Simple vehicle models

Models of vehicles without articulation.

Simple vehicle 3 DOF

The linear version of this model is described by equations ?? and the nonlinear version of this model is described by equations ??.

The simple vehicle model is illustrated in Fig. 62. The vector basis \(\{ {\rm O} {\bf i} {\bf j} {\bf k} \}\) is fixed to the inertial reference frame. The vector basis \(\{ {\rm T} {\bf t}_x {\bf t}_y {\bf t}_z \}\) is fixed to the simple vehicle. The vector basis \(\{ {\rm F} {\bf e}_x {\bf e}_y {\bf e}_z \}\) is fixed to the front axle.

The center of gravity of the vehicle is located at the point \({\rm T}\). The front and rear axles are located at the points \({\rm F}\) and \({\rm R}\), respectively. Point \({\rm O}\) is the origin. The constant \(a\) measures the distance of point \({\rm F}\) to \({\rm T}\) and \(b\) the distance of point \({\rm T}\) to \({\rm R}\). The angles \(\alpha_{\rm F}\) e \(\alpha_{\rm R}\) are the front and rear slip angles, respectively. \(\alpha_{\rm T}\) is the vehicle side slip angle and \(\psi\) is the vehicle yaw angle. \(\delta\) is the steering angle.

Fig. 62 Simple model.

Equations of Motion

The generalized coordinates are

\[\begin{split}q_1 &= x \\ q_2 &= y \\ q_3 &= \psi, \\\end{split}\]

where \(x\) and \(y\) are the coordinates of the CG of the vehicle. \(\psi\) is the yaw angle.

The position of the CG in relation to the origin \({\rm O}\) is

(1)\[{\bf p}_{{\rm T}/{\rm O}} = x \, {\bf i} + y \, {\bf j}.\]

The time derivative of equation (1) gives the velocity

\[{\bf v}_{\rm T} = \dot{x} \, {\bf i} + \dot{y} \, {\bf j}.\]

The velocity of the front axle is

\[{\bf v}_{\rm F} = {\bf v}_{\rm T} + {\bf r} \wedge {\bf p}_{{\rm F}/{\rm T}},\]

where \({\bf p}_{{\rm F}/{\rm T}}\) is the position of point \({\rm F}\) in relation to point \({\rm T}\) and \({\bf r} = \dot{\psi} \, {\bf k}\). Thus,

(2)\[{\bf v}_{\rm F} = \left( \dot{x} - a \dot{\psi} \sin \psi \right) {\bf i} + \left( \dot{y} + a \dot{\psi} \cos \psi \right) {\bf j}.\]

the velocity of the rear axle is

\[{\bf v}_{\rm R} = {\bf v}_{\rm T} + {\bf r} \wedge {\bf p}_{{\rm R}/{\rm T}},\]

where \({\bf p}_{{\rm R}/{\rm T}}\) is the position of point \({\rm R}\) in relation to point \({\rm T}\). Thus,

(3)\[{\bf v}_{\rm R} = \left( \dot{x} + b \dot{\psi} \sin \psi \right) {\bf i} + \left( \dot{y} -b \dot{\psi} \cos \psi \right) {\bf j}.\]

Using equations (2) and (3), the slip angles can be written as

(4)\[\begin{split}\alpha_{\rm F} &= \arctan \left( \frac{\dot{y} + a \dot{\psi} \cos \psi}{ \dot{x} - a \dot{\psi} \sin \psi} \right) - \left( \delta + \psi \right) \\ \alpha_{\rm R} &= \arctan \left( \frac{\dot{y} - b \dot{\psi} \cos \psi}{ \dot{x} + b \dot{\psi} \sin \psi} \right) - \psi\end{split}\]

The force vector at the front axle is

\[{\bf F}_{\rm F} = F_{x,{\rm F}} \, {\bf e}_x + F_{y,{\rm F}} \, {\bf e}_x,\]

It can be also written as

(5)\[{\bf F}_{\rm F} = \left[ F_{x,{\rm F}} \cos \left( \psi + \delta \right) - F_{y,{\rm F}} \sin \left( \psi + \delta \right) \right] {\bf i} + \left[ F_{x,{\rm F}} \sin \left( \psi + \delta \right) + F_{y,{\rm F}} \cos \left( \psi + \delta \right) \right] {\bf j}.\]

The force vector at the rear axle is

\[{\bf F}_{\rm R} = F_{x,{\rm R}} {\bf t}_x + F_{y,{\rm R}} {\bf t}_y\]

or

(6)\[{\bf F}_{\rm R} = \left[ F_{x,{\rm R}} \cos \psi - F_{y,{\rm R}} \sin \psi \right] {\bf i} + \left[ F_{x,{\rm R}} \sin \psi + F_{y,{\rm R}} \cos \psi \right] {\bf j}.\]

