Traction

[oh9mustang]

Instead of High jacking someone’s thread I will attempt to show that a 8" slick and a 10" slick both with the same compound will have the same traction for a given car.

M/T slick
Tread Width: 8.0"
Section Width: 10.0"
Diameter: 22.4"
Circumference: 70.0"
Rim Size: 8.0"
Code: MHR-03
Compound Type: HB-11


M/T slick
Tread Width: 10.0"
Section Width: 11.1"
Diameter: 26.3"
Circumference: 82.0"
Rim Size: 8.0"
Code: MHR-64
Compound Type: HB-11

Both have a compound of HB-11 That compound tells how sticky the tire is.

All Slick manufacturers have various different compounds for different vehicles, weight and road surface.

Here’s the tricky part.
For simplifying the numbers lets take a 4,000 lb vehicle, a 10 inch wide tire and a 5 inch wide tire. And to make it easy lets say that each tire has a 2" wide footprint (cross section)(what’s on the ground). Both tire compounds are the same.
The 10 inch wide tire has 20 sq. inch of surface area on the ground.
The vehicle weighs 4,000 so that would be 1,000 per tire, the 10 inch wide tire has 50lb per sq. inch on the footprint.
Now the 5in tire. It has 10 sq. inch footprint, same 4,000 lb vehicle except now there is 1,000 lbs on a 10 inch foot print or 100 lbs per sq. in.

10 inch slick has 50lb per sq. inch on the foot print but is twice as wide as the 5 in slick that has 100 lb per sq. in.

The actual number for a 8 vs 10 inch slick are different but the outcome is the same, The 10 will have less weight per sq in than the 8, but it's wider so they become the same.
If compounds are the same and the weight of the vehicle is the same the traction between a 8 and 10 inch slick will be no different.

Ain't math fun.

 

[flounderlipps]

Are you taking into account a wrinkle wall slick will elongate the contact patch?

 

[autumn_again]

math is great, but you have to take into effect real world situations. theres too many variables. did you take into account that the contact patch is going to be different if you put the 10 inch tire on a 10 inch rim? on an 8 inch rim its going to have more buldge in the sidewall.

you have to take into account weight transfer and how weight is distributed. how many cars have perfect 50/50 weight distribution front to back, even more so both sides weigh the exact same.

when your launching the tire with the most weight will be the right rear. weight will be distributed from front to back.

also have to take into account air pressure. a slick with 40 psi in it isn't going to have much of a contact patch as a slick with 20 psi.

not sure how you got on the subject of tires, just figured i'd throw that out there.

 

[oh9mustang]

autumn_again
I'm not sure what your trying to get at . but if both tires have the correct rim the traction will be the same. And if a car is set up correctly the left tire will have the most grip. That’s were your sitting. Maybe a re read would be in order.

flounderlipps
Ahh ok, but both 8 and 10" slicks are wrinkle wall. So the math still works.

Gremlins rule

 

[DarkFireGT]

The actual number for a 8 vs 10 inch slick are different but the outcome is the same, The 10 will have less weight per sq in than the 8, but it's wider so they become the same.
If compounds are the same and the weight of the vehicle is the same the traction between a 8 and 10 inch slick will be no different.

That's where you're wrong. You're right, the 5" tire will carry double the weight of the 10" tire per square inch. We all know that.

What you're failing to mention, again, is how you came to your conclusion. All you state is "but it's wider, so they become the same". How do you figure? Yes, there's a bit more rolling resistance and unsprung weight, however, the larger contact patch gives much more increased traction than mass it adds. For example, 100lbs takes approximately 1/10th off your 1/4 mile time. Let's say that the difference in tire and wheel weight between the 5" and 10" is 100lbs. So that'll take 1/10 off the quarter. However, the larger contact patch will get the car out of the hole a lot quicker, and reduce 60ft times by 2/10 to 3/10 of a second, which is over a second off of the 1/4 mile time.

