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Role of rear wheels in steering

  1. May 11, 2009 #1
    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?
  2. jcsd
  3. May 11, 2009 #2
    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.
  4. May 11, 2009 #3
    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)
    square rear wheel.jpg

    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.
  5. May 11, 2009 #4
    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.
  6. May 11, 2009 #5
    Is this a joke? Fit 'massless' square wheels instead of circular ones and see whether your 'massless' honda can be steered.
  7. May 11, 2009 #6
    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.
  8. May 11, 2009 #7
    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.??
  9. May 12, 2009 #8


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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.

    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).
  10. May 12, 2009 #9
    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.
    Last edited: May 12, 2009
  11. May 18, 2009 #10
    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 :smile: )
  12. May 20, 2009 #11
    no replies? :frown:
  13. May 20, 2009 #12


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    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.
  14. May 20, 2009 #13
    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.
  15. May 20, 2009 #14
    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.

    Either ask:
    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.
    Last edited: May 20, 2009
  16. May 20, 2009 #15
    Read up on roll centre. That is not the same as centre of mass.
  17. May 20, 2009 #16
    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.
    Last edited by a moderator: May 4, 2017
  18. May 20, 2009 #17
    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.
  19. May 20, 2009 #18
    Ahh my bad :(. I'm having trouble deciphering exactly what he means.
  20. May 20, 2009 #19


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    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.
  21. May 20, 2009 #20
    Last edited: May 20, 2009
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