Car Theoretical Handling Stability Simulation-MATLAB

introduction

Car handling stability means that the car can drive in the direction given by the driver through the steering system and steering wheels under the condition that the driver does not feel excessive tension and fatigue, and when encountering external disturbances, the car can resist the disturbance and maintain stability ability to drive. In the research of handling stability, the car is often regarded as a control system, so the evaluation index of car handling stability mainly includes time-domain response and frequency-domain response, and the time-domain response mainly includes the steady-state response and transient response under the steering wheel angle step input. State response, frequency domain response includes yaw rate frequency response characteristics.

1. Simulation parameters of vehicle handling stability

Total car mass m/kg	Moment of inertia $I/（kg\cdot m^2)$	Distance from center of mass to front axle a/m	Distance from center of mass to rear axle b/m	Comprehensive front wheel cornering stiffness $K1/(N/rad)$	Integrated rear wheel cornering stiffness $K2/(N/rad)$
3018	10437	1.84	1.88	-23147	-38318

2. Vehicle Stability Factor and Static Reserve Coefficient

1. MATLAB program

m=3018;%质量
I=10437;%转动惯量
a=1.84;%质心至前轴的距离
b=1.88;%质心至后轴的距离  
k1=-23147;%前轮综合侧偏刚度
k2=-38318;%后轮综合侧偏刚度
K=m./(a+b).^2.*(a./k2-b./k1);%计算稳定性因数 
SM=k2/(k1+k2)-a/(a+b);%计算静态储备系数
fprintf('汽车稳定性因数K=%.4f\n',K);%汽车稳定性因数输出
fprintf('汽车静态储备系数S.M.=%.4f\n',SM);%汽车静态储备系数输出

2. Running results

Vehicle stability factor $K=0.0072$
Automobile static reserve coefficient$S.M.=0.1288$

3. Draw the steady-state yaw rate gain curve of the car

1. MATLAB program

m=3018;%质量
I=10437;%转动惯量
a=1.84;%质心至前轴的距离
b=1.88;%质心至后轴的距离  
k1=-23147;%前轮综合侧偏刚度
k2=-38318;%后轮综合侧偏刚度
K=m/(a+b).^2*(a/k2-b/k1);%计算稳定性因数 
u=0:1:150;%定义速度范围 
w1=u/3.6/(a+b)./(1+0.8*K*u.^2/3.6^2);%计算汽车横摆角速度增益 
plot(u,w1,':')%绘制汽车横摆角速度增益曲线 
hold on; 
w2=u/3.6/(a+b)./(1+K.*u.^2/3.6^2);%计算汽车横摆角速度增益 
plot(u,w2)%绘制汽车横摆角速度增益曲线 
hold on;
w3=u/3.6/(a+b)./(1+1.2*K*u.^2/3.6^2);%计算汽车横摆角速度增益 
plot(u,w3,'--');%绘制汽车横摆角速度增益曲线 
hold on;         
axis([0,150,0,2]);%定义坐标轴范围 
xlabel('车速/(km/h)'); 
ylabel('汽车横摆角速度增益/(1/s)'); 
legend('0.8K','1.0K','1.2K');    

2. Running results

4. Draw the relationship curve between the lateral acceleration of the vehicle and the difference between the front and rear wheel slip angles

1. MATLAB program

m=3018;%质量
I=10437;%转动惯量
a=1.84;%质心至前轴的距离
b=1.88;%质心至后轴的距离  
k1=-23147;%前轮综合侧偏刚度
k2=-38318;%后轮综合侧偏刚度
K=m./(a+b).^2.*(a./k2-b./k1);%计算稳定性因数 
ay=0:0.05:1.5;%定义侧向加速度范围 
a1_a2=(a+b).* 0.8*K.*ay*9.8;%计算汽车前后侧偏角之差 
plot(ay,a1_a2);%绘制汽车前后侧偏角之差
hold on;                        
a11_a22=(a+b).*K.*ay*9.8;%计算汽车前后侧偏角之差 
plot(ay,a11_a22,':');%绘制汽车前后侧偏角之差 
hold on;                        
a111_a222=(a+b).* 1.2* K.*ay* 9.8;%计算汽车前后侧偏角之差 
plot(ay,a111_a222,'--');%绘制汽车前后侧偏角之差 
hold on;                       
xlabel('侧向加速度ay/g');
ylabel('侧偏角α/(°)');        
legend('0.8K','1.0K','1.2K'); 

2. Running results

5. Draw the relationship curve between the turning radius of the car and the square of the speed

1. MATLAB program

m=3018;%质量
I=10437;%转动惯量
a=1.84;%质心至前轴的距离
b=1.88;%质心至后轴的距离  
k1=-23147;%前轮综合侧偏刚度
k2=-38318;%后轮综合侧偏刚度
K=m./(a+b).^2.*(a./k2-b./k1);%计算稳定性因数 
u=0:1:20;%定义速度范围 
R1=1+0.8*K*u.^2;%计算汽车转向半径的值 
axis([0 300 0 3]);%设置坐标轴范围 
plot(u.^2,R1);%绘制汽车转向半径与速度平方的曲线
hold on;
R2=1+K*u.^2;%计算汽车转向半径的值 
plot(u.^2,R2,':');%绘制汽车转向半径与速度平方的曲线
hold on; 
R3=1+1.2*K*u.^2;%计算汽车转向半径的值 
plot(u.^2,R3,'-');%绘制汽车转向半径与速度平方的曲线
hold on;
xlabel('速度平方/(m^2/s^2)');
ylabel('转向半径/m'); 
legend('0.8K', '1.0K','1.2K');

2. Running results

Summarize

Stability factor K=0, neutral steering; stability factor K>0, understeer; stability factor K<0, oversteer; stability factor K=0.0072 of this car, that is, understeer. Static reserve coefficient SM=0, with neutral steering characteristics; static reserve coefficient SM>0, with understeer characteristics; static reserve coefficient SM<0, with excessive steering characteristics; the static reserve coefficient of this car is SM=0.1288, with Understeer characteristics.
Understeer, stability factor K>0, steady-state yaw rate gain $w_r/d=and/L/(1+K\ast u^2)$, the larger K is, the larger the understeer amount is;
from the relationship curve of the difference between the lateral acceleration of the vehicle and the side slip angle of the front and rear wheels, it can be known that when the stability factor (greater than zero) is constant, the greater the lateral acceleration, the greater the side slip angle. The greater the angle is; when the lateral acceleration is constant, the greater the stability factor (greater than zero), the greater the slip angle.
According to the relationship curve between the steering radius and the square of the speed, when the stability factor (greater than zero) is constant, the larger the square of the speed is, the larger the turning radius is; when the square of the speed is constant, the larger the stability factor (greater than zero), the turning radius bigger.
Note:Neutral steering, that is, stability factor K=0, steady-state yaw rate gain$w_r/d=u/L$ , the yaw rate gain has a linear relationship with the vehicle speed, and the slope is 1/L;; too much steering, the stability factor K<0, the steady-state yaw rate gain$w_r/d=and/L/(1+K\ast u^2)$, the smaller K is, the larger the amount of excessive steering is.