Why do tires slip easier from a standstill start?

[oiltop1500]

Hello!
Have some questions I've been thinking about for a while...

1.
why is it easier to make the wheels slipping if you rev the engine and relase the clutch fast (this from a standstill start)?
exampel; I can be in first gear and have full throttle and no slip will oucure in my car with low power figures, but if you release cluch fast and rev the engine, then you can make the wheels slipping.

Is this beacase of the angular velocity of the wheels is zero (standstill start) and the engine/clutch have a high angular velocity?

2.
When a burnout is performed what happends? slip is high and the force that normally would be propeling the wheels
is made into kinetic energy? the only thing that is different from normal condition is that the car is held stationary (by the brakes).
where is the breaking point that makes the wheels slipping, what force is it that makes this happen?Tanks!

 

[FactChecker]

Hello!
Have some questions I've been thinking about for a while...

1.
why is it easier to make the wheels slipping if you rev the engine and relase the clutch fast (this from a standstill start)?
exampel; I can be in first gear and have full throttle and no slip will oucure in my car with low power figures, but if you release cluch fast and rev the engine, then you can make the wheels slipping.

Is this beacase of the angular velocity of the wheels is zero (standstill start) and the engine/clutch have a high angular velocity?

Yes, in that case it is because the wheels are stationary and the engine is turning fast. But the wheels do not always have to be stopped.
In general, the wheels will slip when there is a large difference between the wheel rpm and the engine rpm (adjusted for gear ratio) and the clutch is popped. The tires will also slip if you are going fast, put in the clutch, let the engine slow, and then pop the clutch. You can also squeel the tires by down-shifting when the car is going too fast for the new gear. (Don't test this. You may hurt something.) So it isn't always because the wheels are stopped -- it's because of a large difference, one way or another, between the wheels and the engine.

2.
When a burnout is performed what happends? slip is high and the force that normally would be propeling the wheels
is made into kinetic energy?

The force performs work on the wheel that becomes heat and makes the tires burn.

the only thing that is different from normal condition is that the car is held stationary (by the brakes).
where is the breaking point that makes the wheels slipping, what force is it that makes this happen?

When the wheels are not rotating, static friction is stopping it. When the static friction limit is exceeded, the wheels start rotating and dynamic friction provides resistance.

 

[I like Serena]

Additionally, consider that power is force times speed ($P=Fv$).
That is, at low speeds (and stand still as an extreme) we get more force on the wheels from either the flywheel or the engine itself.
And more force means quicker to slip.

 

[rcgldr]

One issue not mentioned yet is the weight transfer to the rear tires that occurs during acceleration. Once a car is accelerating, there's more downforce on the rear tires, so they are less likely to spin.

 

[Ranger Mike]

As in all things in Racing it about tires, Tires, TIRES. Traction is the friction between the tire and the ground and this permits the vehicle to move forward , usually along a linear plane. This traction occurs as the resistance to spinning (tire slippage) when torque is applied to the wheel, and in turn, the tire.When you " drop the clutch" ( rev engine to 7000 RPM with clutch pedal fully depresses then release it) many things happen. The higher the engine's energy output, the higher the rotational acceleration that will occur on the wheel/tire. We have rotational forces overcoming any capability of the tire to grip the pavement. Rotational acceleration is too high and overcome the traction, causing the tire to slide pass the linear plane. Tire spin + no grip = tire burn out.Normal clutch engagement means the engine is accelerated to a steady state of 1500 RPM and we slowly release the clutch. The stationary clutch disc is squeezed between the spinning fly wheel and spinning pressure plate that is attached to the flywheel. The clutch disc begins to spin as does the transmission and drive shaft. The differential changes this rotational force to a 90 degree vector. The rotational force continues to rotate the wheel /tire until the tire meets the asphalt surface. The friction between the pavement and tire is sufficient to cause traction. The vehicle begins to move forward.Many factors can influence the effectiveness of this action. Having enough power to cause the angular acceleration to overcome the traction force, allows for quick acceleration. But to win races, the driver must apply this power better than the other drivers.

