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Discussion Starter #1 (Edited)
I've been avoiding writing this up because a) I'm lazy and b) I'm probably going to come across as sounding preachy. Still, I think it is good to get this all in one place so that the discussions about lightness and acceleration/cornering/braking can be straightened out.

Some equations that describe what we are talking about (remember to use consistent units when using these equations):

F = M * A: Force equals mass times acceleration

F_f = mu * N: Friction force available equals coefficient of friction times normal force

F_r = M * v^2 / r: Force necessary to make you go in a circle of radius 'r' at velocity 'v' equals mass times velocity squared divided by the radius of the circle.
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Case 1: Speeding up

Your engine provides a given force to the ground. You can figure out this force by measuring your torque on a dyno. Torque equals force times moment_arm, where moment_arm is going to have to include things like your gear ratios and the radius of your wheels/tires. Rearranging the first equation gives:

A = F / M

So, you can increase acceleration either with increasing the force put out by your engine or decreasing the mass. Since you have a redline that you can't rev past, you have to shift gears. Everytime you upshift you choose a gear ratio that decreases the amount of torque delivered to the wheels. This is why a low torque engine that revs high (i.e. makes good horsepower) can accelerate a car faster than an engine with more torque but a lower redline.

You can also increase acceleration by decreasing the mass you have to move. This is typically considered to be the way Lotus looks at things. Less is more...
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Case 2: Traction limited acceleration

This can apply to accelerating the car but more typically applies to braking and cornering. Let's look at braking first.

Case 2a: Braking

In most cars braking is limited by the amount of friction force provided by the tires. From the equations above we have:

F = M * A and
F = mu * N

In the case of braking, A is the amount of deceleration you get. Let's first think about this in terms of a car that has one wheel. In that case, N is the mass of the car and setting the two equations equal we get:

M * A = mu * M

Of course, the Ms cancel and we get:

A = mu

So, the deceleration you get is basically equal to the friction coefficient of the tires against the pavement. If you cut the friction coefficient in half, the amount of deceleration you get is half as much.

If you want to translate acceleration into braking distance from a given speed down to zero speed:

distance = (velocity^2) / (2 * A)

So, if you cut the acceleration in half, you double your stopping distance. Note: This all assumes constant braking deceleration.

Of course, you actually have four tires on a car that each carry some fraction of the mass of the car (let's say each tire supports 1/4 of the mass). Now we would have:

M * A = (mu * 1/4 * M) + (mu * 1/4 * M) + (mu * 1/4 * M) + (mu * 1/4 * M)

Clearly, if you add all four terms up, you end up with what we had before...A = mu.

The world isn't quite this simple of course. As some people have pointed out, mu is not constant with respect to the normal force (the amount of downforce on a tire). Typically, as the normal force increases, mu decreases. For the sake of argument (that is to say, I have no reason to believe this is exactly how tires interact with pavement), let's say that:

mu = 10 m/s^2 * (1.0 - (M/5000 kg))

FWIW, 10 m/s^2 is about one 'g'. So, if we load a tire that has this kind of friction coefficient with 1000 kg, we get a friction coefficient of:

mu = 8 m/s^2

If we lighten the car and make the tire bear just 500 kg,

mu = 9 m/s^2

Since A=mu, this means the car will stop a little more than 10% faster when the weight is cut in half.

There is more to this though. The tire is in contact with the ground over an area. A given square centimeter of contact patch only supports a fraction of the car's weight. If you increase the size of the contact patch, you decrease the amount of weight that each square centimeter of contact patch has to support. Since mu increases when the amount of weight being supported decreases, you can expect that increasing the size of the contact patch will increase the amount of friction available to you.

Tire dynamics are tricky though and I'm a bit out of my depth to discuss them in any detail. However, you roughly know the amount of weight per square centimeter that your tires support. Let's say you inflate your tires to 30 lbs/in^2. If you discount the amount of weight supported by the sidewalls of your tires you can say that each square inch of contact patch is supporting 30 lbs of car (I recommend switching to metric units if you actually play with these numbers though). You can use this information to weigh your car...drive each tire over a piece of paper and trace the outline of the contact patch. Measure the area of the contact patches, add the areas together and multiply by the tire pressure. This should give you the weight of the car (assuming the sidewalls don't support any of the weight of the car...which they actually do, so this won't be 100% right)

If you decrease the air pressure in your tires you will increase the size of your contact patch. This decreases the amount of weight supported by each square centimeter and therefore increases mu. This MAY mean that you can brake harder. Remember that the sidewalls of any performance tire are pretty stiff. If you drop your pressure too much you may actually REDUCE the amount of tire that is actually working for you...you may make it so that only the edge of the tire (the part under the sidewall) is the only part working...this would mean that you've made your mu go down and you'll loose grip.

