I'm having issues wrapping my head around that too jimbo. Everythinng I learned from all of my statics and physics classes in college dictate that if you increase the surface area that a force is put upon, then the force should be decreased for any given area compared to the original smaller area. So you would think having a wider tire would distribute the force across a larger area, in essence putting less wear per unit of area compared to the smaller tire. I would think this in turn would permit for longer tire life since it would be putting less force on the area.
However I'm not a mechanical engineer by training, I just had to take a lot of their classes to be ready for the PE-Exam. There might be some "left hand rule" for tires I don't know about that dictates these types of things.
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Put another way to sum up my confusion, it sounds like you are saying, for any given tire compound and tread design, it doesn't matter if you have a 6-inch wide tire or a 12-inch wide tire, your friction and slip upon launch will be identical in each case because youre contact patch is the same in each case.
Well, not exactly. As simply as I can describe it, a 3000 lb. car is being pulled toward the earth's surface by gravity. The only thing stopping it from reaching its objective is the equal and opposite force imparted by the four tires. The pressurized air in those tires is pushing the car away from the earth's surface. Given a fixed vehicle mass of 3000 lbs. and a fixed tire inflation pressure of 30 psi, you end up with a fixed total area of tire contact patch needed to equalize the forces. Does that part make sense?
If you add weight to the car but keep the tire pressure the same, the size of the contact patch must neccessarily increase. If you keep the weight of the car the same but deflate the tires, the size of the contact patch must increase in order to keep the forces equal. Hopefully this makes sense as well.
Thus, the area of the contact patch remains contant so long as inflation pressure and the car's weight remain constant. This is true irrespective of the width of the tire. It then just becomes a question of the contact patch's shape, and it is this which determines the car's ultimate handling characteristics.
To be sure, other factors come into play and must be taken into account in determining which tire is best for a given task, but hopefully now the issue of contact patch size is more comprehensible.
Well, not exactly. As simply as I can describe it, a 3000 lb. car is being pulled toward the earth's surface by gravity. The only thing stopping it from reaching its objective is the equal and opposite force imparted by the four tires. The pressurized air in those tires is pushing the car away from the earth's surface. Given a fixed vehicle mass of 3000 lbs. and a fixed tire inflation pressure of 30 psi, you end up with a fixed total area of tire contact patch needed to equalize the forces. Does that part make sense?
If you add weight to the car but keep the tire pressure the same, the size of the contact patch must neccessarily increase. If you keep the weight of the car the same but deflate the tires, the size of the contact patch must increase in order to keep the forces equal. Hopefully this makes sense as well.
Thus, the area of the contact patch remains contant so long as inflation pressure and the car's weight remain constant. This is true irrespective of the width of the tire. It then just becomes a question of the contact patch's shape, and it is this which determines the car's ultimate handling characteristics.
To be sure, other factors come into play and must be taken into account in determining which tire is best for a given task, but hopefully now the issue of contact patch size is more comprehensible.
I'll challenge your description a little. Take a set of 4 inch wide tires from the fifties (18 inchers were on MG Tseries and race cars of that era) and a set of these new 245/45-18 and on two otherwise identical cars start them around a circular track (NASCAR comes to mind) gradually increase speed until each car loses traction and slides into the wall. You are saying both tires would have the same contact patch - I'm saying the car with the larger patch (wider tire)will stick to the pavement better. Another way to think about it is two people that weigh the same and are the same height. One wears size 9, the other size 13's. One has more traction than the other. What about a woman in heels as opposed to flats. Same weight, different psi applied to the ground.
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I'll challenge your description a little. Take a set of 4 inch wide tires from the fifties (18 inchers were on MG Tseries and race cars of that era) and a set of these new 245/45-18 and on two otherwise identical cars start them around a circular track (NASCAR comes to mind) gradually increase speed until each car loses traction and slides into the wall. You are saying both tires would have the same contact patch - I'm saying the car with the larger patch (wider tire)will stick to the pavement better. Another way to think about it is two people that weigh the same and are the same height. One wears size 9, the other size 13's. One has more traction than the other. What about a woman in heels as opposed to flats. Same weight, different psi applied to the ground.
Correct, I am saying both tires have the same contact patch in total area. This is basic Newtonian Physics. Can't be helped. What you are doing is adding new variables that require different solutions.