If we have more power and want to apply it we need a higher coefficient of friction, which is essentially traction. One way to do this is increase the tire contact patch or area that makes contact with the pavement. A drag slick is a wide tire with no tread and presents the maximum grip to the pavement surface. The surface on which the vehicle accelerates will impact traction so we avoid the section of pavement the previous driver managed to oil with his poorly built internal combustion engine using stock connecting rod bolts. The tire and pavement surface temperature will have an effect on the frictional forces of the tire.Vehicle acceleration depends on the power to weight ratio. The higher power to weight ratio, the faster the vehicle can accelerate. Once the power of a vehicles becomes so high, it becomes too dangerous to drive slowly.One way to increase traction is to apply downforce to the tire. Take a tire off the car and place it on the garage floor and stand it up like it's on the car. Now place each hand at the max diameter and slide it across the floor. Pretty easy. Now get the fat kid next door to come over and sit on the tire and try this again. Putting fat boys extra 100 pounds on it makes the sliding more difficult. Right? DOWN FORCE. We get down force by aerodynamics at speed Huge wings, spoilers etc. We get down force by "weight transfer" during acceleration. There is no such thing a weight transfer. The suspension thinks there is added weight but we are really dealing with transitional forces during acceleration. As the differential rotates the wheel/tire and the tire meets resistance by the pavement, we have rotational torque acting to lift the front of the vehicle. If we are really hooked up we will lift the front end doing a "wheelie" But in most cases the front end lifts a few inches and we have the rear suspension compressing the rear springs. The vehicle thinks that additional weight is being placed on the rear of the vehicle. We have more down force on the tire and thus more traction.Hope this helps

rm

 

[oiltop1500]

So when you try to accelerate fast from a standstill, is it both the large difference between the wheel (0 rpm) and the engine high rpm that will increase the force even more to the tires (or what is it?). And the high torque to the tires in first gear.

I think it is much easier to make the tires slip from a standstill and you give full throttle. Than if you have some speed (in first gear)

Thanks

 

[I like Serena]

At speed 0, we have a theoretical infinite torque, which is only limited by how much the transmissions allow to be transferred without breaking anything.
So that's good for immediate slipping.
Then, when we start accelerating, we get more pressure on the rear as @rcgldr mentioned, so that an engine that drives the rear tires will stop slipping quicker.
At higher speeds, we get less torque (with the engine at max power).
That is because the maximum net force is inversely linearly related to speed ($F_{max}=\frac {P_{max}} v$).

 

[FactChecker]

At speed 0, we have a theoretical infinite torque, which is only limited by how much the transmissions allow to be transferred without breaking anything.

I don't understand this. Does this assume an infinite gear ratio? What about the known torque of an engine at a given RPM and a fixed first gear ratio?

 

[I like Serena]

I don't understand this. Does this assume an infinite gear ratio? What about the known torque of an engine at a given RPM and a fixed first gear ratio?

Whatever the gear ratio, we have a certain non-zero power.
Consequently $F=\frac Pv$ is infinite.
More practically, if we spin up the flywheel to high speed, and release the clutch (if we have one) fast enough, we will either burn through the clutch plates, or the tires will start slipping.

 

[FactChecker]

Whatever the gear ratio, we have a certain non-zero power.
Consequently $F=\frac Pv$ is infinite.

I have stalled a stick-shift car too often to believe that there is any reality to this.

 

[I like Serena]

I have stalled a stick-shift car too often to believe that there is any reality to this.

Once we get to even the lowest non-zero speed, it stalls, although this can take a while or even be avoided if we bring the flywheel to high speed.

 

[tygerdawg]

Lots of interesting comments. I am not a vehicle dynamicist, so correct me if I am wrong. What is NOT being discussed are the concepts of attempting to accelerate the inertia of the vehicle and the associated change in momentum of the vehicle.

The engine + transmission + drive train supplies and abundance of torque to the wheel. This is necessary for drivability, isn't it? The vehicle must accelerate from zero speed. But it must also accelerate from a typical driving speed to a passing speed. So the torque available covers the wide spectrum of performance requirements, and this is much more than start-from-zero torque requirements. Otherwise, to accelerate the car from a zero speed without without the rubber tire slipping would only require a very small amount of torque.

But the vehicle inertia must be accelerated at acceptable rates to be drivable. If one applies all of that excess torque that is available, more than the "minimum" amount, then the friction force at the rubber tire-to-road interface is not sufficient to move that vehicle inertia. Hence: wheel spin.

 

[russ_watters]

I don't understand this. Does this assume an infinite gear ratio? What about the known torque of an engine at a given RPM and a fixed first gear ratio?

It does. The speed of the tire is zero so the ratio is anything / zero = infinite.

I have stalled a stick-shift car too often to believe that there is any reality to this.

I'm wondering if you missed that this situation doesn't *literally* result in infinite torque being applied to the wheels because of the caveat provided: we all know infinite torque is impossible, so *something* in the system will fail when you try to reconcile the discrepancy in speed. What you get though is a really, really high torque.

But if a car stalls, it means it didn't get enough gas. Pretty much any car can produce enough power to spin the tires when the clutch is let out.

