Why do fat tires offer better grip on the road than thin tires? *SOLVED*

[notfred]

Originally posted by: Skoorb
...The reason being that the friction coefficient is equal...

No, it's not. The coefficient of friction is different for a skinny tire on asphalt than it is for a wide tire on asphalt. Changing the area of contact between the two surfaces changes the coefficient of friction.

That's straight from my physics professor.

 

[LAUST]

Contact Patch

 

[StageLeft]

Originally posted by: notfred

Originally posted by: Skoorb
...The reason being that the friction coefficient is equal...

No, it's not. The coefficient of friction is different for a skinny tire on asphalt than it is for a wide tire on asphalt. Changing the area of contact between the two surfaces changes the coefficient of friction.

That's straight from my physics professor.

Yes, it is. Either you didn't follow what your professor was saying properly or he was wrong. In a very simple physics environment - something we've identified that tires and grip are NOT constrained by - the simple equation f = uF states that a coefficient of friction multiplied by the normal force results in the force required to successfully push an object (as mentioned later there may be a <= sign thrown in in either the static or kinetic case). Since we know that a brick on its end and a brick on its side require the same pushing force (f), and since the normal force obviously doesn't change (F), then naturally u is the same as well.

 

[Kev]

Originally posted by: Skoorb

Originally posted by: Roger
What ever Skoorb, think what you want

It's true! The force required to move something when it's not moving is: f = uF where f is force required, u is the coefficient of friction, F is the normal force (force of gravity pulling it down). The equation is the same (actually I just gave the kinetic equation - static is a touch different but irrelevant to this) for a moving object except in that case f is the force required to keep the object in motion, and when something is moving u (kinetic friction now) is likely to be lower than static friction.

So I still have no answer!

the fat tire weighs more than the thin tire so the equations are not equal.

 

[StageLeft]

Originally posted by: Kev

Originally posted by: Skoorb

Originally posted by: Roger
What ever Skoorb, think what you want

It's true! The force required to move something when it's not moving is: f = uF where f is force required, u is the coefficient of friction, F is the normal force (force of gravity pulling it down). The equation is the same (actually I just gave the kinetic equation - static is a touch different but irrelevant to this) for a moving object except in that case f is the force required to keep the object in motion, and when something is moving u (kinetic friction now) is likely to be lower than static friction.

So I still have no answer!

the fat tire weighs more than the thin tire so the equations are not equal.

If that was the answer, then you could simply take the additional weight of the fat vs. the small and throw it in the backseat

 

[WinkOsmosis]

Originally posted by: Kev

Originally posted by: Skoorb

Originally posted by: Roger
What ever Skoorb, think what you want

It's true! The force required to move something when it's not moving is: f = uF where f is force required, u is the coefficient of friction, F is the normal force (force of gravity pulling it down). The equation is the same (actually I just gave the kinetic equation - static is a touch different but irrelevant to this) for a moving object except in that case f is the force required to keep the object in motion, and when something is moving u (kinetic friction now) is likely to be lower than static friction.

So I still have no answer!

the fat tire weighs more than the thin tire so the equations are not equal.

Dude... Are you stupid? Talking about the weight of the damn tire which probably changes the weight of the whole car by less than .1%! Hahahahahha.

 

[notfred]

Originally posted by: Skoorb

Originally posted by: notfred

Originally posted by: Skoorb
...The reason being that the friction coefficient is equal...

No, it's not. The coefficient of friction is different for a skinny tire on asphalt than it is for a wide tire on asphalt. Changing the area of contact between the two surfaces changes the coefficient of friction.

That's straight from my physics professor.

Yes, it is. Either you didn't follow what your professor was saying properly or he was wrong. In a very simple physics environment - something we've identified that tires and grip are NOT constrained by - the simple equation f = uF states that a coefficient of friction multiplied by the normal force results in the force required to successfully push an object (as mentioned later there may be a <= sign thrown in in either the static or kinetic case). Since we know that a brick on its end and a brick on its side require the same pushing force (f), and since the normal force obviously doesn't change (F), then naturally u is the same as well.

I didn't misunderstand him, I was specifically talking about a tire, and he said "the coefficient of friction is different". I pushed a physics book across a desk and said "so the coefficent of friction for this book on this table is different like this (pushing with face of book on desk) than like this (stand the book on it's spine and psuh across desk)?" He said "yes".

