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

In summary: Sorry, I'm getting confused here. In the case of a car turning, there is a centripetal force that is created due to the friction of the wheels against the ground. However, this force is created even if the car is not being driven by an engine. Is that correct?Yes, that is correct.
  • #1
alkaspeltzar
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TL;DR 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.

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  • #2
alkaspeltzar said:
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

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.
Yes, this is correct.
 
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  • #3
Please, see:
https://en.m.wikipedia.org/wiki/Camber_thrust

When cornering forces are strong and wheels are made of a deformable material, there is certain amount of radial movement or “crabbing”, since each landing contact patch will deform a little respect to the already deformed patch in contact with the rolling surface.
 
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  • #4
Dale said:
Yes, this is correct.
Thank you Dale for confirming
 
  • #5
Lnewqban said:
Please, see:
https://en.m.wikipedia.org/wiki/Camber_thrust

When cornering forces are strong and wheels are made of a deformable material, there is certain amount of radial movement or “crabbing”, since each landing contact patch will deform a little respect to the already deformed patch in contact with the rolling surface.
Interesting to read about the camber thrust and how it really makes sense/supports the outer push on the ground due to friction i was referring to. From there yes the tire deforms and you get the true cornering force as it come back to the vehicle, thru the hub and such.

This is helpful. Thank you
 
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  • #6
The force pushing the ground to the right may or may not exist and isn't relevant here.

Centripetal force doesn't need to be created by engine power, it happens on its own as an effect of the turn. I think it would be best to ignore/discard the propusive force on the front wheels and focus solely on the centripetal force you are asking about.
 
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  • #7
russ_watters said:
Centripetal force doesn't need to be created by engine power, it happens on its own as an effect of the turn.
That is exactly what we were saying and I agree with you. The example above ignores the engine.

I am not concerned with propulsive force. I am assuming the car is coasting and you turn the front wheels. Due to the inertia the car wants to continue traveling forward, but the tires don't. There is a static friction force that pushes on the road to avoid that change, which is where the lateral/perpendicular force comes into play, which is the centripetal force turning the car.

If it was a front wheel drive, then you have this effect plus the engine helping drive.
 
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  • #8
alkaspeltzar said:
That is exactly what we were saying and I agree with you. The example above ignores the engine.

I am not concerned with propulsive force. I am assuming the car is coasting and you turn the front wheels. Due to the inertia the car wants to continue traveling forward, but the tires don't. There is a static friction force that pushes on the road to avoid that change, which is where the lateral/perpendicular force comes into play, which is the centripetal force turning the car.

If it was a front wheel drive, then you have this effect plus the engine helping drive.
Sorry, you're right. I got tripped up by the switch from "inward" to "right", vs the diagram.
 
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  • #9
russ_watters said:
Sorry, you're right. I got tripped up by the switch from "inward" to "right", vs the diagram.
No issues, it was a good point as we don't need an engine to make things turn. Any kid with his bike on a big hill can tell you that.

As for the verbage confusion, yeah that is my bad. Sorry the friction from the wheel when turning a car to the left(like the picture), creates a rightward force against the ground. The ground pushes back with an equal and opposite friction force inward, which is the centripetal force. Hope that is clearer.

I got it. THank you
 
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  • #10
russ_watters said:
Centripetal force doesn't need to be created by engine power,
True, but the wheels are always at an angle that's steeper than the tangent so there will be a force which slows the vehicle down because of the hysteresis of the tyre deformation. That does mean that you would need to compensate by increasing the engine power. You do this unconsciously . Even with rails, there will be more losses around a curve than on the straight.

I was interested to learn (on my one and only hands-on joy ride) that turning an aeroplane (by banking) will cause loss of height as you 'slip down sideways' and the rudder deals with this by adjusting your heading to compensate so you are pointing uphill slightly..
 
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  • #11
sophiecentaur said:
I was interested to learn (on my one and only hands-on joy ride) that turning an aeroplane (by banking) will cause loss of height as you 'slip down sideways' and the rudder deals with this by adjusting your heading to compensate so you are pointing uphill slightly..
or use a bit more elevator. The rudder will result in a more coordinated turn, but at higher speeds, you don't need much rudder.
 
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  • #12
rcgldr said:
or use a bit more elevator. The rudder will result in a more coordinated turn, but at higher speeds, you don't need much rudder.
Yeah, to clarify a bit more: in a turn the plane tends to yaw away from the turn not toward it. The rudder input is to avoid sliding(slipping) through the turn. The plane descends because the lift vector is no longer vertical, so more elevator means higher angle of attack and more lift to counteract that.

I don't like a plane as an example here, very different dynamics. Cars are complicated enough. It's not clear to me that turning a car or other wheeled vehicle necessarily requires energy loss.
 
  • #13
To avoid any energy loss when steering with wheels, I’d say there has to be friction. Even with rails there will be rubbing and pneumatic tyres get hot, even when going straight. Only with a rigid bar or wire can the centripetal force be the only force involved.
 
