How is static friction the centripetal force during a car turning?

[bob012345]

The power that would have otherwise gone to fight the 30% of aerodynamic drag that is no longer present due to the speed reduction.

That complicates the problem because it would be a non-linear relationship over the time the speed is reduced. What simplifying assumptions do you want to make?

 

[rcgldr]

We are told that the car slows by 20 mph. However, we are not told that this is a one-time slowdown.

The 20 mph is the reduction in speed over a period of time from corner entry to corner exit and an extreme case (Radillion at Spa). Power is being consumed as soon as the tires deform, but the consumed power takes time to slow the car down. The worst is Spa, at Radillion, a 5 g dip (including gravity), around 3 g's of aerodynamic downforce, allowing for a 5 g right turn at the same time, so a lot of tire deformation due to both vertical and lateral loads. The drivers were running at full throttle, but you could hear the engine rpm's drop in videos of the old 3 liter V8's that spun at 18,000 to 20,000 rpm. Depending on the year, which affects power and downforce, the maximum drop was a bit over 20 mph.

Also at Spa, a video example of the corner name Pouhon, taken flat out, corner entry speed 309 kph, corner exit speed onto the short straight, 291 kph, around 4 g, an 18 kph (11 mph) drop, not quite as bad as Radillion, which is an extreme case, but you get an idea of what is going on. You have to go to youtube to watch this video:



Formula 1 tires dissipate heat fairly quickly, to the the air, road, and as black body radiation as seen in the infrared video from my prior post. Note how quickly the temperature drops once they exit a corner onto a straight.

 

[bob012345]

The 20 mph is the reduction in speed over a period of time from corner entry to corner exit and an extreme case (Radillion at Spa). Power is being consumed as soon as the tires deform, but the consumed power takes time to slow the car down...Formula 1 tires dissipate heat fairly quickly, to the the air, road, and as black body radiation as seen in the infrared video from my prior post. Note how quickly the temperature drops once they exit a corner onto a straight.

Do you have data on how hot the tires get above their normal level in the turn? That alone could be used to estimate how much energy goes into the tires on the turns.

 

[rcgldr]

Do you have data on how hot the tires get above their normal level in the turn? That alone could be used to estimate how much energy goes into the tires on the turns.

Doing a quick web search, my guess from 100 C to 130 C, but this is mentioned as the operating temperature range. Even if you knew the temperature increase, keep in mind the tires are constantly dissipating heat at a fairly rapid rate, which would also need to be taken into account.

 

[bob012345]

Doing a quick web search, my guess from 100 C to 130 C, but this is mentioned as the operating temperature range. Even if you knew the temperature increase, keep in mind the tires are constantly dissipating heat at a fairly rapid rate, which would also need to be taken into account.

We are talking about a very short window of a couple of seconds. We can assume it rises in the turn and falls after the turn. Knowing the temperature range and time should allow an estimate of the excess energy gain and loss. A different rise and fall rate could help account for the extra dissipation in the turn at least to first order.

 

[Richard R Richard]

When there is no slip in pure rolling, the heating of the tire comes from the load applied to the ground, not from the sliding of the tire on the ground, when the car is static it supports its weight in one direction only, but at a straight line speed. it has the weight plus the downforce applied cyclically to each point of the wheel in constantly varying direction, and in curves the resultant is the vector sum of the weight, downforce, and centripetal acceleration. This charge of greater magnitude up to 10g cyclically requests the chemical bonds at higher frequency the faster it travels, it alters the balance in the bonds between atoms, that extra vibration is translated into heat. There is no point in checking it against any friction, and measuring it is difficult due to the intentional aerodynamic cooling of the chassis, and the loss of the rubber's own material that degrades.
Static friction provides the change of direction, but it is not the cause of the heating, nor of the loss of kinetic energy in the turns. It is clear that if there is no static friction, turns cannot be made, keeping a tire continuously at the limit of its static friction does not make it heat up, what makes it heat up is the continuous change of that stress, If vibrating a tire with the same tension as when it is spinning will heat up, even without rolling.

