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

[sophiecentaur]

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.

So some slip occurs and that, of course, can't be where there's static friction and it is not the major factor in normal cornering. But, in some cases, it's all slip. So what is the argument when I point out that saying it's all static friction is not the whole story? Small contributions to effects are usually examined in Physics and Engineering, especially when those contributions dominate at times.
If you want to exclude slip then how do you deal with drifting round corners? Is there some matter of principle involved?

 

[sophiecentaur]

a steel wheel would bend with the same efficiency as a rubber one of the same dimensions

A steel wheel that's strong enough would have too high a modulus I think. The air in a pneumatic tyre gives lifting strength yet allows the modulus to be low enough (I guess you could call it a composite structure). Steel wouldn't work - but I bet somebody tried!. Steel wheels only work when you have a rail and a flange on the wheel. No good if you want to steer 'anywhere' but pretty efficient.

 

[jbriggs444]

A steel wheel that's strong enough would have too high a modulus I think. The air in a pneumatic tyre gives lifting strength yet allows the modulus to be low enough (I guess you could call it a composite structure). Steel wouldn't work - but I bet somebody tried!. Steel wheels only work when you have a rail and a flange on the wheel. No good if you want to steer 'anywhere' but pretty efficient.

Steel on ice is a well developed technology. However, it uses significantly different operating principles.

 

[sophiecentaur]

Steel on ice is a well developed technology. However, it uses significantly different operating principles.

Yes; good point. Turning always appears to 'damage' the surface of the ice tho' (showers of ice when it's done quickly). So there still must be some slip, if that happens. I wonder if skates have been designed with multiple segments - equivalent to a distorting pneumatic tyre. Keeping as much of the skate on the turning circle would minimise slip due to minimising the speed differential along the skate.

 

[jbriggs444]

Yes; good point. Turning always appears to 'damage' the surface of the ice tho' (showers of ice when it's done quickly). So there still must be some slip, if that happens. I wonder if skates have been designed with multiple segments - equivalent to a distorting pneumatic tyre. Keeping as much of the skate on the turning circle would minimise slip due to minimising the speed differential along the skate.

The curve of the blade (the "rocker") on figure skates keeps the contact patch short and even results in some level of conformance to a curved path.

 

[hutchphd]

Steel wouldn't work - but I bet somebody tried!. Steel wheels only work when you have a rail and a flange on the wheel. No good if you want to steer 'anywhere' but pretty efficient.

Pardon me if I have missed parts of these discussions, but isn't the "slip" of a tire as it corners the result of differential lateral (parallel to road surface) stretching and relaxation of the rubber while the tread maintains static contact across the patch?
This would make steel a non-starter (unless on a rail). I assume everybody understands the function of the taper on the solid axle train wheel.

 

[jbriggs444]

Pardon me if I have missed parts of these discussions, but isn't the "slip" of a tire as it corners the result of differential lateral (parallel to road surface) stretching and relaxation of the rubber while the tread maintains static contact across the patch?
This would make steel a non-starter (unless on a rail). I assume everybody understands the function of the taper on the solid axle train wheel.

Iron-shod wheels were a thing, back in the day. They were more durable than wood.

Not really competitive with rubber on the track though.

 

[sophiecentaur]

but isn't the "slip" of a tire as it corners the result of differential lateral (parallel to road surface) stretching and relaxation of the rubber while the tread maintains static contact across the patch?

I'd put it the there way round. Without a deformable tyre, the slip would be right across the length of the footprint except at the single point on the footprint vertically below the axle. That's the only part of the wheel that's actually traveling at the right speed. Rubber and air will flex and allow the major part of the footprint to stay in contact and give 'static friction' but there must be leading and trailing parts of the footprint which have lower pressure on the road. They are not distorted enough to follow the surface until they get close enough and the pressure is high enough (limiting friction). If you don't acknowledge that this always happens then where do you draw a line between where it does and where it doesn't happen over a range of wheel materials and between a flat (squealing) tyre and a fully inflated tyre.

I feel I'm arguing about what appears to be an article of faith here; slip is not 'allowed' and it all has to be static friction. Why? Of course, tyres are designed to work as well as possible but that doesn't mean slip between tyre and road can be eliminated. (Other losses are involved, of course.)

 

[hutchphd]

Part of my question is the semantic distinction: slip vs slide . From Wikipedia

The lateral slip of a tire is the angle between the direction it is moving and the direction it is pointing. This can occur, for instance, in cornering, and is enabled by deformation in the tire carcass and tread. ... Sliding may occur, starting at the rear of the contact patch, as slip angle increases.

As I read this the slip is the tire interaction angle but its use in this colloquy seems all over the place and I am a little confused. I think slip does not necessitate slide but may include slide beyond elastic deformation limits ?

