Neutral Steer Responses, Stability Factor K, Neutral Steer Point and Static Margin
Neutral Steer Responses
For neutral steer, the static directional stability \( N_\beta (= a C_F \; – b C_R) = 0\).
As a result \( Q = -Y_\beta N_r \).
The simplified derivatives for steady-state responses of the neutral steer car are
| Response to steer angle \(\delta\) (Control) | Response to applied side force \(F_y\) at c.g. | Response to applied yaw moment \(N\) |
| \[ \dfrac{1 / R}{\delta} = \; – \dfrac{1}{V} \dfrac{N_\delta}{N_r} \tag{5.26} \] | \[ \dfrac{1 / R}{F_y} = 0 \tag{5.30} \] | \[ \dfrac{1/R}{N} = – \dfrac{1}{VN_r} \tag{5.34} \] |
| \[ \dfrac{r}{\delta} = \;- \dfrac{N_\delta}{N_r} \tag{5.27} \] | \[ \dfrac{r}{F_y} = 0 \tag{5.31} \] | \[ \dfrac{r}{N} = -\dfrac{1}{N_r} \tag{5.35} \] |
| \[ \dfrac{V^2/R}{\delta} = \; – V \dfrac{N_\delta}{N_r} \tag{5.28} \] | \[ \dfrac{V^2/R}{F_y} = 0 \tag{5.32} \] | \[ \dfrac{V^2/R}{N} = – \dfrac{V}{N_r} \tag{5.36} \] |
| \[ \dfrac{\beta}{\delta} = \; – \dfrac{Y_\delta}{Y_\beta} + \dfrac{1}{Y_\beta} \dfrac{N_\delta}{N_r} (Y_r – mV) \tag{5.29} \] | \[ \dfrac{\beta}{F_y} = – \dfrac{1}{Y_\beta} \tag{5.33}\] | \[ \dfrac{\beta}{N} = \dfrac{Y_r – mV}{Y_\beta N_r} \tag{5.37} \] |
Ackermann steer angle
From the yaw velocity response to a steer angle (5.27), and using \( N_\beta = a C_F \; – b C_R = 0 \) \[ \dfrac{r}{\delta} = \; – \dfrac{N_\delta}{N_r} = \; \dfrac{aC_F}{\dfrac{1}{V} (a^2C_F + b^2 C_R)} = \; \dfrac{VC_F}{aC_F + (\dfrac{b}{a})bC_R} =\dfrac{V}{l} \tag{5.39} \] i.e. \[ \delta = \dfrac{l}{V/r} = \dfrac{l}{R} \tag{5.40} \] This is the Ackermann steer angle.
From equation (5.27), \( \delta N_\delta = -rN_r \), and in a steady cornering, the control moment \( \delta N_\delta\) produced by the steering input balances the yaw damping moment \( r N_r \).
Therefore, the Ackermann steering angle \( \delta_{Acker} = l/R \) is the angle required to balance out the yaw damping moment.
( \(N_\beta = 0 \) for the neutral steer car.)
Stability Factor K for Understeer/Oversteer Vehicles
For a general understeer/oversteer vehicles, the curvature response to a steer angle is \[ \begin{align} \dfrac{1 / R}{\delta} &= \dfrac{Y_\beta N_\delta-N_\beta Y_\delta}{VQ} \\ & = \dfrac{Y_\beta N_\delta-N_\beta Y_\delta}{V(N_\beta Y_r \; – N_\beta mV \; – Y_\beta N_r)} \\ &= \dfrac{1}{ \dfrac{V(N_\beta Y_r – Y_\beta N_r)}{Y_\beta N_\delta – N_\beta Y_\delta} – \dfrac{N_\beta mV^2}{Y_\beta N_\delta – N_\beta Y_\delta} } \\ &= \dfrac{1}{l \; – \dfrac{N_\beta mV^2}{Y_\beta N_\delta – N_\beta Y_\delta} } \qquad \text{(by substituting in the derivatives)} \end{align} \] Finally \[ \dfrac{1 / R}{\delta} = \dfrac{1/l}{1+KV^2} \tag{5.41} \] where the stability factor \[ \begin{align} K &= \dfrac{m N_\beta}{l (N_\beta Y_\delta – Y_\beta N_\delta)} &= \dfrac{m}{l} \dfrac{- \dfrac{N_\beta}{Y_\beta} }{ N_\delta \; – \dfrac{N_\beta}{Y_\beta} Y_\delta} \tag{5.42} \end{align} \]
The stability factor \( K\) takes the same sign as \(N_\beta = aC_F -b C_R\) (The cornering stiffness \(C_F\) and \(C_R\) are both negative).
