matlab汽车理论第一章

m1=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;

L=3.2;

a=1.947;

hg=0.9;

Iw=2*Iw1+4*Iw2;

for i=1:171

n(i)=(i+29)*20;

Ttq(i)=-19.313+295.27*(n(i)/1000)-165.44*(n(i)/1000)^2+40.874*(n(i)/1000)^3-3.8445*(n(i)/1000)^4;

end

for j=1:5

for i=1:171

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)^2/21.15+mz*g*f;

end

end

plot(ua,Ft,ua,Fz)

title('汽车驱动力与行驶阻力平衡图');

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

ylabel('Ft(N)');

for k=1:175

n1(k)=3300+k*0.1;

Ttq(k)=-19.313+295.27*(n1(k)/1000)-165.44*(n1(k)/1000)^2

+40.874*(n1(k)/1000)^33.8445*(n1(k)/1000)^4;

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

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

Fz(k)=CdA*ua(k)^2/21.15+mz*g*f;

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

end

for k=1:175