 

[sophiecentaur]

The lateral slip of a tire is the angle between the direction it is moving and the direction it is pointing. This can occur, for instance, in cornering, and is enabled by deformation in the tire carcass and tread. ... Sliding may occur, starting at the rear of the contact patch, as slip angle increases.

You are right and my statements have related to sliding between tyre and road. The friction force has to be enough for the tyre to become distorted, rather than just to slip over the surface.

There must always be a transition from free to stuck, during which the friction is not a 'static' force. Sometimes there will be no static friction at all. The sideways 'bulge' of the tyre will vary up to a maximum around the centre of the footprint. I think that actually implies that there must be some scuffing by all parts of the footprint as Curl of the surface displacement changes during contact so doesn't that mean there's not even total static friction all over the footprint?

 

[Richard R Richard]

The distance between two points on a circle is a chord. The arc of circumference between these two points is always longer than the chord. When the tire hits the ground, the circumference is elastically compressed, to reduce its length from the arc measure to the chord measure.
If we define a forward direction and turn to the left, the compression on the left side of the tire will be less than on the right side, just enough to absorb the difference in turning radius.
As long as the area or weight differential over the area differential, multiplied by the mean coefficient of friction, does not exceed the maximum static friction force, integral over the tread area of the mean pressure per unit, there will be static frictional force with a component perpendicular to the speed of the car, that will be the maximum centripetal force without slip.
If the tire demands more, with tight curves, adding more load by weight or by aerodynamics, etc., then, there may be "also" a dynamic friction force, which opposes the relative sliding of the tire with respect to the ground, whose direction is a combination of the speed of the car and the tangential speed of the tire (it can be higher if you accelerate, the same if you only turn, zero if you skid, or retrograde as very few drivers know how to achieve).
To pretend that all the centripetal force is caused by the sliding friction of the tire, in my opinion is an error, easily demonstrable.
In the 500 miles more than half of the trip, the vehicle is turning, although there is cant, the tire forces in a centripetal direction during each turn, thus lasting approximately 100 miles. But can you imagine how long a tire lasts while sliding it 100 miles, even though I changed the speed of the vehicle very little? They are different demands.
A city car will be doubling 2% of the route with a duration of 60000km, it would be the equivalent of dragging it 1200km… and it is clear that this last wear will be higher than the previous one.
That is my point of view, static friction is always one or two orders greater than dynamic, and only in severe conditions of use, is it the main cause of wear, together with those derived from acceleration or braking. A vehicle on the road where there are curves that are longer, it wears less than in the city, where the turns are short and slower. But on the road more length is covered by doubling than in the city, so the dynamic friction would not correspond to the wear.

 

[sophiecentaur]

To pretend that all the centripetal force is caused by the sliding friction of the tire, in my opinion is an error, easily demonstrable.

Is anyone suggesting that? Not me (intentionally).

I am suggesting that there must be sliding somewhere on the edge of the footprint because the normal force is not great enough to produce static friction. Good tyres, inflated to a proper value will have little of this effect, compared with the static friction but when there's something wrong - say underinflation, the slip angle will increase and the distortion can be extreme in regions where the contact pressure is low.
In the examples I have already given, there is significant slip yet the car or motorcycle still corners.

 

[hutchphd]

To pretend that all the centripetal force is caused by the sliding friction of the tire, in my opinion is an error, easily demonstrable.

No one is making that claim.
But my instinct here is that any sliding anywhere on the contact patch will likely lead to a cascade where the entire patch will break free. And then you have only sliding friction.
I admit I had not considered the simple fact that the chord is shorter...gives the tire a lot more ways to stretch and conform.

 

[Richard R Richard]

Is anyone suggesting that? Not me (intentionally).

No one is making that claim.

Hi ! Maybe I have gone very to the extreme and it was not the intention.
I'm just trying to give an alternative point of view to

Cornering only happens when there is slip in some form.

which in my opinion is the opposite in meaning.

The lateral slip of a tire is the angle between the direction it is moving and the direction it is pointing. This can occur, for instance, in cornering, and is enabled by deformation in the tire carcass and tread. ... Sliding may occur, starting at the rear of the contact patch, as slip angle increases.

I have already agreed that there may be some dynamic friction, but not as the main cause for the centripetal force in all cases, if in some cases already analyzed very evident, I continue in my attempt to explain to OP how to relate "friction" vs "centripetal force"

So some slip occurs and that, of course, can't be where there's static friction and it is not the major factor in normal cornering.

I anticipated that I disagree, the same slippage in meters occurs for a curve with a wide radius as for a curve with a small radius, it is only a function of the total turning angle and the width of the tire, therefore the same slippage of the same tire in the same speed conditions cannot explain the two different radii of gyration. The different turning radii with their associated centripetal force need something else to be explained, and my point of view is that the cause is static friction, parallel to the wheel axis, which has a component in the direction of the center of the curve.

Regards.