The generalized forces are

\[Q_k = \sum_{j = 1} ^p {\bf F}_j \cdot \frac{\partial {\bf p}_j}{\partial q_k}\]

In the model

(7)\[\begin{split}Q_1 &= {\bf F}_{\rm F} \cdot \frac{\partial {\bf p}_{{\rm F}/{\rm O}}}{\partial q_1} + {\bf F}_{\rm R} \cdot \frac{\partial {\bf p}_{{\rm R}/{\rm O}}}{\partial q_1} \\ Q_2 &= {\bf F}_{\rm F} \cdot \frac{\partial {\bf p}_{{\rm F}/{\rm O}}}{\partial q_2} + {\bf F}_{\rm R} \cdot \frac{\partial {\bf p}_{{\rm R}/{\rm O}}}{\partial q_2} \\ Q_3 &= {\bf F}_{\rm F} \cdot \frac{\partial {\bf p}_{{\rm F}/{\rm O}}}{\partial q_3} + {\bf F}_{\rm R} \cdot \frac{\partial {\bf p}_{{\rm R}/{\rm O}}}{\partial q_3},\end{split}\]

where

(8)\[\begin{split}\frac{\partial {\bf p}_{{\rm F}/{\rm O}}}{\partial q_1} &= \frac{\partial {\bf p}_{{\rm F}/{\rm O}}}{\partial x} = {\bf i} \\ \frac{\partial {\bf p}_{{\rm F}/{\rm O}}}{\partial q_2} &= \frac{\partial {\bf p}_{{\rm F}/{\rm O}}}{\partial y} = {\bf j} \\ \frac{\partial {\bf p}_{{\rm F}/{\rm O}}}{\partial q_3} &= \frac{\partial {\bf p}_{{\rm F}/{\rm O}}}{\partial \psi} = - a \sin \psi \, {\bf i} + a \cos \psi \, {\bf j}\end{split}\]

and

(9)\[\begin{split}\frac{\partial {\bf p}_{{\rm R}/{\rm O}}}{\partial q_1} &= \frac{\partial {\bf p}_{{\rm R}/{\rm O}}}{\partial x} = {\bf i} \\ \frac{\partial {\bf p}_{{\rm R}/{\rm O}}}{\partial q_2} &= \frac{\partial {\bf p}_{{\rm R}/{\rm O}}}{\partial y} = {\bf j} \\ \frac{\partial {\bf p}_{{\rm R}/{\rm O}}}{\partial q_3} &= \frac{\partial {\bf p}_{{\rm R}/{\rm O}}}{\partial \psi} = b \sin \psi \, {\bf i} - b \cos \psi \, {\bf j}\end{split}\]

Substituting equations (5), (6), (8) and (9) in equations (7) we have

\[\begin{split}Q_1 &= F_{x,{\rm F}} \cos \left( \psi + \delta \right) + F_{x,{\rm R}} \cos \psi - F_{y,{\rm F}} \sin \left( \psi + \delta \right) - F_{y,{\rm R}} \sin \psi \\ Q_2 &= F_{x,{\rm F}} \sin \left( \psi + \delta \right)+ F_{x,{\rm R}} \sin \psi + F_{y,{\rm F}} \cos \left( \psi + \delta \right) + F_{y,{\rm R}} \cos \psi \\ Q_3 &= F_{x,{\rm F}} a \sin \delta + F_{y,{\rm F}} a \cos \delta - F_{y,{\rm R}} b.\end{split}\]

The kinetic energy of the system is given by

\[T = \frac{1}{2} m_{\rm T} {\bf v}_{\rm T} \cdot {\bf v}_{\rm T} + \frac{1}{2} I_{\rm T} \dot{\psi}^2.\]

or

\[T = \frac{1}{2} m_{\rm T} \left( \dot{x}^2 + \dot{y}^2 \right) + \frac{1}{2} I_{\rm T} \dot{\psi}^2.\]

The Lagrange formulation is given by

(10)\[\frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_k} \right) - \frac{\partial T}{\partial q_k} = Q_k.\]

For the three generalized coordinates of the system

(11)\[\begin{split}\frac{\partial T}{\partial q_1} &= \frac{\partial T}{\partial x} = 0 \\ \frac{\partial T}{\partial q_2} &= \frac{\partial T}{\partial y} = 0 \\ \frac{\partial T}{\partial q_3} &= \frac{\partial T}{\partial \psi} = 0\end{split}\]

and

\[\begin{split}\frac{\partial T}{\partial \dot{q}_1} &= \frac{\partial T}{\partial \dot{x}} = m_{\rm T} \dot{x} \\ \frac{\partial T}{\partial \dot{q}_2} &= \frac{\partial T}{\partial \dot{y}} = m_{\rm T} \dot{y} \\ \frac{\partial T}{\partial \dot{q}_3} &= \frac{\partial T}{\partial \dot{\psi}} = I_{\rm T} \dot{\psi}.\end{split}\]

