汽车运用工程综合大作业汽运大作业授课教师：许洪国班级： 441002 学号： 44100221 姓名：衡威威（一）汽车动力性计算（1）绘制汽车驱动力与行驶阻力平衡图riiTg0tqTFt；4433221)/()/()/()/(cnacnacnacnaaTtq；ruiiag377.0n0；其他参数已知； 又以上条件可求出Ft，并画出autF图形。15.212aDwfAuCfGFF；参数均已知，可求出awfuFF）（图形。数据表格见：《附表一》（2）求最高车速m axau令Ft=Ff+Fw,得到方程058.62968.17316.4042.01065.12344-aaaauuuu使用 MATLAB求解>> p=[0.000165,-0.042,4.16,-173.68,629.58] p = 0.0002 -0.0420 4.1600 -173.6800 629.5800 >> px=poly2str(p,'x') px = 0.000165 x^4 - 0.042 x^3 + 4.16 x^2 - 173.68 x + 629.58 >> format rat% >> r=roots(p) r = 17290/159 8296/117 +11840/193i 8296/117 -11840/193i 2782/697 m axau=108.7km/h. 这与从驱动力—行驶阻力平衡图观察结果一致。最大爬坡度maxi的求解见（ 4）。（3）绘制加速度倒数曲线)(1wftFFFmdudta，其中22202267.4844.401gTgfwiriiIrIm12.61gi，43242.192.5817.96208.650556.201735.18631aaaauuuua；11.32gi，432048.093.364.12185.167954.120118.5111aaaauuuua；69.13gi，4320023.034.062.1905.49663.81645.1791aaaauuuua；00.14gi，432000165.0042.016.468.17358.62911.891aaaauuuua；由以上条件绘出aua1图形如下：数据表格见：《附表二》（4）绘制汽车动力因数特性曲线GFFwtD，N36211G，22123.015.21FaaDwuuAC12.61gi，3621142.192.5817.92608.650507.1659-D432aaaauuuu；11.32gi，36211048.093.364.12185.167909.843-D432aaaauuuu；69.13gi，362110023.034.062.1905.49614.458-D432aaaauuuu；69.13gi36211000165.0042.016.468.17309.271-D432aaaauuuu; 由以上条件求出D，并画出auD图像如下：数据表格见：《附表三》求解maxi如下：欲求m axi ，只需求m ax1D，用 MAITLAB求解如下：先令43242.192.5817.92608.650507.1659uauauauay, 求解 y’=0 的根>> y=[-5.68,176.76,-1852.34,6505.08] y = -142/25 4419/25 -92617/50 84566/13 >> yx=poly2str(y,'x') yx = -5.68 x^3 + 176.76 x^2 - 1852.34 x + 6505.08>> format rat% >> r=roots(y) r = 2593/237 + 1397/676i 2593/237 - 1397/676i 1903/206 得 到maxi对 应 的 速 度1au =1903/206=9.24km/h, ...
