# Role of rear wheels in steering

## Main Question or Discussion Point

Rear wheels' rigid positioning parallel to the longitudinal axis of the vehicle somehow makes the vehicle (two as well as a four wheeler) go around a curve when steered. If the rear wheel is hinged to the frame making deflection about the vertical axis possible (Like the wheels in TV stands) then the vehicle would deviate from the straight line path only initially and continue along that direction. How does the positioning of rear wheels accomplish this?

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Assume that both the front and rear wheel are purely rolling, no lateral sliding. Then the direction of the wheel establishes the direction vector for that point on the system. If the front and rear wheel are not aligned (because either has been steered), then the direction of motion at the front is different from that at the rear. This defines motion on a circular arc, such that these two motions are tangent to that arc. There is a well defined center of curvature and a radius of curvature for this arc. Motion in a straight line is simply the limiting case where the center of curvature moves to infinity and the radius of curvature becomes infinite.

I'm afraid i didn't get the hold of what you said. It would be great if you explain it in terms of forces. In case of a square rear wheel(weird, but just hypothetically) , when the front wheels are steered, the net frictional force act in the direction perpendicular to the plane of the wheel (the one parallel to it go to zero after altering the angular momentum) and as soon as the COM's velocity vector matches the direction of the wheel, forces cease to act (Even though the steering wheel is held at a position different from the neutral position)

But in case of a circular rear wheel, things are different. Forces (centripetal) continue to remain at a constant position of the steering wheel different from the neutral position.

Steering is a purely geometrical effect and does not require any forces at all, provided only massless objects are steered. You complicate it by wanting to think in terms of forces when it is basically a kinematic problem.

Steering is a purely geometrical effect and does not require any forces at all, provided only massless objects are steered. You complicate it by wanting to think in terms of forces when it is basically a kinematic problem.
Is this a joke? Fit 'massless' square wheels instead of circular ones and see whether your 'massless' honda can be steered.

Why would it be a joke? You are the one who made the foolish proposal to talk about square wheels, not I. I think we are done.

I did not 'propose' to use a square wheel in an actual design. I used it to see what would happen in that case. Is the first law, a foolish proposition to talk about "no forces" acting on a body.??

rcgldr
Homework Helper
Going by the title of this thread, role of rear wheels in steering, in the normal case, it's because they are pointed straight ahead while the front wheels are yawed inwards. If you extended the axis of the front and rear tires with imaginary lines, the point at where those lines cross would be the center of the circle that the car would tend to follow.

Once the car starts turning, you have an inwards centripetal force from the pavement to the tires, coexisting with an outwards (centrifugal) reaction force (due to centripetal acceleration times mass of car) from the tires to the pavement. This cause the contact patches to deform and yaw outwards. For all 4 tire surfaces, the actual path is a bit to the outside of the direction the tire is pointed, called slip angle. Since the rear tires are pointed straight ahead relative to the car, the car ends up pointed a bit inwards from the actual path the cars is taking.

For racing cars, the expected maximum average of the slip angle at front and rear is called working slip angle, and is related to stiffness of the tires. The stiffest tires are used on Indy Racing League cars on oval tracks, with a working slip angle around 2 degrees. Formula 1 cars are around 3 degrees, and most other cars are higher. Bias ply (not radials) racing tires have the highest slip angles, up to 8 or so degrees for the tires used in the Formula 1 cars of 1967 (last year before downforce was used), a bit less for modern bias ply racing tires. (Bias ply racing tires have about the same grip when sliding as when not "sliding", when near the limits, making them easier to control).

Steering is a purely geometrical effect and does not require any forces at all, provided only massless objects are steered. You complicate it by wanting to think in terms of forces when it is basically a kinematic problem.
Further to the post above:

When turning the wheel the contact patch distorts (which gives the slip angle described), the angle between the direction of travel and direction of contact patch describes a lateral force that causes the car to turn.

Its the force that causes the turn not the actual amount of slip angle, as in the post above different tyres are desigend to give max cornering at different levels of slip. The variation between the relative slip angles between the front at rear tyres determines over/understeer.

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Going by the title of this thread, role of rear wheels in steering, in the normal case, it's because they are pointed straight ahead while the front wheels are yawed inwards. If you extended the axis of the front and rear tires with imaginary lines, the point at where those lines cross would be the center of the circle that the car would tend to follow.
Yes. The positioning and nature of the rear wheels is the key for a circular turn of the vehicle. That's why, i, for a thought experiment, included square rear wheels. In this hypothetical case, when we turn the steering wheel, the vehicle takes an instant turn of, say an angle 'theta' from the straight line path, and continues along with it- the square rear wheels just dragged along. (see the picture i attached in my previous post).
The explanation i got in this thread is that the turn center lies on the intersection point of the axis of the rear and front wheels for pure rolling condition. Why is this so? Anyway i don't think this is an explanation for the change of scenario created by introducing circular rear wheels. What exactly happens then? I think this is intimately related to why the vehicle also rotates about the COM at the same rate it revolves about the turn center. (like the moon )

no replies?

russ_watters
Mentor
Sorry, but post #10 doesn't seem to add anything to the conversation, which is probably why you aren't getting any replies. It seems pointless to me.