You're error lies in the assumption that it takes the same amount of force to push a brick that's laying on it's side rather thanstood on it's end. It likely takes a larger force to push the brick that's on it's side.

 

[StageLeft]

I didn't misunderstand him, I was specifically talking about a tire, and he said "the coefficient of friction is different".

He's right in regards to a tire, because of reasons mentioned in this thread.

I pushed a physics book across a desk and said "so the coefficent of friction for this book on this table is different like this (pushing with face of book on desk) than like this (stand the book on it's spine and psuh across desk)?" He said "yes".

I'd recommend finding a better physics teacher. Unless the book is using similar forces acting on its contact with the desk as a tire on the road there is NO difference in a coefficient of friction whether a book is on its end, on its tip, or flat on its back. In reality? Perhaps there is a difference, but from a simple physics equation point of view there is no difference.

You're error lies in the assumption that it takes the same amount of force to push a brick that's laying on it's side rather thanstood on it's end. It likely takes a larger force to push the brick that's on it's side.

I made no error - it takes the exact same amount of force regardless of its orientation on the ground. I will qualify that again by mentioning that there could be unidentified factors working with the brick and its surface similar to how they will with a tire on ashphalt, but from a basic mathmatical view there is no difference, and if you answered on a physics exam that there was a difference (without being given the additional factors at work, which are difficult to identify) you'd get the answer wrong.

Given two "pure" materials - the kind one would only find IN a physics question the flat brick vs.the brick on its end are exactly the same in regards to the forces needed to push them because their coefficient of friction is the same.

 

[Kev]

Originally posted by: WinkOsmosis

Originally posted by: Kev

Originally posted by: Skoorb

Originally posted by: Roger
What ever Skoorb, think what you want

It's true! The force required to move something when it's not moving is: f = uF where f is force required, u is the coefficient of friction, F is the normal force (force of gravity pulling it down). The equation is the same (actually I just gave the kinetic equation - static is a touch different but irrelevant to this) for a moving object except in that case f is the force required to keep the object in motion, and when something is moving u (kinetic friction now) is likely to be lower than static friction.

So I still have no answer!

the fat tire weighs more than the thin tire so the equations are not equal.

Dude... Are you stupid? Talking about the weight of the damn tire which probably changes the weight of the whole car by less than .1%! Hahahahahha.

actually i was thinking about bikes, he wasn't specific that he was talking about car tires. in other words, go to hell

 

[notfred]

I'd recommend finding a better physics teacher. Unless the book is using similar forces acting on its contact with the desk as a tire on the road there is NO difference in a coefficient of friction whether a book is on its end, on its tip, or flat on its back. In reality? Perhaps there is a difference, but from a simple physics equation point of view there is no difference.

Come on Skoorb, you really think I'm going to believe you over a physics professor? We are not talking about a "simple physics equation", we're talking about a tire, which exists in reality. Yes, there is a difference in the coefficients of friction for the different sized tires. You say "but from a simple physics equation point of view there is no difference". What is this "simple physics equation you're talking about? One that already assumes you know the coefficient of friction? Of course it doesn't matter in that equation, as it's already given.

I made no error - it takes the exact same amount of force regardless of its orientation on the ground. I will qualify that again by mentioning that there could be unidentified factors working with the brick and its surface similar to how they will with a tire on ashphalt, but from a basic mathmatical view there is no difference, and if you answered on a physics exam that there was a difference (without being given the additional factors at work, which are difficult to identify) you'd get the answer wrong.

What is this "basic mathematical view" you're talking about? You keep referring to equations, but I've yet to see your "simple equation" that computes the coefficient of friction of different sized tires. All your equation assume that the coefficient is given to you. In such a case, you should be given different coefficients for the wide and narrow tire.

Given two "pure" materials - the kind one would only find IN a physics question the flat brick vs.the brick on its end are exactly the same in regards to the forces needed to push them because their coefficient of friction is the same.

No they're not, give me any shred of evidence showing that the coefficient of friction is the same for all objects made of the same material.

 

[WinkOsmosis]

Originally posted by: notfred

I'd recommend finding a better physics teacher. Unless the book is using similar forces acting on its contact with the desk as a tire on the road there is NO difference in a coefficient of friction whether a book is on its end, on its tip, or flat on its back. In reality? Perhaps there is a difference, but from a simple physics equation point of view there is no difference.

