How to prove a car must turn in a curve?

[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!

 

[CWatters]

http://auto.howstuffworks.com/steering1.htm

Turning the Car

You might be surprised to learn that when you turn your car, your front wheels are not pointing in the same direction.

Continues...

 

[hihiip201]

http://auto.howstuffworks.com/steering1.htm

Oh my gosh, so you are telling my i was stresssing about a false example/scenraio the whole night?

 

[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?

 

[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.

 

[hihiip201]

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.

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?

 

[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.

 

[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).

 

[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?

 

[sophiecentaur]

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?

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.

 

[sophiecentaur]

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).

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...

 

[hihiip201]

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...

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.

 

[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.

 

[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.

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.

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.

 

[CWatters]

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..

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

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

 

[rcgldr]

... if it was turning than the velocity of the outer wheel would be moving faster ...

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.

 

[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!

 

[rcgldr]

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!

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

 

[sophiecentaur]

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

Yebbut you pay loads extra for those!

 

[hihiip201]

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.

Sorry about the confusion:

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 velocitysince the back wheels are driving wheel, I am 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.

 

[sophiecentaur]

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.

This is simply not a feasible explanation. A car with just one front and one back wheel (kept upright with castors / stabilisers) would never turn a corner by that argument. You are confusing cause and effect. The steering wheel (the direction controller) has no direct connection with the distribution of the drive between the two back wheels. It controls the angle of the front wheel(s) and that's what determines the radius of the turn.
You might also remember 1. that half the cars in the world happen to have front wheel drive and 2. that they still go round corners when there is no motive force.

I have no idea what this means, I'm afraid.

Also, not allowing the wheels to slip in your model is missing the major part of the way the car steers. You may as well propose that the steering is achieved by curved rails - which is relatively easy to analyse.

 

[xxChrisxx]

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)

The reason why you are having such problems is that you are eliminating the mechanism by which cars turn at anything above a crawling pace. Cars turn by slipping, by assuming no slip you have created a situation that is irreverent to real world vehicles.

Also only consider a front wheel drive case, it's simpler as you can ignore the rear wheels as they are free to spin at whatever speed they need to.

 

[hihiip201]

This is simply not a feasible explanation. A car with just one front and one back wheel (kept upright with castors / stabilisers) would never turn a corner by that argument. You are confusing cause and effect. The steering wheel (the direction controller) has no direct connection with the distribution of the drive between the two back wheels. It controls the angle of the front wheel(s) and that's what determines the radius of the turn.
You might also remember 1. that half the cars in the world happen to have front wheel drive and 2. that they still go round corners when there is no motive force.

I have no idea what this means, I'm afraid.

Also, not allowing the wheels to slip in your model is missing the major part of the way the car steers. You may as well propose that the steering is achieved by curved rails - which is relatively easy to analyse.

Ok...I guess I am very confused the...

so how should I go about understanding the rotation of the car? what is the cause and what is the effect here? I am very confused...

 

[hihiip201]

The reason why you are having such problems is that you are eliminating the mechanism by which cars turn at anything above a crawling pace. Cars turn by slipping, by assuming no slip you have created a situation that is irreverent to real world vehicles.

Also only consider a front wheel drive case, it's simpler as you can ignore the rear wheels as they are free to spin at whatever speed they need to.

what do you mean by "crawling"? I'm confused as in, if wheels can slip, then why can't the front wheels just move at the same angle so that backwheels "slip" and the car turn translationally without rotating? if not, what are the restriction on this slipping of the wheels?


thanks!

 

[sophiecentaur]

Ok...I guess I am very confused the...

so how should I go about understanding the rotation of the car? what is the cause and what is the effect here? I am very confused...

The cause is the 'inwards' force, approximately along the axis of the wheels - in particular, the front wheel, which initiates the turn because it is at an angle to the instantaneous direction of travel. The back wheel is pointing, roughly, along a tangent to the curve. (I assume that you don't actually mean 'rotation' but 'motion in a curve'?)

I'm sorry but some of your choices of words make it very difficult to understand what you mean (I also found your diagram a bit confusing, I'm afraid.)

I just looked in an old book of mine on sports cars by Colin Chapman - designer of early Lotus cars. It reminded me of the term 'Ackerman Steering' Google this term and you will find loads of stuff to help you.

 

[xxChrisxx]

what do you mean by "crawling"? I'm confused as in, if wheels can slip, then why can't the front wheels just move at the same angle so that backwheels "slip" and the car turn translationally without rotating? if not, what are the restriction on this slipping of the wheels?thanks!

Seriously you need to read up on slip angles. As at higher speeds, the car both translates and rotates. So much so that at the limit of grip on a balanced car, all four wheels are actually drifting and the car is purely translating round a corner.