The time derivative is given by

(12)\[\begin{split}\frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_1} \right) &= \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{x}} \right) = m_{\rm T} \ddot{x} \\ \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_2} \right) &= \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{y}} \right) = m_{\rm T} \ddot{y} \\ \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_3} \right) &= \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{\psi}} \right) = I_{\rm T} \ddot{\psi}.\end{split}\]

Substituting equations (11) and (12) in (10) the equations of motion become

\[\begin{split}m_{\rm T} \ddot{x} &= F_{x,{\rm F}} \cos \left( \psi + \delta \right) + F_{x,{\rm R}} \cos \psi - F_{y,{\rm F}} \sin \left( \psi + \delta \right) - F_{y,{\rm R}} \sin \psi \\ m_{\rm T} \ddot{y} &= F_{x,{\rm F}} \sin \left( \psi + \delta \right)+ F_{x,{\rm R}} \sin \psi + F_{y,{\rm F}} \cos \left( \psi + \delta \right) + F_{y,{\rm R}} \cos \psi \\ I_{\rm T} \ddot{\psi} &= F_{x,{\rm F}} a \sin \delta + F_{y,{\rm F}} a \cos \delta - F_{y,{\rm R}} b,\end{split}\]

this is,

(13)\[\begin{split}\ddot{x} &= \frac{F_{x,{\rm F}} \cos \left( \psi + \delta \right) + F_{x,{\rm R}} \cos \psi - F_{y,{\rm F}} \sin \left( \psi + \delta \right) - F_{y,{\rm R}} \sin \psi}{m_{\rm T}} \\ \ddot{y} &= \frac{F_{x,{\rm F}} \sin \left( \psi + \delta \right)+ F_{x,{\rm R}} \sin \psi + F_{y,{\rm F}} \cos \left( \psi + \delta \right) + F_{y,{\rm R}} \cos \psi}{m_{\rm T}} \\ \ddot{\psi} &= \frac{F_{x,{\rm F}} a \sin \delta + F_{y,{\rm F}} a \cos \delta - F_{y,{\rm R}} b}{I_{\rm T}}.\end{split}\]

State Equations

The chosen state vector is

\[\begin{split}{\rm z}_1 &= x \\ {\rm z}_2 &= y \\ {\rm z}_3 &= \psi \\ {\rm z}_4 &= \dot{x} \\ {\rm z}_5 &= \dot{y} \\ {\rm z}_6 &= \dot{\psi}\end{split}\]

Thus, considering the equation of motion in equation (13), the state equation is given by

\[\begin{split}\dot{{\rm z}}_1 &= {\rm z}_4 \\ \dot{{\rm z}}_2 &= {\rm z}_5 \\ \dot{{\rm z}}_3 &= {\rm z}_6 \\ \dot{{\rm z}}_4 &= \frac{F_{x,{\rm F}} \cos \left( {\rm z}_3 + \delta \right) + F_{x,{\rm R}} \cos {\rm z}_3 - F_{y,{\rm F}} \sin \left( {\rm z}_3 + \delta \right) - F_{y,{\rm R}} \sin {\rm z}_3}{m_{\rm T}} \\ \dot{{\rm z}}_5 &= \frac{F_{x,{\rm F}} \sin \left( {\rm z}_3 + \delta \right)+ F_{x,{\rm R}} \sin {\rm z}_3 + F_{y,{\rm F}} \cos \left( {\rm z}_3 + \delta \right) + F_{y,{\rm R}} \cos {\rm z}_3}{m_{\rm T}} \\ \dot{{\rm z}}_6 &= \frac{F_{x,{\rm F}} a \sin \delta + F_{y,{\rm F}} a \cos \delta - F_{y,{\rm R}} b}{I_{\rm T}}\end{split}\]

With the slip angles

\[\begin{split}\alpha_{\rm F} &= \arctan \left( \frac{{\rm z}_5 + a {\rm z}_6 \cos {\rm z}_3}{ {\rm z}_4 - a {\rm z}_6 \sin {\rm z}_3} \right) - \left( \delta + {\rm z}_3 \right) \\ \alpha_{\rm R} &= \arctan \left( \frac{{\rm z}_5 - b {\rm z}_6 \cos {\rm z}_3}{ {\rm z}_4 + b {\rm z}_6 \sin {\rm z}_3} \right) - {\rm z}_3,\end{split}\]

see equation (4).