Come on Skoorb, you really think I'm going to believe you over a physics professor? We are not talking about a "simple physics equation", we're talking about a tire, which exists in reality. Yes, there is a difference in the coefficients of friction for the different sized tires. You say "but from a simple physics equation point of view there is no difference". What is this "simple physics equation you're talking about? One that already assumes you know the coefficient of friction? Of course it doesn't matter in that equation, as it's already given.

I made no error - it takes the exact same amount of force regardless of its orientation on the ground. I will qualify that again by mentioning that there could be unidentified factors working with the brick and its surface similar to how they will with a tire on ashphalt, but from a basic mathmatical view there is no difference, and if you answered on a physics exam that there was a difference (without being given the additional factors at work, which are difficult to identify) you'd get the answer wrong.

What is this "basic mathematical view" you're talking about? You keep referring to equations, but I've yet to see your "simple equation" that computes the coefficient of friction of different sized tires. All your equation assume that the coefficient is given to you. In such a case, you should be given different coefficients for the wide and narrow tire.

Given two "pure" materials - the kind one would only find IN a physics question the flat brick vs.the brick on its end are exactly the same in regards to the forces needed to push them because their coefficient of friction is the same.

No they're not, give me any shred of evidence showing that the coefficient of friction is the same for all objects made of the same material.

Coefficient of friction is a property of the substance. That's the way it works.

 

[notfred]

Originally posted by: WinkOsmosis
Coefficient of friction is a property of the substance. That's the way it works.

Hmm, this is the definition I found
coefficient of friction
noun
(Mechanical engineering) the force required to move two sliding surfaces over each other, divided by the force holding them together. It is reduced once the motion has started

Doesn't say it's a property of a substance.

here's another:
Friction is a resistive force that prevents two objects from sliding freely against each other. The coefficient of friction (u) is a number that is the ratio of the resistive force of friction (Fr) divided by the normal or perpendicular force (Fn) pushing the objects together. It is represented by the equation:

u = Fr / Fn.

Hmm, by that definition there is a higher coefficient of friction for two objects that require more force to slide past each other than two objects that require less force to slide past each other (assuming the same normal force). For example, a roadway and two different tires.

 

[TechnoKid]

When we did a lab in HS physics, using a newton meter, it did require more force to pull the brick face down rather than pulling it on its end.

edit: i think it could have been vice versa. ill have to double check my notes.

 

[StageLeft]

notfred I never said fat tires have the same real world grip as thin tires. Nobody did.

Their REAL WORLD coefficient of friction is obviously larger for the fat tires - that is why they grip better. Read FrustratedUser's long post for a good explanation of why. Also follow the link in the thread where two physics professors answer (near beginning).

Given two "pure" materials - the kind one would only find IN a physics question the flat brick vs.the brick on its end are exactly the same in regards to the forces needed to push them because their coefficient of friction is the same.

No they're not, give me any shred of evidence showing that the coefficient of friction is the same for all objects made of the same material.

Yes, notfred, they are the same. Strictly speaking and going with the basic, controlled, environment surface area has nothing whatsoever to do with coefficient of friction. Zip. You can have a brick on its end or on the flat surface and the force to move it is the same. Friction is not dependent on surface area. and The coefficient of static friction does not depend upon contact surface area.

You can choose to believe your physics teacher or you can follow the answers of the two physics profs on that link near the beginning, as well as the obvious understanding of Dr.Pizza, another physics teacher who answered in this thread.

When your professor told you that the book was easier to move on its spine than on its back he erred in explaining the situation to you. The only reason it might be easier on its spine is for the same reason fat tires grip better than thin. It's a move involved explanation of a coefficient of friction - one that takes into account difficult to identify asperities. The most basic explanation of a coefficient of friction, which anything else builds upon, states that surface area is totally unrelated to the coefficient of friction. I can't believe that your physics professor never mentioned that?

Feel free to plug in "coefficient of friction surface area" into google and you'll have a multitude of documents on the matter.

 

[StageLeft]

Originally posted by: TechnoKid
When we did a lab in HS physics, using a newton meter, it did require more force to pull the brick face down rather than pulling it on its end.

edit: i think it could have been vice versa. ill have to double check my notes.

