02_Equations of Motion and Steady-State Responses

Equations of Motion and Steady-State Responses

Equations of motion

From Newton’s second law \[ \left\{ \begin{array}{l} \begin{align} \Sigma F &= ma \\ \Sigma T &= I\alpha \end{align} \end{array} \right. \] where \(\alpha\) is angular acceleration.
For the bicycle model the equations become \[ \left\{ \begin{array}{l} Y = Ma_y \\ N = I_z \dfrac{dr}{dt} = I_z \dot{r} \end{array} \right. \] where \(Y\) is the total tyre lateral force and \(N\) is the resultant yaw moment.\( I_z\) is the moment of inertia about z-axis. The yaw acceleration \(\dot{r}\) is the time derivative of the yaw velocity \(r\).

About the lateral acceleration \(a_y\) term

Uniform circular motion

The formulae for the uniform circular motion \[ \left\{ \begin{array}{l} a_y = \dfrac{V^2}{R} \\ V = R\omega \end{array} \right. \] From the equations, delete the turn radius \(R\) \[ a_y = V \omega \] In our notation, the yaw velocity \( \omega \) is \(r\). Therefore \[ a_y = V r \]

Lateral acceleration \(a_y\) term for the bicycle model

\[ \begin{align} a_y &= \text{(centripetal acceleration)} + \text{(direct lateral acceleration)} \\ &= Vr + \dot{v} \\ &= Vr + V\dot{\beta} \qquad \text{(V constant and} \; \beta \; \text{small angle (} \tan{\beta} \approx \beta \text{))} \\ &= V(r + \dot{\beta}) \end{align} \] In steady-state, the vehicle slip angle \( \beta \) is constant and \( \dot{\beta} = 0 \).

About the total lateral tyre force \(Y\) and resulting yaw moment \(N\)

When the vehicle experiences the forward velocity \(V\) and lateral velocity \(v\) at c.g., the tyres also experience them.
For a small angle \(\beta\), \( V \approx u\).

A positive yaw velocity \(r\) at c.g. produces lateral velocity \( -br \) experienced at the rear tyre. The rear slip angle is \[ \begin{align} \alpha_R &= \dfrac{v-br}{V} = \dfrac{v}{V} – \dfrac{br}{V} \\ &= \beta \; – \dfrac{br}{V} \end{align} \]

The front slip angle is \[ \begin{align} \alpha_F &= \dfrac{v+ar}{V} – \delta = \dfrac{v}{V} + \dfrac{ar}{V} – \delta \\ &= \beta \; + \dfrac{ar}{V} – \delta \end{align} \]

Therefore, the front and rear lateral tyre forces are \[ \left\{ \begin{array}{l} Y_F = C_F \alpha_F = C_F (\beta \; + \dfrac{ar}{V} – \delta) \\ Y_R = C_R \alpha_R = C_R (\beta \; – \dfrac{br}{V}) \end{array} \right. \]

Then, the total lateral tyre force \(Y\) and tyre yaw moment about z-axis through the c.g. \(N\) are \[ \left\{ \begin{array}{l} Y = Y_F + Y_R \\ N = N_F + N_R = Y_Fa \; – Y_Rb \end{array} \right. \] i.e. \[ \left\{ \begin{array}{l} Y = (C_F +C_R)\beta + \dfrac{1}{V}(aC_F -bC_R)r \; – C_F \delta \\ N = (aC_F \;- bC_R)\beta + \dfrac{1}{V}(a^2C_F +b^2C_R)r \; – aC_F \delta \end{array} \right. \]

Derivative notation – linear analysis

The principle of superposition

Linear systems are linear in terms of variables, and the systems can be described by linear differential equations. Suppose that the response of a linear system to the input \(u_1\), applied to the system is \(y_1\), and similarly, the input \(u_2\) produces the response \(y_2\).
Then, the response of the system to the input \( u(t) = \alpha u_1(t) + \beta u_2(t) \) is \[ y(t) = \alpha y_1(t) + \beta y_2(t) \]