You also have to take into account weight transfer from the rear to the front of the car under braking as well if you want to accurately calculate braking distances.

This is pretty much the extent of my tire knowledge...my impression is that there are many more details and variables that effect things (e.g. deformation of the tire).

All of the above applies to acceleration (if it is traction limited) as well since acceleration is just like deceleration, just in the opposite direction. You'll note that dragsters quite often have very low tire pressures on the driven tires in order to maximize grip.

Case 2b: Cornering

All of the above applies to cornering as well, but radial acceleration (physics speak for cornering) is slightly different.

F_r = M * v^2 / r
F_f = mu * M

So,

M * v^2 / r = mu * M

yielding,

v = sqrt(mu * r)

So, the velocity you can go around a corner is equal to the square root of the friction coefficient times the radius of the corner. All the same arguments about the friction coefficient stated in the braking example apply here. Deformation of the tire can be quite obvious under cornering...most of us have seen pictures of sidewalls rolling over. Clearly you have to have enough inflation to avoid that since you aren't really even using the tire if its sidewall is sufficiently rolled over.
-----------------------------------

Again, I'm sure there are many more details concerning tire dynamics (contact patch shapes, camber, etc), however I believe the above points are useful for understanding the basics.

I'll get down off my soapbox now...
 

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Discussion Starter #5
shay2nak said:
damn, that is one big ass post! :p Thanks for taking the time & effort!
My pleasure.

Physics: It's not just a career, it's a lifestyle. :)

Most of my physics friends can't help but lecture about stuff like this. :) We lecture eachother all the time. *chuckle*
 

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shay2nak
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originally posted by Nochmal
Physics: It's not just a career, it's a lifestyle.
Most of my physics friends can't help but lecture about stuff like this. We lecture eachother all the time. *chuckle*

I'm an engineer, so I know where you're coming from. :)
 

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Nochmal said:
I've been avoiding writing this up because a) I'm lazy and b) I'm probably going to come across as sounding preachy. Still, I think it is good to get this all in one place so that the discussions about lightness and acceleration/cornering/braking can be straightened out.

Some equations that describe what we are talking about (remember to use consistent units when using these equations):

F = M * A: Force equals mass times acceleration

F_f = mu * N: Friction force available equals coefficient of friction times normal force

F_r = M * v^2 / r: Force necessary to make you go in a circle of radius 'r' at velocity 'v' equals mass times velocity squared divided by the radius of the circle.
--------------------------------------
Case 1: Speeding up

Your engine provides a given force to the ground. You can figure out this force by measuring your torque on a dyno. Torque equals force times moment_arm, where moment_arm is going to have to include things like your gear ratios and the radius of your wheels/tires. Rearranging the first equation gives:

A = F / M

So, you can increase acceleration either with increasing the force put out by your engine or decreasing the mass. Since you have a redline that you can't rev past, you have to shift gears. Everytime you upshift you choose a gear ratio that decreases the amount of torque delivered to the wheels. This is why a low torque engine that revs high (i.e. makes good horsepower) can accelerate a car faster than an engine with more torque but a lower redline.

You can also increase acceleration by decreasing the mass you have to move. This is typically considered to be the way Lotus looks at things. Less is more...
--------------------------------------------
Case 2: Traction limited acceleration

This can apply to accelerating the car but more typically applies to braking and cornering. Let's look at braking first.

Case 2a: Braking

In most cars braking is limited by the amount of friction force provided by the tires. From the equations above we have:

F = M * A and
F = mu * N

In the case of braking, A is the amount of deceleration you get. Let's first think about this in terms of a car that has one wheel. In that case, N is the mass of the car and setting the two equations equal we get:

M * A = mu * M

Of course, the Ms cancel and we get:

A = mu

So, the deceleration you get is basically equal to the friction coefficient of the tires against the pavement. If you cut the friction coefficient in half, the amount of deceleration you get is half as much.