 

[rcgldr]

When there is no slip in pure rolling, the heating of the tire comes from the load applied to the ground, not from the sliding of the tire on the ground

Even when pure rolling, there is still significant deformation at the contact patch for typical tires. Formula 1 tires are setup with negative camber, so that involves some deformation even when going straight. As posted earlier, the hysteresis related to deformation and recovery is a contributor to heat generation.

 

[sophiecentaur]

Discussing car tyre behaviour is, of course, interesting and relevant but my point is that loss of energy is fundamental when steering. To produce a centripetal force requires an angle between the wheel plane and the tangent. There will always be slip because of the finite length of contact footprint. For point contact there can be no 'steering' and also there will be infinite pressure on the surface.

 

[cmb]

Summary:: I understand for a car to turn, there must be a centripetal force. As the car turns it is friction of the front wheels creates an inward force. How, looking for a conceptual answer

Hello, as you can see i am trying to understand conceptually how the tires during turning create a centripetal force. It was explained to me that as we turn the car tires, the tires similar to a ski or a wedge, now want to push the ground to the side and forward. If the ground was loose, this makes sense as the tires would slide forward and scrap the ground to the side. Is it this force of friction that causes the ground to push back, have a perpendicular component, pushing the car inward and turn?

As you negotiate a turn, if you are turning left, your wheels are pushing to the right against the floor. Static friction allows the floor to "push back" against your wheels, allowing you to turn left.

View attachment 285559

At the heart of your question is the question of how a wheel works.

A ('perfect') wheel provides no restricted movement in one direction whilst preventing movement in the perpendicular direction, subject to the static friction between the wheel and the surface. A rolling wheel is essentially static with the ground at the moment it rolls over it (nothwithstanding a more complex discussion about 'tyre slip', which isn't a necessary part of that discussion).

A vehicle rolling along in the direction of the unrestricted rolling direction of its wheels has an incipient restriction perpendicular to those wheels. If perturbed, either through forces acting on the vehicle or by perturbations in the angles of those wheels, that friction force, perpendicular to the free rolling direction of the wheel, will be conducted into the wheel hub and thus the vehicle that retains the hub on its chassis.

If some, but not all, the wheels are perturbed, then there may be a yawing torque resultant on the vehicle, about the cross-section of the projected rolling axes of the wheels.

 

[Richard R Richard]

Discussing car tyre behaviour is, of course, interesting and relevant but my point is that loss of energy is fundamental when steering. To produce a centripetal force requires an angle between the wheel plane and the tangent. There will always be slip because of the finite length of contact footprint. For point contact there can be no 'steering' and also there will be infinite pressure on the surface.

I agree that the axis of rotation of the wheel is not perpendicular to the direction of advance or tangential. I just want to clarify that there is not necessarily slip, it is possible that the vehicle receives centripetal force, up to the maximum value that allows static friction in a direction perpendicular to the tangential speed.
In ideal conditions a wheel has a single point of contact with the ground, in practice the contact is superficial, a rectangular area in the best of cases, where the force is transformed into pressure that is not uniform on the surface, nor is it it is constant in time.
When the tire hits the ground, it can apply static force without sliding for the short time that rolling contact lasts.

 

[cmb]

There is a theory expounded by certain engineers that say a tyre can only generate grip if it is slipping. This is a misunderstanding of incipient forces at a boundary of surface energies. There is no need to revert to a discussion of tyres when discussing how wheels work.

Imagine a surface of fine needle like structures and very narrow metal wheels passing through and between them. You would still expect a wheeled vehicle to turn, but there would not be any slip.

The OP's question is fundamentally about the functioning of wheels, they allow a hub to move in one direction but not in the other.

I'd also like to dispel one other massive, humongous myth; that only the front wheels steer.

The OPs picture shows only forces on the front wheels, this is very wrong. All 4 wheels steer and apply the necessary force, it is the relative directions of the wheels to the vehicle motion (rotation rate) that induces yaw. If the car is already in a yawing motion, a lateral steer can still be achieved with all wheels pointing in the same direction, at that point the wheels are opposing an existing yaw, rather than creating one, and this is a '4-wheel drift'.