Sorry, but post #10 doesn't seem to add anything to the conversation, which is probably why you aren't getting any replies. It seems pointless to me.
Not exactly pointless. It says what i gathered from the thread- that it has taken me nowhere near the answer-. It also has an additional point as regards the rotation of the entire vehicle about it's COM.

Rotation comes from the differnce in slip angle between front and rear wheels. No slip angle at rear (sliding frictionless blocks) means no lateral thrust, you'll get translation but no rotation. It the car will literally go sideways.

NOTE: This is a highly simplified view, because in reality you would get some rotation as the thrust at the front wheels will create a moment about the centre of the car.

Hence the thing about over and understeer when the relative slip angle is larger or smaller at rear.

The original question is quite ill defined and semi-confusing to be honest with you. One thing that makes these type of discussions very difficult is that you assert something, or talk about something hypothetical and then ask why its true. Then when you dont get an answer that mathes what you think already you have a go at people.

if a hypothetical scenario is true.
why does a car do what it does.

As they are easy to answer and we can go from there. Asserting something and then attempting to decribe the geometry over the interweb by text always leads to disaster.

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Yes. The positioning and nature of the rear wheels is the key for a circular turn of the vehicle. That's why, i, for a thought experiment, included square rear wheels. In this hypothetical case, when we turn the steering wheel, the vehicle takes an instant turn of, say an angle 'theta' from the straight line path, and continues along with it- the square rear wheels just dragged along. (see the picture i attached in my previous post).
The explanation i got in this thread is that the turn center lies on the intersection point of the axis of the rear and front wheels for pure rolling condition. Why is this so? Anyway i don't think this is an explanation for the change of scenario created by introducing circular rear wheels. What exactly happens then? I think this is intimately related to why the vehicle also rotates about the COM at the same rate it revolves about the turn center. (like the moon )
Read up on roll centre. That is not the same as centre of mass.

I honestly find the OP's questions pretty confusing - what is the point of these hypothetical "square wheels"? Is it just to imagine the condition where the rear wheels slide on the ground with no preferential direction, and if so, is it the same as just having the brakes locked?

As for the rest, I don't really follow your reasoning for the forces involved. Are you familiar with a http://en.wikipedia.org/wiki/Tractrix" [Broken]? It might apply to your locked rear wheel scenario.

Otherwise, I agree with the earlier statement that basic steering is purely geometric in nature. You can get a good approximation of the direction of a vehicle knowing only the steering angle (note the relatively small values of typical slip angles compared to angles to which a wheel can be steered).

I'll also mention as a matter of interest that the assertion in the OP that rear wheel must be rigidly fixed in order to steer a car is contradicted by cars with four-wheel steering, such as the 4th generation Honda Prelude made in the mid-90's. It had rear wheels that steered in the same direction as the front wheel for small steering angles (such as for high-speed lane changes), but in the opposite direction for large angles, such as for parallel parking. It was a successful system, but ultimately it was dropped because the benefits were outweighed by the costs of the additional complexity of the linkage, etc.; the point is just that it worked.

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... I think this is intimately related to why the vehicle also rotates about the COM at the same rate it revolves about the turn center. (like the moon )
Read up on roll centre. That is not the same as centre of mass.
I think he meant that that car rotates about its COM at the same angular rate that it (and its COM) move about the turn center. The is true only if there is no understeer or oversteer, however.

Ahh my bad :(. I'm having trouble deciphering exactly what he means.

rcgldr
Homework Helper
Hence the thing about over and understeer when the relative slip angle is larger or smaller at rear.
Oversteer and understeer are more commonly used to refer to slippage than slip angles. Oversteer when a car tends to yaw inwards during a turn, understeer when it tends to yaw outwards.

The slip angles are dependent on the side loads on the tires, which are affected by mass distrubition between front and back on a car. A tail heavy car will have higher slip angles at the rear than a front heavy car. The slip angles are also related to tire stiffness, but to prevent yaw oscillation the stiffness between fron and rear are kept nearly equal.

Admittedly it's been a while since i've read it but the slip angle is the explination given in Milliken & Milliken - Race Car Vehichle Dynamics. (im 99.8% sure of this)

The tendancy to yaw one way or the other is a product of the relative (ratio of) slip angles.

Rajamani - Vehichle Dynamics and Control- Top of Pg 63

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rcgldr
Homework Helper
The tendancy to yaw one way or the other is a product of the relative (ratio of) slip angles.
Depends on what relative slip angles means. The front tires are almost always turned inwards during a turn so they always have a higher absolute slip angle, which is why rear wheel drive makes sense in a race car, because the grip left over from a lower slip angle can be used to propel the car.

For the slip angles to be the same, once in a turn at the limits, the steering wheel would be pointed straight ahead (I've done this with setups in racing games, but it's not a good setup).

