I know they do but I cannot fid any thoery to prove this. my understanding of traction is the irregularities between two surfaces haveing a pressure exerted on them. If this is true then a tire with 1 cubic foot of contact patch would have as much traction as a tire with 2 cubic feet of contact patch. WTF!!! I cant disprove this!

its way too late for me to delve into physics (i am half asleep as i type this), however, the size of the contact patch on the surface, and the gummier the material, the more traction.

true there has to be a limmit, but i believe that that limit is acheived be rotational mass and energy exerted to move that mass. if you look at F1 cars, from 20 or 30 years ago... they had cars dumping out 1000 hp, and huge tires.

hope that helps, some.

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hoop
Red '86 GT
KYB's, Eibach's
and spellcheck crash's windows
Poly Bushings
Accel Coil

I guess that I'm in the same boat as bHooper (phone rang and woke me up!). I can add that the surface must also be considered for the amount of traction gained, for instance, the difference between (extreme) dry pavement and wet pavement. Lots of variables will come into play.

Okay, frictional physics isn't my speciality, but I'll tackle this one. (Strangely not a lot of call for nuclear physics or advanced theoretical physics on this forum.)

Okay, basically everything has adhesive properties. In other words, to objects when placed together, tend to resist movement in relationship to each other. Things that are considered 'slippery' simply have a very low resistance to this movement. The more surface area the two objects have in contact with each other, the more they resist movement in relationship to each other because there's more 'sticking them together'.

Example: Which is easier to do? Place your finger on the surface of your hood. Now drag it across the hood. Place your entire palm on the surface of your hood and drag it across while applying the same amount of pressure as you did with your finger.

With tires, the surface area is determined by the width and length of the surface area of the tire in contact with the road (called the 'footprint' of the tire.) You can affect the footprint of the tire by raising or lowering the air pressure of the tire, which is why you see people playing with it at the drag strip. Too much pressure and the footprint is too small for maximum traction. Too little and the middle of the footprint doesn't have enough pressure on it to keep it planted on the street.

Lets say you have two tires of equal diameter, one is 5 inches wide and the other is 10 inches wide. Given equal vehicle weight and equal air pressure they will have roughly the same footprint. The wider tire will have less length but more width because it's pressure picks more of the tire off the street. However, this would be way more psi than the tire should have in it. So if both tires have the appropriate amount of air pressure (the smaller tire requires more, think of your bicycle. Mine takes 45psi. My car is around 30-35psi.)

So now we have a 10 inch wide tire with a very similar length of footprint to the 5 inch wide tire, thus giving it a large footprint. Given the same vehicle, the 10 inch tire has roughly half the pounds per square inch of pressure on it's footprint, but the actual weight applied to the tires is not linear with respect to friction. (ie doubling the weight on a tire does not double it's friction. It actually ends up less than double). So while adding weight over the drive tires does improve grip, it doesn't do it as efficiently as increasing the footprint. This also demonstrates why the smaller footprint doesn't have as much traction. While it has more psi on the ground (weight the vehicle puts on the wheel divided by the area of the footprint) that increased psi doesn't overcome the traction gained by having more square inches on the ground.

Let's say both our tires have a 5 inch long footprint. That means our 10 inch wide tire has 50 square inches of rubber on the road (5 x 10) while the 5 inch wide tire has only 25 square inches (5 x 5). The total weight on the tire is the same (same vehicle) which means the 5 inch tire is putting twice as much weight per inch on the ground. (Like the difference of having a woman stand on your back in heels or barefoot).

Static (and kinetic) friction is a function of the friction coefficients of the two object in contact, the surface area shared by the objects, and the force pushing the two objects together. 'Stickier' objects make for more friction, having more surface area in contact makes for more friction, and pushing them together harder makes for more friction. If you parked a Mack truck on the back of a Top Fuel dragster, it would have trouble spinning the tires, assuming the suspension didn't break or the tires blow under the weight. The friction would be increased.

I don't have the exact equation handy, but basically speaking, increasing the surface area increases friction MORE than increasing the weight. That means that while a narrower tire puts more weight per square inch on the road, the friction per square inch is lower than the wider tire.

And that's what it really boils down to: Friction per square inch. A wider tire has more friction per square inch, and thus resists the tendancy to slip against the road better (traction). It's much easier to drag your finger down a pane of glass than it is to drag your entire hand, even if you apply less force with your hand. Your hand generates more friction per square inch, and there's more square inches in contact. That makes for a much higher total friction and more resistance to slippage.

As for putting more traction under your car, the methods are myriad. Wider tires is the most common. Hot tires make more friction, hot rubber is 'stickier' than cold rubber. Adjusting tire pressure to give more usable footprint is the same as wider tires, but since it has a maximum point, wider tires are eventually needed as vehicle horsepower is increased. Trak-bite is also a good method, making the tire and the street stick to each other more. Slicks is another one, again by eliminating the recesses in a treaded tire, more footprint is acheived. Larger diameter tires also have larger footprints.

So when the concern is launching a car or improving cornering, it normally boils down to how much rubber can you put on the road, since the other variables are either harder to control or have a much smaller limit to how effective they can be.

Hope that helps (and you haven't fallen asleep reading.)

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85 GT (to be modified...)
--Um, no the paint isn't oxidized, that's the ultra-rare Whirlpool White textured finish... yeah.
89 Lincoln Town Car (has Fiero Envy)
--Hey! Can I have an anti-sway bar too?

