What happens if two wheels turn toward each other....

[Jibran96]

Does anyone know what will happen if the two front wheels of a car turn away from each other (left goes left and right goes right) while the car is moving or if they turn inwards towards each other? Other than the stress on the steering rod, will this cause braking and if yes, how effective can this braking be considering that it is an average sedan? Will this braking be substantial?

 

[Dr.D]

If the wheels turn only slightly in (or slightly out), the effect should be increased friction and a lot of tire wear. If wheels turn into a greater degree, at some point they will cease to roll and go into sliding only. This would be a braking action, but it would be difficult to control; not a good idea.

 

[BvU]

Turns out front-wheel drive cars are aligned with toe-out and rear-wheel drives with toe-in.
But not too much in modern cars, or they will be pulled apart...

 

[Jibran96]

Thank you for your replies. Dr.D, just to be clear, for instance if a car is moving forward and the front two wheels tun inwards by a slight angle (not too much), I now understand that the friction will increase and this will cause a braking effect, but will this be becasue of increased surface area (wheel to ground) or the restriction of the wheel movement?
Thanks in advance!

 

[BvU]

The restriction of the wheel movement will force skidding instead of rolling, with a correspondingly higher friction coefficient.

 

[jbriggs444]

One might (or might not) get some better insight to the effect by considering the behavior of a small patch on the surface of a tire as it rotates into contact with the road and then back off. Let us assume a toe-in configuration. Toe-out will be the same, just with some signs reversed.

As the patch first rotates into contact with the road on the front side of the contact area, it is under no lateral force. As it rotates further back, the geometry of the wheel tends to force the patch away from the centerline of the vehicle. The tire is not rigid. It is made of rubber. It can flex. The patch continues to track straight back while the body of the tire above it moves outward, away from the centerline. Of course, this results in an outward force on the patch and a corresponding inward force on the body of the tire.

An inward force on a tire with toe-in has a component that retards rotation. [Squeeze together on the bottom of a pair of tires with toe-in and you'll tend to make your car move backward] Of course, in an ideal world, energy is conserved. This squeezing effect that retards rotation, effectively consuming energy, must put that energy somewhere. And it does. The rubber in the tire acts like a spring, storing potential energy in its deflection.

If we continue to follow the patch backward, it may undergo so much lateral stress that it skids across the pavement laterally, draining energy into friction. In any case, it will eventually rotate up off the pavement and snap back to its unstressed position, draining potential energy into vibration which will ultimately damp into heat.

 

[quickquestion]

Does anyone know what will happen if the two front wheels of a car turn away from each other (left goes left and right goes right) while the car is moving or if they turn inwards towards each other? Other than the stress on the steering rod, will this cause braking and if yes, how effective can this braking be considering that it is an average sedan? Will this braking be substantial?

According to what I know so far, if the 2 front tires are suddenly turned 90 degrees, if the tires have a uS of 1.7 and the car is not accelerating, and has an even weight distribution, the car will stop instantly if the car is traveling at a forwards velocity less than .85.

I could be wrong, and if I am, someone please correct me because I have been basing my car physics on this principle so far and not having much progress, and if I'm wrong about this scenario this may help my life greatly.
Turns out i was wrong and i suck. But, when I did carphysics, I did it the supposed right way, but got bad results, so ironically me being wrong with this post didn't help me much.
The correct answer is, according to Newtonian model, a car with tires of 1.7 us can go no faster than 8.33 m/s if you don't want it to slip when you turn the tires 90 in one frame of simulated time.

 

[jbriggs444]

According to what I know so far, if the 2 front tires are suddenly turned 90 degrees, if the tires have a uS of 1.7 and the car is not accelerating, and has an even weight distribution, the car will stop instantly if the car is traveling at a forwards velocity less than .85.

In what units? Clearly this must be wrong since velocity is not unitless.

 

[quickquestion]

In what units? Clearly this must be wrong since velocity is not unitless.

