# How to prove a car must turn in a curve?

1. Jan 20, 2013

### hihiip201

Hi

I have been having a hard time visualizing how a car must turn in a curve

we know 2 things:

1. if the steering wheel is held at an angle, the front wheels must be at a certain angle relative to the car frame as well as the back wheel at all time.

2. the wheels cannot have any velocity component that is perpendicular to the direction of the tires or or the car would be slipping.

so assuming we have a car with front wheel inclined angle , initially at rest, how is its motion going to be like?

I imagine that just after t = 0, both wheel front and back would have to move toward there they were oriented at t=0 because as lim t goes to zero their instantaneous velocity must be along the tires(perpendicular to tire's normal) , but if that is the case then the backwheel would have violated (2)

so I thought, could it be that the outter wheel to whatever direction is moving faster? but then it would violate (1) as @ t= t+dt, the relative length between the two back wheel would then increased by a small amount.

thank you guys!

2. Jan 20, 2013

### CWatters

3. Jan 20, 2013

### hihiip201

4. Jan 20, 2013

### hihiip201

but still, i think the unanswered question for me is that, mathematically the back wheels both move in the vertical direction, but if it was turning than the velocity of the outter wheel would be moving faster, then at t = dt, wouldn't there be a slight distortion between the length of the back wheel?

Also, if the front wheels angle were fixed (like in op). what would happen? it is still possible for the car to turn?

Last edited: Jan 20, 2013
5. Jan 20, 2013

### 256bits

If you have a solid axle connecting both front wheels, and/or a solid axle connecting both rear wheels, then your vehicle would have trouble turning, since outer and inner wheels would be rotating at the same angular velocity.

Last edited: Jan 20, 2013
6. Jan 20, 2013

### hihiip201

I see, but it will still turn tho won't it? just slipping?

also, what of the initial velocity of the back wheels in the normal case?

7. Jan 20, 2013

### sophiecentaur

Front and back wheels will have a 'slip angle' - (the angle between the plane of the wheel and the line of its path). The pneumatic tyre distorts in contact with the road and tends 'crawl' up into the curve on the way round - the elliptical footprint has an axis that is even further pointed into the curve.
In order to make a car easier to control (by boy racers and their grannies alike) it is normal to arrange that the front wheels (the steered ones) have a slightly greater slip angle than the back wheels. This gives the car 'understeer' which makes it much less likely that the rear wheels will fly out of the curve and make you do a 180.
However, understeer means that cornering is not, in fact, as good and high performance cars do not have it built in. Hence you see can the drivers doing 'reverse lock' on the way round a corner to counteract the oversteer.

8. Jan 20, 2013

### hihiip201

I have always just pictured that:

the back wheel is pushing the front wheel forward, but the front wheel's motion is restricted by its orientation, so static friction that pushes the front wheel in their plane
s normal cause a motion that is to whatever direction they are turning.

so it seems that this is still true, but it is just that the front wheels directions are not the same which cleared a lot of my confusions.

But I still have some questions:

1. So pretty much tires velocity are different from their plane's normal ?

2. Even the back wheels don't have the same velocity(in both direction and magnitude) as lim t goes to zero? (t = zero is when car starts to move).

Last edited: Jan 21, 2013
9. Jan 21, 2013

### hihiip201

So I just read up wiki and it seems that the slip angle is a result of tire deformation, so assuming we have a rigid wheel that has an ideal contact point with the ground, will the wheel turns in the direction of which they are oriented?

10. Jan 21, 2013

### sophiecentaur

That will depend upon how simplified your model is. If you are talking 'train tracks' there will be no slip. I can't actually think of a satisfactory half way house model to describe what would happen with a rolling wheel at an angle to the direction of the car. Presumably it would have to involve no slip normal to the plane of the wheel and no slip in the plane either. That looks too simple to be worth considering. Any real wheel on a real surface will have some deformation and, hence some slip - imo.

11. Jan 21, 2013

### sophiecentaur

I think that with any form of tyre, it will not be simply static friction that is at work. This would be true except at zero speed, which is not very interesting because neither wheel would be turning.
I don't understand what you mean here.

Not the same as what? Do you mean not equal to each other? Toe in or toe out are not (first order) relevant to the basics of steering but because of weight distribution and the action of suspension etc...

12. Jan 21, 2013

### hihiip201

Yes I meant that they are not equal to each other.

I'm really just trying to think of a simplified model of the turning wheels, as well as trying to understand why it is designed that way.

sorry for all the questions so just let me make a well organized question:

first, math question
1. for a rigid body, not just cars or wheels, point A with r1 from instantaneous center of zero velocity , point B with r2 , where r1 > r2. meaning that at any time vA > vB.
if that is true, if we take t = 0 at some point in time where the body is rotating about its i.c.z.v.

