汽车理论matlab作业

、确定一轻型货车的动力性能

1)绘制汽车驱动力与行驶阻力平衡图；

2)求汽车最高车速与最大爬坡度；

3)绘制汽车行驶加速度倒数曲线；用计算机求汽车用H档起步加速行驶至70km/h所需

的加速时间。

已知数据略。(参见《汽车理论》习题第一章第3题)

解题程序如下:用Matlab语言

(1)绘制汽车驱动力与行驶阻力平衡图—

m仁2000; m2=1800; mz=3880;

g=9.81; r=0.367; CdA=2.77; f=0.013; nT=0.85;

ig=[5.56 2.769 1.644 1.00 0.793]; i0=5.83;

If=0.218; Iw1=1.798; Iw2=3.598;

Iw=2*lw1+4*lw2;

for i=1:69

n (i)=(i+11)*50;

Ttq(i)=-19.313+295.27*( n(i)/1000)-165.44*( n(i)/1000)A2+40.874*( n( i)/1000)A3-3.8

445*( n( i)/1000)A4;

end

for j=1:5

for i=1:69

Ft(i,j)=Ttq(i)*ig(j)*i0* nT/r;

ua(i,j)=0.377*r* n( i)/(ig(j)*i0);

Fz(i,j)=CdA*ua(i,j)A2/21.15+mz*g*f;

end

end

plot(ua,Ft,ua,Ff,ua,Ff+Fw)

title(' 汽车驱动力与行驶阻力平衡图');

xlabel('ua(km/h)');

ylabel('Ft(N)');

gtext( 'Ft1')

gtext( 'Ft2' )

gtext( 'Ft3' )

gtext( 'Ft4' )

gtext( 'Ft5' )

gtext( 'Ff+Fw')

(2)求最大速度和最大爬坡度

for k=1:175

n1(k)=3300+k*0.1;

Ttq(k)=-19.313+295.27*( n1(k)/1000)-165.44*( n1(k)/1000)A2

+40.874* (n 1(k)/1000)A33.8445* (n 1(k)/1000)A4;

Ft(k)=Ttq(k)*ig (5)*i0* nT/r;

ua(k)=0.377*r* n1(k)/(ig(5)*i0);

Fz(k)=CdA*ua(k)A2/21.15+mz*g*f;

E(k)=abs((Ft(k)-Fz(k)));

end

for k=1:175

if(E(k)==mi n(E))

disp(' 汽车最高车速=');

disp(ua(k));

disp('km/h');

end

end

for p=1:150

n2(p)=2000+p*0.5;

Ttq(p)=-19.313+295.27*( n2(p)/1000)-165.44*( n2(p)/1000)A2+40.874*( n2(p)/1000) 人3-

3.8445*( n2(p)/1000)M;

Ft(p)=Ttq(p)*ig(1)*iO* nT/r;

ua(p)=0.377*r* n2(p)/(ig(1)*i0);

Fz(p)=CdA*ua(p)A2/21.15+mz*g*f;

af(p)=as in ((Ft(p)-Fz(p))/(mz*g));

end

for p=1:150

if(af(p)==max(af))

i=ta n(af(p));

disp(' 汽车最大爬坡度=');

disp(i);

end

end

汽车最高车速=99.0679km/h

汽车最大爬坡度=0.3518

(3)计算2档起步加速到70km/h所需时间

for i=1:69

n(i)=(i+11)*50;

Ttq(i)=-19.313+295.27*( n(i)/1000)-165.44*( n(i)/1000)A2+40.874*( n( i)/1000)A3-3.8 445*( n( i)/1000)A4;

end

for j=1:5

for i=1:69

deta=1+lw/(mz*「A2)+lf*ig(j)A2*i0A2* nT/(mz*「A2);

ua(i,j)=0.377*r* n( i)/(ig(j)*i0);

a(i,j)=(Ttq(i)*ig(j)*i0* nT/r-CdA*ua(i,j)A2/21.15

-mz*g*f)/(deta*mz);

if(a(i,j)<=0)

a(i,j)=a(i-1,j);

end

if(a(i,j)>0.05)

b1(i,j)=a(i,j);

u1(i,j)=ua(i,j);

else

b1(i,j)=a(i-1,j);

u1(i,j)=ua(i-1,j);

end

b(i,j)=1/b1(i,j);

end

end

x1=u1(:,1);y 1= b(:,1);