If K > 0, the vehicle understeers. If K = 0, the vehicle is neutral steer. If K < 0, the vehicle oversteers.
With the simple bicycle model, the stability factor \(K\) is independent of the forward velocity \(V\).
The path curvature response to steering \(\dfrac{1/R}{\delta} \) and the yaw velocity response to steering \( \dfrac{r}{\delta} \) for a single value of \(K\) are illustrated below.
Steer angle required to cancel the lateral force \(Y_0\) in terms of the stability factor \( K\)
Consider the side force \( Y_0\) applied at the c.g. and the steer input is made, so that the yaw velocity \(r=0\).
The equations of motion (5.10) in steady-state ( \(\dot{\beta} = \dot{r} = 0\) ) are
\[
\left\{
\begin{array}{l} \begin{align}
0 &= Y_0 + Y_\beta \beta + Y_\delta \delta \\
0 &= N_\beta \beta + N_\delta \delta
\end{align} \end{array}
\right.
\]
Then, from the equations, delete \( \beta \)
\[
\dfrac{\delta}{Y_0} = \dfrac{-N_\beta}{N_\beta Y_\delta – Y_\beta N_\delta}
\]
i.e.
\[
\dfrac{\delta}{Y_0} = \; – \dfrac{l}{m} K
\]
With the side force \(Y_0\) applied at c.g., the steer angle required to prevent the car from turning is proportional to the negative of the stability factor \(K\).
Neutral Steer Point
The Neutral Steer Point for the bicycle model is the point on the chassis at which an external lateral force can be applied without producing a steady-state yaw velocity.
Suppose no steering \( \delta = 0\) and an applied lateral force \(Y_0\) at the neutral steer point \(d\) from the front axle. This arrangement will produce no yaw velocity \(r= 0\).
From the equations of motion (5.10), in steady state ( \(\dot{\beta} = \dot{r} = 0\) ) \[ \left\{ \begin{array}{l} \begin{align} 0 &= Y_0 + Y_\beta \beta \\ 0 &= -(d-a)Y_0 + N_\beta \beta \end{align} \end{array} \right. \] Delete \(\beta\) to find \[ Y_0 \{ 1+ \dfrac{Y_\beta}{N_\beta}(d-a)\} = 0 \] i.e. \[ d = a \; – \dfrac{N_\beta}{Y_\beta} \] Normalising by the wheelbase \(l\) \[ \begin{align} \text{NSP} &= \dfrac{d}{l} \\ &= \dfrac{a}{l} \; – \dfrac{N_\beta}{Y_\beta} \dfrac{1}{l} \tag{5.45} \end{align} \] For the bicycle model, by substituting in the derivatives (5.9) \[ \text{NSP} = \dfrac{d}{l} = \dfrac{C_R}{C_F + C_R} \]
Static Margin
The static margin for the bicycle model is the distance between the neutral steer point and the c.g, normalised by the wheelbase.
\[
\begin{align}
\text{SM} &= \dfrac{d-a}{l} \tag{5.47}\\
&= \; – \dfrac{N_\beta}{Y_\beta} \dfrac{1}{l} \tag{5.48}
\end{align}
\]
For the bicycle model, by substituting in the derivatives (5.9)
\[
\text{SM} = \dfrac{-(a/l)C_F +(b/l)C_R}{C_F + C_R} \tag{5.48a}
\]
From (5.48), since \( Y_\beta = C_F + C_R \) is always negative, the static margin has the same sign as the \( N_\beta \).
If \(\text{SM} >0 \), the neutral steer point exist to the rear of c.g, and the vehicle understeers.
Olley Definition of Oversteer/Understeer
The earliest definition of oversteer/understeer by the path taken by a vehicle initially running on a straight path and subject to a side force \(F_y\) applied at c.g.
From the equations of motion (5.10) with no steering input \( \delta = 0 \)
\[
\left\{
\begin{array}{l}
mV(r + \dot{\beta}) &= Y_\beta \beta + Y_r r + F_y \\
I_z \dot{r} &= N_\beta \beta + N_r r
\end{array}
\right.
