BRZ vs Mustang - Here is the review the haters have been waiting for.

[carbonBLUE]

The actual grip that a tyre can generate is dictated by the coefficient of friction of the rubber compound used in the tyre. The higher the coefficient, the more grip which can be generated. The relation that is used is called Armonton's Law, and
the equation is: F=µN, where F is the force generated, µ is the coefficient of friction, and N is the weight on the surface considered (in our case, the weight on the tyre).
So, if you increase the weight on the tyre, then the frictional force will increase as
well, in proportion to the increase in weight on the tyre - but the coefficient of
friction will remain the same. The level of grip of the tyre (forgetting about
suspension niceties - we are only discussing tyres here) is totally dictated by the
coefficient of grip of the tyre and the weight acting on it - not the area of the
contact between the tyre and the road.

stole this :P


also what is important is tire compound if your running a soft compound tire your tires will heat up more quickly giving more grip this also helps out on lighter cars

heavier cars or cars that create a lot of down force can use harder compound tires because their weight causes more friction and they can heat up the tires more effectively to create more grip, this is a practice used in endurance races because tires that are harder usually last longer, in short lap races a soft compound will be used to get better lap times....

forza fans and gran trismo guys probably have seen the difference running soft and hard compound tires, yes i know its a game but they try to make it as real as possible, not a good reference but its an easy one you can test yourself if you have the game

[Dimman]

Originally Posted by carbonBLUE

The actual grip that a tyre can generate is dictated by the coefficient of friction of the rubber compound used in the tyre. The higher the coefficient, the more grip which can be generated. The relation that is used is called Armonton's Law, and
the equation is: F=µN, where F is the force generated, µ is the coefficient of friction, and N is the weight on the surface considered (in our case, the weight on the tyre).
So, if you increase the weight on the tyre, then the frictional force will increase as
well, in proportion to the increase in weight on the tyre - but the coefficient of
friction will remain the same. The level of grip of the tyre (forgetting about
suspension niceties - we are only discussing tyres here) is totally dictated by the
coefficient of grip of the tyre and the weight acting on it - not the area of the
contact between the tyre and the road.

stole this :P


also what is important is tire compound if your running a soft compound tire your tires will heat up more quickly giving more grip this also helps out on lighter cars

heavier cars or cars that create a lot of down force can use harder compound tires because their weight causes more friction and they can heat up the tires more effectively to create more grip, this is a practice used in endurance races because tires that are harder usually last longer, in short lap races a soft compound will be used to get better lap times....

forza fans and gran trismo guys probably have seen the difference running soft and hard compound tires, yes i know its a game but they try to make it as real as possible, not a good reference but its an easy one you can test yourself if you have the game

That doesn't sound right...

[fatoni]

Originally Posted by carbonBLUE

ohh wow your special, get two balls of the same size, set one ball on a table, that would be the tire off the car standing upright unloaded, now take your hand and press on the second ball, the ball will absorb the pressure and compress, this is replicating a tire when its on a car and the car's weight is on the tire thus giving a larger contact patch

now on test two get a volley ball(smaller ball) and a basket ball(larger ball)
inflate both balls to the same PSI and dip the bottoms in paint
press both balls onto a large piece of paper, now press down on both balls with equal force
measure the diameter of each paint print

the result should show that the volley ball has a smaller contact patch than the basket ball, its 3rd grade science project i hope you can complete it...


you can even further the test by putting more pressure on the basket ball to resemble a higher load for larger diameter tires that are usually used on heavier cars the measure the difference in diameter vs the volleyball with less load

thats not right. do you know what psi means? if the pressure is the same, and the pressure per area is the same...the area is the same. third grade project or not its clearly not that intuitive

[Spaceywilly]

Originally Posted by Dimman

That doesn't sound right...

It is right. This is why a formula one car could drive on the ceiling at top speed. They create massive downforce to press the tires into the road that ends up being greater than the weight of the car. The formula one car uses aero downforce but as far as the tire's ability to hold onto the road downforce from weight works just a well. The problem is that more weight makes the car roll more so you end up with less grip in turns and it means there is more inertia to overcome in order to accelerate, and more momentum to overcome in order to turn or decelerate.

This part

The level of grip of the tyre (forgetting about
suspension niceties - we are only discussing tyres here) is totally dictated by the
coefficient of grip of the tyre and the weight acting on it - not the area of the
contact between the tyre and the road.