 

[sophiecentaur]

I just want to clarify that there is not necessarily slip,

How can there not be slip if the inner and outer extremes of the footprint are traveling at different speeds?

 

[cmb]

How can there not be slip if the inner and outer extremes of the footprint are traveling at different speeds?

"Consider a wheel of negligible width"

Yes, there is slip of the tyre in the real world but slip is not 'necessary' for a wheel to impart a centripetal force to accelerate a vehicle around a corner.

Consider a train on a track, where the front bogey is turned by virtue of the tracks it is riding on.

 

[sophiecentaur]

"Consider a wheel of negligible width"

. . . . and no deformation of wheel or surface? I feel you would have to explain what "negligible" means. Zero width would involve infinite pressure because the footprint would be a single point and a finite length of footprint would imply that there would be a range of tangential speeds over the length. If the wheel (and ground) were totally rigid, there would have to be slip or, if not totally rigid, there would be deformation (=hysteresis).
One arrangement that may not involve basic slip could be a wheel with a vertical axis, running around the inside of a vertical cylinder but then you would need a ball bearing to support the weight of the vehicle. I feel that would be too extreme to consider valid to this topic.

 

[hutchphd]

How can there not be slip if the inner and outer extremes of the footprint are traveling at different speeds?

If there is any natural "crown" on the inflated tire, then the camber could make the tire roll slightly like a cone depending upon location of the contact patch relative to the center line .That's how it can happen for a finite width tire: I do not know if it does.

 

[sophiecentaur]

If there is any natural "crown" on the inflated tire, then the camber could make the tire roll slightly like a cone depending upon location of the contact patch relative to the center line .That's how it can happen for a finite width tire: I do not know if it does.

I see. You mean that the difference in tangential speed would match the difference in the surface speeds of the tyre. That way there would be no difference in the speeds of the two surfaces. The angle of the cone would, I suppose, need to be a function of the wheel radius and the turning radius. OK for one special radius for the wheel and that's along the same lines as the design of bevel gears (?).

 

[hutchphd]

But the suspension changes all the angles all the time in response to lateral force etc. I really have no idea whether this is a real effect or not. I do find F1 entertaining, though, and their attention to excruciating detail is legendary and usually well tested. Maybe somebody actually knows...

 

[sophiecentaur]

But the suspension changes all the angles all the time in response to lateral force etc. I really have no idea whether this is a real effect or not. I do find F1 entertaining, though, and their attention to excruciating detail is legendary and usually well tested. Maybe somebody actually knows...

Petrol heads use tyres and tyres use rubber and are designed with slip in mind; that's good Engineering. Some contributors are still of the opinion that slip is not a fundamental factor in cornering. Personally, I can't see how a 'constrained' wheel can have the same tangential velocity all across its contract footprint.
We are suffering here from an insistence to use the term 'static' because the car is traveling on a constant radius, rather than considering what actually goes on over the line of contact with the ground. Some parts of the (definitely rigid) rim are going faster and some are going slower, instantaneously. Can they be considered 'static'?

 

[Richard R Richard]

Some parts of the (definitely rigid) rim are going faster and some are going slower, instantaneously. Can they be considered 'static'?

Undoubtedly, its exposition is logical, but it has a drawback, the width of the wheel is limited, let's call it $d$, and if you observe it will have the same longitudinal slip between the inside and outside of the cover regardless of the radius of the curve, it is Say the longitudinal slip is? $d$ for a curve of plane angle $\theta \cdot d$. A $10m$ radius curve at $180$ degrees has the same slip as a $1km$ radius curve.

So the heating comes from making turns at a greater angle on the "zigzag" track, but obtaining the zigzag does not depend on the fact that there is slip, it depends on the static friction or the static grip or static gear, of the wheel and pavement, on a small scale. there is slippage, but that is not what prompts the car to turn. On the inner side we can think that the sliding will be in the opposite direction to the advance and on the outside towards the advance, there will be an internal line where there is pure rolling.
One thing is to turn with grip as in F1 and another thing to do drifting, I think that in "mythbusters" tests were made on whether it was better to turn while drifting or maintaining grip. The results were….