\]
In steady state \(\dot{\beta} = \dot{r} = 0\)
\[
\left\{
\begin{array}{l}
r(mV – Y_r) &= Y_\beta \beta + F_y \\
\beta &= \; – \dfrac{rN_r}{N_\beta}
\end{array}
\right.
\]
\(Y_r \) is negligible compared to \(mV\)
By deleting \(\beta \) and using \(V = rR\)
\[
\dfrac{1/R}{F_y \; – \dfrac{mV^2}{R}} = \dfrac{N_\beta}{VN_r Y_\beta} \tag{5.49a}
\]
\( \dfrac{mV^2}{R} \) is the centrifugal force, and \( F_y \; – \dfrac{mV^2}{R} \) is the total side force.
Since \( VN_r = a^2C_F + b^2 C_R < 0 \) and \( Y_\beta = C_F + C_R <0 \), the sign of the response is the same as the static directional stability \(N_\beta = aC_F - bC_R \).
If \(N_\beta > 0 \), the car understeers.
If \(N_\beta = 0 \), the car is neutral steer, and there is no change in the path.
If \(N_\beta < 0 \), the car oversteers.
Effective Slip Angles
From the turning geometry \[ \left\{ \begin{array}{l} \begin{align} \alpha_F &= \beta + \dfrac{ar}{V} – \delta \tag{5.4} \\ \alpha_R &= \beta \; – \dfrac{br}{V} \tag{5.3} \end{align} \end{array} \right. \] Then \[ -\alpha_F + \alpha_R = -(\beta + \dfrac{ar}{V} – \delta) + (\beta \; – \dfrac{br}{V}) \] i.e. \[ \begin{array}{l} \delta &= \dfrac{r(a+b)}{V} + (-\alpha_F + \alpha_R) \\ \delta &= \dfrac{l}{R} + (-\alpha_F + \alpha_R) \end{array} \tag{5.50} \] From the general curvature response in terms of the stability factor \(K\) (5.41) \[ \begin{array}{l} \delta = \dfrac{l}{R} + \dfrac{lKV^2}{R} \\ \text{where the understeer gradient} \qquad lK = \dfrac{mN_\beta}{N_\beta Y_\delta – Y_\beta N_\delta} \end{array} \] From a comparison to (5.50) \[ -\alpha_{Fe} + \alpha_{Re} = \dfrac{lKV^2}{R} \tag{5.50a} \] i.e., the stability factor \[ K = \dfrac{1}{l} \biggl(\dfrac{-\alpha_{Fe} + \alpha_{Re}}{V^2/R}\biggr) \tag{5.51} \] where \( \alpha_{Fe} \), \(\alpha_{Re}\) are the effective slip angles.
Bundorf Cornering Compliances \(D_F\), \( D_R\)
From (5.50a), the required adjustment to the steering angle at speed due to the understeer
\[
\begin{align}
\dfrac{lKV^2}{R} &= \; – \alpha_F + \alpha_R \\
&= -\dfrac{F_F}{C_F} + \dfrac{F_R}{C_R} \\
&= (-\dfrac{W_F}{C_F} + \dfrac{W_R}{C_R} ) \dfrac{a_y}{g} \\
&= (-\dfrac{W_F/C_F}{g} + \dfrac{W_R/C_R}{g} ) a_y \\
&= (D_F \; – D_R )\dfrac{\pi}{180} a_y
\end{align}
\]
where
\[
\left\{
\begin{array}{l}
\text{the front and rear tyre lateral force}\qquad &F_F = (b/l)F, \qquad F_R = (a/l)F \\
\text{the total lateral force}\qquad &F = F_F + F_R = Ma_y = M\dfrac{V^2}{R} \\
\text{the front and rear static weight}\qquad &W_F = (b/l)W, \qquad W_R = (a/l)W \\
\text{the total weight of the complete vehicle}\qquad &W = Mg
\end{array}
\right.
\]
\(D_F\) and \(D_R\) are the Bundorf cornering compliances at the front and rear in deg /g.
The contributions of various effects to \(D_F\) and \(D_R\), called the understeer budgets, can be summed with the linear system assumption. The totla understeer is \(D_F \; – D_R \).
Reference
[1] W. Milliken and D. Milliken, Race Car Vehicle Dynamics.