However, in many occasions it is more convenint to use the states \(v_{\rm T}\) e \(\alpha_{\rm T}\) instead of \(\dot{x}\) e \(\dot{y}\).

The relation between this pair of states is given by

(14)\[\begin{split}\dot{x} &= v_{\rm T} \cos \left( \psi + \alpha_{\rm T} \right) \\ \dot{y} &= v_{\rm T} \sin \left( \psi + \alpha_{\rm T} \right).\end{split}\]

The time derivative o equation (14) is given by

(15)\[\begin{split}\ddot{x} &= \dot{v}_{\rm T} \cos \left( \psi + \alpha_{\rm T} \right) - v_{\rm T} \left( \dot{\psi} + \dot{\alpha}_{\rm T} \right) \sin \left( \psi + \alpha_{\rm T} \right) \\ \ddot{y} &= \dot{v}_{\rm T} \sin \left( \psi + \alpha_{\rm T} \right) + v_{\rm T} \left( \dot{\psi} + \dot{\alpha}_{\rm T} \right) \cos \left( \psi + \alpha_{\rm T} \right).\end{split}\]

Substituting equation (15) in the equations of motion (13) we have

\[\begin{split}\dot{v}_{\rm T} \cos \left( \psi + \alpha_{\rm T} \right) - v_{\rm T} \left( \dot{\psi} + \dot{\alpha}_{\rm T} \right) \sin \left( \psi + \alpha_{\rm T} \right) &= \frac{F_{x,{\rm F}} \cos \left( \psi + \delta \right) + F_{x,{\rm R}} \cos \psi - F_{y,{\rm F}} \sin \left( \psi + \delta \right) - F_{y,{\rm R}} \sin \psi}{m_{\rm T}} \\ \dot{v}_{\rm T} \sin \left( \psi + \alpha_{\rm T} \right) + v_{\rm T} \left( \dot{\psi} + \dot{\alpha}_{\rm T} \right) \cos \left( \psi + \alpha_{\rm T} \right) &= \frac{F_{x,{\rm F}} \sin \left( \psi + \delta \right)+ F_{x,{\rm R}} \sin \psi + F_{y,{\rm F}} \cos \left( \psi + \delta \right) + F_{y,{\rm R}} \cos \psi}{m_{\rm T}} \\ \ddot{\psi} &= \frac{F_{x,{\rm F}} a \sin \delta + F_{y,{\rm F}} a \cos \delta - F_{y,{\rm R}} b}{I_{\rm T}}.\end{split}\]

Manipulating and simplifying

\[\begin{split}\dot{v}_{\rm T} &= \frac{F_{x,{\rm F}} \cos \left( \alpha_{\rm T} - \delta \right) + F_{x,{\rm R}} \cos \alpha_{\rm T} + F_{y,{\rm F}} \sin \left( \alpha_{\rm T} - \delta \right) + F_{y,{\rm R}} \sin \alpha_{\rm T}}{m_{\rm T}} \\ \dot{\alpha}_{\rm T} &= \frac{- F_{x,{\rm F}} \sin \left( \alpha_{\rm T} - \delta \right) - F_{x,{\rm R}} \sin \alpha_{\rm T} + F_{y,{\rm F}} \cos \left( \alpha_{\rm T} - \delta \right) + F_{y,{\rm R}} \cos \alpha_{\rm T} - m_{\rm T} v \dot{\psi}}{m_{\rm T} v_{\rm T}} \\ \ddot{\psi} &= \frac{F_{x,{\rm F}} a \sin \delta + F_{y,{\rm F}} a \cos \delta - F_{y,{\rm R}} b}{I_{\rm T}}.\end{split}\]

The slip angles become

\[\begin{split}\alpha_{\rm F} &= \arctan \left( \frac{v_{\rm T} \sin \left( \psi + \alpha_{\rm T} \right) + a \dot{\psi} \cos \psi}{ v_{\rm T} \cos \left( \psi + \alpha_{\rm T} \right) - a \dot{\psi} \sin \psi} \right) - \left( \delta + \psi \right) \\ \alpha_{\rm R} &= \arctan \left( \frac{v_{\rm T} \sin \left( \psi + \alpha_{\rm T} \right) - b \dot{\psi} \cos \psi}{ v_{\rm T} \cos \left( \psi + \alpha_{\rm T} \right) + b \dot{\psi} \sin \psi} \right) - \psi.\end{split}\]

Simplifying

(16)\[\begin{split}\alpha_{\rm F} &= \arctan \left( \frac{v_{\rm T} \sin \alpha_{\rm T} + a \dot{\psi}}{ v_{\rm T} \cos \alpha_{\rm T}} \right) - \delta \\ \alpha_{\rm R} &= \arctan \left( \frac{v_{\rm T} \sin \alpha_{\rm T} - b \dot{\psi}}{ v_{\rm T} \cos \alpha_{\rm T}} \right)\end{split}\]