Finally, I've never seen a professional driver using skinny tires. There's a reason. Auto racing has been going on since cars were invented. Drag racing has been around since the 20's, and the NHRA has been around for over 50 years. Over this amount of time, there has been millions (if not billions) put into research to see what would make a car go the fastest. If fat tires were no better than skinny tires, they'd use skinny tires.

Your initial math and thinking is correct. Your conclusion is incorrect, and your logic for the conclusion is missing.

 

[autumn_again]

autumn_again
I'm not sure what your trying to get at . but if both tires have the correct rim the traction will be the same. And if a car is set up correctly the left tire will have the most grip. That’s were your sitting. Maybe a re read would be in order.

flounderlipps
Ahh ok, but both 8 and 10" slicks are wrinkle wall. So the math still works.

Gremlins rule

M/T slick
Tread Width: 8.0"
Section Width: 10.0"
Diameter: 22.4"
Circumference: 70.0"
Rim Size: 8.0"
Code: MHR-03
Compound Type: HB-11


M/T slick
Tread Width: 10.0"
Section Width: 11.1"
Diameter: 26.3"
Circumference: 82.0"
Rim Size: 8.0"
Code: MHR-64
Compound Type: HB-11



as stated in your first post.

View attachment 248197

see how all these cars are leaving? the drivers side front tire is always a little bit higher then the passenger side? think of how your transfering the weight. think of if you lift the car up so the car looks like that, where is all the weight going to? the right rear tire.

its shocking the right rear tire the most. if you stand on 1 foot, are you going to have your weight equally distributed through both your feet? when you launch the car the front end is going to lift up. not that drastically in most of our cases, but thats the best example i could find to show you.

matt dasilvas s197. n/a. bolt ons. stock head cams i think he might have a ported stock intake, but he didn't see any power gain from it. hes currently running mild bolt ons, 12.04 @116.

if your into racing, you should recognize the last name. joe dasilva is an extremely popular racer.

My S197 Project build thread - WWW.S197FORUM.COM heres his build thread if your curious as to what exactly he has done. i would think his set up is pretty damn good since hes pulling better times then most s/c s197s i've seen run.

just figured i'd throw that in there so it doesn't seem like it has to be a 600 horsepower car to transfer weight like that.


and like i said in my original post, what 4,000 lb car weighs a thousand pounds at each corner. i sure as hell don't have a second motor in my trunk, so my rear end is lighter then my front end. and if you cut my car in half, all the components are not symmetrical. i don't have a steering shaft, master cylinder, etc. all on the passenger side also.

 

[lostprophet894]

opcorn:

 

[GreyDiesel]

opcorn:

wait wait, I did that on Oh9's other post, but it looks like the actions is over here now...

opcorn:

 

[J.Cagle]

Couldn't you just picture this car running skinnies on all four corners!!
I don't think it would be doing this

 

[oh9mustang]

You guys are showing a lack of reason. Math is math, 1000 lbs on a 10 in. tire that has a 50 lbs per sq in. is the same as 1000 lbs on a 5 in tire that has 100 lbs per sq in.
If you feel like saying the wider tire would have more traction, then go right ahead. keep in mind that just by saying it, don't make it true.
Compound of tire, weight of vehicle = traction, The size of the tire matters little.
The cars leaving in the pictures that lean to the passengers side are just set up wrong. If you set up a vehicle to leave straight, and then you get in the vehicle, it will lean left or passenger side. This is not rocket science.
There is no difference between 8 and 10 inch slicks when it comes to traction, if the slicks have the same compound.

J.Cagle
Just why doesn’t that GM have 18 in wide tires on it? The 8s that it has are doing the same job that a 10 would do. Not skinnies. a comparison between
8s and 10s.
The numbers I put in the first post are just to make the math easier for some, not that you would compare a 5 in tire to a 10 in tire in real life. But, unless you create some kind of (special) math the figures work out the same.
This is like telling Columbus people that the world is not flat, and the earth rotates around the sun. people can believe what they will.
Math is math, and the saying of a subject doesn't make that subject true.
If your weight was 200 lbs and you stood on one foot , your weight would not change, just the psi on the foot that was on the ground would double. And sliding that foot would be twice as difficult as sliding a foot that the weight was distributed between two feet rather than one.

