汽车理论课后习题1.3

解：1） >> clear all

>> f=0.013; G=38800;Ff=G*f; CdA=2.77;u1=0:1:120; Fw=CdA.*u1.^2/21.15;F=Ff+Fw;Ig=[5.56 2.769 1.644 1.000 0.793];k=1:5; >> ngk=[600 600 600 600 600];

>> ngm=[4000 4000 4000 4000 4000]; >> r=0.367;I0=5.83;eta=0.85;

>> ugk=0.377*r*ngk(k)./(Ig(k).*I0); >> ukm=0.377.*r.*ngm(k)./(Ig(k).*I0); >> for k=1:5 u=ugk(k):ukm(k);

n=Ig(k)*I0.*u./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(k)*I0*eta/r; plot(u,Ft,u1,F) hold on,grid on end

2）最高车速求解程序： >>

[n,u]=solve('n=u*0.793*5.83/(0.377*0.367)','(-19.313+295.27*(n/1000)-165.44*(n/1000)^2+40.874*(n/1000)^3-3.8445*(n/1000)^4)*0.793*5.83*0.85/0.367-(2.77*u^2/21.15+3880*10*0.013)') n =

3299.9224534468163066899978278864 264.13689338171544956471411061266 3533.8761660967238474546442908623 + 2707.8751259314497120563837339829*i 3533.8761660967238474546442908623 - 2707.8751259314497120563837339829*i u =

98.757345195946534184690962185964 7.9048701073070256438355939579073 105.75891807712350450878660177051 + 81.039043290184796798403093329529*i 105.75891807712350450878660177051 - 81.039043290184796798403093329529*i 分析：u的后三个解明显不符合题意，所以最高车速是98.757km/h

求最大爬坡度和最大附着系数

>>I=0.218; Iw1=1.798; Iw2=3.598; L=3.2; m=3880; r=0.367; >>eta=0.85;Cd=2.77; f=0.013; i0=5.83; a=1.947; hg=0.9; >>g=10;ig=[5.56;2.769;1.644;1.00;0.793]; >>for i=1:5

th(i)=1+(Iw1+Iw2)/(m*r^2)+(I*i0^2+(ig(i)^2*eta))/(m*r^2); end

>>for i=1:5

ual(i,1)=(600*0.377*r)/(ig(i)*i0); ual(i,2)=(4000*0.377*r)/(ig(i)*i0); end

>>ua=ual(1,1):.01:ual(1,2); >>[x,y]=size(ua); >>for i=1:y

n=ua(i)*ig(1)*i0/(0.377*r);

Tq=-19.313+295.27*(n/1000)-165.44*(n/1000)^2+40.874*(n/1000)^3-3.8445*(n/1000)^4; Ft(i)=Tq*ig(1)*i0*eta/r; Fw(i)=Cd*ua(i)^2/21.15; Fc(i)=Ft(i)-Fw(i) end

>>plot(ua,Ft/1000,':',ua,Fc/1000,'-',ua,Fw/1000,'--')

>>D=max(Fc)/m/g;

>>af=asin((D-f*sqrt(1+f^2-D^2))/(1+f^2)); >>i=tan(af); >>q=i;

>>cf2=q/(a/L+q*hg/L); >> i i =

0.3448 >> cf2

cf2 =

0.4888 3）

>> clc clear

I=0.218; Iw1=1.798; Iw2=3.598;L=3.2; m=3880; r=0.367; eta=0.85; Cd=2.77; f=0.013; i0=5.83; a=1.947; hg=0.9; g=10; ig=[5.56;2.769;1.644;1.00;0.793]; umax=96; for i=1:5

th(i)=1+(Iw1+Iw2)/(m*r^2)+(I*i0^2+(ig(i)^2*eta))/(m*r^2); end

for i=1:5

ual(i,1)=(600*0.377*r)/(ig(i)*i0); if (4000*0.377*r)/(ig(i)*i0)

ual(i,2)=umax; end end

for j=1:5

ua=ual(j,1):ual(j,2); [x,y]=size(ua); for i=1:y

n=ua(i)*ig(j)*i0/(0.377*r);

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(j)*i0*eta/r;

Fz=Cd*ua(i)^2/21.15+m*g*f;

a(i)=(th(j)*m)/(Ft-Fz); end

plot(ua,a,'-r') hold on end

>> clc clear

%定义个变量值：

I=0.218; Iw1=1.798; Iw2=3.598; L=3.2; m=3880; r=0.367; eta=0.85;Cd=2.77; f=0.013; i0=5.83; a=1.947; hg=0.9; g=10;ig=[5.56;2.769;1.644;1.00;0.793]; %求出各档所对应加速质量系数 for i=1:5

th(i)=1+(Iw1+Iw2)/(m*r^2)+(I*i0^2+(ig(i)^2*eta))/(m*r^2); end

umax=99;

%求出各档在最大车速范围内稳定转速所对应的速度

for i=1:5

ual(i,1)=(600*0.377*r)/(ig(i)*i0); if (4000*0.377*r)/(ig(i)*i0)

ual(i,2)=umax; end end

%从二档到五档按照A方法选取合适加速度 for u=1:98 if(u

else if(u<=ual(2,2)) %当在二档ua内的加速度倒数 n=u*ig(2)*i0/(0.377*r);

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(2)*i0*eta/r; Fz=Cd*u^2/21.15+m*g*f; a(u)=(th(2)*m)/(Ft-Fz);

else if(u<=ual(3,2)) %在二档最大速度到三档最大速度内加速度倒数 n=u*ig(3)*i0/(0.377*r);

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*eta/r; Fz=Cd*u^2/21.15+m*g*f; a(u)=(th(3)*m)/(Ft-Fz);

else if(u<=ual(4,2)) %在三档最大速度到四档最大速度内加速度倒数 n=u*ig(4)*i0/(0.377*r);

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(4)*i0*eta/r; Fz=Cd*u^2/21.15+m*g*f; a(u)=(th(4)*m)/(Ft-Fz);

else if(u<=ual(5,2)) %在四档最大速度到五档最大速度内加速度倒数 n=u*ig(5)*i0/(0.377*r);

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(5)*i0*eta/r; Fz=Cd*u^2/21.15+m*g*f; a(u)=(th(5)*m)/(Ft-Fz); end end end end end end

ux=1:ual(2,1);

%给低速无意义阶段的速度赋予5km/h时同样的加速度

[x,y]=size(ux); for i=1:y a(i)=a(y+1) end t=0; u=1:98; for i=1:98

t=t+1/3.6*a(i);%图解积分法求解时间 ts(i)=t; end

plot(ts,u)

title('速度时间图') ylabel('ua(km/h)') xlabel('t(s)')

text(33,70,'\\leftarrow 70km/h加速时间')

>> ts(70) ans =27.4830

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