 

[rcgldr]

How can there not be slip if the inner and outer extremes of the footprint are traveling at different speeds?

Via deformation. There is some actual slippage, more so during at what could be called partial contact. To answer your question though, the inside would get compressed more in the direction of rolling than the outside. The contact patch is also deformed at an outwards angle, which correlates to slip angle, the difference between a tire's actual path and the direction it is pointed. For racing cars, the term working slip angle refers to the slip angle for maximum lateral acceleration. Formula 1 cars in the late 1960's before downforce, using bias play racing tires, had working slip angles between 8 to 10 degrees, enough to make the cars appear to be drifting in turns. Modern Formula 1 radial slicks have working slip angles around 3 degrees. The tires used on Indy race cars for high speed ovals have the relatively stiffest sidewalls, with working slip angles around 2 degrees.

 

[sophiecentaur]

I would agree with most of that. The strength of the pneumatic rubber tyre is that it allows a controlled amount of deformation and that’s been a success story for over a hundred years.
but there is still some slip and non static friction. This will be at the beginning and end of the contact time when the local contact pressure is low and the tyre distortion is starting and relaxing.
Without a deforming tyre, the wheel or road surface will get chewed up permanently and the only place you could say there’s static friction would be where the wheel and road speeds are equal; a small part of the footprint.
Road surface is usually much more rigid but you get loud squealing at low speed on the surface of many indoor car parks because of the latex floor coating (anti dust measure, I believe).

 

[cmb]

. . . . and no deformation of wheel or surface? I feel you would have to explain what "negligible" means. Zero width would involve infinite pressure because the footprint would be a single point and a finite length of footprint would imply that there would be a range of tangential speeds over the length. If the wheel (and ground) were totally rigid, there would have to be slip or, if not totally rigid, there would be deformation (=hysteresis).
One arrangement that may not involve basic slip could be a wheel with a vertical axis, running around the inside of a vertical cylinder but then you would need a ball bearing to support the weight of the vehicle. I feel that would be too extreme to consider valid to this topic.

I'm saying that in the real world yes of course there may be a small amount of slip, but it could be truly paltry.

I mean, imagine a motorcycle tyre rolling down a highway surface and cornering around a long curve of several hundred metres radius. The contact patch would be a cm or so across for a motorcycle tyre, meaning that the slip would be literally unmeasurable and only discernible with some 'perfect' mathematics being applied to an imprecise world.

Nonetheless, despite the slip on one side of the motorcycle tyre being no more than a few parts in a million to the other, it can corner as effectively around that highway bend as could a sports car with a huge excessively wide tyre.

My point is that the cornering ability of a tyre is not a function of the 'slip' it experiences. If you thought that, then you'll end up thinking that a motorcycle with a contact patch a couple of cm across would be an order of magnitude worse at cornering than a sports car with a 275(mm) wide tyre.

 

[rcgldr]

I mean, imagine a motorcycle tyre rolling down a highway surface and cornering around a long curve of several hundred metres radius. The contact patch would be a cm or so across for a motorcycle tyre, meaning that the slip would be literally unmeasurable and only discernible with some 'perfect' mathematics being applied to an imprecise world.

Other than for something like a 10 speed bicycle, motorcycle tires have a reasonably large contact patch, due to a combination of tire size, profile, and soft compound rubber. The slip angle in higher g turns is significant.

 

[cmb]

Other than for something like a 10 speed bicycle, motorcycle tires have a reasonably large contact patch, due to a combination of tire size, profile, and soft compound rubber. The slip angle in higher g turns is significant.

Yes, but my point is that the ability of a tyre to grip a corner is not a function of its ability to slip, so one is not a function of another and there is no technical reason to discuss them as if they are one concept.

A 120psi bicycle racing tyre is another good example, a contact patch of a few mm across.

If the argument is that tyres 'must slip' to generate centripetal forces, then the same argument may apply to any scenario of friction, that in fact no friction force is created until the contact patch slips.