Therefore, the new set of states is

\[\begin{split}{\rm x}_1 &= x \\ {\rm x}_2 &= y \\ {\rm x}_3 &= \psi \\ {\rm x}_4 &= v_{\rm T} \\ {\rm x}_5 &= \alpha_{\rm T} \\ {\rm x}_6 &= \dot{\psi}\end{split}\]

and the state equation is

(17)\[\begin{split}\dot{{\rm x}}_1 &= {\rm x}_4 \cos \left( {\rm x}_3 + {\rm x}_5 \right) \\ \dot{{\rm x}}_2 &= {\rm x}_4 \sin \left( {\rm x}_3 + {\rm x}_5 \right) \\ \dot{{\rm x}}_3 &= {\rm x}_6 \\ \dot{{\rm x}}_4 &= \frac{F_{x,{\rm F}} \cos \left( {\rm x}_5 - \delta \right) + F_{x,{\rm R}} \cos {\rm x}_5 + F_{y,{\rm F}} \sin \left( {\rm x}_5 - \delta \right) + F_{y,{\rm R}} \sin {\rm x}_5}{m_{\rm T}} \\ \dot{{\rm x}}_5 &= \frac{- F_{x,{\rm F}} \sin \left( {\rm x}_5 - \delta \right) - F_{x,{\rm R}} \sin {\rm x}_5 + F_{y,{\rm F}} \cos \left( {\rm x}_5 - \delta \right) + F_{y,{\rm R}} \cos \alpha_{\rm T} - m_{\rm T} {\rm x}_4 {\rm x}_6}{m_{\rm T} {\rm x}_4} \\ \dot{{\rm x}}_6 &= \frac{F_{x,{\rm F}} a \sin \delta + F_{y,{\rm F}} a \cos \delta - F_{y,{\rm R}} b}{I_{\rm T}}\end{split}\]

The nonlinear model of the package uses the equations (16) and (17)

Linearization

The state equation (17) describes the nonlinear version of the model. It can be written as

\[\dot{{\bf x}} = {\bf f} \left( {\bf x}, {\bf u} \right),\]

where the state vector is given by

\[\begin{split}{\bf x} = \left[ \begin{array}{c} {\rm x}_1 \\ {\rm x}_2 \\ {\rm x}_3 \\ {\rm x}_4 \\ {\rm x}_5 \\ {\rm x}_6 \end{array} \right] = \left[ \begin{array}{c} x \\ y \\ \psi \\ v_{\rm T} \\ \alpha_{\rm T} \\ \dot{\psi} \end{array} \right],\end{split}\]

the input vector is

\[\begin{split}{\bf u} = \left[ \begin{array}{c} \delta \\ F_{x,{\rm F}} \\ F_{x,{\rm R}} \\ F_{y,{\rm F}} \\ F_{y,{\rm R}} \end{array} \right]\end{split}\]

and the vector function \({\bf f}\) is

\[\begin{split}{\bf f} = \left[ \begin{array}{c} {\rm f}_1 \\ {\rm f}_2 \\ {\rm f}_3 \\ {\rm f}_4 \\ {\rm f}_5 \\ {\rm f}_6 \end{array} \right] = \left[ \begin{array}{c} \dot{{\rm x}}_1 \\ \dot{{\rm x}}_2 \\ \dot{{\rm x}}_3 \\ \dot{{\rm x}}_4 \\ \dot{{\rm x}}_5 \\ \dot{{\rm x}}_6 \end{array} \right].\end{split}\]

The linearization can be made choosing an operating point that represents the vehicle traveling in a straight line with a constant forward velocity

A linearização deste sistema pode ser feita para um veículo se movimentando em linha reta com uma determinada velocidade \(v_{{\rm T},0} > 0\). In this case, the operating point of each state are

\[\begin{split}{\rm x}_{1,op} &= x_{op} \\ {\rm x}_{2,op} &= y_{op} \\ {\rm x}_{3,op} &= \psi_{op} = 0 \\ {\rm x}_{4,op} &= v_{{\rm T},op} = v_{{\rm T},0} \\ {\rm x}_{5,op} &= \alpha_{{\rm T},op} = 0 \\ {\rm x}_{6,op} &= \dot{\psi}_{op} = 0,\end{split}\]

As we are going to see below, the operating point of the states \(x\) e \(y\) do not have to be defined. They do not influence the dynamics of the linear system.

