汽車理論matlab作業(yè) 一、確定一輕型貨車的動力性能。 歐陽學(xué)文 1〕繪制汽車驅(qū)動力與行駛阻力平衡圖； 2〕求汽車最高車速與最大爬坡度； 3〕繪制汽車行駛加速度倒數(shù)曲線；用計算機求汽車用Ⅱ檔起步加速行駛至70km/h所需 的加速時間。 已知數(shù)據(jù)略?！矃⒁姡ㄆ嚴碚摚┝?xí)題第一章第3題〕 解題程序如下:用Matlab語言 (1)繪制汽車驅(qū)動力與行駛阻力平衡圖 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; Iw=2*Iw1+4*Iw2; fori=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)^33.8445*(n(i)/1000)^4;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)^2/21.15+mz*g*f;end end plot(ua,Ft,ua,Ff,ua,Ff+Fw) title('汽車驅(qū)動力與行駛阻力平衡圖');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)^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 fork=1:175 if(E(k)==min(E)) disp('汽車最高車速='); 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)^2+40.874*(n2(p)/1000) ^33.8445*(n2(p)/1000)^4; Ft(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 forp=1:150 if(af(p)==max(af)) i=tan(af(p)); disp('汽車最大爬坡度='); disp(i); end end 汽車最高車速=99.0679km/h 汽車最大爬坡度=0.3518 〔3〕計算2檔起步加速到70km/h所需時間 fori=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)^33.8445*(n(i)/1000)^4; end forj=1:5 fori=1:69 deta=1+Iw/(mz*r^2)+If*ig(j)^2*i0^2*nT/(mz*r^2); ua(i,j)=0.377*r*n(i)/(ig(j)*i0); a(i,j)=(Ttq(i)*ig(j)*i0*nT/rCdA*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 b1(i,j)=a(i1,j); u1(i,j)=ua(i1,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('加速度倒數(shù)時間曲線');axis([0120030]); xlabel('ua(km/h)'); ylabel('1/aj'); gtext('1/a1') gtext('1/a2') gtext('1/a3') gtext('1/a4') gtext('1/a5') fori=1:69A=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) t=ua(1,2)*b(1,2); forp1=2:69 t1(p1)=(ua(p1,2)ua(p11,2))*(b(p1,2)+b(p11,2))*0.5;t=t+t1(p1); end forp2=j:69 t2(p2)=(ua(p2,3)ua(p21,3))*(b(p2,3)+b(p21,3))*0.5;t=t+t2(p2); end forp3=k:m t3(p3)=(ua(p3,4)ua(p31,4))*(b(p3,4)+b(p31,4))*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)+(70ua(m,4))*b(m,4); tz=t/3.6; disp('加速時間='); disp(tz); disp('s'); 加速時間=29.0585s 二、計算與繪制題1中貨車的1〕汽車功率平衡圖； 2〕最高檔與次高檔的等速百公里油耗曲線。 已知數(shù)據(jù)略?！矃⒁姡ㄆ嚴碚摚┝?xí)題第二章第7題〕 解題程序如下:用Matlab語言 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=[815120716142603300634033804]; Iw=2*Iw1+4*Iw2; nd=400;Qid=0.299; forj=1:5 fori=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)^33.8445*(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('汽車功率平衡圖)'); xlabel('ua(km/h)'); ylabel('Pe,Pz(kw)'); gtext('I') gtext('II') gtext('III') gtext('IV') gtext('V') gtext('P阻') forj=1:5 fori=1:8 Td(i)=19.313+295.27*(n1(i)/1000.0)165.44*(n1(i)/1000.0)^2+40.874*(n1(i)/10 00.0)^33.8445*(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)^45.8629*Pd(1)^3+72.379*Pd(1)^2416.46*Pd(1)+1326.8; b(2)=0.043072*Pd(2)^42.0553*Pd(2)^3+36.657*Pd(2)^2303.98*Pd(2)+1354.7; b(3)=0.0068164*Pd(3)^40.51184*Pd(3)^3+14.524*Pd(3)^2189.75*Pd(3)+1284.4; b(4)=0.0018555*Pd(4)^40.18517*Pd(4)^3+7.0035*Pd(4)^2121.59*Pd(4)+1122.9; b(5)=0.00068906*Pd(5)^40.091077*Pd(5)^3+4.4763*Pd(5)^298.893*Pd(5)+1141.0; b(6)=0.00035032*Pd(6)^40.05138*Pd(6)^3+2.8593*Pd(6)^273.71