Force and moment derivatives of the bicycle model

In the above expressions for the total lateral force \(Y\) and tyre yaw moment \( N \), \(\beta\), \(r\) and \(\delta\) are variables and others are constants.
By taking partial devrivatives \[ \left\{ \begin{array}{l} \begin{align} Y = f(\beta, r, \delta) &= \dfrac{\partial Y}{\partial \beta} \beta + \dfrac{\partial Y}{\partial r}r + \dfrac{\partial Y}{\partial \delta}\delta = (C_F + C_R)\beta + \dfrac{1}{V}(aC_F \; – bC_R) r + (-C_F) \delta \\ N = f(\beta, r, \delta) &= \dfrac{\partial N}{\partial \beta} \beta + \dfrac{\partial N}{\partial r}r + \dfrac{\partial N}{\partial \delta}\delta = (a C_F \; – b C_R)\beta + \dfrac{1}{V}(a^2C_F + b^2C_R) r + (-a C_F)\delta \end{align} \end{array} \right. \] i.e. \[ \left\{ \begin{array}{l} Y = Y_\beta \beta + Y_r r + Y_\delta \delta \\ N = N_\beta \beta + N_r r + N_\delta \delta \end{array} \right. \] where

Damping-in-Sideslip Derivative \(Y_\beta = C_F + C_R <0\)	Lateral Force/Yaw Coupling Derivative \(Y_r = \dfrac{1}{V}(aC_F – bC_R) \)	Control Force Derivative\(Y_\delta = -C_F >0 \)
Static Directional Stability Derivative\( N_\beta = a C_F – b C_R \)	Yaw Damping Derivative \(N_r = \dfrac{1}{V}(a^2C_F + b^2C_R) <0\)	Control Moment Derivative \( N_\delta = -a C_F >0 \)

Remarks on the derivatives

Control moment derivative \( N_\delta = -a C_F >0 \)

Describes the relationship between the tyre induced yaw moment \(N_\delta \delta\) and the steering input \(\delta\). The yaw moment linearly increase with the steer input \(\delta\).

Yaw damping derivative \(N_r = \dfrac{1}{V}(a^2C_F + b^2C_R) <0 \)

Describes the relationship between the tyre induced yaw moment \(N_r r\) and the vehicle yawing velocity \(r \) [rad/s]. The derivative is always negative and this is in analogous to a yaw velocity damper trying to reduce the yaw velocity.
\(N_r \to \; – \infty\) with \(V \to 0 \), and \(N_r \to 0 \) with \(V \to \infty \)

Static directional stability derivative \( N_\beta = a C_F – b C_R \)

Describes the relationship between the tyre induced yaw moment \(N_\beta \beta\) and the vehicle slip angle \(\beta\). It is the difference between the yaw moment about c.g. produced by the front tyres and the rear tyres per unit \(\beta\).
If \(N_\beta >0 \), the car understeers. The yaw moment \(N_\beta \beta \) tries to reduce the vehicle slip angle (Restoring).
If \(N_\beta =0 \), the car is neutral steer.
If \(N_\beta <0 \), the car oversteers. The yaw moment \(N_\beta \beta\) tries to increase the vehicle slip angle (Upsetting).

Control force derivative \(Y_\delta = -C_F >0 \)

Describes the relationship between the part of the tyre lateral force \(Y_\delta \delta \) and the steering input \(\delta\). The lateral tyre force linearly increase with the steer input \(\delta\).

Damping-in-sideslip derivative \(Y_\beta = C_F + C_R <0\)

Describes the relationship between the part of the tyre lateral force \(Y_\beta \beta\) and the vehicle slip angle \(\beta\). The derivative is always negative. Therefore, for a positive sideslip velocity \(v\) (and positive sideslip angle \(\beta\)), negative lateral forces are generated.

Lateral force/yaw coupling derivative \(Y_r = \dfrac{1}{V}(aC_F – bC_R) \)

Describes the relationship between the part of the tyre lateral force \(Y_r r\) and the vehicle yaw velocity \(r\).

Complete equations of motion

\[ \left\{ \begin{array}{l} \mathrm{mV}(\mathrm{r}+\dot{\beta}) &=\mathrm{Y}_{\beta} \beta+\mathrm{Y}_{\mathrm{r}} \mathrm{r}+\mathrm{Y}_{\delta} \delta &= (C_F + C_R)\beta + \dfrac{1}{V}(aC_F \; – bC_R) r + (-C_F) \delta \\ \mathrm{I}_{\mathrm{z}} \dot{\mathrm{r}} &=\mathrm{N}_{\beta} \beta+\mathrm{N}_{\mathrm{r}} \mathrm{r}+\mathrm{N}_{\delta} \delta &= (a C_F \; – b C_R)\beta + \dfrac{1}{V}(a^2C_F + b^2C_R) r + (-a C_F)\delta \end{array} \right. \]