  • #14
russ_watters said:
Yeah, to clarify a bit more: in a turn the plane tends to yaw away from the turn not toward it. The rudder input is to avoid sliding(slipping) through the turn. The plane descends because the lift vector is no longer vertical, so more elevator means higher angle of attack and more lift to counteract that.

I don't like a plane as an example here, very different dynamics. Cars are complicated enough. It's not clear to me that turning a car or other wheeled vehicle necessarily requires energy loss.
Not to mention that the centripetal force is provided by a horizontal component of the Lift vector and not due to any kind of friction with the air or drag.

Also, for cars, static friction is for flat turns. Banked turns do not need static friction if the car is traveling at the proper designed speed limit. In this case a component of the Normal force acts as the centripetal force.

https://en.m.wikipedia.org/wiki/Banked_turn
 
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  • #15
bob012345 said:
Not to mention that the centripetal force is provided by a horizontal component of the Lift vector and not due to any kind of friction with the air or drag.

Also, for cars, static friction is for flat turns. Banked turns do not need static friction if the car is traveling at the proper designed speed limit. In this case a component of the Normal force acts as the centripetal force.

https://en.m.wikipedia.org/wiki/Banked_turn
My suggestion of the flight analogy has rather taken over here but, afaics, there's still an energy cost involved in a banked turn. If you are trimmed for optimum fuel consumption straight and level then are you implying that doing a banked turn would not cause you to drop in height, slow down or use more fuel? Clearly you can't change altitude in a car but isn't the energy consideration still there in both situations?
I know that analogies can disrupt the flow of an argument and it might have been better if I hadn't introduced it, despite the fact that it's valid, imo.
 
  • #16
sophiecentaur said:
My suggestion of the flight analogy has rather taken over here but, afaics, there's still an energy cost involved in a banked turn. If you are trimmed for optimum fuel consumption straight and level then are you implying that doing a banked turn would not cause you to drop in height, slow down or use more fuel? Clearly you can't change altitude in a car but isn't the energy consideration still there in both situations?
I know that analogies can disrupt the flow of an argument and it might have been better if I hadn't introduced it, despite the fact that it's valid, imo.
I believe you would add some power, increasing speed to increase lift to maintain altitude during a turn. Of course in many situations you are either in a climbing turn or decending turn when maintaining altitude is not an issue. In the old war movies doesn't the engine always rev up just as the plane is banking?
 
  • #17
sophiecentaur said:
My suggestion of the flight analogy has rather taken over here but, afaics, there's still an energy cost involved in a banked turn. If you are trimmed for optimum fuel consumption straight and level then are you implying that doing a banked turn would not cause you to drop in height, slow down or use more fuel? Clearly you can't change altitude in a car but isn't the energy consideration still there in both situations?
I know that analogies can disrupt the flow of an argument and it might have been better if I hadn't introduced it, despite the fact that it's valid, imo.
Lift and drag are inexorably coupled so yes, engine power in a plane has to increase in a turn to avoid slowing down or descending. I consider that different from how a wheeled vehicle works (loss is not required to create the force).
 
  • #18
russ_watters said:
Lift and drag are inexorably coupled so yes, engine power in a plane has to increase in a turn to avoid slowing down or descending. I consider that different from how a wheeled vehicle works (loss is not required to create the force).
I used to be a private pilot and sometimes I would circle a tree rather low in some field to practice banked turns without losing altitude.
 
  • #19
bob012345 said:
I used to be a private pilot and sometimes I would circle a tree rather low in some field to practice banked turns without losing altitude.
I'm taking lessons right now. That's one of the exercises. My usual reference is a grain silo.
 
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  • #20
russ_watters said:
I'm taking lessons right now. That's one of the exercises. My usual reference is a grain silo.
I take it you have passed the right of passage.. the solo flight?
 
  • #21
russ_watters said:
Lift and drag are inexorably coupled so yes, engine power in a plane has to increase in a turn to avoid slowing down or descending. I consider that different from how a wheeled vehicle works (loss is not required to create the force).
Different yes but the common effect of losing energy is there. The drag in flight is equivalent to friction.
Perhaps using the concept of static friction is not appropriate in this case. Hanging on a rope as you’re being swung round would be a justified use of the static friction term in circular motion but PF is always having to deal with the ‘proper’ terms to use in many situations. Tyres always have a slip angle so no static friction, IMO,
 
  • #22
bob012345 said:
I take it you have passed the right of passage.. the solo flight?
Yes, I'm basically finished and waiting for my checkride.
 
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  • #23
russ_watters said:
Yes, I'm basically finished and waiting for my checkride.
My only advice is make sure you let your Examiner know that you see all the planes flying around you.
 