 

[stangdude2000]

A car will have the same contact patch if the tires are filled to the same PSI no matter what the size of the tire is. It is simple physics/statics - forces must balance. The "cross section" as you called it is completely dependant upon how much air pressure is in the tire. In your example, you used a 4000 lb car with a even weight distribution. Let's say each of the tires were filled to 20 psi. Each tire would have a contact patch of 50 square inches. In an 8" wide slick, the dimensions would be 8 wide by 6.25" while the 10" slick would be 10" wide by 5" deep. Same area, just different dimensions. Funny thing is, in a theoretical world, there would be absolutely no traction difference since friction isn't dependant upon area anyway. Theoretically, any two tires with the same compound would grip the same no matter how big or small the tire and contact patch.

Unfortunately, we do not live in a theoretical world. When you get into friction, everything you think you know and everything you are taught about in basic physics class goes out with the trash. First, let's be clear: humans don't fully understand real world friction yet. There are several known things that affect friction in racing:

1. A wider tire means the middle section (main area of grip) becomes "softer" or can twist more. That allows it to grab the ground harder than a narrower tire. It is the same principle as any object. Is it easier to bend a 2x4" that is 6 inches long or 6 feet long? Same basic principle here.

2. A properly heated slick becomes "sticky." It will actually be sticky to the touch. Rules of friction are now gone. A wide contact patch between two sticky objects will provide MUCH more grip than a narrow one of the same area. No really sure how to explain why on that one without getting WAY too scientific. Imagine if the contact patches were rows of dots. The front row "sticks" and provides all the force. The ones behind it in the direction of motion stick some, but don't do very much since the lead dot is taking the brunt of the hit. Each dot in that front row, no matter how wide, will be sticking just about as much as the one to its left or right. As you get wider, there are more guys up front to do the "heavy sticking" and provides more force.

3. Abrasion factor. If there is too much force, materials will simply shear - or in the case of tires, shred. There is a theory that the leading edge of the tire (front line of the area of the contact patch) is pressed down upon by the rest of the rotating tire. In a narrow tire, this will create an intense local pressure that will begin shredding the outer layer of the tire. These shreads will be extremely small, but will act almost as small ball bearings and destroy the traction. A wider tire will provide a larger area on that leading edge that will reduce the pressure and the abrasion factor.

4. Perhaps just as important as anything related to friction is the tire diameter. Yes, I know we are talking about two slicks with the same diameter but different widths. Take a look at the pictures above. You can see that in every one, the tire elongates as they are being launched and the diameter increases. The diameter increases more as the tire gets wider. This effectively increases the final drive ratio over a narrower tire, allowing it to accellerate quicker. Obviously this only has an advantage for a while - using a 5 foot wide tire would do more harm than good, even if it got really tall on the launch! Along these lines, we all have somewhat of a CVT on our cars, especially when we use slicks. The diamater gets closer to original as you increase your speed and the rate of accelleration decreases, effectively dropping the final drive ratio and increasing your top speed!


Friction is complicated of a topic enough. When you throw weird compounds like rubber into the mix, it really gets crazy.

 

[Steel Horse]

You guys are showing a lack of reason. Math is math, 1000 lbs on a 10 in. tire that has a 50 lbs per sq in. is the same as 1000 lbs on a 5 in tire that has 100 lbs per sq in.
If you feel like saying the wider tire would have more traction, then go right ahead. keep in mind that just by saying it, don't make it true.
Compound of tire, weight of vehicle = traction, The size of the tire matters little.
The cars leaving in the pictures that lean to the passengers side are just set up wrong. If you set up a vehicle to leave straight, and then you get in the vehicle, it will lean left or passenger side. This is not rocket science.
There is no difference between 8 and 10 inch slicks when it comes to traction, if the slicks have the same compound.