If you want to translate acceleration into braking distance from a given speed down to zero speed:

distance = (velocity^2) / (2 * A)

So, if you cut the acceleration in half, you double your stopping distance. Note: This all assumes constant braking deceleration.

Of course, you actually have four tires on a car that each carry some fraction of the mass of the car (let's say each tire supports 1/4 of the mass). Now we would have:

M * A = (mu * 1/4 * M) + (mu * 1/4 * M) + (mu * 1/4 * M) + (mu * 1/4 * M)

Clearly, if you add all four terms up, you end up with what we had before...A = mu.

The world isn't quite this simple of course. As some people have pointed out, mu is not constant with respect to the normal force (the amount of downforce on a tire). Typically, as the normal force increases, mu decreases. For the sake of argument (that is to say, I have no reason to believe this is exactly how tires interact with pavement), let's say that:

mu = 10 m/s^2 * (1.0 - (M/5000 kg))

FWIW, 10 m/s^2 is about one 'g'. So, if we load a tire that has this kind of friction coefficient with 1000 kg, we get a friction coefficient of:

mu = 8 m/s^2

If we lighten the car and make the tire bear just 500 kg,

mu = 9 m/s^2

Since A=mu, this means the car will stop a little more than 10% faster when the weight is cut in half.

There is more to this though. The tire is in contact with the ground over an area. A given square centimeter of contact patch only supports a fraction of the car's weight. If you increase the size of the contact patch, you decrease the amount of weight that each square centimeter of contact patch has to support. Since mu increases when the amount of weight being supported decreases, you can expect that increasing the size of the contact patch will increase the amount of friction available to you.

Tire dynamics are tricky though and I'm a bit out of my depth to discuss them in any detail. However, you roughly know the amount of weight per square centimeter that your tires support. Let's say you inflate your tires to 30 lbs/in^2. If you discount the amount of weight supported by the sidewalls of your tires you can say that each square inch of contact patch is supporting 30 lbs of car (I recommend switching to metric units if you actually play with these numbers though). You can use this information to weigh your car...drive each tire over a piece of paper and trace the outline of the contact patch. Measure the area of the contact patches, add the areas together and multiply by the tire pressure. This should give you the weight of the car (assuming the sidewalls don't support any of the weight of the car...which they actually do, so this won't be 100% right)

If you decrease the air pressure in your tires you will increase the size of your contact patch. This decreases the amount of weight supported by each square centimeter and therefore increases mu. This MAY mean that you can brake harder. Remember that the sidewalls of any performance tire are pretty stiff. If you drop your pressure too much you may actually REDUCE the amount of tire that is actually working for you...you may make it so that only the edge of the tire (the part under the sidewall) is the only part working...this would mean that you've made your mu go down and you'll loose grip.

You also have to take into account weight transfer from the rear to the front of the car under braking as well if you want to accurately calculate braking distances.

This is pretty much the extent of my tire knowledge...my impression is that there are many more details and variables that effect things (e.g. deformation of the tire).

All of the above applies to acceleration (if it is traction limited) as well since acceleration is just like deceleration, just in the opposite direction. You'll note that dragsters quite often have very low tire pressures on the driven tires in order to maximize grip.

Case 2b: Cornering

All of the above applies to cornering as well, but radial acceleration (physics speak for cornering) is slightly different.

F_r = M * v^2 / r
F_f = mu * M

So,

M * v^2 / r = mu * M

yielding,

v = sqrt(mu * r)

So, the velocity you can go around a corner is equal to the square root of the friction coefficient times the radius of the corner. All the same arguments about the friction coefficient stated in the braking example apply here. Deformation of the tire can be quite obvious under cornering...most of us have seen pictures of sidewalls rolling over. Clearly you have to have enough inflation to avoid that since you aren't really even using the tire if its sidewall is sufficiently rolled over.
-----------------------------------

Again, I'm sure there are many more details concerning tire dynamics (contact patch shapes, camber, etc), however I believe the above points are useful for understanding the basics.

