Car testdrive (lane change manoeuvre)

Car testdrive (lane change manoeuvre)
State dimension:	1
Differential states:	7
Continuous control functions:	3
Discrete control functions:	1
Interior point inequalities:	7

The mathematical equations form a small-scale ODE model.

Mathematical formulation

We consider a single-track model, derived under the simplifying assumption that rolling and pitching of the car body can be neglected. Consequentially, only a single front and rear wheel is modeled, located in the virtual center of the original two wheels. Motion of the car body is considered on the horizontal plane only.

Four controls represent the driver's choice on steering and velocity. We denote with $w_{\delta }$ the steering wheel's angular velocity. The force $F_{\text{B}}$ controls the total braking force, while the accelerator pedal position $\varphi$ is translated into an accelerating force. Finally, the selected gear $\mu$ influences the effective engine torque's transmission.

Resulting MIOCP

For $t\in [t_0,t_f]$ almost everywhere the mixed-integer optimal control problem is given by

$\begin{array}{ll}\underset{x(\cdot ),u(\cdot ),\mu (\cdot )}{\min }& t_{\text{f}}\\ \text{s.t.}& \dot{x}(t)& =& f(t,x(t),u(t),\mu (t)),\\ & x(t_0)& =& x_0,\\ & r(t,x(t),u(t))& \ge & 0,\\ & \mu (t)& \in & \{1,2,3,4,5\}.\end{array}$

Parameters

These fixed values are used within the model.

$\begin{array}{llcl}\text{Parameter}& \text{Value}& \text{Unit}& \text{Description}\\ m& 1.239\cdot 10^3& \text{kg}& \text{Mass of the car}\\ g& 9.81& \frac{\text{m}}{{\text{s}}^2}& \text{Gravity constant}\\ l_{\text{f}}& 1.19016& \text{m}& \text{Front wheel distance to center of gravity}\\ l_{\text{r}}& 1.37484& \text{m}& \text{Rear wheel distance to center of gravity}\\ e_{\text{SP}}& 0.5& \text{m}& \text{Drag mount point distance to center of gravity}\\ R& 0.302& \text{m}& \text{Wheel radius}\\ I_{\text{zz}}& 1.752\cdot 10^3& {\text{kg m}}^2& \text{Moment of inertia}\\ c_{\text{w}}& 0.3& -& \text{Air drag coefficient}\\ \rho & 1.249512& \frac{\text{kg}}{{\text{m}}^3}& \text{Air density}\\ A& 1.4378946874& {\text{m}}^2& \text{Effective flow surface}\\ i_{\text{g}}^1& 3.09& -& \text{Transmission ratio of first gear}\\ i_{\text{g}}^2& 2.002& -& \text{Transmission ratio of second gear}\\ i_{\text{g}}^3& 1.33& -& \text{Transmission ratio of third gear}\\ i_{\text{g}}^4& 1.0& -& \text{Transmission ratio of fourth gear}\\ i_{\text{g}}^5& 0.805& -& \text{Transmission ratio of fifth gear}\\ i_{\text{t}}& 3.91& -& \text{Engine torque transmission ratio}\\ B_{\text{f}}& 1.096\cdot 10^1& -& \text{Pacejka coefficients (stiffness)}\\ B_{\text{r}}& 1.267\cdot 10^1& -& \\ C_{\text{f}}& 1.3& -& \text{Pacejka coefficients (shape)}\\ C_{\text{r}}& 1.3& --& \\ D_{\text{f}}& 4.5604\cdot 10^3& --& \text{Pacejka coefficients (peak)}\\ D_{\text{r}}& 3.94781\cdot 10^3& --& \\ E_{\text{f}}& -0.5& --& \text{Pacejka coefficients (curvature)}\\ E_{\text{r}}& -0.5& --& \\ \end{array}$

Test course

The double-lane change manoeuvre presented in <bibref>Gerdts2005</bibref> is realized by constraining the car's position onto a prescribed track at any time $t\in [t_0,t_{\text{f}}]$. Starting in the left position with an initial prescribed velocity, the driver is asked to manage a change of lanes modeled by an offset of 3.5 meters in the track. Afterwards he is asked to return to the starting lane. This manoeuvre can be regarded as an overtaking move or as an evasive action taken to avoid hitting an obstacle suddenly appearing on the starting lane.

From a mathematical point of view, the test track is described by setting up piecewise cubic spline functions $P_{\text{l}}(x)$ and $P_{\text{r}}(x)$ modeling the top and bottom track boundary, given a horizontal position $x$.