At a more convienient time, could you post that equation? Please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

I could look it up, but the variables involved pretty much make it impossible to get any real information from just the equation.

Anyone know the traction co-efficient of the rubber in their particular brand of tires? Asphalt? Blacktop? I sure as heck don't.

Especially since the co-efficient changes based on temperature, humidity. Add to that the need to know exactly what the weight distribution at each drive wheel... pretty much the equation becomes academic instead of practical. If you knew all the numbers, you could in theory caluclate the exact amount of shaft horsepower it would take to spin the tires, and perhaps in some way calculate the precise launch RPM. But once again, it would only be good for THAT launch and would require precise measurements of track temperature, tire temperature, etc.

The basic equation boils down to what I said before. The more rubber you put on the street, the better your traction gets. The other ways to get it is with softer rubber (it's stickier, costs more, and wears out faster), Trak-bite which is only useful on drag runs, and using slicks instead of street tires. In practical terms you simply want to mount the biggest, widest tires you can on your drive wheels.

I can look up the equation if you like, but it just isn't really useful.

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85 GT (to be modified...)
--Um, no the paint isn't oxidized, that's the ultra-rare Whirlpool White textured finish... yeah.
89 Lincoln Town Car (has Fiero Envy)
--Hey! Can I have an anti-sway bar too?

Ah yes, I agree with what you say as I have seen it work in the real world.

But can you explain why larger pads on the disk brakes wouldn't stop faster? Everybody seems to say so. Pressure on them would be the same yet area would be greater so friction should be greater.

My head hurts now.

That explanation was WAY more confusing than it had to be.
The friction formula you were taught in high school physics (you DID take physics, right?) --F=uN where F is frictional force available, u is coefficient of friction, and N is "downforce" or force perpendicular to the interface of the materials--does not apply to tires. That formula applies to non-deforming materials which do not get pieces torn off as they slide.
When a tire is pushed against pavement, the rubber gets squeezed into and interlocks with irregularities in the pavement. When the tire slides or spins, these little interlocking pieces get torn off. This does two things: it leaves rubber on the pavement and it makes frictional force a function of both the shear strength and the "softness" or conformability of the tire.
In the above tires example, if the 50 square inch tire had a sideways force of 50 lbs applied to it, the rubber at the tires-pavement interface would be under 1 psi of shearing stress. For the 100 square inch tire the same lateral force would generate 1/2 psi of shearing stress at the interface. If the rubber tears at a shear stress of 1.5 (for this example; it's really much higher) psi, then the 50 square inch tire would slide at 75 pounds of lateral force, while the 100 square inch tire would slide at 150 pounds of lateral force.
And that's how the leopard got his spots.

Brake pads are much closer to the ideal F=uN formula than tires, so larger brake pads do not exert more force than smaller ones (provided the geometric center is in the same place relative to the rotor), they only wear more slowly.

Ditto Will's reply. (You beat me to it!)

You got it Will, I just couldn't remember the difference between F=uN and the deforming equation (which is the really nasty complex formula I was thinking of).

Your example is a lot better. Guess that's what happens when an ex-nuclear physicist tries to explain why bigger tires stick better...

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85 GT (to be modified...)
--Um, no the paint isn't oxidized, that's the ultra-rare Whirlpool White textured finish... yeah.
89 Lincoln Town Car (has Fiero Envy)
--Hey! Can I have an anti-sway bar too?

N in the EQ has to be in newtons doesn't it?
anyways if you are bored here are some numbers for you EQs

Friction coefficients
Us= Mu static Uk= Mu kinetic
tire on dry concrete= Us=1 Uk=.7-.8
tire on wet concrete= Us=.7 Uk=.5
tire on ice concrete= Us=.3 Uk=.02

Have fun... these are out of an old physics book from 1998

It F and N could be in grains Troy as long as they're in the same units.

not to mention wider tires ride harder, hydroplane lots easier. Where does that all figure in? ie/ little cars like vw with little 6 inch wide tires get thru a foot of snow, and the mustang or camaro with foot wide shoes, just sits n spins-even on a few inches of snow.

I heard somewhere the vettes get about the best friction on their tires for a standard car, approaching 1 G of acceleration in a turn or straight ahead or breaking. And that is the best a tire can do, 1 G, under 'normal' circumstances.

except for drag racers. You know that giant wings they put on the back? Once they hit about 90mph or more that pushes down on the tires with so much force that drag racers can accelerate with several Gs, because the downward force is so great.

Look at the 1/4 mile times that drag racers acheive and do the math, they are accelerating faster than 1 G, quite a bit faster!

Depends how the tire is compounded. Soft or hard.
A top fueler accelerates at about 4 Gs for the first half of the quarter, then it tapers off as the clutches hook up and the torque multiplication drops off.
Formula 1 cars can hit 4 Gs, but it occurs mostly under braking.
It must be difficult to formulate a rubber compound that moulds into the granules of the pavement, but stays intact as it flows reluctantly sideways over and around all the little hills and valleys of the pavement.
By reluctance, I mean once the tire is pressed down into the grains of the tarmac, it doesn't want to be dislodged sideways or forward, etc.
The tires resistance to moving(flowing) sideways across the tarmac while cornering defines it's cornering power most of the time.
Ripping bits out of the tire, with the signature black streaks shows the limit of the rubbers physical strength has been exceeded.