I wouldn't say it "must be wrong", just could be more right. Unless, it is wrong for some reason besides that one.
Let's see, the carmass is KG, and the grav accel is 9.8m/s, so if we plug it in, our max Fforce from the front tires is carmass*gravaccel/2*1.7.
And force=mass*acceleration, so (in a simulation of one frame duration) the force needed to stop the car instantly would be force=carmass*carvelocity. Thus carmass*velocity cannot exceed carmass*gravaccel/2*1.7.
And we can cancel out carmass since it is on both sides of the equation. so velocity cannot exceed gravaccel/2. Thus our units are m/s. So our car cannot go faster than gravaccel/2*1.7. (8.33 meters per second.)
And the type of units are kg and meters per second.
So yeah, the equation was absolutely wrong, but for a different reason than was listed (the actual reason was because I forgot to put in gravaccel).
And no, me being wrong on this will not help me with my carphysics, because I used gravaccel in my car physics, just forgot to put it in the first post.

 

[jbriggs444]

And force=mass*acceleration, so (in a simulation of one frame duration) the force needed to stop the car instantly would be force=carmass*carvelocity.

The force needed to stop the car instantly would be infinite.

Force is equal to the rate of change of momentum. $F=\frac{dp}{dt}$

If you have a non-zero change of momentum and a zero interval, that quotient is infinite (actually it's division by zero).

 

[quickquestion]

The force needed to stop the car instantly would be infinite.

Force is equal to the rate of change of momentum. $F=\frac{dp}{dt}$

If you have a non-zero change of momentum and a zero interval, that quotient is infinite (actually it's division by zero).

true, but my latest post says this.
(in a simulation of one frame duration)

 

[jbriggs444]

true, but my latest post says this.
(in a simulation of one frame duration)

Could you slow down and explain what you are talking about? None of this makes any sense.

 

[quickquestion]

Could you slow down and explain what you are talking about? None of this makes any sense.

Sometimes it feels that way to me as well.

Basically, if you are in a car simulation (and you want to follow the rules of the Newtonian model), and you turn the steering wheel 90 degrees in one frame (the physics engine updates at the end of each frame) if you car has a velocity (not accel, but a velocity) higher than 8.33 meters per second, the tires will not make the car stop instantly, but instead slip.

 

[jbriggs444]

Sometimes it feels that way to me as well.

Basically, if you are in a car simulation (and you want to follow the rules of the Newtonian model), and you turn the steering wheel 90 degrees in one frame (the physics engine updates at the end of each frame) if you car has a velocity (not accel, but a velocity) higher than 8.33 meters per second, the tires will not make the car stop instantly, but instead slip.

Who said that the model updates one frame per second?

 

[Windadct]

Isn't this a vector problem - the portion of the direction of travel relative to the rolling direction and the portion normal to it, let's call this the Scrub. Depending on the tire composition and the surface you will have different results regarding tire and system distortion and finally breakaway and skidding/scrub. -- Also - as a real world question for the group, is a skidding tire really sliding friction? I have always considered more of a tearing, material issue. (i.e. the surface area does matter).

 

[quickquestion]

Who said that the model updates one frame per second?

Not me, my simulations run at 30 or 60 frames per second.

8.33 meters per second is based on the Pixels To Meters conversion in the physics engine.

 

[rcgldr]

rear-wheel drives with toe-in

This is combined with caster, resulting in a slight outwards steering torque on each front tire. If the car starts to wander into a left or right turn, then the "outer" front tire will experience more outwards steering torque than the "inner" tire, causing the steering to self correct (although after a correction, the car's direction has slightly changed).

For rear wheel drive, the rear tires also have a bit of toe in. If a car starts to wander and turn, the "outside" tire gets additional downforce and would tend to steer the rear end inwards to correct the turn, due to interaction between slip angle and downforce. For front wheel drive, with a bit of toe out, the "outside" tire would tend to steer the front end outwards to correct the turn.