I'll get down off my soapbox now...
For those that don't get it, he's saying that Collin Chapman Da Man and performance through light weight (Acceleration, cornering, braking) even applies here in the USA:D
 

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Discussion Starter #8 (Edited)
The following does not apply directly to lightness and performance, but it does amplify a bit on tire dynamics:

The coefficient of friction is different for a sliding tire and a tire that is not sliding. In your physics textbooks you'll see talk about static coefficients of friction and kinetic coefficients of friction. The static coefficient (non-sliding) of friction is generally larger than the kinetic (sliding) coefficient. This is why locking up your brakes makes your braking distance longer.

This also is what makes some tires less forgiving than others. Typically, race tires have a larger difference between their static and kinetic friction coefficients than more forgiving street tires. The book referenced above may explain why this is in detail but in general it seems that in order to get the highest static friction coefficients you end up with a larger static/kinetic difference. This means that once the race tires break loose (under braking, acceleration or cornering) your grip drops a lot and it takes more to gather things up. Street tires, obviously, are engineered to be more forgiving and have more similar static/kinetic coefficients.

All this goes out the window when you aren't on pavement. Sometimes on snow it is better to lock up your brakes and pile up snow in front of your tires as you slide. Also, you'll notice rally drivers nearly always are sliding. I understand this to be due to the fact that on loose surfaces you get more longitudinal grip (good for accelerating and braking) than you get transverse grip (good for cornering). So, they pitch their cars sideways so they can use the longitudinal grip to give them acceleration transverse to the direction they are traveling (i.e. to make the car go around a corner).

OK OK...lecture over, I promise....
 

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The equations can be simplified considerably:

ME + ELISE = FUN



I was forced to take a year of physics in college, but haven't used it much.
 

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I am hoping the delete option will be a reality and available by the time my order is required to be taken.
 

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Discussion Starter #11
jml1952 said:
The equations can be simplified considerably:

ME + ELISE = FUN



I was forced to take a year of physics in college, but haven't used it much.
Hah. As I like to say, physics is everything. :) The more you understand physics, the more you understand the world around you...even if you never write down an equation. The equations are just a compact language...you don't have to think in terms of the equations but sometimes it helps to train your intuition. The problem usually is that people have had a bad experience with a poor teacher who didn't really understand what they were talking about AND couldn't communicate very well.

Careful...that soapbox is looking mighty comfy right about now. :)
 

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Nochmal said:
This also is what makes some tires less forgiving than others. Typically, race tires have a larger difference between their static and kinetic friction coefficients than more forgiving street tires. The book referenced above may explain why this is in detail but in general it seems that in order to get the highest static friction coefficients you end up with a larger static/kinetic difference.
Nicely stated...

Paul's book does a wonderful job of explaining where the "friction" between tire & tarmac is generated. Most comes from mechanical "interlocking", not some friction coefficient. A sliding tire has LOST nearly all of its mechanical interlocking. A race tire is designed to optimize that mechanical interlocking with the tarmac, thus as you correctly point out the difference in the force between a "hooked-up" race tire and a sliding one is one hella alot BIGGER than with a street tire (which is designed to wear).
 

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Nochmal said:

All this goes out the window when you aren't on pavement. Sometimes on snow it is better to lock up your brakes and pile up snow in front of your tires as you slide. Also, you'll notice rally drivers nearly always are sliding. I understand this to be due to the fact that on loose surfaces you get more longitudinal grip (good for accelerating and braking) than you get transverse grip (good for cornering). So, they pitch their cars sideways so they can use the longitudinal grip to give them acceleration transverse to the direction they are traveling (i.e. to make the car go around a corner).
Interesting theory... I've always wondered why when the driving instructors say that skidding isn't the fastest way around a track, the rally guys don't seem to listen. :)
 

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One thing about the coefficient of friction is based on 2 surfaces - the tire and the type of surface - sliding or not. Starting friction is greater than sliding friction yes. But my question is since the contact area of a tire is not a function of friction why is it that a wider tire is better than a skinnier tire (aside from the heat aspect)?

Since heat is a waste product of friction i understand why a hotter tire would have less friction (hold) than a skinnier tire but at the same temp a 12' tire should have the same grip as a 1" tire of the same compound - yet this is obviously wrong in the "real world" (examples include the scored landing way of the Space Shuttle).

