汽车理论matlab课后复习.docx

1、汽车理论matlab课后复习 一、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;

2、Iw1=1.798; Iw2=3.598; Iw=2*Iw1+4*Iw2; for i=1:69 n(i)=(i+11)*50; 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=1:69 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)2/21.15+mz*g*f; end end plot(ua,F

3、t,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)2 +40.874*(n1(k)/1000)33.8445*(n1(k)/1000)4; Ft(k)=Ttq(k)*ig(5)*i0*n

4、T/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 if(E(k)=min(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)2+40.874*(n2(p)/1000)3-3.8445*(n2(p)/1000)4; Ft(

5、p)=Ttq(p)*ig(1)*i0*nT/r; ua(p)=0.377*r*n2(p)/(ig(1)*i0); Fz(p)=CdA*ua(p)2/21.15+mz*g*f; af(p)=asin(Ft(p)-Fz(p)/(mz*g); end for p=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 for i=1:69 n(i)=(i+11)*50; Ttq(i)=-19.313+29

6、5.27*(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=1:69 deta=1+Iw/(mz*r2)+If*ig(j)2*i02*nT/(mz*r2); 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)2/21.15 -mz*g*f)/(deta*mz); if(a(i,j)0.05) b1(i,j)=a(i,j); u1(i,j)=ua(i,j); else

7、 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(0 120 0 30); xlabel(ua(km/h); ylabel(1/aj); gtext(1/a1)gtext(1

8、/a2)gtext(1/a3)gtext(1/a4)gtext(1/a5) for i=1:69 A=ua(i,3)-ua(69,2); if (A0) j=i; end B=ua(i,4)-ua(69,3); if(B0) k=i; end if(ua(i,4)=70) m=i; end end t=ua(1,2)*b(1,2); for p1=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 for p2=j:69 t2(p2)=(ua(p2,3)-ua(p2-1,3)*(b(p2,3)+b(p

9、2-1,3)*0.5; t=t+t2(p2); end for p3=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

10、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=5.56 2.769 1.644 1.00 0.793; 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; nd=400; Qid=0.299; for j=1:5 for i=1:69 n(i)=(i+11)*50; Ttq(i)=-19.313

11、+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; end end plot(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)gt

12、ext(IV)gtext(V)gtext(P阻) for j=1:5 for i=1:8 Td(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; u(i,j)=0.377*n1(i)*r/(ig(j)*i0); end 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*P

13、d(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)4-0.05

14、138*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)40.00075215*Pd(8)3+0.71113*Pd(8)245.291*Pd(8) +1129.7; u1=u(:,1); u2=u(:,2); u3=u(:,3); u4=u(:,4); u5=u(:,5); B1=polyfit(u1,b,3); B2=polyfit(u2,b,3); B3=poly

15、fit(u3,b,3); B4=polyfit(u4,b,3); B5=polyfit(u5,b,3); for q=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,5)=polyval(B5,ua(q,5); end for i=1:5 for q=1:69 Q(q,i)=Pz(q,i)*bh(q,i)/(1.02*ua(q,i)*7.05); end end plot(ua(:,4),Q(:,4)

16、,ua(:,5),Q(:,5); title(四档五档等速百公里油耗图); xlabel(ua(km/h); ylabel(Qs(L/100km); 三、改变1.3题中轻型货车的主减速器传动比，做出为5.17、5.43、5.83、6.17、6.33时的燃油经济性加速时间曲线，讨论不同值对汽车性能的影响。Matlab程序：m1=2000 ; m2=1800 ; m=3880 ; r0=0.367 ; gt=0.85 ; f=0.013 ; CDA=2.77 ; i0=5.83 ; If=0.218 ; Iw1=1.798 ; Iw2=3.598 ; Ig5=5.56 2.769 1.644 1.

