Automotive Why do tires slip easier from a standstill start?

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Tires are more likely to slip from a standstill due to the significant difference between the stationary wheels and the high RPM of the engine, which creates a large torque that can exceed the tires' grip. When the clutch is released quickly after revving the engine, the high angular velocity of the engine can overwhelm the static friction of the tires, leading to slippage. During burnouts, the force that would normally propel the vehicle is converted into heat, causing tire wear while the vehicle remains stationary. Factors such as weight transfer during acceleration and the power-to-weight ratio also influence traction, as increased downforce on the rear tires enhances grip. Overall, effective power application and tire traction are crucial for optimal acceleration and performance.
  • #31
I could have missed it, but I don't believe anyone pointed out that technically, the traction is actually higher when the tire is still stopped (at rest with respect to the pavement). That is called stiction. Once the tire begins to slip, it slips more easily, that is with less torque. That is called kinetic friction. These frictional changes are part of Dry friction. The above lesser kinetic friction-after-start occurs even before the tire rubber begins to liquefy from heat and "lubricate" the contact surfaces. This exceptionally slippery liquid condition is then known as Lubricated friction. All tires, once they are rolling, have less friction than they did parked because there is always a tiny bit of slip inherently included. These processes are why a vehicle can cling to a tilted, critical slope when parked, but speedily slip away immediately once it starts sliding. Of course the tires may soon began to melt too.

The heuristic reasoning for Dry friction is that the rubber and pavement surfaces are generally both bumpy in a macro sense, so at mutual rest it is as though the two "toothed" surfaces are mechanically engaged like fine gears meshing. Once the tire starts slipping, it thereafter begins to ride higher (away from the pavement) as the "gear teeth bounce" from tooth tip to tooth tip rather than fully re-meshing. This sort of bouncing kinetic frictional loss also occurs in a micro sense between two sliding surfaces even if they are deemed to be very smooth (asperity). Therefore starting friction is always more than continuous friction. Even at the most extreme case, the rounded outer electron valance shell of the near-perfect smooth surfaces finally supply unavoidable small, gear-like, electrostatic bumps and dips to each "flat" surface, and these engage the rounded "shell" bumps and dips of the adjacent surface, thereby maintaining the initial gear-like "quasi-mechanical" mesh... until the objects move ever so slightly at different relative speeds. This smooth, near impossible-to-achieve condition, gets down to the nitty-gritty and is along the lines of various van der Waals forces.

A whole new category of friction arises in automotive braking. Certainly the anti-lock brakes help prevent tire skid by delaying heat liquefaction, but the brake material friction also comes into account. In auto road-racing, most brake-pad compounds are especially chosen to avoid gassing when hot. That sort of hot gas can lubricate the brakes similar to any other fluid (all gas is a fluid) and cause extreme brake fade during frequent use. Unfortunately there is a tendency for these low-gassing compounds to not stop well when they are cold, so these super low-fade brakes are largely useless on the streets. This fact is well accentuated by heavy freight train brakes that must not fade from heat while descending long grades, but do not work well at all until they are initially well heated. Consequently the train operators must learn to always apply them 15 or more seconds in advance, taking into account a preliminary heating delay. At 60 miles an hour, with a standard application, such a train will typically go up to a quarter mile before the brakes even begin to work to stop the 30 plus million pounds of freight. A hard application emergency stop is just a little quicker. It still takes some planning ahead, but is still much more user-friendly than stopping a ship or aircraft that are entirely supported by pure fluid.

More friction definitions here.

Wes
 
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  • #32
Wes Tausend said:
All tires, once they are rolling, have less friction than they did parked because there is always a tiny bit of slip inherently included.
That's where I left off, wondering.
Wes Tausend said:
the traction is actually higher when the tire is still stopped
jerromyjon said:
Is that before or after the tires regain traction?
Yeah, it got complicated. I'm back on it, now, but whew.

Lubricated friction is a case of fluid friction where a lubricant fluid separates two solid surfaces, as in spinning, melting tires.
 
  • #33
Don't ever forget the driver in any of these discussions about vehicle performance .

Many cases of slipping at start off are just caused by ham footed drivers with no feel for the vehicle's response to the accelerator pedal .
 
  • #34
The way I look at launch, in a manual transmission car, is to think of the sudden clutch engagement as a rotational collision between the engine/flywheel assembly and the (stationary) drivetrain/wheels/tires. Without the traction force, conservation of angular momentum would be conserved, and the tires would rotate. With a high engine speed and sudden enough engagement, the tires don't have enough traction to hold the wheels stationary during this collision. So for a short time, no power at all would be required from the engine in order to break the tires loose. (A similar jolt happens when engaging an A/C compressor clutch, and it can cause belt slip and squeal.)

If you shift hard enough in the lower gears, you can get this to happen again. In the higher gears, there is not usually enough angular momentum relative to the wheels to get high enough torque during the collision.

While it's true that P=Fv, that only applies to the power applied to the road, not necessarily the power the engine is producing. It would take an infinite gear reduction to actually produce a large amount of power (actually any power at all) at the tires when stationary.

I agree that a slipping clutch is not a gear reduction. If you examine it as a free body, the torque in equals the torque out. However, the power in is not equal to the power out, due to the shaft speed differences. The rest is lost to heat. Even the common torque converter can only double the torque, and that's only because it has a stator contacting the transmission housing.

For a reference, I discuss this very thing in the book mentioned in my signature.
 
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  • #35
Randy Beikmann said:
While it's true that P=Fv, that only applies to the power applied to the road, not necessarily the power the engine is producing. It would take an infinite gear reduction to actually produce a large amount of power (actually any power at all) at the tires when stationary.
I agree. And it is a mistake to treat P = Fv as though P is constant as v decreases and therefore F increases. In fact, engine torque decreases as the engine RPM goes below the point of highest torque. So F decreases and P decreases even more. The wheel spin from popping the clutch is primarily due to inertia of the mismatch between engine RPM and wheel RPM through the gear ratio of first gear.

PS. I disagree with a prior comment (I forget where) that most cars can spin the wheels without popping the clutch. I think that most can not.
 
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  • #36
Randy Beikmann said:
It would take an infinite gear reduction to actually produce a large amount of power ...
Gear reduction does not produce power. Though, it can produce torque or rpm (one at the expense of the other). The output power is always the same (in principle) as the input power.
Randy Beikmann said:
... at the tires when stationary.
Again, if it is stationary, there cannot be any power, only torque.

I know you understand those principles @Randy Beikmann , I just wanted to make that clear to anyone who would misread your post.
 
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  • #37
jack action said:
Gear reduction does not produce power. Though, it can produce torque or rpm (one at the expense of the other). The output power is always the same (in principle) as the input power.

Again, if it is stationary, there cannot be any power, only torque.

I know you understand those principles @Randy Beikmann , I just wanted to make that clear to anyone who would misread your post.

You're right, that could have been said more clearly. Gear reductions certainly do not increase power - at best they break even, and if there's any friction they reduce power.
 
  • #38
They only seem to slip easier.

There are other things going on like contact patch size change with suspension compression.

Most cars with independent suspension on the drive wheels encounter camber change during a hard acceleration event which normally peaks just as the power is applied.

We can break traction at 260mph on our car and hoping that doesn't happen when we test at KSC on the 15th of this month.
 

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