic; in the meantime, any other suggestions will be greatly appreciated.

 

Thanks Griemann & if my brain hurts too much from reading your suggested topic...I promise not to ask any more questions about traction & friction

 

Kevin,

(Yea,Still an Inliner)

[260DET]

Tyre pressure comes into it too, a lower pressure increases the size of the tyre contact patch so giving more grip.

[Guest Anonymous]

I'm by no means an engineer (not even close) but I do believe that there are far too many variables in actual applications to make a general theory applicable. Most of the above posts address theory and physics, but if you apply this to two differnt cars you would get different results. For example, two cars running the same 10.5" drag radail tires. One a street car with stock suspension and the other a race car modified for straight line acceleration. The srteet car with 450 horsepower may heavily spin the tires and run 12 second quater mile times. The race car with 1200 horsepower may hook hard and run a 8 second quater mile. Both of the above examples are based on actual cars. Just a thought, I may be wrong (it wouldn't be the first time).

[Scottie-GNZ]

Just as important or maybe even more so than tread width is compound. Put a pair of 235 DRs on a 400hp Z-car, heat them up and with the proper launch technique the car will rocket off the line, squatting heavily with no wheelspin. Replace those with a pair of gonzo 295 standard radials and with the same launch technique you could probably lay down 200' of rubber. Pretty but not effective.

[Guest greimann]

Both of the above posts are absolutely correct. The complexities and unpredictibilities of the dynamics of a vehicle accelerating does not lend itself to a simple formula that calculates the coeffecient of friction of a tire on pavement. Which, going back to Kevin's original inquiry, tire width is one of many variables that have an effect on the outcome

[johnc]

Race Car Vehicle Dynamics, page 32, section 2.3 Logitudinal Force.

[Kevin Shasteen]

Yea JohnC, I came across that book last night while searching the web.

 

In my search for a common approach to a tire (prior to actully paying for one) I'm not looking for absolutes; only attempting to try & figure a way to know what to look for; after all, one can theorize all day long-but eventually you have to apply your knowledge (Think we call this R&D..or at the track we call it Test & Tune).

 

I also found out that traction/friction or the loss of it is actually "Rocket Science" (oh man-where's my slide ruler); however, once an individual can acknowledge the variables & assign a numerical substanct to that variable...then I'm confidant one can get pretty close, in theory, to the actual requirement.

 

John, from reading last night on the web, the book reviews on "Race Car Vehicle Dynamics" illuded to the PaceJKA's Mystery Formula, does this book give the novice/enthusiast something they can use-or would I actually have to be a Rocket Scientist to understand their explanation.

 

BTW; how much does the book run for; the website I found advertising the book was for Seminars or Schools...so I had no idea what the book was selling for.

 

How in detail does the book get in regards to tire traction/friction forces?

 

Thanks for all inputs from everyone-good info all the way round.

 

Kevin,

(Yea,Still an Inliner)