datsun2401972 wrote:What I don't understand I guess(unless I'm right), is how you can say that by taking, for example, a 5" wide wheel with a 5" wide tire and comparing it to a 10" wide wheel with a 10" wide tire, all other factors constant, that the 10"wheel/tire will not have more contact patch than a 5"wheel/tire.
Now, I'm completely dumbfounded as to how anyone could say 10" is not greater than 5". But let me apologize before hand if someone should point out the error in my statements. That would actually be appreciated.
Alright, let's do some math (ack!). First, let me make it clear, contact patch geometries are complicated, rely on a large number of variables, and are not fully understood.
To get rid of all the crap that'll make this impossible to explain, let's just say that all contact patches are rectangular.
Now, think about the tire as a separate entity. At rest, a tire only has two forces acting on it. One of them is a normal (vertical) force, pointed downward, that is equal to whatever percent of the car's weight rests on that tire. There is another equal and opposite force that acts (pointing upwards) on the tire, and that is the reaction force that the road applies. This force is EQUAL in magnitude and opposite in direction to the force that the car applies downwards on the tire. These forces MUST be equal when the car is at rest -- if they weren't equal, they would not cancel out, and the wheel would be experiencing a vertical acceleration, which it obviously does not when it is at rest.
Now, lets say a tire is just like a balloon (which it is to an extent). Say you have a tire (not installed on a vehicle), inflated to 45psi with no load on it (since it is not on the vehicle). You want this tire to support 500lbs (one corner of a 2000lb car with 50/50 static weight distribution and 50/50 static cross weighting). So you load the tire onto the hub and put the car on the ground. The tire will deform slightly when it hits the ground, forming a flat in the circular profile of the tire when looking at it from a side view. This is the side view of the contact patch.
Because you introduce a flat in a circle, you reduce the area inside the circle. Because this circle has a depth (the width of the tire) and defines a volume, the volume inside the tire will decrease as well when the car is on the ground. Because the volume in the tire has decreased but the amount of gas inside has remained the same, the pressure inside the tire MUST increase. All this deformation is related to a number of factors (sidewall stiffness, inflation pressure, tire construction and compound, temperature, etc. etc. etc.). Once again to simplify this explanation, lets just say that a particular type of tire we're looking at will deform until the internal pressure is 50psi (up from 45psi unloaded). This property will remain nearly the same across a wide range of tire sizes.
Remember that the up and down forces on the tire must be equal in magnitude and opposite in direction. Let's say we have two tires, one is 5" wide at the tread, and the other is 10" wide at the tread. Since the tire is now at 50psi, and has 500lbs acting on it, the area of the contact patch must ALWAYS be 10 square inches in order for the forces to cancel out ( 50 lbs/in^2 * 10 in^2 = 500 lbs).
Say you have the 5" wide tire. 10 in^2 / 5 in = 2 in. This means with the first tire, the contact patch will be 2" long (front to back) and 5" wide (side to side). With the 2nd tire, 10" wide, 10 in^2 / 10 in = 1 in. This means with the wider tire, the contact patch will now be only 1" long front to back, and 10" long side to side.
Therefore, as you can see, the area remains constant despite the fact that the tread section width changes. Since frinctional force depends on contact area, we can deduce that over a variety of widths, frictional forces will remain nearly the same at constant operating conditions.
"Constant operating conditions" is the catch. Changing the width of the tire changes operating conditions, such as: the ability of the tire to absorb and shed heat / stay at operating temperature, rolling resistance, rotational inertia, cornering stiffness (slip angle vs. lateral force), and so on.
The largest thing that increased tread width changes is breakaway characteristics, because changing the tire width will change the cornering stiffness of the carcass. A narrower tire has less cornering stiffness, that is, it will produce a given amount of lateral force at a higher slip angle (a wider tire will produce the same lateral force at a lower slip angle, because it's cornering stiffness is higher). Because the narrower tire will travel through a wider range of slip angles before reaching it's "limit" (maximum lateral force), the driver has a better "feel" of when the tire will let go, as opposed to a wide tire which operates in a narrow range of slip angles before reaching it's ultimate grip and will therefore not give as much warning before breaking away.
These breakaway characteristics are very involving to quantify, and doing so would require an analysis of the vehicle's steering system (scrub radius, caster, mechanical trail), and would require quantitative knowledge of a tire's performance (self-centering torque vs. slip angle). I have done the best i can without diving into a full-depth analysis of the system.
It is possible to "over tire" a car, which is the point at which rolling resistance, rotational inertia, and the inability of the tire to stay at the compound's operating temperature, become a detriment to the car's handling.
Quote »AZhitman: Tire compound, internal construction, and suspension setup is NOTHING without width. So tell me again, how important is it?
Let me spell that out for the rest of the crowd. If a tire manufacturer has no dimension(width being part of dimension) to start with, there is no point in testing different compounds or internal construction. Why? Because the tire doesn't have dimension, without dimension there's no tire!!!
TREAD WIDTH IS IMPORTANT![/quote]I suppose that's true, but thats a seedy argument.
The truth of the matter is that vehicle dynamics is the study of compromises in almost every context (roll centers vs. camber curves, tire size, anti-dive and anti-squat geometries). The job of the vehicle dynamicist or car setup expert is to find the best compromise; the setup which best accounts for all operating conditions the chassis is to endure.
It's the same with tires. Without proper width, the tire wouldn't operate in it's most efficient range; without compound and construction, that range of efficiency would be much lower.