@ t = dt, the displacement of A would be va times dt, and vb times dt for point B, we also know that va and vb @ t = 0 is along the same direction, meaning A would have a greater displacement than point B @ t = dt.

hence, wouldn't there be a infinitesimal alongation of the distance vector AB?

my guess is that this "difference" goes to zero as we take dt = 0.

question 2:

back to the car, i
I am aware that the front wheels angle are different from each other when rotating, assuming slip angle are zero, with rigid tires.

is it okay for me to think of the car turning design like this:

the driving force of turning is the difference in speed of the back wheels, one is greater than the other, and since the distance between the two back tires don't change, therefore the only possible motion for the back wheels is rotation + translation.

and the plane of the front wheels are just oriented normal to the point of rotation to "fit" the current rotating motion about rotation point O. in order to minimize resistance.

13. Jan 21, 2013

### sophiecentaur

I think I'd need a diagram and some more explanation of your model before I could understand what you mean.
Also, could you tell us where, exactly, you are planning to take this? Are you after a very simple equation to describe what goes on?
It might be easier to imagine just one wheel at the front and one at the back (magic stabilisers included, of course) to make the model easier to start with.

14. Jan 21, 2013

### rcgldr

If you extend the axis of the front and rear tires of a car with imaginary lines, the point at where these lines cross is the radius the car will tend to turn. The actual radius will be somewhat larger due to 'slip angle' (the deformation and some slippage at the contact patch) depending on the load on the tires and how stiff the tires are.

High downforce cars like Formula 1 cars have the wings set to produce understeer at high speed, to prevent snap oversteer when pulling 4 g turns. Non-downforce cars have to compromise on the setup to get the cars to turn well at moderate speed, and excessive oversteer at high speed. On some non-downforce race cars, an alternate method to 'reverse lock' is 'induced understeer' where the driver steers inwards excessively to increase slip angle at the front enough just enough to wash out the front end to match the rear end oversteer, or at least steer inwards just enough to get the car more stable in higher speed turns.

15. Jan 21, 2013

### CWatters

Not sure if I understand your question but have you heard of a differential?..

http://en.wikipedia.org/wiki/Differential_(mechanical_device [Broken])

Last edited by a moderator: May 6, 2017
16. Jan 21, 2013

### rcgldr

Even in the case of a go kart with a solid rear axle, there's enough flex in a tire's tread to allow it to deform at the contact patch, allowing the contact patch speed to be slightly slower or faster than the average surface speed of a tire. Also for high g turns, some go-kart frames are setup to allow the inner rear tire to lift a bit.

As already mentioned, cars use a differential that allow the tires to rotate at different speeds. A performance differential will limit the difference in speed to prevent a single tire from spinning during heavy acceleration.

17. Jan 21, 2013

### sophiecentaur

The differential is such a damn smart invention!! Not only does it divide the power between the wheels in just the right proportions but it also gears down the drive from the engine to a useful rotation rate.
Hopeless on an icy road, though. The last thing you want then ( I speak from this morning's experience) is one wheel gripping and the other one spinning wildly!

18. Jan 21, 2013

### rcgldr

That's an "open" differential. Some type of limited slip differential will eliminate this problem.

19. Jan 21, 2013

### sophiecentaur

Yebbut you pay loads extra for those!

20. Jan 21, 2013

### hihiip201

thank you so much for you guys help, I did not expected this much responds from such badly expressed question:

for the math:

http://tinypic.com/view.php?pic=ychgp&s=6
(I'm aware of the car differential btw)

for the car physics:

my goal here is to try to understand why cars turn the way they do, or why they are designed the way they are for turning, and I would like to know if the following is true:

each velocity are "designed" to move at a certain direction, velocity, such that they "resemble" a rotating rigid body. where direction of wheels are normal to their instantaneous center of velocity

since the back wheels are driving wheel, Im imaginging that when a car turns, it is DUE to the difference in speed between the left and right rear wheels, and we simply orient the front wheels at an angle so that their planes are normal to that center of rotation.

in other word , we have 2 restriction of motion :

1. rigid body (if we hold front wheels the same angle relative to the back at all time)

2. no slipping, the wheel's plane are oriented in the same direction of its motion (assume rigid wheels)

hence, we are simply "matching" these criteria together, since we know the front wheels position will rotate according to the difference in left and right rear wheels, we make their planes's normal intersect the center of rotation so that it will not slip when that rotation occur.

Last edited: Jan 21, 2013