Does traction increase as speed increases?
The car would perform worse in the fact that it takes a larger force to divert the vehicle from its path due to the increased speed. Since "Traction Circle" hasn't changed you can no longer turn the wheel as far as you could at lower speeds.
Some jackass will get in here with a situation caused by another completely differenct reaction and argue with me some. But the prior statement is unargueably true.
Even in a radial tire, the tire will "grow" in diameter as speed increases.
Therefor the width will decrease, so less tread in contact with the road.
Traction should be a constant, given tire compound and surface don't
change with speed.
Generally the traction force at the tire contact patch increases with increasing download on the tire. Most street cars lose load on the tire at higher speeds (negative downforce aka "lift"). A wing or other aero device that works may add enought downforce @ 100 mph to overcome lift or a bit more.
Good production based race cars can pickup quite a bit more downforce and the drag that goes with it. A Top Fuel wing produces thousands of lbs of downforce @ 300+ mph. The whole Formula 1 car is a aero machine which can create downforce @180-200 mph equal to many time the weight of the vehicle. It's not for drive traction at high speeds but for better braking and cornering, both in the 4-5g+ area. Ouch!
Generally the traction force at the tire contact patch increases with increasing download on the tire. Most street cars lose load on the tire at higher speeds (negative downforce aka "lift"). A wing or other aero device that works may add enought downforce @ 100 mph to overcome lift or a bit more.
Good production based race cars can pickup quite a bit more downforce and the drag that goes with it. A Top Fuel wing produces thousands of lbs of downforce @ 300+ mph. The whole Formula 1 car is a aero machine which can create downforce @180-200 mph equal to many time the weight of the vehicle. It's not for drive traction at high speeds but for better braking and cornering, both in the 4-5g+ area. Ouch!
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For this arguement the coefficient of friction will basically remain the same.
No type of racing was mentioned though. This is not 100% on topic but just keep in mind [addressing the general public not you] that the coefficient of friction will vary with the angle you are turning, weight on tyre, and temp.
When the tyre is loaded the CoF goes down but the max tractive force goes up.
It doesn't seem to me that the contact patch will change much... The amount of force being put down is so much greater than the amount of PSI that would be lost from increased area. I am not going to lie, the contact patch will grow when more weight is applied but I don't think that is the main attraction when you have vertical loading and shearing to consider. Much more so when we are considering the average street car. I am sure professional racing will always have expections though!
I would love to hear you guys converse about this more. Especially, if some more racers drop us some hints and not dumb post ****** who come in here and muck the thread up.
Listening in!
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... take a tethered ball for example, yet it can rotate infinately. From a stand still you can quickly accelerate the ball around the pole to a good speed. Now, try to spin it faster around the pole with the same acceleration as when it was stopped. You would need a butt load more force to accelerate it just as fast from that point. Speed vs Traction is similar, once up to speed you need more force to accomplish the same thing as when standing still. As Old SStroker said, the transmission is multiplying the engine torque on the lower (1st, 2nd) gears with lower speed (0-60), as compared to higher (3rd, 4th) gears once up to speed (70-120). Torque Converter equipped cars will be even worse, since they multiply torque from a dead stop and decrease that torque multiplication higher in the rpms. 
So it is not that traction increases with speed (it remains constant), but rather the amount of force required to break traction at speeds increases substantially.
As you can see, for a street F-Body (both A4 and M6), torque at the pavement decreases on higher gears (which equals higher MPH).
Eng TQ trans gears RWTQ
350 3.06 3.23 3459
350 1.63 3.23 1843
350 1.00 3.23 1131
350 0.70 3.23 791
Eng TQ trans gears RWTQ
350 2.66 3.42 3184
350 1.78 3.42 2131
350 1.30 3.42 1556
350 1.00 3.42 1197
350 0.74 3.42 886
350 0.50 3.42 599
Last edited by 12secSS; Aug 13, 2007 at 03:45 PM.
You are thinking in terms of total traction over a given distance (1/4 mile). You have to remember that distance traveled per unit time (velocity) increases at the same rate as surface area applied per unit time.
I think the correct way to think about traction is at a minute instance (instantaneous traction) as opposed to over a given distance (average traction).
Please do yourself a favor this Sunday. Try and find two programs. One is NASCAR the other is NHRA. Both groups have determined that tires grow the faster the tire turns. If you watch the NHRA during the water box burn out you can actually see the tire grow as it rotates. There are other factors like side wall height that also effects the diameter as well as the tire surface as it contacts the pavement.
Let me desribe the thought experiment I was trying to run. Let's assume we're discussing straightline acceleration only. Let's assume that we are using a CVT, so that the exact same torque will be delivered at 20 mph and at 40 mph. Let's assume that any tire growth or deformation will be the same at both speeds. Let's assume we're doing all this in a vacuum so that downforce and air resistance are not factors (the engine is being fed air from a really long hose).
Now run the experiment: starting at a steady-state 20 mph, mash the throttle and apply X000 ft-lbs of torque. Do the same at 40. Keep increasing X until the tires break loose at either 20 or 40. Will they break loose at 20 first, 40 first, or will X be the same at 20 and 40?
I think 20 first because rubber-on-pavement fricton is not simple friction. I don't know the technical terms, sorry, but down at the molecular level the rubber is deforming around the surface irregularities of the pavement. It has to be able to deform, yet not tear, and recover When you exceed the rubber's capability to do that, the tires spin and smoke. At 20 you have fewer rubber "particles" per unit time, and hence they have a higher unit loading and you are closer to their point of not being able to recover.