Steady-state response

In steady-state, \( \dot{r} = 0 \) and \( \dot{\beta} = 0 \) (or \( \dot{v} =0 \)), and the equations of motion become \[ \left\{ \begin{array}{l} \begin{align} \mathrm{mVr}&=\mathrm{Y}_{\beta} \beta+\mathrm{Y}_{\mathrm{r}} \mathrm{r}+\mathrm{Y}_{\delta} \delta\ \tag{1}\\ 0&=\mathrm{N}_{\beta} \beta+\mathrm{N}_{\mathrm{r}} \mathrm{r}+\mathrm{N}_{\delta} \delta \tag{2} \end {align}\end{array} \right. \] Plug in \(r = \dfrac{V}{R} \) to equations (1) and (2) to find \[ \left\{ \begin{array}{l} \begin{align} -Y_{\delta} \delta&=Y_{\beta} \beta+\left(V Y_{r}-m V^{2}\right)\left(\frac{1}{R}\right) \tag{3} \\ -\mathrm{N}_{\delta} \delta&=\mathrm{N}_{\beta} \beta+\mathrm{VN}_{\mathrm{r}}\left(\frac{1}{R}\right) \tag{4} \end{align} \end{array} \right. \]

[1] Response to steer angle \(\delta\) (Control)

From equations (3) and (4), delete \( \beta\) to find \[ \left\{ \begin{align} \begin{array}{l} \dfrac{1 / \mathrm{R}}{\delta}&=\dfrac{\mathrm{Y}_{\beta} \mathrm{N}_{\delta}-\mathrm{N}_{\beta} \mathrm{Y}_{\delta}}{\mathrm{VQ}} \\ \dfrac{\mathrm{r}}{\delta}&=\mathrm{V} \dfrac{1 / \mathrm{R}}{\delta}=\dfrac{\mathrm{Y}_{\mathrm{\beta}} \mathrm{N}_{\mathrm{\delta}}-\mathrm{N}_{\mathrm{\beta}} \mathrm{Y}_{\mathrm{\delta}}}{\mathrm{Q}} \\ \dfrac{\mathrm{v}^{2} / \mathrm{R}}{\delta}&=\dfrac{\mathrm{V}\left(\mathrm{Y}_{\beta} \mathrm{N}_{\delta}-\mathrm{N}_{\beta} \mathrm{Y}_{\delta}\right)}{\mathrm{Q}} \end{array} \end{align} \right. \] where \[ \mathrm{Q}=\mathrm{N}_{\beta} \mathrm{Y}_{\mathrm{r}}-\mathrm{N}_{\beta} \mathrm{mV}-\mathrm{Y}_{\mathrm{\beta}} \mathrm{N}_{\mathrm{r}} \]

From equations (1) and (2), delete \( r\) then \[\frac{\beta}{\delta}=\dfrac{\mathrm{Y}_{\delta} \mathrm{N}_{\mathrm{r}}-\mathrm{N}_{\delta}\left(\mathrm{Y}_{\mathrm{r}}-\mathrm{mV}\right)}{\mathrm{Q}}\]

[2] Response to applied side force \(F_y\) at c.g.

Assuming the steer angle \( \delta = 0 \), from equations (3) and (4) \[ \left\{ \begin{array}{l} \begin{align} -\mathrm{F}_{\mathrm{y}}&=\mathrm{Y}_{\beta} \beta+\left(\mathrm{V} \mathrm{Y}_{\mathrm{r}}-\mathrm{m} \mathrm{V}^{2}\right)\left(\dfrac{1}{\mathrm{R}}\right) \\ 0&=\mathrm{N}_{\mathrm{\beta}} \beta+\mathrm{VN}_{\mathrm{r}}\left(\dfrac{1}{\mathrm{R}}\right) \end{align} \end{array} \right. \] then, the responses are \[ \left\{ \begin{array}{l} \begin{align} \dfrac{1/R}{F_y} &= -\dfrac{N_\beta}{VQ} \\ \dfrac{r}{F_y} &= \; – \dfrac{N_\beta}{Q} \\ \dfrac{V^2/R}{F_y} &= -\dfrac{VN_\beta}{Q} \\ \dfrac{\beta}{F_y} &= + \dfrac{N_r}{Q} \end{align} \end{array} \right. \]

