汽车理论matlab作业.docx
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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=[5.562.7691.6441.000.793];i0=5.83;
If=0.218;Iw1=1.798;Iw2=3.598;
Iw=2*lw1+4*lw2;
fori=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
forj=1:
5
fori=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
fork=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*(n1(k)/1000)A33.8445*(n1(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
fork=1:
175
if(E(k)==min(E))
disp('汽车最高车速=');
disp(ua(k));
disp('km/h');
end
end
forp=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)=asin((Ft(p)-Fz(p))/(mz*g));
end
forp=1:
150
if(af(p)==max(af))
i=tan(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
fori=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)A3-3.8445*(n(i)/1000)A4;
end
forj=1:
5
fori=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);y1=b(:
1);
x2=u1(:
2);y2=b(:
2);
x3=u1(:
3);y3=b(:
3);
x4=u1(:
4);y4=b(:
4);
x5=u1(:
5);y5=b(:
5);
Plot(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5);
title('加速度倒数时间曲线');
axis([O120030]);
xlabel('ua(km/h)');
ylabel('1/aj');
gtext('1/a1')
gtext('1/a2')
gtext('1/a3')
gtext('1/a4')
gtext('1/a5')
加速度倒載时间曲线
fori=1:
69
A=ua(i,3)-ua(69,2);if(A<1&A>0)
j=i;
end
B=ua(i,4)-ua(69,3);
if(B<2&B>0)
k=i;
end
if(ua(i,4)<=70)
m=i;
end
endt=ua(1,2)*b(1,2);forp1=2:
69
t1(p1)=(ua(p1,2)-ua(p1-1,2))*(b(p1,2)+b(p1-1,2))*0.5;t=t+t1(p1);
end
forp2=j:
69
t2(p2)=(ua(p2,3)-ua(p2-1,3))*(b(p2,3)+b(p2-1,3))*0.5;t=t+t2(p2);
end
forp3=k:
m
t3(p3)=(ua(p3,4)-ua(p3-1,4))*(b(p3,4)+b(p3-1,4))*0.5;t=t+t3(p3);
end
t=t+(ua(j,3)-ua(69,2))*b(69,2)+(ua(k,4)-ua(69,3))*b(69,3)+(70-ua(m,4))*b(m,4);
tz=t/3.6;
disp('加速时间=');
disp(tz);
disp('s');
加速时间=29.0585s
二、计算与绘制题1中货车的1)汽车功率平衡图;
2)最高档与次高档的等速百公里油耗曲线。
已知数据略。
(参见《汽车理论》习题第二章第7题)解题程序如下:
用Matlab语言
m1=2000;m2=1800;mz=3880;g=9.81;
r=0.367;CdA=2.77;f=0.013;nT=0.85;
ig=[5.562.7691.6441.000.793];
i0=5.83;If=0.218;Iw1=1.798;Iw2=3.598;
n1=[8151207161420122603300634033804];
Iw=2*Iw1+4*Iw2;
nd=400;Qid=0.299;
forj=1:
5
fori=1:
69n(i)=(i+11)*50;
Ttq(i)=-19.313+295.27*(n(i)/1000)-165.44*(n(i)/1000)人2+40.874*(n(。
/1000)人3-3.8445*(n(0/1000)人4;
Pe(i)=n(i)*Ttq(i)/9549;
ua(i,j)=0.377*r*n(i)/(ig(j)*i0);
Pz(i,j)=(mz*g*f*ua(i,j)/3600.+CdA*ua(i,j)A3/76140.)/nT;
end
endplot(ua,Pe,ua,Pz);
title('汽车功率平衡图)');
xlabel('ua(km/h)');
ylabel('Pe,Pz(kw)');
gtext('I')
gtext('II')
gtext('III')
gtext('IV')
gtext('V')
gtext('P阻')
汽车功率平衡图
40
60
ua/(krrVh)
80
m120
wo
90
70
Gn
5n
40
3n
°020
MWOJd
forj=1:
5
fori=1:
8
Td(i)=-19.313+295.27*(n1(i)/1000.0)-165.44*(n1(i)/1000.0)A2+40.874*(n1(i)/1000.0)人3-3.8445*(n1(i)/1000.0)A4;
Pd(i)=n1(i)*Td(i)/9549;
u(i,j)=0.377*n1(i)*r/(ig(j)*i0);
endendb
(1)=0.17768*Pd
(1)A4-5.8629*Pd
(1)A3+72.379*Pd
(1)A2-416.46*Pd
(1)+1326.8;b
(2)=0.043072*Pd
(2)A4-2.0553*Pd
(2)A3+36.657*Pd
(2)A2-303.98*Pd
(2)+1354.7;b(3)=0.0068164*Pd(3)A4-0.51184*Pd(3)A3+14.524*Pd(3)A2-189.75*Pd(3)+1284.4;b(4)=0.0018555*Pd(4)A4-0.18517*Pd(4)A3+7.0035*Pd(4)A2-121.59*Pd(4)+1122.9;b(5)=0.00068906*Pd(5)A4-0.091077*Pd(5)^3+4.4763*Pd(5)A2-98.893*Pd(5)+1141.0;b(6)=0.00035032*Pd(6)A4-0.05138*Pd(6)A3+2.8593*Pd(6)A2-73.714*Pd(6)+1051.2;b(7)=0.00028230*Pd(7)A4-0.047449*Pd(7)A3+2.9788*Pd(7)A2-84.478*Pd(7)+1233.9;b(8)=-0.000038568*Pd(8)A40.00075215*Pd(8)A3+0.71113*Pd(8)A245.291*Pd(8)+1129.7;
u1=u(:
1)';u2=u(:
2):
u3=u(:
3)';u4=u(:
4)';u5=u(:
5)';
B仁polyfit(u1,b,3);
B2=polyfit(u2,b,3);
B3=polyfit(u3,b,3);
B4=polyfit(u4,b,3);
B5=polyfit(u5,b,3);
forq=1:
69
bh(q,1)=polyval(B1,ua(q,1));
bh(q,2)=polyval(B2,ua(q,2));
bh(q,3)=polyval(B3,ua(q,3));
bh(q,4)=polyval(B4,ua(q,4));
bh(q,