Once the car is in motion, the shape of the contact patch plays a very important role in how the vehicle handles certain tasks, as I said in my previous post. A wider tire has a wider contact patch side to side, but shorter measured front to back. The converse is true of a narrower tire.
When the car is in motion there are (basically) two forces acting against the tire that affect handling. The first is coming from in front of the car trying to keep it from going forward. The other is from the side trying to keep the car from turning. These are vector forces. In your example, the wider tire works better at keeping the car from sliding sideways because thats the way the contact patch is oriented. Trying to keep it simple, does that make sense?
Basically, this is why dragsters use very narrow tires up front. They don't have to make the vehicle change direction and are better suited to overcoming the force acting against the car's forward motion. Lets leave the dragster's rear tires alone for this discussion, please.
This is also why road racing cars use wider tires. They are better suited to overcoming the forces trying to keep the car from changing direction. All because of the shape of the contact patch.
Racers fine tune the handling of their car by adjusting tire pressures, thus altering the size of the contact patch, as described in my previous post. And I won't complicate the illustration any further by introducing factors such as sidewall deflection, slip angles, coefficients of friction and tire compounds.
I'll challenge your description a little. Take a set of 4 inch wide tires from the fifties (18 inchers were on MG Tseries and race cars of that era) and a set of these new 245/45-18 and on two otherwise identical cars start them around a circular track (NASCAR comes to mind) gradually increase speed until each car loses traction and slides into the wall. You are saying both tires would have the same contact patch - I'm saying the car with the larger patch (wider tire)will stick to the pavement better. Another way to think about it is two people that weigh the same and are the same height. One wears size 9, the other size 13's. One has more traction than the other. What about a woman in heels as opposed to flats. Same weight, different psi applied to the ground.
Yeah you're hitting on what I was thinking. Because say I have a 7" wide tire now but replace it with a 10" wide tire. That's ~3 extra inches of tire that is making contact with the ground my previous tire didn't have at all no matter how low the psi was to make contact with. If both tires are pretty much already flat slicks then you're pretty much making contact with the entire tire already. I could understand rounded tires that need to be pressed down on to make the rounded edges meet the road, but not on something flat. You still end up with the same amount of fource as before, it's just smaller per unit area then before.
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Breeze, I'm impressed. Do you work in the tire industry?
I'll throw another factor that gets put into the mix, that of contact patch pressure. (Tire Technology - Advanced Basics 102)
AGAIN, NOBODY IS FORCING YOU TO READ THIS, SKIP IT IF THIS BORES YOU!!!
First, most tires act like balloons - (in fact this is one of the fundamental models that is still used in one form or another yet today). Essentially, except for the belt and construction influences (which are not insignificant, but in general can be set aside for this discussion), the tire patch is determined by tire pressure and the load that is on it, just like Breeze said.
If you measure the amount of turning friction on a tire (like, as an example, the friction when you are stationary and turning your steering wheel), it turns out that the amount of torque at the tire patch is mostly independent of the tread pattern, or tire width (maybe only a few percent influence from an all-season tread to a summer-only tread). So a space saver spare tire at 32 PSI (WAY below the recommended pressure) and a P245/45R18 at 32 PSI on the same vehicle will have pretty much the same turning friction. It sounds incredible (CALL RIPLEY'S!), but I have personally seen such measurements.
Now, the part that starts to get tricky is the actual parts of rubber that contact the ground. The most ideal traction for rubber to the road inside the tire patch is the LOWEST pressure. The higher the pressure, the less traction available at that point in the tire patch.
If you have a tire patch with many localized areas of high pressure, like in a snow tire (large voids of no contact, with many blocks and a net area much less than an average all-season tire), even if it is the same size, inflation pressure, and load as a comparison all-season, the net traction (on dry conditions) will be much lower - even though you have the same size and shape tire patch.
The more distributed and lower pressure you get, the more grip you have (weather notwithstanding). Summer-only tread patterns have less grooves, and larger, flatter blocks, which increases the net area of rubber in contact with the road for the same load and pressure (same tire patch). Ergo, it has more grip, because it has less points at which the localized pressure is high.