17、00 0.793; Ig0=5.17 5.43 5.83 6.17 6.33; B=1326.8 -416.46 72.379 -5.8629 0.17768; 1354.7 -303.98 36.657 -2.0553 0.043072; 1284.4 -189.75 14.524 -0.51184 0.0068164; 1122.9 -121.59 7.0035 -0.18517 0.0018555; 1141.0 -98.893 4.4763 -0.091077 0.00068906; 1051.2 -73.714 2.8593 -0.05138 0.00035032; 1233.9 -

18、84.478 2.9788 -0.047449 0.00028230; 1129.7 -45.291 0.71113 -0.00075215 -0.000038568;n=815 1207 1614 2012 2603 3006 3403 3804;for i=1:5 for k=1:8ua(i,k)=0.377*0.367*n(k)/(Ig0(i)*Ig5(5); 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; F5(i,k)=0.013

19、*3880*9.8+2.77.*ua(i,k)2/21.15;Pe(i,k)=F5(i,k)*ua(i,k)/(3600*0.85);b5(i,k)=B(k,1)+B(k,2)*Pe(i ,k)+B(k,3)*Pe(i,k)2+B(k,4)*Pe(i,k)3+B(k,5)*Pe(i,k)4; endend ua1=25; s1=50;Fa5=0.013*3880*9.8+2.77.*ua1.2/21.15;Pe5=Fa5.*ua1/(3600*0.85);d1=polyfit(Pe(1,:),b5(1,:),3);ba1=polyval(d1,Pe5);d2=polyfit(Pe(2,:),b

20、5(2,:),3);ba2=polyval(d2,Pe5);d3=polyfit(Pe(3,:),b5(3,:),3);ba3=polyval(d3,Pe5);d4=polyfit(Pe(4,:),b5(4,:),3);ba4=polyval(d4,Pe5);d5=polyfit(Pe(5,:),b5(5,:),3);ba5=polyval(d5,Pe5) ;ba=ba1 ba2 ba3 ba4 ba5;Qa1=Pe5.*ba*50/(ua1*102*7)ua2=25:40;Q2=1+(2*Iw1+4*Iw2)/(m*r02)+If*Ig5(2).2*Ig0(2)2*gt/(m*r02)Fb5

21、=0.013*3880*9.8+2.77.*ua2.2/21.15+Q2*m*0.25;Pb5=Fb5.*ua2/(3600*0.85);db1=polyfit(Pe(1,:),b5(1,:),3);bb1=polyval(db1,Pb5);db2=polyfit(Pe(2,:),b5(2,:),3);bb2=polyval(db2,Pb5);db3=polyfit(Pe(3,:),b5(3,:),3);bb3=polyval(db3,Pb5);db4=polyfit(Pe(4,:),b5(4,:),3);bb4=polyval(db4,Pb5);db5=polyfit(Pe(5,:),b5(

22、5,:),3);bb5=polyval(db5,Pb5);bb=bb1 bb2 bb3 bb4 bb5;Pb=Pb5 Pb5 Pb5 Pb5 Pb5;Qb=Pb.*bb/(367.1*7);for i=1:5 for j=1:15 qb(i,j)=Qb(i,j)+Qb(i,j+1); end end Qb2=sum(qb)ua3=40; s2=250;Fc5=0.013*3880*9.8+2.77.*ua3.2/21.15;Pc5=Fc5.*ua3/(3600*0.85);dc1=polyfit(Pe(1,:),b5(1,:),3);bc1=polyval(dc1,Pc5);dc2=polyf

23、it(Pe(2,:),b5(2,:),3);bc2=polyval(dc2,Pc5);dc3=polyfit(Pe(3,:),b5(3,:),3);bc3=polyval(dc3,Pc5);dc4=polyfit(Pe(4,:),b5(4,:),3);bc4=polyval(dc4,Pc5);dc5=polyfit(Pe(5,:),b5(5,:),3);bc5=polyval(dc5,Pc5) ;bc=bc1 bc2 bc3 bc4 bc5;Qc3=Pc5.*bc*250/(ua3*102*7)ua4=40:50;Q4=1+(2*Iw1+4*Iw2)/(m*r02)+If*Ig5(4).2*I