Does not seem right, but maybe it was part of a larger point

[Dimman]

Originally Posted by carbonBLUE

So, if you increase the weight on the tyre, then the frictional force will increase as well, in proportion to the increase in weight on the tyre - but the coefficient of friction will remain the same.

Originally Posted by Spaceywilly

It is right. This is why a formula one car could drive on the ceiling at top speed. They create massive downforce to press the tires into the road that ends up being greater than the weight of the car. The formula one car uses aero downforce but as far as the tire's ability to hold onto the road downforce from weight works just a well. The problem is that more weight makes the car roll more so you end up with less grip in turns and it means there is more inertia to overcome in order to accelerate, turn, or decelerate.

Actually, I've been going through some of my books and the section I quoted of carbon's is incorrect. A tire's Cf has a curve based on load, and Cf decreases with load. Frictional force will increase with vertical load but it is not proportional.

For example at 500 lbs vertical load a tire may provide 700 lbs of available traction.

But the same tire at 2000 lbs vertical load may only provide 1500 lbs of traction.

[ZDan]

Originally Posted by Deslock

I agree with most of that (and yes tire width, tire PSI, and vehicle weight have a larger impact). I think you're reading a bit too much into my comment... again, my point was simply that you can't only look at tire width per weight.

I never said or implied that you *should* only look at tire width per weight.

You said:

The Mustang's 255/40R19 tires have a larger diameter, so in addition to a wider contact patch, the Mustang also has a longer contact patch.

The Mustang *does* have a larger contact patch, but not because it's tires are wider and larger diameter.

It has a larger contact patch because it weighs a LOT more than the FR-S/BRZ and runs the same tire pressure.

It would have about the same contact patch area if it were on 255/40-17s, the patch would just be *wider* and *shorter*.

BTW, an interesting page: http://www.performancesimulations.co...on-tires-1.htm

I've seen that before, and their model *assumes* that contact patch width is constan with load, which is totally untrue! They TOTALLY botch the area calculation. They go through a calculation to estimate the contact patch length based on radial deflection, then they calculate area as that length by a *constant* width. But contact patch width ALSO increases with load.

In addition to the contact patch width changing with load, which they don't account for, the SHAPE will also change. As width of the contact patch gets larger with load, it will also become more rectangular as the contact patch width approaches the width of the tire. In this way, contact patch area gets larger at a greater rate than c.p. width * c.p. length would indicate.

I don't know why they used the method they did, but it isn't even close to measuring actual contact patch area.

Keeping it in the thread instead of PM'ing because I figure it's a conversation worth having in the open, since there's clearly a lot of confusion and IMO unnecessary and misguided fixation on "contact patch". Short version: wider tires will give slightly more lateral grip, *not* due to increased "contact patch. Tire make/model selection are infinitely more important.

[Spaceywilly]

Originally Posted by Dimman

Actually, I've been going through some of my books and the section I quoted of carbon's is incorrect. A tire's Cf has a curve based on load, and Cf decreases with load. Frictional force will increase with vertical load but it is not proportional.

For example at 500 lbs vertical load a tire may provide 700 lbs of available traction.

But the same tire at 2000 lbs vertical load may only provide 1500 lbs of traction.

Yes load in that case refers to cornering load or acceleration/deceleration load. More downforce on the tire would shift this whole curve upwards, but not in a linear way.

[Dimman]

Originally Posted by Spaceywilly

Yes load in that case refers to cornering load. More downforce on the tire would shift this whole curve upwards, but not in a linear way.

I don't think so. Vertical load is vertical load. It doesn't matter if it's generated by weight transfer or by wings.

In the 500/700 lbs example the Cf is 1.4

Adding downforce to get the 2000/1500 lbs example the Cf is .75

It is an ultimate gain in traction, but the Cf still decreases. They talk about it as a tire's efficiency.

Remember cornering g is based on the available traction (second number) over vehicle weight.

So in the first case the car has 2800 lbs of available traction. If it is properly balanced 50/50 a 2800 lbs car will corner at 1.0g.

With the (ridiculous amount) downforce, even with the decreasing Cf, we still have 6000 lbs of traction and the same 2800 lbs car would corner at ~2.14g

[Dimman]

We need to make a tire performance thread...

