完整版汽车理论习题Matlab程序Word文件下载.docx
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飞轮转动惯量If=0.218kg•m2
二前轮转动惯量Iw1=1.798kg•m2
四后轮转动惯量Iw2=3.598kg•m2
变速器传动比ig(数据如下表)
Ⅰ档
Ⅱ档
Ⅲ档
Ⅳ档
Ⅴ档
四档变速器
6.09
3.09
1.71
1.00
-
五档变速器
5.56
2.769
1.644
0.793
轴距L=3.2m
质心至前轴距离(满载)a=1.974m
质心高(满载)hg=0.9m
解:
Matlab程序:
(1)求汽车驱动力与行驶阻力平衡图和汽车最高车速程序:
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=[5.562.7691.6441.000.793];
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)/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;
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'
Ft3'
Ft4'
Ft5'
Ff+Fw'
zoomon;
[x,y]=ginput
(1);
zoomoff;
disp('
汽车最高车速='
disp(x);
km/h'
汽车最高车速=
99.3006
km/h
(2)求汽车最大爬坡度程序:
n=[600:
Fw1=CDA*ua1.^2/21.15;
Fz1=Ff+Fw1;
Fi1=Ft1-Fz1;
Zoomon;
imax=100*tan(asin(max(Fi1/G)));
汽车最大爬坡度='
disp(imax);
%'
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:
clear
Fw2=CDA*ua2.^2/21.15;
Fw3=CDA*ua3.^2/21.15;
Fw4=CDA*ua4.^2/21.15;
Fw5=CDA*ua5.^2/21.15;
deta1=1+(Iw1+Iw2)/(m*r^2)+(If*ig
(1)^2*i0^2*nT)/(m*r^2);
deta2=1+(Iw1+Iw2)/(m*r^2)+(If*ig
(2)^2*i0^2*nT)/(m*r^2);
deta3=1+(Iw1+Iw2)/(m*r^2)+(If*ig(3)^2*i0^2*nT)/(m*r^2);
deta4=1+(Iw1+Iw2)/(m*r^2)+(If*ig(4)^2*i0^2*nT)/(m*r^2);
deta5=1+(Iw1+Iw2)/(m*r^2)+(If*ig(5)^2*i0^2*nT)/(m*r^2);
a1=(Ft1-Ff-Fw1)/(deta1*m);
ad1=1./a1;
a2=(Ft2-Ff-Fw2)/(deta2*m);
ad2=1./a2;
a3=(Ft3-Ff-Fw3)/(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([099010]);
汽车的加速度倒数曲线'
ua(km/h)'
1/a'
1/a1'
1/a2'
1/a3'
1/a4'
1/a5'
a=max(a1);
af=asin(max(Ft1-Ff-Fw1)/G);
C=tan(af)/(a/L+hg*tan(af)/L);
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)>
>
ig=[5.562.7691.6441.000.793];
u1=0.377*r*nmin./ig/i0;
u2=0.377*r*nmax./ig/i0;
deta=0*ig;
fori=1:
5
deta(i)=1+(Iw1+Iw2)/(m*r^2)+(If*(ig(i))^2*i0^2*nT)/(m*r^2);
end
ua=[6:
0.01:
99];
N=length(ua);
n=0;
Tq=0;
Ft=0;
inv_a=0*ua;
delta=0*ua;
N
k=i;
ifua(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/1000)^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;
elseifua(i)<
=u2(3)
n=ua(i)*(ig(3)*i0/r)/0.377;
Ft=Tq*ig(3)*i0*nT/r;
inv_a(i)=(deta(3)*m)/(Ft-Ff-Fw(i));
=u2(4)
n=ua(i)*(ig(4)*i0/r)/0.377;
Ft=Tq*ig(4)*i0*nT/r;
inv_a(i)=(deta(4)*m)/(Ft-Ff-Fw(i));
else
n=ua(i)*(ig(5)*i0/r)/0.377;
Ft=Tq*ig(5)*i0*nT/r;
inv_a(i)=(deta(5)*m)/(Ft-Ff-Fw(i));
end
a=delta(1:
k);
t(i)=sum(a);
plot(t,ua)