J.Cagle
Just why doesn’t that GM have 18 in wide tires on it? The 8s that it has are doing the same job that a 10 would do. Not skinnies. a comparison between
8s and 10s.
The numbers I put in the first post are just to make the math easier for some, not that you would compare a 5 in tire to a 10 in tire in real life. But, unless you create some kind of (special) math the figures work out the same.
This is like telling Columbus people that the world is not flat, and the earth rotates around the sun. people can believe what they will.
Math is math, and the saying of a subject doesn't make that subject true.
If your weight was 200 lbs and you stood on one foot , your weight would not change, just the psi on the foot that was on the ground would double. And sliding that foot would be twice as difficult as sliding a foot that the weight was distributed between two feet rather than one.

Those cars lean to the passenger side because the torque from the engine is actually twisting the chassis. They are not setup wrong at all.

See even the top fuelers do it.

http://images.google.com/imgres?img...v=/images?q=top+fuel+dragster&hl=en&sa=N&um=1

 

[DarkFireGT]

You guys are showing a lack of reason. Math is math, 1000 lbs on a 10 in. tire that has a 50 lbs per sq in. is the same as 1000 lbs on a 5 in tire that has 100 lbs per sq in.
If you feel like saying the wider tire would have more traction, then go right ahead. keep in mind that just by saying it, don't make it true.
Compound of tire, weight of vehicle = traction, The size of the tire matters little.
The cars leaving in the pictures that lean to the passengers side are just set up wrong. If you set up a vehicle to leave straight, and then you get in the vehicle, it will lean left or passenger side. This is not rocket science.
There is no difference between 8 and 10 inch slicks when it comes to traction, if the slicks have the same compound.

J.Cagle
Just why doesn’t that GM have 18 in wide tires on it? The 8s that it has are doing the same job that a 10 would do. Not skinnies. a comparison between
8s and 10s.
The numbers I put in the first post are just to make the math easier for some, not that you would compare a 5 in tire to a 10 in tire in real life. But, unless you create some kind of (special) math the figures work out the same.
This is like telling Columbus people that the world is not flat, and the earth rotates around the sun. people can believe what they will.
Math is math, and the saying of a subject doesn't make that subject true.
If your weight was 200 lbs and you stood on one foot , your weight would not change, just the psi on the foot that was on the ground would double. And sliding that foot would be twice as difficult as sliding a foot that the weight was distributed between two feet rather than one.

Again, you don't state how you got to your final conclusion. You just, again, stated we were wrong and stated your conclusion. You said you could "prove it" to us. Prove it.

 

[GreyDiesel]

A car will have the same contact patch if the tires are filled to the same PSI no matter what the size of the tire is. It is simple physics/statics - forces must balance. The "cross section" as you called it is completely dependant upon how much air pressure is in the tire. In your example, you used a 4000 lb car with a even weight distribution. Let's say each of the tires were filled to 20 psi. Each tire would have a contact patch of 50 square inches. In an 8" wide slick, the dimensions would be 8 wide by 6.25" while the 10" slick would be 10" wide by 5" deep. Same area, just different dimensions. Funny thing is, in a theoretical world, there would be absolutely no traction difference since friction isn't dependant upon area anyway. Theoretically, any two tires with the same compound would grip the same no matter how big or small the tire and contact patch.

Unfortunately, we do not live in a theoretical world. ...

1. A wider tire means the middle section (main area of grip) becomes "softer" or can twist more. ...

2. A properly heated slick becomes "sticky." ... As you get wider, there are more guys up front to do the "heavy sticking" and provides more force.

3. Abrasion factor. ...A wider tire will provide a larger area on that leading edge that will reduce the pressure and the abrasion factor....

4. ...The diameter increases more as the tire gets wider....

...Friction is complicated of a topic enough. When you throw weird compounds like rubber into the mix, it really gets crazy.

effing brilliant stangdude, thanks for posting. Are you an engineer? I am currently finishing up a degree in Physics. I think Oh9 posed a great question that illustrates the difference between the theoretical and real worlds. Math is math, but when and why does math actually apply to the real world? This is a question that boggles even the best mathematicians and physicists.