0:07-0:10


At very very low speeds, the car turns in the direction that the wheels are pointed. So your turn centre is in line with the rear wheels. Ie, draw a line perpendicular through the rear wheels and perpendicular to the front wheels.

Unless you have a geometry that allows the inner and outer wheels to take different paths (because the inner wheel is taking a shorter radius) you will scrub the tyres.
Reference: Ackermann steering angle

At higher speeds the tyre tread distorts meaning the tyre is traveling in a different direction than the contract patch. The angle between these two is the slip angle. This slip angle induces a cornering force which acts to turn the car into the turn.

This is the crucial thing. The lateral force caused by the turn induces a slip angle in the rear wheels. So the rear wheels are actually contributing to the turn. The centre of the turn (instantaneous centre) is now perpendicular to the slip angles front and rear.

When the front slip angle exeeds the rear, the car will understeer.
When the rear slip angle exeeds the front, the car will oversteer.

In your scenario, the car will permanently and terminally understeer as you have no slip angle on the rear.


I've bolded several words/concepts above. You should try to read up on them before approaching this again. If you have any questions on them, just ask.

 

[rcgldr]

ackermann steeing

Wiki article:

wiki_ackermann_steering

slip angle

Because of deformation, and some slippage at the contact patch, the actual radius will be somewhat larger than idealized radius. In race cars, because the downforce on the tires is greater on the outside tires, ackermann is reduced or eliminated and toe in may be used, so that the inside front tire has less slip angle than the outside front tire.

Even in the idealized case of no tire slipping and perfect ackermann steering, you still have some of the turning force provided by the rear tires, due to the centripetal acceleration of the car resulting in a reactive outwards force on all 4 tires, with an equal but opposing inwards (centripetal) force from the tires (a Newton third law pair of forces, you also have outwards force of tires on pavement, and equal and opposing inwards force of pavement on tires). The ratio of force between front and rear depend on weight distribution of the car (assuming constant speed). Image of an idealized example of a car turning, no tire slipping, and perfect ackerman steering setup:

 

[CWatters]

I have another drawing that shows how all four wheels can be tangents to the curve they are following if the front two wheels are at different angles...

 

[sophiecentaur]

I have another drawing that shows how all four wheels can be tangents to the curve they are following if the front two wheels are at different angles...

I imagine that the steering linkage can be made to do almost anything you want. It must be a matter of what makes vehicles corner best at speed.

PS No one has yet mentioned the technique of additional rear (four) wheel steering which has been used for high end cars.

 

[hihiip201]

The cause is the 'inwards' force, approximately along the axis of the wheels - in particular, the front wheel, which initiates the turn because it is at an angle to the instantaneous direction of travel. The back wheel is pointing, roughly, along a tangent to the curve. (I assume that you don't actually mean 'rotation' but 'motion in a curve'?)

I'm sorry but some of your choices of words make it very difficult to understand what you mean (I also found your diagram a bit confusing, I'm afraid.)

I just looked in an old book of mine on sports cars by Colin Chapman - designer of early Lotus cars. It reminded me of the term 'Ackerman Steering' Google this term and you will find loads of stuff to help you.

And are these "inward" forces reaction forces normal to the planes of the front wheels caused by the "forward" forces on front wheels by the back wheels?

 

[CWatters]

The cars inertia will try to make it go in a straight line (a tangent to the curve). The front wheels point in a different direction so they provide a sideways force on the front of the car.

In effect inertia is trying to drag the froont wheels sideways and it's friction that provides the sideways force making the car turn. On ice you go straight on!

The same applies to an un powered soap box cart so makes no difference if the car is front or rear wheel drive.

 

[sophiecentaur]

And are these "inward" forces reaction forces normal to the planes of the front wheels caused by the "forward" forces on front wheels by the back wheels?

That would imply it would only go around a bend if it were driven! If the wheels are not being braked, then there are no tangential forces on any of the wheels. Have you looked at all diagrams, read the google hits and tried to understand what they are trying to tell you? I have a feeling that you are hanging on to some idea that is hindering you from actually getting this. Try stepping back and starting again with this problem.

 

[hihiip201]

That would imply it would only go around a bend if it were driven! If the wheels are not being braked, then there are no tangential forces on any of the wheels. Have you looked at all diagrams, read the google hits and tried to understand what they are trying to tell you? I have a feeling that you are hanging on to some idea that is hindering you from actually getting this. Try stepping back and starting again with this problem.

No I haven't yet, and I am going to just discard al preassumptions I have at this moment and read them...guess that's my best bet.I thought I knew, but I don't, not one bit, my arrogance has cost me 2 days of mental torment. And it is time I liberate myself , I will not ask anymore questions until I have squeeze every last bit of all the resources made available to me by you all good folks.