$\begin{array}{rlr}{P}_{\text{l}}(x)& :=& \{\begin{array}{llrcl}0& \text{if }& & x& \le 44,\\ 4h_2(x-44{)}^3& \text{if }& 44<& x& \le 44.5,\\ 4h_2(x-45{)}^3+h_2& \text{if }& 44.5<& x& \le 45,\\ h_2& \text{if }& 45<& x& \le 70,\\ 4h_2(70-x{)}^3+h_2& \text{if }& 70<& x& \le 70.5,\\ 4h_2(71-x{)}^3& \text{if }& 70.5<& x& \le 71,\\ 0& \text{if }& 71<& x.& \\ \end{array}\\ P_{\text{u}}(x)& :=& \{\begin{array}{llrcl}{h}_1& \text{if }& & x& \le 15,\\ 4(h_3-h_1)(x-15{)}^3+h_1& \text{if }& 15<& x& \le 15.5,\\ 4(h_3-h_1)(x-16{)}^3+h_3& \text{if }& 15.5<& x& \le 16,\\ h_3& \text{if }& 16<& x& \le 94,\\ 4(h_3-h_4)(94-x{)}^3+h_3& \text{if }& 94<& x& \le 94.5,\\ 4(h_3-h_4)(95-x{)}^3+h_4& \text{if }& 94.5<& x& \le 95,\\ h_4& \text{if }& 95<& x.& \\ \end{array}\end{array}$

where $B=1.5\text{m}$ is the car's width and

$h_1:=1.1B+0.25,h_2:=3.5,h_3:=1.2B+3.75,h_4:=1.3B+0.25.$

Reference Solutions

Reference solutions for the case of a fixed time-grid are given in <bibref>Gerdts2005</bibref>. Solutions for a non-fixed time grid are given in <bibref>Gerdts2006</bibref>.

Source Code

C

The differential equations in C code:

// Controls
double C_steer = u[0];
double C_brake = u[1];
double C_acc   = u[2];

// Differential states
double X_v     = xd[2];
double X_beta  = xd[3];
double X_psi   = xd[4];
double X_wz    = xd[5];
double X_delta = xd[6];

// Intermediate values
double alpha_f, alpha_r, v_km_h, v_km_h2;
double F_Ax, F_Ay, F_Bf, F_Br, F_Rf, F_Rr, F_sf, F_sr, F_lr, F_lf;
double f_R, f_1, w_mot, f_2, f_3, M_mot, M_wheel;
double X_v_cos_X_beta, X_v_sin_X_beta;

X_v_cos_X_beta = X_v * cos ( X_beta );
X_v_sin_X_beta = X_v * sin ( X_beta );
alpha_f        = X_delta - atan( ( P_l_f * X_wz - X_v_sin_X_beta ) / X_v_cos_X_beta );
alpha_r        =           atan( ( P_l_r * X_wz + X_v_sin_X_beta ) / X_v_cos_X_beta );

F_sf    = P_D_f * sin( P_C_f * atan( P_B_f*alpha_f - P_E_f*(P_B_f*alpha_f - atan(P_B_f*alpha_f)) ) );
F_sr    = P_D_r * sin( P_C_r * atan( P_B_r*alpha_r - P_E_r*(P_B_r*alpha_r - atan(P_B_r*alpha_r)) ) );

F_Ax    = 0.5 * P_c_w * P_rho * P_A * X_v*X_v;
F_Ay    = 0.0;
F_Bf    = 2.0/3.0 * C_brake;
F_Br    = 1.0/3.0 * C_brake;

v_km_h  = X_v / 100.0 * 3.6;
v_km_h2 = v_km_h * v_km_h;
f_R     = P_f_R0 + P_f_R1 * v_km_h + P_f_R4 * v_km_h2 * v_km_h2;
F_Rf    = f_R * P_F_zf;
F_Rr    = f_R * P_F_zr;

f_1     = 1.0 - exp( -3.0 * C_acc );
w_mot   = X_v * P_ig * P_i_t / P_R;
f_2     = -37.8 + (1.54 - 0.0019 * w_mot_i) * w_mot;
f_3     = -34.9 - 0.04775 * w_mot;
M_mot   = f_1 * f_2_i + ( 1.0 - f_1 ) * f_3_i;
M_wheel = P_ig * P_i_t * M_mot_i;
	
F_lr    = M_wheel / P_R - F_Br - F_Rr;
F_lf    = - F_Bf - F_Rf;

// 0 Horizontal position x
rhs[0] = X_v * cos( X_psi - X_beta );
// 1 Vertical position y
rhs[1] = X_v * sin( X_psi - X_beta );
// 2 Velocity v
rhs[2] = 1.0 / P_m * (
	  (F_lr - F_Ax) * cos(X_beta) + F_lf * cos(X_delta + X_beta)
	- (F_sr - F_Ay) * sin(X_beta) - F_sf * sin(X_delta + X_beta) );
// 3 Side slip angle beta
rhs[3] = X_wz - 1.0 / (P_m * X_v) * (
	  (F_lr - F_Ax) * sin(X_beta) + F_lf * sin(X_delta + X_beta)
	+ (F_sr - F_Ay) * cos(X_beta) + F_sf * cos(X_delta + X_beta) );
// 4 Yaw angle psi
rhs[4] = X_wz;
// 5 Velocity of yaw angle w_z
rhs[5] = 1.0 / P_I_zz * (
	  F_sf * P_l_f * cos(X_delta) - F_sr * P_l_r
	+ F_lf * P_l_f * sin(X_delta) - F_Ay * P_e_SP );
// 6 Steering angle delta
rhs[6] = C_steer;

Variants

See testdrive overview page.

References

<bibreferences/>