The vector of the states operating points is

\[\begin{split}{\bf x}_{op} = \left[ \begin{array}{c} {\rm x}_{1,op} \\ {\rm x}_{2,op} \\ {\rm x}_{3,op} \\ {\rm x}_{4,op} \\ {\rm x}_{5,op} \\ {\rm x}_{6,op} \end{array} \right].\end{split}\]

The operating point of the inputs is

\[\begin{split}\delta_{op} &= 0 \\ F_{x,{\rm F},op} &= 0 \\ F_{x,{\rm R},op} &= 0 \\ F_{y,{\rm F},op} &= 0 \\ F_{y,{\rm R},op} &= 0.\end{split}\]

The vector of the input operating points is

\[\begin{split}{\bf u}_{op} = \left[ \begin{array}{c} \delta_{op} \\ F_{x,{\rm F},op} \\ F_{x,{\rm R},op} \\ F_{y,{\rm F},op} \\ F_{y,{\rm R},op} \end{array} \right].\end{split}\]

Expanding the system in a Taylor series and neglecting the higher order terms, we have

(18)\[\begin{split}{\bf f}_{lin}\left( {\bf x}, {\bf u} \right) = {\bf f} \left( {\bf x}_{op}, {\bf u}_{op} \right) + \nabla{\bf f}\left( {\bf x}_{op}, {\bf u}_{op} \right) \left[ \begin{array}{c} {\bf x} - {\bf x}_{op} \\ {\bf u} - {\bf u}_{op} \end{array} \right].\end{split}\]

where

(19)\[\begin{split}{\bf f} \left( {\bf x}_{op}, {\bf u}_{op} \right) = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ v_{{\rm T},0} \\0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right].\end{split}\]

The gradient of \({\bf f}\) is

\[\begin{split}\nabla {\bf f} = \left[ \begin{array}{ccccccccccc} \frac{ \partial f_1 }{ \partial x } & \frac{ \partial f_1 }{ \partial y } & \frac{ \partial f_1 }{ \partial \psi } & \frac{ \partial f_1 }{ \partial v } & \frac{ \partial f_1 }{ \partial \alpha_{\rm T} } & \frac{ \partial f_1 }{ \partial \dot{\psi} } & \frac{ \partial f_1 }{ \partial \delta } & \frac{ \partial f_1 }{ \partial F_{x, {\rm F}} } & \frac{ \partial f_1 }{ \partial F_{x, {\rm R}} } & \frac{\partial f_1}{F_{y,{\rm F}}} & \frac{\partial f_1}{\partial F_{y,{\rm R}}} \\ \frac{ \partial f_2 }{ \partial x } & \frac{ \partial f_2 }{ \partial y } & \frac{ \partial f_2 }{ \partial \psi } & \frac{ \partial f_2 }{ \partial v } & \frac{ \partial f_2 }{ \partial \alpha_{\rm T} } & \frac{ \partial f_2 }{ \partial \dot{\psi} } & \frac{ \partial f_2 }{ \partial \delta } & \frac{ \partial f_2 }{ \partial F_{x, {\rm F}} } & \frac{ \partial f_2 }{ \partial F_{x, {\rm R}} } & \frac{\partial f_2}{F_{y,{\rm F}}} & \frac{\partial f_2}{\partial F_{y,{\rm R}}} \\ \frac{ \partial f_3 }{ \partial x } & \frac{ \partial f_3 }{ \partial y } & \frac{ \partial f_3 }{ \partial \psi } & \frac{ \partial f_3 }{ \partial v } & \frac{ \partial f_3 }{ \partial \alpha_{\rm T} } & \frac{ \partial f_3 }{ \partial \dot{\psi} } & \frac{ \partial f_3 }{ \partial \delta } & \frac{ \partial f_3 }{ \partial F_{x, {\rm F}} } & \frac{ \partial f_3 }{ \partial F_{x, {\rm R}} } & \frac{\partial f_3}{F_{y,{\rm F}}} & \frac{\partial f_3}{\partial F_{y,{\rm R}}} \\ \frac{ \partial f_4 }{ \partial x } & \frac{ \partial f_4 }{ \partial y } & \frac{ \partial f_4 }{ \partial \psi } & \frac{ \partial f_4 }{ \partial v } & \frac{ \partial f_4 }{ \partial \alpha_{\rm T} } & \frac{ \partial f_4 }{ \partial \dot{\psi} } & \frac{ \partial f_4 }{ \partial \delta } & \frac{ \partial f_4 }{ \partial F_{x, {\rm F}} } & \frac{ \partial f_4 }{ \partial F_{x, {\rm R}} } & \frac{\partial f_4}{F_{y,{\rm F}}} & \frac{\partial f_4}{\partial F_{y,{\rm R}}} \\ \frac{ \partial f_5 }{ \partial x } & \frac{ \partial f_5 }{ \partial y } & \frac{ \partial f_5 }{ \partial \psi } & \frac{ \partial f_5 }{ \partial v } & \frac{ \partial f_5 }{ \partial \alpha_{\rm T} } & \frac{ \partial f_5 }{ \partial \dot{\psi} } & \frac{ \partial f_5 }{ \partial \delta } & \frac{ \partial f_5 }{ \partial F_{x, {\rm F}} } & \frac{ \partial f_5 }{ \partial F_{x, {\rm R}} } & \frac{\partial f_5}{F_{y,{\rm F}}} & \frac{\partial f_5}{\partial F_{y,{\rm R}}} \\ \frac{ \partial f_6 }{ \partial x } & \frac{ \partial f_6 }{ \partial y } & \frac{ \partial f_6 }{ \partial \psi } & \frac{ \partial f_6 }{ \partial v } & \frac{ \partial f_6 }{ \partial \alpha_{\rm T} } & \frac{ \partial f_6 }{ \partial \dot{\psi} } & \frac{ \partial f_6 }{ \partial \delta } & \frac{ \partial f_6 }{ \partial F_{x, {\rm F}} } & \frac{ \partial f_6 }{ \partial F_{x, {\rm R}} } & \frac{\partial f_6}{F_{y,{\rm F}}} & \frac{\partial f_6}{\partial F_{y,{\rm R}}} \end{array} \right].\end{split}\]