Any differences you'd have found would have been an experiment showing real world influences on basic formulas as opposed to experimentally proving the strict relationship of the coefficient of friction and surface area (that relationship being non-existent, strictly speaking).

 

[notfred]

Bah, this is fruitless. I keep saying that the two tires have different coefficients of friction, and you keep talking about surface area. The first link you posted says, "Friction is not dependent on surface area." It also says F=uN, where u is the coefficient of friction, and N is the normal force. That shows that friction isn't dependent on surface area, only u and N.

However, it does NOT say that two different objects automatically have the same value for u solely because they're made out of the same substance. I'm saying that the wider tire has a higher coefficient of friction. You're saying that friction is not dependent on surface area. Those are not two contradictory statements...

Obviously, the wider tire DOES have a higher coefficient of friction, it's pretty apparent from the experimental data (cars with wider tires stick better). You're also correct in saying that F=uN can't account for that, as that equation assumes that u is known, even though we have no idea what it is for two different tires.

You're right insofar as you say that this problem can't be explained solely with the equation F=uN. However, you haven't shown that two different tires will have the same value for u.

BTW, your second link wouldn't load for me.

 

[StageLeft]

Originally posted by: notfred
Bah, this is fruitless. I keep saying that the two tires have different coefficients of friction, and you keep talking about surface area. The first link you posted says, "Friction is not dependent on surface area." It also says F=uN, where u is the coefficient of friction, and N is the normal force. That shows that friction isn't dependent on surface area, only u and N.

However, it does NOT say that two different objects automatically have the same value for u solely because they're made out of the same substance. I'm saying that the wider tire has a higher coefficient of friction. You're saying that friction is not dependent on surface area. Those are not two contradictory statements...

Obviously, the wider tire DOES have a higher coefficient of friction, it's pretty apparent from the experimental data (cars with wider tires stick better). You're also correct in saying that F=uN can't account for that, as that equation assumes that u is known, even though we have no idea what it is for two different tires.

You're right insofar as you say that this problem can't be explained solely with the equation F=uN. However, you haven't shown that two different tires will have the same value for u.

BTW, your second link wouldn't load for me.

I don't know why you're trying to dispute something I'm not even claiming. I didn't say two tires have the same coefficient of friction.My problem with your explanation - and the one your physics proffessor wrongly indicated to you - is that because an object has more surface area it has a greater coefficient of friction. Tires aside he used the same statement with the textbook. Strictly speaking, and factual, is that surface area is independent of the coefficient of friction. Therefore the two contrasting materials are responsible for u, not the area between them.

Built upon this fact, and not in spite of it is that asperties between the two materials can increase the coefficient of friction as surface area increases. Saying a wider tire has better grip than a small one because of a greater surface area is like answering "How does a rocket work" with "Because it has power come out one side." It's not a functional answer.

In regards to your statement

Given two "pure" materials - the kind one would only find IN a physics question the flat brick vs.the brick on its end are exactly the same in regards to the forces needed to push them because their coefficient of friction is the same.

No they're not, give me any shred of evidence showing that the coefficient of friction is the same for all objects made of the same material. "

I have done that but not surprisingly you will not achknowledge it. I have given you the answer. Coefficient of friction is irrelevant with surface area between "two "pure" materials - the kind one would only find IN a physics question". In the real world there is more to the answer as irregularities in the materials affect the coefficient of friction, but strictly speaking there is no difference.

I'll again reiterate that your physics professor, not clearly stating this to begin with, erred in his introduction of friction to the class. Only when one has grasped Amonton's law should one jump beyond it to practical application. You can find additional reading about it here..

And as I said earlier if you're taking a physics test and argue that a brick on its side will require more force than a brick on its end you're gonna get a zero, unless you have the almost intangible ability to calculate the asperities influencing the contact points of the brick and what its lying upon.

I think if you'd read the entire thread before quoting my first post we could have both saved a lot of effort here.

 

[StageLeft]

This is a good page.

In general frictional forces are independent of the area of contact although this is an empirical observation not a theory. Consider a metallic brick and a metallic table. The reason that friction is nearly independent of surface area is if the "microscopic" area of contact of the brick to the table is independent of the orientation of the brick. If this is not the case, then friction will have a small dependence on area. In normal circumstances, with the largest surface area of the brick in contact with the table there are a large number of "contact" points that support the load. With the smallest area in contact (brick standing on end) there are fewer contacts but as long as the area of each contact is larger due to the higher pressure (same force, smaller unit area) then there will be no difference in the amount of static friction. Over wide limits, most materials follow this and hence friction is largely independent of surface area. iF you have a situation where the microscopic contact area does not scale in accordance with the pressure, then static friction will depend upon orientation.

