精选优质文档-倾情为你奉上第二次作业 车辆工程 八班 胡文贵 解:汽车功率平衡图程序 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); pz=F.*u1./3600; for k=1:5u=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).