Calculating the partial derivatives,

(20)\[\begin{split}\nabla {\bf f}\left( {\bf x}_{op}, {\bf u}_{op} \right) = \left[ \begin{array}{ccccccccccc} 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & v_{{\rm T},0} & 0 & v_{{\rm T},0} & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{m_{\rm T}} & \frac{1}{m_{\rm T}} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & \frac{1}{m_{\rm T} v_{{\rm T},0}} & \frac{1}{m_{\rm T} v_{{\rm T},0}} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{a}{I_{{\rm T}}} & - \frac{b}{I_{{\rm T}}} \end{array} \right]\end{split}\]

Substituting equations (19) and (20) in (18) the linearized system becomes

(21)\[\begin{split}f_{1,lin} &= \dot{x} = v_{\rm T} \\ f_{2,lin} &= \dot{y} = v_{{\rm T},0} \left( \psi + \alpha_{{\rm T}}\right) \\ f_{3,lin} &= \dot{\psi} = \dot{\psi} \\ f_{4,lin} &= \dot{v}_{\rm T} = \frac{F_{x,{\rm F}} + F_{x,{\rm R}}}{m_{\rm T}} \\ f_{5,lin} &= \dot{\alpha}_{\rm T} = \frac{F_{y,{\rm F}} + F_{y,{\rm R}}}{m_{\rm T} v_{{\rm T},0}} - \dot{\psi} \\ f_{6,lin} &= \ddot{\psi} = \frac{a F_{y,{\rm F}} - b F_{y,{\rm R}}}{I_{\rm T}}\end{split}\]

It is important to note that if the longitudinal forces of the tire, \(F_{x,{\rm F}}\) and \(F_{x,{\rm R}}\), are zero, the state \(v_{\rm T}\) is constant.

In the same operating point, the linear slip angles from equation (16) become

(22)\[\begin{split}\alpha_{{\rm F},lin} &= \alpha_{{\rm T}} + \frac{a}{v_{{\rm T},0}} \dot{\psi} - \delta \\ \alpha_{{\rm F},lin} &= \alpha_{{\rm T}} - \frac{b}{v_{{\rm T},0}} \dot{\psi}.\end{split}\]

The linear model of the package uses equations (21) and (22).

Finally, a linear tire model

(23)\[\begin{split}F_{y,{\rm F}} &= - K_{\rm F} \alpha_{\rm F} = - K_{\rm F} \alpha_{{\rm T}} - \frac{a K_{\rm F}}{v_{{\rm T},0}} \dot{\psi} + K_{\rm F} \delta \\ F_{y,{\rm R}} &= - K_{\rm R} \alpha_{\rm R} = - K_{\rm R} \alpha_{{\rm T}} + \frac{b K_{\rm R}}{v_{{\rm T},0}} \dot{\psi}\end{split}\]

