汽车理论习题答案.doc

1.3、1）绘制汽车驱动力与行驶阻力平衡图。2）求汽车的最高车速、最大爬坡度及克服该坡度时相应的附着率。3）绘制汽车行驶加速倒数曲线，用图解积分法求汽车有档起步加速行驶至70km/h的车速时间曲线，或者用计算机求汽车用档起步加速至70km/h的加速时间。n_min=600;n_max=4000;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;i0=5.83;If=0.218;Iw1=1.798;Iw2=3.598;Iw=2*Iw1+4*Iw2;L=3.2;a=1.947;b=L-a;hg=0.9;g=9.8;1）绘制汽车驱动力与行驶阻力平衡图：for i=1:3401 n(i)=599+i; 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;endfor j=1:4 for i=1:3401 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 endplot(ua,Ft,ua,Fz) ,gridtitle(汽车驱动力与行驶阻力平衡图);xlabel(ua(km/h);ylabel(Ft(N); 2）汽车的最高车速： 最大爬坡度： for k=1:1000 n1(k)=1800+k*0.5; Ttq(k)=-19.313+295.27*(n1(k)/1000)-165.44*(n1(k)/1000)2+40.874*(n1(k)/1000)3-3.8445*(n1(k)/1000)4;Ft(k)=Ttq(k)*ig(1)*i0*nT/r;ua(k)=0.377*r*n1(k)/(ig(1)*i0); Fz(k)=CdA*ua(k)2/21.15+mz*g*f; af(k)=asin(Ft(k)-Fz(k)/(mz*g);endfor k=1:1000 if(af(k)=max(af) i=tan(af(k); disp(汽车最大爬坡度=); disp(i); endend汽车最大爬坡度= 0.3522相应的附着率： 3）绘制汽车行驶加速倒数曲线:n=(n_min:1:n_max);N=n./1000;Tq=-19.313+295.27*N-165.44*N.2+40.874*N.3-3.8445*N.4;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;ua1=0.377*n*r/ig(1) /i0;ua2=0.377*n*r/ig(2)/i0;ua3=0.377*n*r/ig(3)/i0;ua4=0.377*n*r/ig(4)/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;Ff= mz*g*f;deta1=1+Iw/(mz*r2)+If*ig(1)2*i02*nT/(mz*r2);deta2=1+Iw/(mz*r2)+If*ig(2)2*i02*nT/(mz*r2);deta3=1+Iw/(mz*r2)+If*ig(3)2*i02*nT/(mz*r2);deta4=1+Iw/(mz*r2)+If*ig(4)2*i02*nT/(mz*r2);a1=( Ft1- Fw1- Ff)/(deta1*mz);a2=( Ft2- Fw2- Ff)/(deta2*mz);a3=( Ft3- Fw3- Ff)/(deta3*mz);a4=( Ft4- Fw4- Ff)/(deta4*mz);x1=1./a1;x2=1./a2;x3=1./a3;x4=1./a4;plot(ua1,x1, k, ua2,x2, b,ua3,x3, r, ua4,x4,y),gridtitle(加速度倒数曲线);xlabel(ua(km/h);ylabel(1/a);汽车用档起步加速至70km/h的加速时间2.7、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。 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;i0=5.83; If=0.218;Iw1=1.798;Iw2=3.598;n1=815 1207 1614 2012 2603 3006 3403 3804;Iw=2*Iw1+4*Iw2;Qid=0.299;L=3.2;a=1.947;b=L-a;hg=0.9;g=9.8;1）汽车功率平衡图：for j=1:4 for i=1:3401 n(i)=599+i; 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; 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)3/76140.)/nT; endend plot(ua,Pe,ua,Pz),grid;title(功率平衡图);xlabel(ua(km/h);ylabel(Pe,Pz(kw);for i=1:8Td(i)=-19.313+295.27*(n1(i)/1000.0)-165.44*(n1(i)/1000.0)2+40.874*(n1(i)/1000.0)3-3.8445*(n1(i)/1000.0)4;Pd(i)=n1(i)*Td(i)/9549;u3(i)=0.377*n1(i)*r/(ig(3)*i0);u4(i)=0.377*n1(i)*r/(ig(4)*i0);end b(1)=0.17768*Pd(1)4-5.8629*Pd(1)3+72.379*Pd(1)2-416.46*Pd(1)+1326.8;b(2)=0.043072*Pd(2)4-2.0553*Pd(2)3+36.657*Pd(2)2-303.98*Pd(2)+1354.7;b(3)=0.0068164*Pd(3)4-0.51184*Pd(3)3+14.524*Pd(3)2-189.75*Pd(3)+1284.4;b(4)=0.0018555*Pd(4)4-0.18517*Pd(4)3+7.0035*Pd(4)2-121.59*Pd(4)+1122.9;b(5)=0.00068906*Pd(5)4-0.091077*Pd(5)3+4.4763*Pd(5)2-98.893*Pd(5)+1141.0;b(6)=0.00035032*Pd(6)4-0.05138*Pd(6)3+2.8593*Pd(6)2-73.714*Pd(6)+1051.2;b(7)=0.00028230*Pd(7)4-0.047449*Pd(7)3+2.9788*Pd(7)2-84.478*Pd(7)+1233.9;b(8)=-0.000038568*Pd(8)4-0.00075215*Pd(8)3+0.71113*Pd(8)2-45.291*Pd(8)+1129.7;for i=1:8Qs3(i)=Pd(i)*b(i)/(1.02*u3(i)*7.15);Qs4(i)=Pd(i)*b(i)/(1.02*u4(i)*7.15);endplot(u3,Qs3, *) ,grid;hold on;plot(u4,Qs4, );title(3档4档等速百公里油耗图);xlabel(u (km/h);ylabel(Qs(L/100km);3、改变1.3题中轻型货车的主减速器传动比，作出i0为5.17、5.43、5.83、6.17、6.33时的燃油经济性加速时间曲线，讨论不同i0值对汽车性能的影响和采用不同变速器对汽车性能的影响。（以最高档，较高转速（n取3403r/min），最经济负荷（即90%负荷大约60Kw）行驶时的油耗）：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;i0=5.17 5.43 5.83 6.17 6.33;If=0.218;Iw1=1.798;Iw2=3.598;Iw=2*Iw1+4*Iw2;n=3403;Qid=0.299;Pe=60;b=0.00028230*Pe4-0.047449*Pe3+2.9788*Pe2-84.478*Pe+1233.9;for i=1:5ua(i)=0.377*n*r/ig(4) /i0(i);Qs(i)=Pe*b/(1.02*ua(i)*7.15);enddisp(百公里消耗燃油=);disp(Qs);disp(L);百公里消耗燃油= 26.9678 28.3240 30.4105 32.1840 33.0186 Lfor q=1:5for i=1:3401 n(i)=599+i; 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 i=1:3401 deta=1+Iw/(mz*r2)+If*ig(4)2*i0(q)2*nT/(mz*r2); ua(i)=0.377*r*n(i)/(ig(4)*i0(q); a(i)=(Ttq(i)*ig(4)*i0(q)*nT/r-CdA*ua(i)2/21.15-mz*g*f)/(deta*mz); b1(i)=a(i);b(i)=1/b1(i);endfor i=1:3401 if(ua(i)=96.6) m=i; endend t=ua(1)*b(1); for p=2:m t1(p)=(ua(p)-ua(p-1)*(b(p)+b(p-1)*0.5; t=t+t1(p);endtz=t/3.6;disp(加速时间=);disp(tz);disp(s); end加速