(damn science geeks)
 

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Nochmal said:
I think it is good to get this all in one place so that the discussions about lightness and acceleration/cornering/braking can be straightened out.
Great post. For me, the critical element was when I learned that friction does not increase linearly with weight (for rubber, I think it might for others materials?) I had to noodle on that one for a few hours - then drive on it for a few dozen laps - before the implications started to sort themselves out. You did a good job of explaining the physics.
 

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Evl said:
Interesting theory... I've always wondered why when the driving instructors say that skidding isn't the fastest way around a track, the rally guys don't seem to listen. :)
Speaking of that, I regularly run into a situation that I can't explain - maybe one of you can. I understand that rubber provides more traction when it can interlock (on a molecular level) with the surface of the road versus sliding across the road. That makes sense to me. What I don't get is why most cars corner considerably faster with fairly high slip angles. In the Spec RX7s we run pretty high slip angles - 10 - 15%. Fast corners are a long series of toss-and-catch corrections as the car hurtles toward the track out point. Every road car, and almost every race car, I've driven responds well to some drift - although most don't like as much as the Spec RX7s.

Is it just a matter of degree? IE sliding the tires within their range of surface deformation/tearing but not so much that they can't interlock with the road surface? I need to study Paul Haney's book some more....
 

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Discussion Starter #20 (Edited)
That's right...get us our Elisi or we will noodle ourselves to death.

A couple of extra points:

1) "Mechanical interlocking" and a "friction coefficient" are not exclusive ideas. That is to say, "friction coefficient" is a concept that includes things like "mechanical interlocking" in it. The coefficient is meant to include all sources of grip. Once upon a time I read that there is even chemical bonding that goes on between a race tire and the road...that one still wierds me out a bit.

2) Patch size and shape. As noted here and in other threads, you can increase your grip by having a larger contact patch. The question of why a wide tire might grip better than a narrow tire with the same overall contact patch size is one that I don't really know the answer. However, I have been told that a tire tends to grip more in the direction parallel to the longest part of the contact patch. That is to say, for drag racing a long narrow contact patch is desired, while for pure cornering a short wide contact patch is desired. This makes some intuitive sense, but I'd be happier if I understood why it should be this way.

3) Slip angles. I believe that a slip angle does not necessarily mean that the tire is sliding. As most of us probably know, slip angle is the difference between the direction that the car is going and the direction the wheels are pointed. A small slip angle (which is typically where the best grip is) can be generated by having a slight twist of the tire while not making the contact patch slide. Again, I don't have solid physical arguments for why this would create more grip but it does make some sense that twisting the tire a bit might make it grab onto the surface more.

This also is part of what makes some tires more forgiving than others. As you increase the slip angle you will eventually start getting less grip. In less forgiving tires the drop off is more sudden than in more forgiving tires. As with the overall static/kinetic friction coefficients, race tires tend to be less forgiving. I understand that many race tires like larger slip angles but if pushed just a bit too far will lose grip quickly.

JeffersonRaley, when you say that a car likes to drift, are you sure that it is sliding or are you just sensing a slip angle that might not actually be due to sliding? What do you mean by a 10-15% slip angle? Do you mean degrees...how do you actually measure the slip angles? I'm not real sure about the question of a bit of sliding giving more grip...I suppose it makes some sense but my gut feeling is that it might be that when a car is at "good" slip angles it might feel like it is sliding even when the contact patches are hooked up.

4) Heat and grip. The reasons that temperature makes a difference to how much a tire grips come down to the materials involved. As you heat up a rubber compound you will change its properties. If you make it very very hard (cold) it isn't hard to understand that it will have less grip than if you make it a little softer (warm). If you make it very very gooey (hot) the molecules of rubber won't hold onto eachother very well and the tire will just ooze around instead of gripping. Different compounds perform well at different temperatures. High end racing tires need to be "hot" in order to have maximum grip, but if you got your street tires to those temperatures you'd have very little grip. Obviously, street tires need to work well when cold since that is primarily where they are used. Race tires are going to get very hot so they need to work best when hot and the rubber compounds are tuned to those temperatures (or at least the tire manufacturers try to tune them). A tire gets heat in it a number of ways and I'm not sure I know them all...sliding clearly puts heat into the tire as does compressing and twisting it (which constantly goes on as a tire is being used). Street tires also get heat from the tread blocks squirming around (which is what makes street tires squeal). I've probably over simplified a few things, but I think the above contains the basics of why temperature matters in a tire.
 
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