The most ideal traction is a slick - very distributed area, no localized pressure zones in the tire patch and maximum grip. Unless you have the wrong tire pressure - then you might have a localized high pressure zone along the center of the tire patch (if your tire pressure is too high - you've effectively got a smaller tire patch handling the same load), or two localized zones of high pressure on the outsides of the tire (if your tire pressure is too low - the middle of the tire doesn't have enough pressure to support the center of the tread).
Longitudinally, the tire can gain more grip by temporarily increasing the net area of rubber in contact with the road. This can be done by shaping the blocks so that under driving or braking torque the forward or rearward blocks that are coming into contact with the road actually touch earlier. Or you can fool with the sidewall stiffness, making it so the tire "wraps" and the shape of the patch becomes longer. You can actually almost double the patch area (and effectively lower the pressure within the patch) by doing this - just look at the pics of the rear tires of a dragster on launch with the front wheels just in the air, and you'll see what I mean.
The size and shape of the tire patch is important. Narrower tire patches tend to have higher localized patch pressure points, especially when operating at slip angles. Wider tire patches tend to have more distributed patch pressure under more driving conditions, with less high pressure points. The amount of resisting torque, known as "aligning torque", though, will be less - meaning it would be like lowering your caster in the car. Steering feedback may be greatly reduced, causing you to need to re-tune your steering or just deal with numb or "video-game-like" steering (snappy response but no feel).
Of course, there are limits - wider is better for handling only up to a point. If you go TOO wide, and you aren't putting enough load on the tire, then it becomes a loosing proposition. You can reach a point where it almost doesn't matter how low your tire pressure is, you just can't put enough load on the tire to create the correct tire patch shape (footprint). It would be like you are always overinflated - and from the above that means higher patch pressures and less overall grip.
Breeze, I'm impressed. Do you work in the tire industry?
....Essentially, except for the belt and construction influences (which are not insignificant, but in general can be set aside for this discussion), the tire patch is determined by tire pressure and the load that is on it, just like Breeze said.
Nope, just have a little experience in suspension tuning and tires are an integral part of that. And thanks for the endorsement solsticeman.
Yep - not only are they an integral part of the tuning of a car, they are prolly absolutely the most important. Tires are the most complex single component on a vehicle.
Four pieces of rubber each the size of a man's hand, temporarily in contact with the road for a fraction of a second - the only thing you have to control the trajectory of 1 1/2 to 3 tons of steel and plastic at deadly velocities.
Yep - not only are they an integral part of the tuning of a car, they are prolly absolutely the most important. Tires are the most complex single component on a vehicle.
Four pieces of rubber each the size of a man's hand, temporarily in contact with the road for a fraction of a second - the only thing you have to control the trajectory of 1 1/2 to 3 tons of steel and plastic at deadly velocities.
Really quite amazing, when you think of it...
When you mentioned balloon tires first picture popped into my head was Roger Rabbit and his cartoon car. "I'm not really a bad car, I'm just drawn that way." Another amazing thing is how a set of 235/60-16 can be fully adequate on a 3500# sedan with 240hp, 0-60 of 6.8 (my current ride) and then you have to have 245/45-18 on 2500# 177hp which by the way tires are almost exactly double in price, wheels have to up there, too - just an aside here, they could have shaved $700-800 from the price of teh Solstice with "normal" sized tires/wheels and I don't think any of us would feel the handling difference. Not that it wouldn't be there, just that we couldn't tell by the seat of our pants. Plus a little drift in the corners couldn't be all bad. I wonder what it will take to break these puppies loose.
__________________
When more than one friend wants to ride shotgun
Pontiac 1926-2010
"We hardly knew 'ya"
Confusion say: "If it ain't broke...give government a crack at it."
When you mentioned balloon tires first picture popped into my head was Roger Rabbit and his cartoon car. "I'm not really a bad car, I'm just drawn that way." Another amazing thing is how a set of 235/60-16 can be fully adequate on a 3500# sedan with 240hp, 0-60 of 6.8 (my current ride) and then you have to have 245/45-18 on 2500# 177hp which by the way tires are almost exactly double in price, wheels have to up there, too - just an aside here, they could have shaved $700-800 from the price of teh Solstice with "normal" sized tires/wheels and I don't think any of us would feel the handling difference. Not that it wouldn't be there, just that we couldn't tell by the seat of our pants. Plus a little drift in the corners couldn't be all bad. I wonder what it will take to break these puppies loose.