Clearly if, as the pure textbook physics answer states, u is independent of surface area the "coefficient of friction is the same for all objects made of the same material.".

Pure textbook physics at the basic level can't answer the tire question. It's not that the basic physics is wrong, but rather it's not involved enough and taking into account everything.

 

[CTho9305]

Originally posted by: notfred
Obviously, the wider tire DOES have a higher coefficient of friction, it's pretty apparent from the experimental data (cars with wider tires stick better). You're also correct in saying that F=uN can't account for that, as that equation assumes that u is known, even though we have no idea what it is for two different tires.

Your answer doesn't help. WHY does u change?

 

[Triumph]

Originally posted by: CTho9305

Originally posted by: notfred
Obviously, the wider tire DOES have a higher coefficient of friction, it's pretty apparent from the experimental data (cars with wider tires stick better). You're also correct in saying that F=uN can't account for that, as that equation assumes that u is known, even though we have no idea what it is for two different tires.

Your answer doesn't help. WHY does u change?

Because Bridgestone designed them that way.

 

[Carapace]

Originally posted by: Skoorb

Originally posted by: DurocShark
It's the side-to-side torque that's affected. In a straight line, there's no difference (other than rolling resistance).

Well then why do drag cars have fat ass tires?

Because the tire expands as it goes faster, in all actuality about 1/4 of the tires width is making contact with the surface. Also, fat tires are not always better. For snow and ice you want more pounds per sqaure inch spread out over a smaller area, thus giving you the pentration through the snow and ice.

 

[LOLyourFace]

nice sig.

 

[Triumph]

I've thought about this long and hard over the years as well. I've done more engineering calculations like this than you can shake a stick at. And I've pretty much come to the conclusion that mu has to change with a wider tire. But this does NOT mean that mu is changing BECAUSE of the increased surface area directly. F=uN still applies as a physics law, but the calculation of u depends on many things, and as was explained above, is really only determined by the ratio of the friction force over the normal force. That is how it is calculated. There is no equation that says u = f(surface area, temperature, velocity, acceleration, Reynold's number). Any table of mu that you find will have come from someone doing a test and dividing friction by normal force. It is in reality a variable created by humans in order to match empirical results.

In summary, mu does change. Why it changes, I don't know.

 

[StageLeft]

Originally posted by: Triumph
I've thought about this long and hard over the years as well. I've done more engineering calculations like this than you can shake a stick at. And I've pretty much come to the conclusion that mu has to change with a wider tire. But this does NOT mean that mu is changing BECAUSE of the increased surface area directly. F=uN still applies as a physics law, but the calculation of u depends on many things, and as was explained above, is really only determined by the ratio of the friction force over the normal force. That is how it is calculated. There is no equation that says u = f(surface area, temperature, velocity, acceleration, Reynold's number). Any table of mu that you find will have come from someone doing a test and dividing friction by normal force. It is in reality a variable created by humans in order to match empirical results.

In summary, mu does change. Why it changes, I don't know.

You're quite right and it's what I think most of us agree upon. It doesn't change simply because as surface area expands so does u. It changes in this case because of what exactly that surface area expansion is doing behind the scenes, and in this case it's introducing the asperities mentioned throughout the thread.

 

[Triumph]

Originally posted by: Carapace

Originally posted by: Skoorb

Originally posted by: DurocShark
It's the side-to-side torque that's affected. In a straight line, there's no difference (other than rolling resistance).

Well then why do drag cars have fat ass tires?

Because the tire expands as it goes faster, in all actuality about 1/4 of the tires width is making contact with the surface. Also, fat tires are not always better. For snow and ice you want more pounds per sqaure inch spread out over a smaller area, thus giving you the pentration through the snow and ice.

Tire width absolutely does have an affect on straight line acceleration. There is no disputing that. The top fuel dragster's don't have fat tires *because* they expand. That's some kind of circular reasoning. The fat tires will provide more width when expanded than a thin tire would provide when expanded. More width = more grip.