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1、汽车理论课后习题MATLAB编程1.3 确定一轻型货车的动力性能（货车可装用4挡或5挡变速器，任选其中的一种进行整车性能计算）：1）绘制汽车驱动力与行驶阻力平衡图。2）求汽车最高车速，最大爬坡度及克服该坡度时相应的附着率。3）绘制汽车行驶加速度倒数曲线，用图解积分法求汽车用2档起步加速行驶至70km/h的车速时间曲线，或者用计算机求汽车用2档起步加速行驶至70km/h的加速时间。解：(1) 求汽车驱动力与行驶阻力平衡图和汽车最高车速：n=600:10:4000;Tq=-19.313+295.27*(n/1000)-165.44*(n/1000).2+40.874*(n/1000).3-3.84
2、45*(n/1000).4;m=3880;g=9.8;nmin=600;nmax=4000;G=m*g;ig=5.56 2.769 1.644 1.00 0.793;nT=0.85;r=0.367;f=0.013;CDA=2.77;i0=5.83;L=3.2;a=1.947;hg=0.9;If=0.218;Iw1=1.798;Iw2=3.598;Ft1=Tq*ig(1)*i0*nT/r;Ft2=Tq*ig(2)*i0*nT/r;Ft3=Tq*ig(3)*i0*nT/r;Ft4=Tq*ig(4)*i0*nT/r;Ft5=Tq*ig(5)*i0*nT/r;ua1=0.377*r*n/ig(1)/i
3、0;ua2=0.377*r*n/ig(2)/i0;ua3=0.377*r*n/ig(3)/i0;ua4=0.377*r*n/ig(4)/i0;ua5=0.377*r*n/ig(5)/i0;ua=0:5:120;Ff=G*f;Fw=CDA*ua.2/21.15;Fz=Ff+Fw;plot(ua1,Ft1,ua2,Ft2,ua3,Ft3,ua4,Ft4,ua5,Ft5,ua,Fz);title(驱动力-行驶阻力平衡图);xlabel(ua(km/s);ylabel(Ft(N);gtext(Ft1),gtext(Ft2),gtext(Ft3),gtext(Ft4),gtext(Ft5),gtext(
4、Ff+Fw);zoom on;x,y=ginput(1);zoom off;disp(汽车最高车速=);disp(x);disp(km/h);汽车最高车速= 99.3006km/h(2)求汽车最大爬坡度程序：n=600:10:4000;Tq=-19.313+295.27*(n/1000)-165.44*(n/1000).2+40.874*(n/1000).3-3.8445*(n/1000).4;m=3880;g=9.8;nmin=600;nmax=4000;G=m*g;ig=5.56 2.769 1.644 1.00 0.793;nT=0.85;r=0.367;f=0.013;CDA=2.77
5、;i0=5.83;L=3.2;a=1.947;hg=0.9;If=0.218;Iw1=1.798;Iw2=3.598;Ft1=Tq*ig(1)*i0*nT/r;ua1=0.377*r*n/ig(1)/i0;Ff=G*f;Fw1=CDA*ua1.2/21.15;Fz1=Ff+Fw1;Fi1=Ft1-Fz1;Zoom on;imax=100*tan(asin(max(Fi1/G);disp(汽车最大爬坡度=);disp(imax);disp(%);汽车最大爬坡度=35.2197%(3)求最大爬坡度相应的附着率和求汽车行驶加速度倒数曲线程序：clearn=600:10:4000;Tq=-19.313
6、+295.27*(n/1000)-165.44*(n/1000).2+40.874*(n/1000).3-3.8445*(n/1000).4;m=3880;g=9.8;nmin=600;nmax=4000;G=m*g;ig=5.56 2.769 1.644 1.00 0.793;nT=0.85;r=0.367;f=0.013;CDA=2.77;i0=5.83;L=3.2;a=1.947;hg=0.9;If=0.218;Iw1=1.798;Iw2=3.598;Ft1=Tq*ig(1)*i0*nT/r;Ft2=Tq*ig(2)*i0*nT/r;Ft3=Tq*ig(3)*i0*nT/r;Ft4=Tq