 

[jerromyjon]

An important point that I don't think anyone mentioned is the power graph of engine performance. When you rev the engine with the clutch disengaged you bring the RPM up which dramatically increases the available torque output of the engine. Revving the engine up, releasing the throttle, and then popping the clutch would only chirp the tires and then the engine would likely stall. (Unless you have a large rotational mass compared to vehicle weight) Flooring the throttle and dumping the clutch as the engine reaches peak output produces optimal difference between torque at the flywheel and static friction at the tire/road contact patch(es).

 

[FactChecker]

An important point that I don't think anyone mentioned is the power graph of engine performance. When you rev the engine with the clutch disengaged you bring the RPM up which dramatically increases the available torque output of the engine. Revving the engine up, releasing the throttle, and then popping the clutch would only chirp the tires and then the engine would likely stall. (Unless you have a large rotational mass compared to vehicle weight) Flooring the throttle and dumping the clutch as the engine reaches peak output produces optimal difference between torque at the flywheel and static friction at the tire/road contact patch(es).

I agree. And if they are not considering this, they are not talking about a real car -- they are talking about some fictional vehicle.
The engine torque curve does not increase with low RPM. It decreases significantly. At a fixed gear ratio, that gives decreased torque available at the tires. The slipping of the clutch allows the engine RPM to be higher than the first gear ratio would otherwise allow.

 

[I like Serena]

The slipping of the clutch allows the engine RPM to be higher than the first gear ratio would otherwise allow.

That's exactly it - with a slipping clutch we can effectively modify the otherwise fixed gear ratio.
And we can get close to delivery of the maximum power of the engine at any speed, which is how we obtain maximum sustained acceleration.
And yes, when fully engaged, an engine can provide limited power at low speed in a given gear ratio, which leads to stalling.
But that's not what we're doing at extremely low speeds, which is why such graphs always begin at a certain minimum speed.

 

[FactChecker]

The engine and clutch work together to allow the engine to stay at a high-torque RPM while the clutch slips and turns the wheels through the fixed gear ratio. That is more complicated than saying that the gear ratio has been effectively changed (It has not been changed.). And it is very different from the F = P/v equation.

 

[jerromyjon]

In my mind it might help simplify the problem by saying the gear ratio from the clutch disk to the drive wheels is constant and the percent of torque transfer from the flywheel to the clutch disk is variable...

 

[jerromyjon]

The engine and clutch work together to allow the engine to stay at a high-torque RPM while the clutch slips and turns the wheels through the fixed gear ratio.

As opposed to "slips" I think you meant the clutch "grabs" 100% = 1:1 through the flywheel to input shaft.
Never mind, I think I interpreted what you were saying incorrectly...

 

[jack action]

I agree with @I like Serena when he states:

At higher speeds, we get less torque (with the engine at max power).
That is because the maximum net force is inversely linearly related to speed ($F_{max}=\frac {P_{max}} v$).

That is the true reason why you can more easily slip the tires at low speed than at high speed: For a given power output, you have more force at a lower car speed.

That being said, the tire force is NOT infinite at $v = 0$. Because, by definition, at $v=0$, the WHEEL power $P$ is also zero. If the car speed is zero, then the wheel rpm is also zero and therefore the wheel power is zero, leading to $F=\frac {0} {0}$, an undetermined form where $F$ can be anything.

$F$ actually depends on the smallest value between the tire friction force, the clutch friction torque and the engine torque (the last two combined with the gear ratio and tire radius as well).

As soon as the clutch engages, the drivetrain will accelerate, but the engine will also decelerate. While decelerating, the engine power will necessarily decrease, leaving less potential torque to the now rotating drivetrain ($T=\frac{P}{\omega}$), leading to a smaller drivertrain acceleration. If the engine decelerate faster than the drivetrain accelerate (depends on the drivetrain and car inertia), then the engine can stall. We usually increase the engine power (opening the throttle) during the process to ensure that the engine doesn't decelerate.

But once the engine and drivetrain rpms are synch, the $F=\frac{P}{v}$ certainly applies.

 

[jerromyjon]

But once the engine and drivetrain rpms are synch, the F=PvF=PvF=\frac{P}{v} certainly applies.

Is that before or after the tires regain traction?

 

[FactChecker]

But once the engine and drivetrain rpms are synch, the $F=\frac{P}{v}$ certainly applies.

If we are talking about different gears between the low and high speed, then I agree that F=P/v implies higher torque at the wheel at the lower speed and the associated lower gear. But if we are talking only about first gear, I think there is a danger of confusing cause and effect. When the engine gets below the ideal RPM for torque, the torque decreases. Period. In that case, F=P/v would imply reduced power due to both decreased engine torque and decreased speed.

