Gannet

All else being equal, does traction increase as speed increases? I would think so, because the surface area of the tire applied to the pavement per unit time would increase, but I'm interested to hear what others think.

2002_Z28_Six_Speed

iTrader: (19)

I would be lead to think that maxium potential force is the same regardless of speed. The rate of speed between the pavement and the tyres are the same. The situation is actually static.

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.

2002_Z28_Six_Speed

iTrader: (19)

By the way, with modern tyres more weight means less coefficient of friction.

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.

FAST LS1

iTrader: (2)

It probably decreases slightly as speed increases.

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.

edcmat-l1

iTrader: (10)

Frictional coefficient will stay the same. What will change is surface area, due to growth, downforce due to airflow, and the resistance against the driveline. Its easy to break em loose trying to move a vehicle from a dead stand still than say accelerating from 75 to 125. But thats just common sense, and the whole basis for roll racing.

Old SStroker

Being in 1st gear (especially with a torque converter) multiplies the torque at the drive wheels by many times what it sees @ 75 mph. That might explain why the tires break loose at launch.

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!

SScam68

iTrader: (20)

Those machines are just unbelievably amazing.

2002_Z28_Six_Speed

iTrader: (19)

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!

12secSS

iTrader: (30)

I will simplify it ... 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

tee-boy

iTrader: (8)

"because the surface area of the tire applied to the pavement per unit time would increase"

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).

Jakes Dad

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.

Gannet

Many good responses; thank you all.

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.

aaronloveszztop

Traction would decrease as your tires spin the centrifugal force flexes the tire taller making the contact patch thinner. You can watch top fuel dragsters as they take off the tires get taller and skinnier.

Old SStroker

CVTs don't have constant output torque with fixed (constant) input torque. If vehicle speed is increasing. You'd have to vary the input torque (engine) to achieve the same torque at the wheels, wouldn't you?

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

Doc99SS

iTrader: (2)

Maximum traction occurs at about 8% to 10% slip. As you pass this 8% to 10% slip zone, the tire loses a little grip but levels out. So from 14% slip on up to whatever, the tire is making smoke and forward grip is nearly constant.

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.

Gannet

The reason I asked the question in the first place is that I was trying to spreadsheet-game different power/trans/gear combos, in an effort to maximize overall acceleration performance with limited traction (street tires).

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.

Doc99SS

iTrader: (2)

Gannet: remember there are two rear tires, you are trying to feed 3300/2 thru each tire, so 1650 ft=lbf of torque thru each tire. I dont know where I'm taking this, but the tire is about 12 inches in radius loaded and squashed from accel (conveniently 1 ft). So you need 1650 lbf thru the contact patch.

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?

hammertime

iTrader: (1)

Three Things...

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)

z28241

iTrader: (13)

Like I said, think about this little throery, slightly off topic, a decent number of us are 12 second guys with drag radial tires, the usual set up, like mine, is bolt ons, stall converter and gears, see my sig., now I want to ask "RUNNING DAMN NEAR HALF INFLATED TIRES" okay 19lbs... BFG's, going to affect trap speed and or E.T.? ie..drag, because all the stuff i have done and my e.t./ mph hasnt changed much, okay a little but stock with free mods and 3.23's I was 13.12/105mph now I am c-sig....

Doc99SS

iTrader: (2)

The choice of rear tires does not effect the trap speed provided that the OD is the same. However, the rear tire can have a big impact on the ET.