Now you've got me curious, what weighs 3500#, has 240 hp, has 16" wheels with 45 series tires and does 0-60 in 6.8. My two guesses are a BMW 3 series or a WRX. Amiright?
Also, from my Enginnering Mechanics book:
"The magnitude of the maximum static frictional force that can be developed is independent of the area of contact, provided the normal pressure is not very low nor great enough to to severely deform or crush the contacting surfaces of the bodies."
Or in other words, "Contact patch doesn't matter, only the coeffecient of friction and the normal force, exept with tires. You need a masters degree to understand that case."
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..Then again, I could be (f)lying.
Yep - not only are they an integral part of the tuning of a car, they are prolly absolutely the most important. Tires are the most complex single component on a vehicle.
Four pieces of rubber each the size of a man's hand, temporarily in contact with the road for a fraction of a second - the only thing you have to control the trajectory of 1 1/2 to 3 tons of steel and plastic at deadly velocities.
Also, from my Enginnering Mechanics book:
"The magnitude of the maximum static frictional force that can be developed is independent of the area of contact, provided the normal pressure is not very low nor great enough to to severely deform or crush the contacting surfaces of the bodies."
Or in other words, "Contact patch doesn't matter, only the coeffecient of friction and the normal force, exept with tires. You need a masters degree to understand that case."
yeah, seems like we were always talking about moving blocks or bricks in Engine Phys. Which is ok for static friction but there are a whole lot of things that deform when a force is applied.
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... what weighs 3500#, has 240 hp, has 16" wheels with 45 series tires and does 0-60 in 6.8. My two guesses are a BMW 3 series or a WRX. Amiright?
Sounds like VQ power to me
What kind of lateral acceleration do you expect from Solstice? I was thinking that 0.95 g would be nice, but then I realised that GM probably didn't install hard enough springs for that. ( Then again, how much stroke is available in a car only 3½ inches off the ground? ) Or does the high load capacity of the tire somehow nullify the loss of grip resulting from weight transfer?
Now you've got me curious, what weighs 3500#, has 240 hp, has 16" wheels with 45 series tires and does 0-60 in 6.8. My two guesses are a BMW 3 series or a WRX. Amiright?
."
It's 235/60 tires and both cars are Grand Prix GTP, 1997 and 2003. The '97 did run just a little faster with 5 less hp but it had a sport-shift selector that would keep the tranny in gear longer. These are stock out of the box figures with no funny smaller pulleys on the s/c, or removing left front headlamp for colder air induction (usually good for a half second in the 1/4 mile, but frowned on for nighttime trips to the market)
__________________
When more than one friend wants to ride shotgun
Pontiac 1926-2010
"We hardly knew 'ya"
Confusion say: "If it ain't broke...give government a crack at it."
...Also, from my Enginnering Mechanics book:
"The magnitude of the maximum static frictional force that can be developed is independent of the area of contact, provided the normal pressure is not very low nor great enough to to severely deform or crush the contacting surfaces of the bodies."
Or in other words, "Contact patch doesn't matter, only the coeffecient of friction and the normal force, exept with tires. You need a masters degree to understand that case."
That's part of the key - if tires are plastic or metal, then deforming or crushing the contact surfaces approximates EXACTLY the simple example from the mech book.
However, rubber is not "non-deformable". Good thing, too. The "equivalent" coefficient of friction of rubber for a piece of rubber on a given surface with uniform force (like the air pressure behind a tire patch) is the integration (OOOOHHH! Clear out those cobwebs, engineers!!!) of the pressure multiplied by the pressure coefficient of the compound of rubber used for the surface to which it is applied. That's a simplified approach - but rubber on some surfaces can have localized areas in excess of 1 for an equivalent coefficient of friction.
Change the surface (like lay down some partially liquified rubber first), and you can actually generate multiple g's of acceleration. It implies the equivalent coefficient of friction, for that period of time for the rear tire patches, is over 2.0 or 3.0!!!
The other part of tread rubber that fits into this is the adhesiveness of the compound. So there's so many things that make this case NOT like the simple box on a ramp coefficient of friction examples from first year physics that it almost doesn't apply.
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