[Spaceywilly]

Vertical load refers to the load pressing the tire into the pavement (downforce). Cornering load or acceleration/deceleration load is in the plane orthogonal to the downforce and attempts to break the tire free from the pavement. At higher vertical loads, it will take more cornering load to break the tire free. It is not because of increased cf but because of increased weight on the tire. The cf stays the same but as mentioned in carbon's post the force generated by the friction against the pavement increases. As cornering load increases, this friction begins to break away in a manner that is describes by the curves in your book.

Think about taking your hand and trying to move it across a table. If you lightly press your hand against the table it moves easily. If you put a book on top of your hand it is hard to move. The cf between your hand and the table doesn't change but there is now more force required to break the friction between your hand and the table.

[Dimman]

Originally Posted by Spaceywilly

Vertical load refers to the load pressing the tire into the pavement (downforce). Cornering load or acceleration/deceleration load is in the plane orthogonal to the downforce and attempts to break the tire free from the pavement. At higher vertical loads, it will take more cornering load to break the tire free. It is not because of increased cf but because of increased weight on the tire. The cf stays the same but as mentioned in carbon's post the force generated by the friction against the earth increases. As cornering load increases, this friction begins to break away in a manner that is describes by the curves in your book.

^ Re-read that.

The curves are represented as either lateral vs vertical load, or Cf vs vertical load, which are two ways of saying the same thing.

Lateral/Vertical = Cf

[carbonBLUE]

Originally Posted by Dimman

I don't think so. Vertical load is vertical load. It doesn't matter if it's generated by weight transfer or by wings.

In the 500/700 lbs example the Cf is 1.4

Adding downforce to get the 2000/1500 lbs example the Cf is .75

It is an ultimate gain in traction, but the Cf still decreases. They talk about it as a tire's efficiency.

Remember cornering g is based on the available traction (second number) over vehicle weight.

So in the first case the car has 2800 lbs of available traction. If it is properly balanced 50/50 a 2800 lbs car will corner at 1.0g.

With the (ridiculous amount) downforce, even with the decreasing Cf, we still have 6000 lbs of traction and the same 2800 lbs car would corner at ~2.14g

this is also correct, and there are tires that give a good amount of Cf
my car for example, the highest recorded lateral g in a test was 1.04g this was due to tires and suspension, sway bars
stock a celica does .86gs car also has no fancy traction control or anything else only ABS

Originally Posted by Dimman

Actually, I've been going through some of my books and the section I quoted of carbon's is incorrect. A tire's Cf has a curve based on load, and Cf decreases with load. Frictional force will increase with vertical load but it is not proportional.

For example at 500 lbs vertical load a tire may provide 700 lbs of available traction.

But the same tire at 2000 lbs vertical load may only provide 1500 lbs of traction.

aslo i was talking about the size of contact patches not Cf :P but still you brought up a good subject

that is correct 2000lbs of vertical load only provides 1500 lbs of traction because the load has surpassed the coefficient of friction the tires can give

ie the tire are overloaded and need either A a larger tire with the same Cf or B tires of the same size that offer a higher amount of friction or grip

[Spaceywilly]

Originally Posted by Dimman

^ Re-read that.

The curves are represented as either lateral vs vertical load, or Cf vs vertical load, which are two ways of saying the same thing.

Lateral/Vertical = Cf

cf is a property between two materials and doesn't change as long as traction is maintained. Once the tire breaks away cf begins to decrease but as long as there is no slipping the cf is constant. Increasing the downforce increases the amount of lateral force required to break traction, so higher cornering or accelaration/deceleration loads can be acheived without spinning the tires.

[Lighting Red]

Originally Posted by Spaceywilly

Vertical load refers to the load pressing the tire into the pavement (downforce). Cornering load or acceleration/deceleration load is in the plane orthogonal to the downforce and attempts to break the tire free from the pavement. At higher vertical loads, it will take more cornering load to break the tire free. It is not because of increased cf but because of increased weight on the tire. The cf stays the same but as mentioned in carbon's post the force generated by the friction against the pavement increases. As cornering load increases, this friction begins to break away in a manner that is describes by the curves in your book.

Think about taking your hand and trying to move it across a table. If you lightly press your hand against the table it moves easily. If you put a book on top of your hand it is hard to move. The cf between your hand and the table doesn't change but there is now more force required to break the friction between your hand and the table.

That's right. There is also the matter of dynamic loading of the tires as the body of the car moves about the suspension system during maneuvers, but you hit the nail on the head. the coefficient of friction does not change, just the loading on the friction point.