If the tires are either end are actually sliding or at least "drifting", then the "slip angle" is a slip and during the slide the car would be understeering or oversteering.

BobG
Homework Helper
I'll also mention as a matter of interest that the assertion in the OP that rear wheel must be rigidly fixed in order to steer a car is contradicted by cars with four-wheel steering, such as the 4th generation Honda Prelude made in the mid-90's. It had rear wheels that steered in the same direction as the front wheel for small steering angles (such as for high-speed lane changes), but in the opposite direction for large angles, such as for parallel parking. It was a successful system, but ultimately it was dropped because the benefits were outweighed by the costs of the additional complexity of the linkage, etc.; the point is just that it worked.

However, while the rear wheels may not be rigidly fixed in 4-wheel steering, they are controlled. Like was mentioned in other posts, the intersection between the axes of the wheels establishes the turning radius. If you angle the rear wheels in addition to the front wheels, you can move that intersection closer to the car and reduce the turning radius.

Letting the rear wheels run free is where you run into problems. The rear wheels tend to keep rolling straight ahead, causing the rear and front to whip around. Carts with all four wheels free to pivot can be a pain to use if you want to move the cart anywhere fast.

Only allowing the front wheels to turn is just the simplest way to control the turn, with simple control often being better than the shortest turning radius. As soon as the front wheels change direction, the entire frame changes direction, including the rigid rear wheels.

In other words, the rigid rear wheels aren't the important part - controlling the angle of all of the wheels is the important part.

First, Thanks a lot for these inputs. I should think about this with the help of these posts to reach a clear picture in my mind. But i'll address a few posts with the thoughts i presently have, some of which are bound to change while i think more about this.
By Chris: One thing that makes these type of discussions very difficult is that you assert something, or talk about something hypothetical and then ask why its true.
No.. I intended to make it clear by doing this. What i've understood till now-what i need to know. I know that if we have a 'square rear wheel' or a locked up one as Belliott suggested, then the path, the vehicle follows, is the one i gave in that picture. This i know. (Now after reading through these posts, i think, i can generalize this as, "if the steering angles of both the front and rear wheels are the same, then the vehicle will momentarily deflect from the original path and move on"- same as the 'square wheel' case.)
What exactly happens when the rear wheels are not locked, to make the vehicle traverse an entirely different path- a curve!.
This was the pattern of the first post.

By Belliott: I think he meant that that car rotates about its COM at the same angular rate that it (and its COM) move about the turn center. The is true only if there is no understeer or oversteer, however.
Ya.. This is still an amazing thing for me. What i think about this is that the frictional forces perpendicular to the planes of the front and rear wheels exert a torque about the COM such that the longitudinal axis of the vehicle always remain tangential to the curve the vehicle traces.

@Jeff and Chris
Oh oh! No slip angles and stuff. Let's restrict ourselves to rigid body dynamics. My peanut brain can't handle this!

@BobG
Your post really helped me a lot.
Like was mentioned in other posts, the intersection between the axes of the wheels establishes the turning radius. If you angle the rear wheels in addition to the front wheels, you can move that intersection closer to the car and reduce the turning radius.
Why should the turn center lie on the intersection? I have read this in several books and in this thread especially. But why?

I know you intended to make things more clear, next time maybe you could draw a diagram with the axis labelled or something. It's nothing you've said or described badly, its just text descriptions are sometimes inadequate, especially to idiots like me who dont tend to read them thoroughly enough.

I've reread it now and doodled a few diagrams of each situation. I'll deal with this last post first.

Why should the turn center lie on the intersection? I have read this in several books and in this thread especially. But why?
This is purely a geometry thing. To turn and follow a circular path you much be turning about a central point. The point of a circle can be found by drawing two lines normal to the curvature of the circle, where they meet is the centre.

Further reading on http://www.carbibles.com/steering_bible.html (jus the top bit is really relevent)

What i think about this is that the frictional forces perpendicular to the planes of the front and rear wheels exert a torque about the COM such that the longitudinal axis of the vehicle always remain tangential to the curve the vehicle traces.
This is sort of the right thinking. Its not friction force however, the force you attribute to friction is the 'side force'.

The side force comes from the slip angle (sorry, i cant help it =, this is where the force comes from :P), the forllowing description isnt slip angle but will suffice for what we want to discuss.

All this basically is, is the direction the car wants to travel in (stright forward) and the direction the type is pointing in. There is an angle between the two. The line that connects these two is the side force.

As you say this force acts perpendicular to the wheel travel and creates a moment (not torque) around centre of the car. Causing yaw. So long as th is yaw rate is the same as the curvature you move tangential to the circle path.

rcgldr
Homework Helper
The longitudinal axis of the vehicle always remain tangential to the curve the vehicle traces.
The front tires follow a larger radius curve than the rear tires, and if you pick the average radius, then ignoring slip angle, the car points a bit outwards of the tangent, but the rate of yaw follows the rate of change in turn direction which I think is your point, but this would have to happen unless the tire contact patches had lateral (perpendicular) movement with respect to the pavement.