 

[I like Serena]

If we are talking about different gears between the low and high speed, then I agree that F=P/v implies higher torque at the wheel at the lower speed and the associated lower gear. But if we are talking only about first gear, I think there is a danger of confusing cause and effect. When the engine gets below the ideal RPM for torque, the torque decreases. Period. In that case, F=P/v would imply reduced power due to both decreased engine torque and decreased speed.

You do realize that we can get more torque than the maximum torque that the engine can provide in first gear?
That's by revving up the engine to the ideal RPM for power, and slipping through the clutch.

 

[FactChecker]

You do realize that we can get more torque than the maximum torque that the engine can provide in first gear?
That's by revving up the engine to the ideal RPM for power, and slipping through the clutch.

Absolutely. If you pop the clutch, you get a lot of torque from the inertia of the engine/flywheel. That is what I said in my first post. But the way several posts were applying F=P/v has nothing to do with that.

 

[I like Serena]

Absolutely. If you pop the clutch, you get a lot of torque from the inertia of the engine/flywheel. That is what I said in my first post. But the way several posts were applying F=P/v has nothing to do with that.

But... with an ideal clutch we still have F=P/v (ignoring the flywheel for now).
And we can get pretty close to an ideal clutch as the Variomatic car proved.

 

[FactChecker]

But... with an ideal clutch we still have F=P/v (ignoring the flywheel for now).
And we can get pretty close to an ideal clutch as the Variomatic car proved.

That is a continuously variable transmission. If we allow the gear ratio to change and the OP was asking why one gear ratio makes it easier to spin the wheels than another, then the answer is obvious.
Suppose we are talking about a fixed gear ratio, not popping the clutch, and the engine is below the max-torque RPM. Then F=P/v still applies, but in the opposite way than several comments here have implied. At a lower steady-state speed (engine inertia not involved), the engine torque will be lower, there will be no added torque by popping the clutch, the gear ratio will not change, and there will be less torque at the wheels.

 

[jack action]

My comment about $F=\frac{P}{v}$ has nothing to do with the powertrain, no matter what it is. It has to do with the wheel power, car velocity and available friction force at the tire-road contact patch.

Given a certain amount of wheel power, if the car velocity is higher, the available tire friction force will necessarily be less. You can use a combustion engine in any gear you want, and electric motor or a nuclear reactor if you wish, the statement will always be true no matter the power source.

So, given a maximum power that a powertrain can produce, and that it is delivered optimally to the wheel (for example, with the best suited gear ratio to link the engine and wheel rpms), the maximum tire friction force will always be at low vehicle speed. Thus, why it is easier to do a burn out while standing still than once you are cruising on the highway.

With a wheel rpm of zero, the engine power doesn't matter, as it is not transferred to the wheel (again, the wheel rpm is zero). The engine power is completely transformed into heat within the slipping clutch. BUT, there is still friction in the clutch, producing a torque, which IS transferred to the wheel. As soon as the wheel began to accelerate then power is transferred to the wheel. If the tire-road friction can handle the wheel torque, the car accelerates and the power is used to propel the car; If it cannot, then only the wheel and powertrain accelerate and all the power is converted into friction heat (It is acting just like a slipping clutch).

 

[I like Serena]

With a wheel rpm of zero, the engine power doesn't matter, as it is not transferred to the wheel (again, the wheel rpm is zero). The engine power is completely transformed into heat within the slipping clutch. BUT, there is still friction in the clutch, producing a torque, which IS transferred to the wheel. As soon as the wheel began to accelerate then power is transferred to the wheel. If the tire-road friction can handle the wheel torque, the car accelerates and the power is used to propel the car; If it cannot, then only the wheel and powertrain accelerate and all the power is converted into friction heat (It is acting just like a slipping clutch).

We don't have all that much heat by friction in the clutch. If we had, the clutch would burn out and cause a fire or some such.
The clutch is really just a variable gear.
If not all possible power is transmitted to the wheel, typically when the wheel is slipping, the engine simply produces less power, and consumes less fuel.

 

[jack action]

The clutch is really just a variable gear.

I don't like that definition. A variable gear would be something that can transform the input torque & rpm into a different output torque & rpm, while keeping the output power equals to the input power. A clutch will split the input power into output power and wasted heat. There is no other way.

 

[FactChecker]

My comment about $F=\frac{P}{v}$ has nothing to do with the powertrain, no matter what it is. It has to do with the wheel power, car velocity and available friction force at the tire-road contact patch.

Oh, sorry. I had completely misinterpreted your post.

 

[Wes Tausend]

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

 

[jerromyjon]

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.

the traction is actually higher when the tire is still stopped

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.

 

[Nidum]

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 .

 

[Randy Beikmann]

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.

 

[FactChecker]

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.