can be used in the linearized equations (21). In this case

\[\begin{split}f_{1,lin} &= \dot{x} = v_{\rm T} \\ f_{2,lin} &= \dot{y} = v_{{\rm T},0} \left( \psi + \alpha_{{\rm T}}\right) \\ f_{3,lin} &= \dot{\psi} = \dot{\psi} \\ f_{4,lin} &= \dot{v}_{\rm T} = \frac{F_{x,{\rm F}} + F_{x,{\rm R}}}{m_{\rm T}} \\ f_{5,lin} &= \dot{\alpha}_{{\rm T}} = - \frac{K_{\rm F} + K_{\rm R}}{m_{\rm T} v_{{\rm T},0}} \alpha_{{\rm T}} - \frac{m_{\rm T} v_{{\rm T},0} + \frac{a K_{\rm F} - b K_{\rm R}}{v_{{\rm T},0}}}{m_{\rm T} v_{{\rm T},0}} \dot{\psi} + \frac{K_{\rm F}}{m_{\rm T} v_{{\rm T},0}} \delta \\ f_{6,lin} &= \ddot{\psi} = - \frac{a K_{\rm F} - b K_{\rm R}}{I_{\rm T}} \alpha_{{\rm T}} - \frac{a^2 K_{\rm F} + b^2 K_{\rm R}}{I_{\rm T} v_{{\rm T},0}} \dot{\psi} + \frac{a K_{\rm F}}{I_{\rm T}} \delta\end{split}\]

In the matrix form

\[\dot{{\bf x}} = {\bf A} {\bf x} + {\bf B} {\bf \hat{u}}\]

or

(24)\[\begin{split}\left[ \begin{array}{c} \dot{x} \\ \dot{y} \\ \dot{\psi} \\ \dot{v}_{\rm T} \\ \dot{\alpha}_{\rm T} \\ \ddot{\psi} \end{array} \right] = \left[ \begin{array}{cccccc} 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & v_{{\rm T},0} & 0 & v_{{\rm T},0} & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & - \frac{K_{\rm F} + K_{\rm R}}{m_{\rm T} v_{{\rm T},0}} & - \frac{m_{\rm T} v_{{\rm T},0} + \frac{a K_{\rm F} - b K_{\rm R}}{v_{{\rm T},0}}}{m_{\rm T} v_{{\rm T},0}} \\ 0 & 0 & 0 & 0 & - \frac{a K_{\rm F} - b K_{\rm R}}{I_{\rm T}} & - \frac{a^2 K_{\rm F} + b^2 K_{\rm R}}{I_{\rm T} v_{{\rm T},0}} \end{array} \right] \left[ \begin{array}{c} x \\ y \\ \psi \\ v_{\rm T} \\ \alpha_{\rm T} \\ \dot{\psi} \end{array} \right] + \left[ \begin{array}{ccccc} 0 & 0 & 0 & \\ 0 & 0 & 0 & \\ 0 & 0 & 0 & \\ 0 & \frac{1}{m_{\rm T}} & \frac{1}{m_{\rm T}} & \\ \frac{K_{\rm F}}{m_{\rm T} v_{{\rm T},0}} & 0 & 0 & \\ \frac{a K_{\rm F}}{I_{\rm T}} & 0 & 0 \end{array} \right] \left[ \begin{array}{c} \delta \\ F_{x,{\rm F}} \\ F_{x,{\rm R}} \end{array} \right]\end{split}\]

A popular model used in the literature considers just the last two rows of the model described in (24) with only the steering angle as input. This is

\[\begin{split}\left[ \begin{array}{c} \dot{\alpha}_{\rm T} \\ \ddot{\psi} \end{array} \right] = \left[ \begin{array}{cccccc} - \frac{K_{\rm F} + K_{\rm R}}{m_{\rm T} v_{{\rm T},0}} & - \frac{m_{\rm T} v_{{\rm T},0} + \frac{a K_{\rm F} - b K_{\rm R}}{v_{{\rm T},0}}}{m_{\rm T} v_{{\rm T},0}} \\ - \frac{a K_{\rm F} - b K_{\rm R}}{I_{\rm T}} & - \frac{a^2 K_{\rm F} + b^2 K_{\rm R}}{I_{\rm T} v_{{\rm T},0}} \end{array} \right] \left[ \begin{array}{c} \alpha_{\rm T} \\ \dot{\psi} \end{array} \right] + \left[ \begin{array}{ccccc} \frac{K_{\rm F}}{m_{\rm T} v_{{\rm T},0}} \\ \frac{a K_{\rm F}}{I_{\rm T}} \end{array} \right] \left[ \begin{array}{c} \delta \end{array} \right]\end{split}\]

Simple vehicle 4 DOF

The linear version of this model is described by equations ?? and the nonlinear version of this model is described by equations ??.

Nonlinear bicycle model nonlinear with 4 degrees of freedom.

Description

The center of gravity of the vehicle is located at the point \(T\). The front and rear axles are located at the points \(F\) and \(R\), respectively. The constant \(a\) measures the distance of point \(F\) to \(T\) and \(b\) the distance of point \(T\) to \(R\). The angles \(\alpha_F\) e \(\alpha_R\) are the front and rear slip angles, respectively. \(\alpha_T\) is the vehicle side slip angle and \(\psi\) is the vehicle yaw angle. \(\delta\) is the steering angle.