24、g0(4)2*gt/(m*r02)Fd5=0.013*3880*9.8+2.77.*ua4.2/21.15+Q4*m*0.2;Pd5=Fd5.*ua4/(3600*0.85);dd1=polyfit(Pe(1,:),b5(1,:),3);bd1=polyval(dd1,Pd5);dd2=polyfit(Pe(2,:),b5(2,:),3);bd2=polyval(dd2,Pd5);dd3=polyfit(Pe(3,:),b5(3,:),3);bd3=polyval(dd3,Pd5);dd4=polyfit(Pe(4,:),b5(4,:),3);bd4=polyval(dd4,Pd5);dd5=

25、polyfit(Pe(5,:),b5(5,:),3);bd5=polyval(dd5,Pd5);bd=bd1 bd2 bd3 bd4 bd5;Pd=Pd5 Pd5 Pd5 Pd5 Pd5;Qd=Pd.*bd/(367.1*7);for i=1:5 for j=1:10 qd(i,j)=Qd(i,j)+Qd(i,j+1); end end Qd4=sum(qd)ua5=50; s2=250;Ff5=0.013*3880*9.8+2.77.*ua5.2/21.15;Pf5=Ff5.*ua5/(3600*0.85);df1=polyfit(Pe(1,:),b5(1,:),3);bf1=polyval

26、(df1,Pf5);df2=polyfit(Pe(2,:),b5(2,:),3);bf2=polyval(df2,Pf5);df3=polyfit(Pe(3,:),b5(3,:),3);bf3=polyval(df3,Pf5);df4=polyfit(Pe(4,:),b5(4,:),3);bf4=polyval(df4,Pf5);df5=polyfit(Pe(5,:),b5(5,:),3);bf5=polyval(df5,Pf5) ;bf=bf1 bf2 bf3 bf4 bf5;Qf5=Pf5.*bf*250/(ua3*102*7)Qi=0.299;Qg=(50-25)/3.6/0.36*0.

27、299;Qg6=Qg Qg Qg Qg QgQ=Qa1 Qb2 Qc3 Qd4 Qf5 Qg6;Qz=sum(Q)/1075*100for k=1:5for i=1:3401; for j=1:5; n(i)=i+599; ua(i,j)=0.377*r0*n(i)./(Ig5(j)*Ig0(k); Q(j)=1+(2*Iw1+4*Iw2)/(m*r02)+If*Ig5(j).2*Ig0(k)2*gt/(m*(r0)2); endenduamax=max(ua);uamin=min(ua);ua2=uamin(2):uamax(2);n=ua2*Ig0(k)*Ig5(2)/(0.377*0.3

28、67);Ttq=-19.313+295.27*(n/1000)-165.44*(n/1000).2+40.874*(n/1000).3-3.8445*(n/1000).4; Ft2=Ttq*Ig0(k)*Ig5(2)*0.85/0.367;F2=0.013*3880*9.8+2.77*ua2.2/21.15;a2=(Ft2-F2)./(Q(2)*m);t2=trapz(ua2,1./a2)/3.6;ua3=uamax(2):uamax(3);n3=ua3*Ig0(k)*Ig5(3)/(0.377*0.367);Ttq3=-19.313+295.27*(n3/1000)-165.44*(n3/1

29、000).2+40.874*(n3/1000).3-3.8445*(n3/1000).4; Ft3=Ttq3*Ig0(k)*Ig5(3)*0.85/0.367;F3=0.013*3880*9.8+2.77*ua3.2/21.15;a3=(Ft3-F3)./(Q(3)*m);t3=trapz(ua3,1./a3)/3.6;ua4=uamax(3):70;n4=ua4*Ig0(k)*Ig5(4)/(0.377*0.367);Ttq4=-19.313+295.27*(n4/1000)-165.44*(n4/1000).2+40.874*(n4/1000).3-3.8445*(n4/1000).4; Ft4=Ttq4*Ig0(k)*Ig5(4)*0.85/0.367;F4=0.013*3880*9.8+2.77*ua4