汽车理论matlab编程作业答案

孙野2008126814.3（1）利用附着系数空载时前轴的利用附着系数φf1：φf1=βz1L(b+zhg)=0.38z13.95(1.85+0.845z)空载时后轴的利用附着系数φr1：φr1=(1−β)z1L(a−zhg)=(1−0.38)z13.95(2.1−0.845z)满载时前轴的利用附着系数φf2：φf2=βz1L(b+zhg)=0.38z13.95(1.0+1.17z)满载时后轴的利用附着系数φr2：φr2=(1−β)z1L(a−zhg)=(1−0.38)z13.95(2.95−1.17z)Matlab程序：clcclearsymsz;f1=0.38*z/((1/3.95)*(1.85+0.845*z));r1=(1-0.38)*z/((1/3.95)*(2.1-0.845*z));f2=0.38*z/((1/3.95)*(1.0+1.17*z));r2=(1-0.38)*z/((1/3.95)*(2.95-1.17*z));f=z;ezplot(f1);holdon;ezplot(f2);ezplot(r1);ezplot(r2);ezplot(f);axis([01.001.0]);title('利用附着系数曲线');xlabel('制动强度z');ylabel('利用附着系数');text(0.38,0.8,'Ør(空载)');text(0.6,0.9,'Ør(满载)');text(0.8,0.45,'Øf(空载)');text(0.8,0.6,'Øf(满载)');text(0.85,0.9,'Ø=z')孙野200812682制动效率空载时前轴的制动效率𝐸𝑓1：𝐸𝑓1=bL𝛽−𝜑𝑓ℎ𝑔𝐿=1.85/3.950.38−𝜑𝑓∙0.845/3.95空载时后轴的制动效率𝐸𝑟1：𝐸𝑟1=aL(1−𝛽)−𝜑𝑟ℎ𝑔𝐿=2.1/3.95(1−0.38)−𝜑𝑟∙0.845/3.95满载时前轴的制动效率𝐸𝑓2：𝐸𝑓2=bL𝛽−𝜑𝑓ℎ𝑔𝐿=1.0/3.950.38−𝜑𝑓∙1.17/3.95满载时后轴的制动效率𝐸𝑟2：𝐸𝑟2=aL(1−𝛽)−𝜑𝑟ℎ𝑔𝐿=2.95/3.95(1−0.38)−𝜑𝑟∙1.17/3.95Matlab程序：clearsymsx;Er1=2.1/3.95/(1-0.38+x*0.845/3.95);Ef2=1.0/3.95/(0.38-x*1.17/3.95);Er2=2.95/3.95/(1-0.38+x*1.17/3.95);ezplot(Ef2);holdon;ezplot(Er1);ezplot(Er2);axis([01.001.0]);孙野200812683title('前后轴制动效率曲线');xlabel('附着系数');ylabel('制动效率（%）');text(0.35,0.9,'Ef');text(0.8,0.9,'Er');text(0.55,0.78,'Er');text(0.65,0.94,'满载');text(0.55,0.65,'空载');(2)①由图可得：空载时，在φ=0.8时的制动效率为0.7，则其制动减速度为0.8g*0.7=0.56g。制动距离为：max2002292.2526.31baaauusg56.092.253030202.002.06.312=6.57m②由图可得：满载时，在φ=0.8时的制动效率为0.87，则其制动减速度为0.8g*0.87=0.696g。制动距离为：max2002292.2526.31baaauusg696.092.253030202.002.06.312=5.34m（3）①若制动系前部管路损坏GzdtdugGFxb2)(2gzzhaLGF后轴利用附着系数grzhaLz后轴制动效率LhLazEgrrr/1/代入数据得：空载时：rE=0.45满载时：rE=0.60a)空载时其最大动减速度ggab36.045.08.0max孙野200812684代入公式：max2002292.2526.31baaauusg36.092.253030202.002.06.312=10.09mb)满载时其最大动减速度ggab48.06.08.0max代入公式：max2002292.2526.31baaauusg48.092.253030202.002.06.312=7.63mB．若制动系后部管路损坏GzdtdugGFxb1)(1gzzhbLGF前轴利用附着系数gfzhbLz前轴制动效率LhLbzEgfff/1/代入数据：空载时：fE=0.57满载时：fE=0.33a)空载时其最大动减速度ggab456.057.08.0max代入公式：max2002292.2526.31baaauusg456.092.253030202.002.06.312=8.02mb)满载时其最大动减速度ggab264.033.08.0max代入公式：max2002292.2526.31baaauusg264.092.253030202.002.06.312=13.67m4.5（1）同步附着系数8.063.025.165.07.20ghbL（2）因7.00所以前轮先抱死7.0fLhLbzEgfff//=7.2/63.07.065.07.2/25.1=0.951（3）最大制动减速度maxba=2/53.67.0smgEf