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  • #24
Getting back on the side topic of energy losses due to tire deformation, an extreme example is Formula 1 race cars, where despite being at full throttle, in high speed, high g turns, 20+mph is lost due to tire deformation. Even in medium to high speed turns, that energy goes into heating up the tires as shown in this infrared video:

 
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  • #25
We also could compare the energy to heat the tires with the loss in kinetic energy of the car and see how close they come. How much does the tempature rise during a turn? I wish the scale were shown. What to use for the specific heat of the rubber and mass of tread that heats up? Based on the reference below I will estimate it at 1000 J/Kg °C and the tread mass about 10 Kg per tire and the temperature rise about 50 °C. That gives ##2 X 10^6## Joules which seem way too high as the whole kinetic energy of the car assuming 1000Km total mass at 200 Km/hr is about ##1.5 X 10^6## Joules! Obviously I'm overestimating the heating of the tires by a lot. There wouldn't be enough energy or power available to do that. What are more reasonable numbers?
It was found that the specific heats ranged from generally 0.8 to 1.2 J g−1 °C−1, which was considerably lower values than those of rubbers, 1.9 to 2.2 J g−1 °C−1.*

*John Wiley & Sons, Inc. J Appl Polym Sci 72: 1513–1522, 1999
 
  • #26
bob012345 said:
##10^6## Joules which seem way too high as the whole kinetic energy of the car assuming 1000Km total mass at 200 Km/hr is about ##1.5 X 10^6## Joules!
We are told that the car slows by 20 mph. However, we are not told that this is a one-time slowdown. One assumes that it is a 20 mph sustained speed reduction that persists until the cornering is complete. So it would be accompanied by a reduction in air resistance.

Waving my hands like a wild man, I will assume that air resistance goes as the square of velocity and we have a 10 percent reduction in velocity. That translates to a roughly 30 percent reduction in power loss to air resistance. Suddenly, 30% of the engine power is going into heating the tires instead of swirling the air. So what you need to know is the power output of the engine in watts and the Joule to calorie conversion ratio. (4.2 to 1 if I remember my high school physics right).

One assumes that most of the thermal energy is generated and remains in the tires rather than being deposited immediately on the road, though that assumption is questionable.
 
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  • #27
jbriggs444 said:
We are told that the car slows by 20 mph. However, we are not told that this is a one-time slowdown. One assumes that it is a 20 mph sustained speed reduction that persists until the cornering is complete. So it would be accompanied by a reduction in air resistance.

Waving my hands like a wild man, I will assume that air resistance goes as the square of velocity and we have a 10 percent reduction in velocity. That translates to a roughly 30 percent reduction in power loss to air resistance. Suddenly, 30% of the engine power is going into heating the tires instead of swirling the air. So what you need to know is the power output of the engine in watts and the Joule to calorie conversion ratio. (4.2 to 1 if I remember my high school physics right).

One assumes that most of the thermal energy is generated and remains in the tires rather than being deposited immediately on the road, though that assumption is questionable.
I assume the tires are always hot but get momentarily hotter in the turns but cool quickly to their normal level because of the intimate contact with the road and also air. Most turns the engine is throttling down yet the tires still heat up. The question in my mind is how much energy is going into heating the tires in the turns due to compression of the rubber? Maybe I overestimated the amount of rubber that gets hot by 10x.

For a 20mph difference of a 1000 kg car the formula is with 20mph equal to 32.2 kph;

##DeltaE = 1000/2 (0.27775^2)(V^2 - (V - 32.2)^2)##

Which gives about 420kJ for ##V##= ##200 km/hr##

Now all I have to do is use ##DeltaE = C DeltaT M##.
 
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  • #28
bob012345 said:
For a 20mph difference of a 1000 kg car the formula is with 20mph equal to 32.2 kph;

##DeltaE = 1000/2 (0.27775^2)(V^2 - (V - 32.2)^2)##
The scenario I thought we were discussing was:
rcgldr said:
where despite being at full throttle, in high speed, high g turns, 20+mph is lost due to tire deformation
It is not completely clear whether we should be considering the KE lost by the vehicle, the extra power diverted from engine to tires during the turn or both.
 
  • #29
jbriggs444 said:
It is not completely clear whether we should be considering the KE lost by the vehicle, the extra power diverted from engine to tires during the turn or both.
The scenario I was discussing is the power applied doesn't change in the turn but the speed reduces by 20mph and the tires get hotter. However by mentioning the throttle going down I was being somewhat confusing and I apologize.

Constant engine power maintains the speed against rolling friction, engine friction and aerodynamic drag. I assume some available kinetic energy at the start of the turn is converted into heating the tires in the turn causing a 20mph speed reduction. What do you consider as the extra power diverted from engine?
 
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  • #30
bob012345 said:
What do you consider as the extra power diverted from engine?
The power that would have otherwise gone to fight the 30% of aerodynamic drag that is no longer present due to the speed reduction.
 
  • #31
bob012345 said:
jbriggs444 said:
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?
 
  • #32
jbriggs444 said:
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.
 
  • #33
rcgldr said:
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.
 
  • #34
bob012345 said:
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.
 
  • #35
rcgldr said:
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.
 
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