This is mistaking the concept of 'incipient forces', those that 'would be a force' if there was motion but there is no motion else there would be a force. This is a distraction to the basic functional operation of a wheel, which is very simple and is where it allows for rolling motion in one direction but resists motion orthogonal to that axis of freedom. Anything stated further is an unnecessary complication.

 

[sophiecentaur]

I'm saying that in the real world yes of course there may be a small amount of slip, but it could be truly paltry.

Trouble is that "truly paltry" is not a numerical value. You must have read about slip angle. Wiki has a long article about it. Cornering only happens when there is slip in some form. Tyres are designed for optimum cornering.
You suggest a very large turning radius would involve very small slip. No problem there; you possibly couldn't measure it and the heating of the tyre would only be due to hysteresis. How is that relevant to the high slip on a fast, tight curve? Just a different situation.

 

[cmb]

Cornering only happens when there is slip in some form.

Cornering only happens when there is a bend in the road, but that doesn't mean the bend in the road causes the cornering.

It is a consequence of existing in a world of physical dimensions that a tyre will slip in a corner, but it is not the slip which defines the cornering. As I mention, different tyres will slip differently, yet their cornering ability is simply the reaction force and the friction coefficient, which are largely the same irrespective of the width of the tyre and degree of slip.

 

[sophiecentaur]

ability of a tyre to grip a corner is not a function of its ability to slip,

That statement is not really relevant and it's the wrong way round; it's not "its ability" it's 'in spite of slip'. My point is that there will always be some slip. A good tyre will minimise it. A bad, nearly flat tyre will slip a lot because the footprint will be long and a big fraction of it will be way inside or way outside the turning circle of the middle of the print.

For an undriven front wheel, the part of the wheel that's about to land on the ground is outside the turning circle so it is going faster (speed proportional to radius from middle of turn). It has to slow down so that can be in firm contact with the ground and just before contact there is not enough pressure to establish static friction - that's when you get slip. The rubber deforms and reduces the slip greatly and contributes to the cornering. When that same bit of tyre starts to lift, it's going slower than the middle of the footprint but deformation keeps it in contact until pressure is low and it slips.
Most of the energy is stored in the rubber and returned less the hysteresis loss.

How can there not be slip then there is a speed difference and too low a friction force to stick parts of the tyre to the road?.

 

[sophiecentaur]

It is a consequence of existing in a world of physical dimensions that a tyre will slip in a corner, but it is not the slip which defines the cornering.

I see your argument but the portion of the tyre that's actually slipping still contributes a centripetal force. The extreme situation is in speedway racing and in drifting round a corner when there is actually no static friction at all - it's all slip. Where would you draw the line?
It seems you want to hold on to the static category more as a matter of principle than anything else. Whenever there's distortion, there is some slip - even when braking in a straight line when the leading and lagging portions of the footprint are actually going at different speeds as the tyre deforms.

 

[cmb]

I agree the slip occurs at the front and trailing edge of the tyre contact. This is not, however, the region of the footprint in which there is static friction and which is the major contribution to centripetal loads.

What I was disputing was that there is any need to discuss slip in this context. It is a consequence of 'being a tyre' and cornering, not a cause of either.

Any slip because the 'inside edge' of the tyre to the 'outside edge' are at different radii thus at different speeds is accommodated by the tyre's compliance on the road.

The conformal compression of the tyre, as you rightly describe, is a matter of units to several percent. The differential speeds inside edge to out are significantly less (thus wholly accommodated in the rubber compliance) except when at very small turning radii (like in a car park when you hear tyres squealing on a slow car).

 

[Richard R Richard]

If the cause is to slide and the effect to bend, then a steel wheel would bend with the same efficiency as a rubber one of the same dimensions and with the same ground conditions. That is not true, 40 years ago I was certain of that when my toy car with bearings greatly improved the grip in the curves with skateboard wheels. I can understand the point that there is a small slip "always", but it is not the main cause, nor the only one, to have perpendicular force and use it as a centripetal force to make a curve.

The coefficient of friction is of "clear" importance, but it is more so formed by a flat contact surface without appreciable denial produced by a flexible material that does not degrade rapidly under the cyclical demand that produces heat by internal friction.