7、*ig(4)*i0*nT/r;Ft5=Tq*ig(5)*i0*nT/r;ua1=0.377*r*n/ig(1)/i0;ua2=0.377*r*n/ig(2)/i0;ua3=0.377*r*n/ig(3)/i0;ua4=0.377*r*n/ig(4)/i0;ua5=0.377*r*n/ig(5)/i0;Fw1=CDA*ua1.2/21.15;Fw2=CDA*ua2.2/21.15;Fw3=CDA*ua3.2/21.15;Fw4=CDA*ua4.2/21.15;Fw5=CDA*ua5.2/21.15;Ff=G*f;deta1=1+(Iw1+Iw2)/(m*r2)+(If*ig(1)2*i02*nT
8、)/(m*r2);deta2=1+(Iw1+Iw2)/(m*r2)+(If*ig(2)2*i02*nT)/(m*r2);deta3=1+(Iw1+Iw2)/(m*r2)+(If*ig(3)2*i02*nT)/(m*r2);deta4=1+(Iw1+Iw2)/(m*r2)+(If*ig(4)2*i02*nT)/(m*r2);deta5=1+(Iw1+Iw2)/(m*r2)+(If*ig(5)2*i02*nT)/(m*r2);a1=(Ft1-Ff-Fw1)/(deta1*m);ad1=1./a1;a2=(Ft2-Ff-Fw2)/(deta2*m);ad2=1./a2;a3=(Ft3-Ff-Fw3)
9、/(deta3*m);ad3=1./a3;a4=(Ft4-Ff-Fw4)/(deta4*m);ad4=1./a4;a5=(Ft5-Ff-Fw5)/(deta5*m);ad5=1./a5;plot(ua1,ad1,ua2,ad2,ua3,ad3,ua4,ad4,ua5,ad5);axis(0 99 0 10);title(汽车的加速度倒数曲线);xlabel(ua(km/h);ylabel(1/a);gtext(1/a1);gtext(1/a2);gtext(1/a3);gtext(1/a4);gtext(1/a5);a=max(a1);af=asin(max(Ft1-Ff-Fw1)/G);C=
10、tan(af)/(a/L+hg*tan(af)/L);disp(假设后轮驱动，最大爬坡度相应的附着率=);disp(C); 假设后轮驱动，最大爬坡度相应的附着率= 0.4219(4) clearnT=0.85;r=0.367;f=0.013;CDA=2.77;i0=5.83;If=0.218;Iw1=1.798;Iw2=3.598;L=3.2;a=1.947;hg=0.9;m=3880;g=9.8;G=m*g; ig=5.56 2.769 1.644 1.00 0.793;nmin=600;nmax=4000;u1=0.377*r*nmin./ig/i0;u2=0.377*r*nmax./ig
11、/i0;deta=0*ig;for i=1:5 deta(i)=1+(Iw1+Iw2)/(m*r2)+(If*(ig(i)2*i02*nT)/(m*r2);endua=6:0.01:99;N=length(ua);n=0;Tq=0;Ft=0;inv_a=0*ua;delta=0*ua;Ff=G*f;Fw=CDA*ua.2/21.15;for i=1:N k=i; if ua(i)=u2(2) n=ua(i)*(ig(2)*i0/r)/0.377; Tq=-19.313+295.27*(n/1000)-165.44*(n/1000)2+40.874*(n/1000)3-3.8445*(n/100
12、0)4; Ft=Tq*ig(2)*i0*nT/r; inv_a(i)=(deta(2)*m)/(Ft-Ff-Fw(i); delta(i)=0.01*inv_a(i)/3.6; elseif ua(i)=u2(3) n=ua(i)*(ig(3)*i0/r)/0.377; Tq=-19.313+295.27*(n/1000)-165.44*(n/1000)2+40.874*(n/1000)3-3.8445*(n/1000)4; Ft=Tq*ig(3)*i0*nT/r; inv_a(i)=(deta(3)*m)/(Ft-Ff-Fw(i); delta(i)=0.01*inv_a(i)/3.6; elseif ua(i) ginputans = 25.8223 70.0737 25.7467 70.0737所以汽车2档原地起步换挡加速行驶至70km/h的加速时间约为25.8s2.7已知货车装用汽油发动机的负荷特性与万有特性。负荷特性曲线的拟合公式为：其中，b为燃油消耗率g/(kWh)；Pe为发动机净功率（kW）；拟合式中的系数随转速n变化。怠速油耗（怠速转速400r/min）。计算与绘制题1.3中货车的