[3] Response to applied yaw moment \(N\)

Assuming the steer angle \( \delta = 0 \),from equations (3) and (4) \[ \left\{ \begin{array}{l} \begin{align} 0 &=\mathrm{Y}_{\beta} \beta+\left(\mathrm{V} \mathrm{Y}_{\mathrm{r}}-\mathrm{m} \mathrm{V}^{2}\right)\left(\dfrac{1}{\mathrm{R}}\right) \\ -N&=\mathrm{N}_{\mathrm{\beta}} \beta+\mathrm{VN}_{\mathrm{r}}\left(\dfrac{1}{\mathrm{R}}\right) \end{align} \end{array} \right. \] then, the responses are \[ \left\{ \begin{array}{l} \begin{align} \dfrac{1/R}{N} &= \dfrac{Y_\beta}{VQ} \\ \dfrac{r}{N} &= \; \dfrac{Y_\beta}{Q} \\ \dfrac{V^2/R}{N} &= \dfrac{VY_\beta}{Q} \\ \dfrac{\beta}{N} &= \; – \dfrac{Y_r-mV}{Q} \end{align} \end{array} \right. \]

Summary

Complete equations of motion

\[ \left\{ \begin{array}{l} \mathrm{mV}(\mathrm{r}+\dot{\beta}) &=\mathrm{Y}_{\beta} \beta+\mathrm{Y}_{\mathrm{r}} \mathrm{r}+\mathrm{Y}_{\delta} \delta \\ \mathrm{I}_{\mathrm{z}} \dot{\mathrm{r}} &=\mathrm{N}_{\beta} \beta+\mathrm{N}_{\mathrm{r}} \mathrm{r}+\mathrm{N}_{\delta} \delta \end{array} \right. \]

Damping-in-Sideslip Derivative \(Y_\beta = C_F + C_R \)	Lateral Force/Yaw Coupling Derivative \(Y_r = \dfrac{1}{V}(aC_F – bC_R) \)	Control Force Derivative \(Y_\delta = -C_F \)
Static Directional Stability Derivative \( N_\beta = a C_F – b C_R \)	Yaw Damping Derivative \(N_r = \dfrac{1}{V}(a^2C_F + b^2C_R) \)	Control Moment Derivative \( N_\delta = -a C_F \)

Steady-state response – the transients \(\dot{\beta} = \dot{r} = 0\)

Response to steer angle \(\delta\) (Control)	Response to applied side force \(F_y\) at c.g.	Response to applied yaw moment \(N\)
\( \dfrac{1 / R}{\delta} = \dfrac{Y_\beta N_\delta-N_\beta Y_\delta}{VQ} \)	\( \dfrac{1 / R}{F_y} = \; – \dfrac{N_\beta}{VQ} \)	\( \dfrac{1/R}{N} = \dfrac{Y_\beta}{VQ} \)
\( \dfrac{r}{\delta} = V \dfrac{1/R}{\delta} = \dfrac{Y_\beta N_\delta – N_\beta Y_\delta}{Q} \)	\( \dfrac{r}{F_y} = \; – \dfrac{N_\beta}{Q}\)	\( \dfrac{r}{N} = \dfrac{Y_\beta}{Q} \)
\( \dfrac{V^2/R}{\delta} = \dfrac{V( Y_\beta N_\delta – N_\beta Y_\delta )}{Q} \)	\(\dfrac{V^2/R}{F_y} = \; – \dfrac{VN_\beta}{Q}\)	\( \dfrac{V^2/R}{N} = \dfrac{VY_\beta}{Q}\)
\( \dfrac{\beta}{\delta} = \dfrac{Y_\delta N_r – N_\delta (Y_r – mV) }{Q}\)	\( \dfrac{\beta}{F_y} = \dfrac{N_r}{Q}\)	\( \dfrac{\beta}{N} = \; – \dfrac{Y_r – mV}{Q}\)

where \(Q = N_\beta Y_r \; – N_\beta mV \; – Y_\beta N_r \)
The top three equations in each columns are identical, just multiplying \(V = Rr \) to both sides of the top equations will lead to the 2nd row.

Reference

[1] W. Milliken and D. Milliken, Race Car Vehicle Dynamics.