On the other hand, at 40 each one has half as much time to take the load.... Hmmmm...
That's what I was thinking about when I started this anyways. And it actually has a practical application. I'm trying to figure out how to accelerate as quickly as possible with a car that is traction-limited. Ideally, you'd find the overall torque loading the tires could handle and keep the delivered torque just below that. Hence the CVT. For a C5, we have to think of things like turbos with boost controllers, close-ratio gears, staged N2O, etc.
Let me desribe the thought experiment I was trying to run. Let's assume we're discussing straightline acceleration only. Let's assume that we are using a CVT, so that the exact same torque will be delivered at 20 mph and at 40 mph. Let's assume that any tire growth or deformation will be the same at both speeds. Let's assume we're doing all this in a vacuum so that downforce and air resistance are not factors (the engine is being fed air from a really long hose).
Now run the experiment: starting at a steady-state 20 mph, mash the throttle and apply X000 ft-lbs of torque. Do the same at 40. Keep increasing X until the tires break loose at either 20 or 40. Will they break loose at 20 first, 40 first, or will X be the same at 20 and 40?
I think 20 first because rubber-on-pavement fricton is not simple friction. I don't know the technical terms, sorry, but down at the molecular level the rubber is deforming around the surface irregularities of the pavement. It has to be able to deform, yet not tear, and recover When you exceed the rubber's capability to do that, the tires spin and smoke. At 20 you have fewer rubber "particles" per unit time, and hence they have a higher unit loading and you are closer to their point of not being able to recover.
On the other hand, at 40 each one has half as much time to take the load.... Hmmmm...
That's what I was thinking about when I started this anyways. And it actually has a practical application. I'm trying to figure out how to accelerate as quickly as possible with a car that is traction-limited. Ideally, you'd find the overall torque loading the tires could handle and keep the delivered torque just below that. Hence the CVT. For a C5, we have to think of things like turbos with boost controllers, close-ratio gears, staged N2O, etc.
How about just using electronic traction control?
FWIW, best traction comes with a small % of slip. This can be programmed into good TC software.
How about this idea: use wheel speed sensors on the unpowered and powered wheels. With no slip, there is a null output. As the drive wheels start to slip, convert the wheel speed delta to an audio beep, the frequency of which increases with increasing slip. Pipe that to your earpiece in the helmet, or the Bose system in the C5. Using the beeps, control the traction with your right foot. You could do it with pitch (frequency) also: pitch increases with slip and peaks at best traction. Maybe that would be more difficult to tell if you are over or underspeeding the tires. Call it traction advisory.
Back to your original assumption. Except for the ~5% slippage for max "bite", isn't the relative velocity of the tire contact patch zero with respect to the road? Molecular "bonding/debonding" is taking place essentially at zero relative velocity, but each bond lasts for a shorter time with increased vehicle velocity but there are proportionatly more bonds occurring. Do those things cancel each other out and result in the same total tractive effort? Perhaps!
Jon
Also, excluding dragstrip tires, radial steel belted high performance tires grow just a tad in diameter with speed and the footprint narrows just a tad,, the overall area of the footprint does not change significantly. Forward grip (traction) will change only due to where you are located on the tire slip chart.
Taking my current combo as a baseline, my car makes about 364 rwtq. 3.42 rear gears, and 2.66 first gear. Overall gearing in 1st is 8.379, so delivered torque is 3305 ftlbs.
The car hooked "ok" but not great with the 5-year-old runflats. I went to larger Pilot Sport 2s and now it hooks just fine. If we assume that the 3300 number has some more room in it, and that slightly larger tires could be fitted, then perhaps we could call 4000 ftlbs a reasonable limit for street tires on my car.
If that's the case, then we can game different power/gearing combos to try and keep the torque delivered as close as possible to 4000...but this assumes the tires that can hold 4000 ftlbs at 20 mph can do the same at 120 mph.
The longitudinal Mu of a very good street/strip tire might be about 1.2. This means that for a vertical dynamic load @ launch on the rear tire is 1000 lbf, the longitudinal force is 1.2 x 1000 = 1200 lbf. If you have peak torque at launch, this tire would not hook-up. 1650 is higher than 1200.... This is why you want a suspension system that transfers as much weight as possible to the rear tires at launch. Am I making any sense?
Taking my current combo as a baseline, my car makes about 364 rwtq. 3.42 rear gears, and 2.66 first gear. Overall gearing in 1st is 8.379, so delivered torque is 3305 ftlbs.
The car hooked "ok" but not great with the 5-year-old runflats. I went to larger Pilot Sport 2s and now it hooks just fine. If we assume that the 3300 number has some more room in it, and that slightly larger tires could be fitted, then perhaps we could call 4000 ftlbs a reasonable limit for street tires on my car.
If that's the case, then we can game different power/gearing combos to try and keep the torque delivered as close as possible to 4000...but this assumes the tires that can hold 4000 ftlbs at 20 mph can do the same at 120 mph.
1) 3.42 * 2.66 = 9.0972 Overall 1st Gear. Your 8.379 overall must have been figured with a 3.15 gear, not 3.42's.
2) Runflats have never been known for their traction, mostly because of the stiff sidewalls required, basically the same reason 18" tires don't hook as well as 15" tires of the same overall diameter.
3) In order to deliver 4000 ft/lbs to the rear axle at 120 mph, your engine needs to produce 1170 ft/lbs. If you had that much torque available, there wouldn't be any need for a transmission (think locomotive)