I was thinking about this last night, and because I have a largely theoretical background, I was able to come up with the contact patch idea. And then I thought, well, friction is just the normal force times the coefficient of friction... but of course its not just that. Its only that in physics classes so we can move on. I am interested in taking some engineering courses so I can get more into the details of these ideas we gloss over in physics courses.. like friction and air resistance.

 

[oh9mustang]

Dark fire'
Reading comprehension. I explained it.

If yous guys would put both 8 & 10 in slicks with the same compound, on the same car, with the same air pressure, on the same planet, the traction will be the same.

AA/FD are a little different than the rest of the world. 8,000 hp will do that and it will also snap the chassis in half .

The normal car world, cars that pick up the drivers side tire are set up wrong. When setting a 4 link they can't be set the same on both sides dew to torque and the driver sitting on the drivers side. Before the 4 link there was traction bars, and with traction bars the drivers side had to be preloaded. I used to get 2-3 guys to sit on the back left while I tightened it around the leaf. Then I would chain down the drivers side of the engine, so the torque would be distributed better to the frame. Even if set up wrong the traction will still be the same for 8s and 10s on the same car.

Think about it logically.
If you drag any object across a rough surface that has a weight on it, the object will get abrazed, Now 1/2 the surface contact area for whatever it is your dragging and the abrasion will double.

 

[autumn_again]

AA/FD are a little different than the rest of the world. 8,000 hp will do that and it will also snap the chassis in half .

The normal car world, cars that pick up the drivers side tire are set up wrong. When setting a 4 link they can't be set the same on both sides dew to torque and the driver sitting on the drivers side. Before the 4 link there was traction bars, and with traction bars the drivers side had to be preloaded. I used to get 2-3 guys to sit on the back left while I tightened it around the leaf. Then I would chain down the drivers side of the engine, so the torque would be distributed better to the frame. Even if set up wrong the traction will still be the same for 8s and 10s on the same car.

Think about it logically.
If you drag any object across a rough surface that has a weight on it, the object will get abrazed, Now 1/2 the surface contact area for whatever it is your dragging and the abrasion will double.

thats exactly why i showed you matt dasilvas car, making 300 horsepower, lifting the front tire off the ground because i knew you would throw out the "oh thats a drag car" phrase. his car has 300 horsepower, is the same as our chassis with a little weight reduction. 300 lbs i think total. reading comprehension. i hear you talk about it, try it some time.

and like darkfire said again. you are not proving sh¡t, you are just stating if compared the tires would perform the same, with the same amount of traction, you are not stating why, your not stating why it would be the same. your giving no reasoning.


you just keep saying the traction will be the same, your not giving a single damn reason why it would be. stating it over and over again doesn't give any kind of reasoning, it just means your stuck in a rut because of a statement that you made that you have no proof for, and don't wanna look like an ass for calling people out when you can not prove your own statement.

 

[stangdude2000]

Dark fire'
Reading comprehension. I explained it.

If yous guys would put both 8 & 10 in slicks with the same compound, on the same car, with the same air pressure, on the same planet, the traction will be the same.

AA/FD are a little different than the rest of the world. 8,000 hp will do that and it will also snap the chassis in half .

The normal car world, cars that pick up the drivers side tire are set up wrong. When setting a 4 link they can't be set the same on both sides dew to torque and the driver sitting on the drivers side. Before the 4 link there was traction bars, and with traction bars the drivers side had to be preloaded. I used to get 2-3 guys to sit on the back left while I tightened it around the leaf. Then I would chain down the drivers side of the engine, so the torque would be distributed better to the frame. Even if set up wrong the traction will still be the same for 8s and 10s on the same car.

Think about it logically.
If you drag any object across a rough surface that has a weight on it, the object will get abrazed, Now 1/2 the surface contact area for whatever it is your dragging and the abrasion will double.

No. Wrong. Just flat out wrong. On everything. Reread my post on how friction and tires work. Friction does depend on area in the real world.

As someone pointed out before, the chassis is twisting - it isn't because of the torque not being split properly or weight not being distributed properly. All chassis twist. Not just top fuel. You don't notice it 99% of the time in street cars though because of their shape. In most street cars, there also isn't enough torque for the engine to twist the chassis. Anyway, the engine creates this gob of torque which in most RWD cars, twists about the direction of the car. Depending on exactly how the car is set up (which direction the flywheel is spinning relative to looking out the front of the car), there will need to be a force to counteract that. That where the engine mounts come into play. For every action, there is an equal, but opposite, reaction. The engine mounts (part of the chassis) are twisted as they resist the force of the engines torque. As a result, the chassis twists either to the passenger's side or the driver's side depending on the car.

A car with a transverely mounted motor would also twist, but it would try to bring the front wheels to the back wheels (which also happens under accelleration).


GreyDiesel, yes, I got a degree in mechanical engineering. I also took several courses on automotive design/engineering and natures of materials which dealt with this type of thing quite a bit. I also designed and built a race car (formula style - < 600 lbs, single seat, Yamaha R6 motor) for my senior design project, so I got to put all that education to work.

 

[oh9mustang]

The reason the traction would be the same is beacuse the compound is the same. The weight is the same, The car is the same, the tire pressure is the same the psi on the contact patch is different for the 8 and 10. Sheeeez.

I'm done taling to a wall.

 

[cycosarge]

OK... in a vacuum his theory works. However, on the dragstrip it doesn't. Depending on the Torque applied to the slick and the weight of the vehicle, the vehicle will need a certain amount of grip from the tires or it will spin. For example, a 4000 pound car with 500 foot pounds of torque and a 10 inch slick will have the same traction as a 3000 pound car, 500 foot pounds and an 8 inch slick. You need to find the right size width tire necessary to ensure a slip-less launch for YOUR car. Torque is key here. The more flash-torque hitting the rear tires at launch, the more traction you need for a given weight. As you increase engine power, you need to increase tire size accordingly. I have a friend who races his 1965 Chevy II, he is running on the ragged edge with traction. His car make 900 HP and 755 Torque with a SC SBC. He is running a 9 inch slick on an 8 inch rim. How does he do it? The 9 inch slick is the minimum needed to launch at his torque level at 3000 rpm. If he were to up the power output, a wider tire or shorter gear would be needed. I try to maintain mininum standards for rear traction as it pertains to the tire. Anything bigger than you need is excess unsprung weight.

 

[DarkFireGT]

Dark fire'
Reading comprehension. I explained it.

If yous guys would put both 8 & 10 in slicks with the same compound, on the same car, with the same air pressure, on the same planet, the traction will be the same.

AA/FD are a little different than the rest of the world. 8,000 hp will do that and it will also snap the chassis in half .

The normal car world, cars that pick up the drivers side tire are set up wrong. When setting a 4 link they can't be set the same on both sides dew to torque and the driver sitting on the drivers side. Before the 4 link there was traction bars, and with traction bars the drivers side had to be preloaded. I used to get 2-3 guys to sit on the back left while I tightened it around the leaf. Then I would chain down the drivers side of the engine, so the torque would be distributed better to the frame. Even if set up wrong the traction will still be the same for 8s and 10s on the same car.

Think about it logically.
If you drag any object across a rough surface that has a weight on it, the object will get abrazed, Now 1/2 the surface contact area for whatever it is your dragging and the abrasion will double.

Thank you, Oh9, I can read. As can most of the members of this board. Not one of us has been able to determine how you came up with your inferred conclusion on traction. In any event, your conclusion is wrong. You can make a point without insulting people, especially moderators. I'm not the first moderator you've insulted. If you would like to continue behavior of that nature, we can take care of that. You offer little to nothing positive to this community. In the majority of your posts, you are instigating arguments and/or insulting members and staff.

You can argue the logic all you want, but it has been proven time and time again that wider tires will hook up better and cause a decrease in 1/4 mile times.