赛车道路况分析问题
数学10-03班
王玉刚10104477 吴曦10104478 徐晓10104479
一、题目。
现要举行一场山地自行车赛，为了了解环行赛道的路况，现对一选手比赛情况进行监测，该选手从A地出发向东到B，再经C、D回到A地（如下图）。现从选手出发开始计时，每隔15min观测其位置，所得相应各点坐标如下表（假设其体力是均衡分配的）： 45 40 35 30 25 20 15 10 5 0-5051015 2025303540 由A→B各点的位置坐标（单位：km） 横坐标x 0.3 4.56 6.45 9.71 13.17 16.23 18.36 20.53 23.15 26.49
纵坐标y 横坐标x 纵坐标y 横坐标x 纵坐标y 横坐标x 纵坐标y 6.56 5.28 4.68 5.19 2.34 6.94 5.55 9.86 5.28 3.87 28.23 29.1 30.65 30.92 31.67 33.03 34.35 35.01 37.5 3.04 2.88 3.68 2.38 2.06 2.58 2.16 1.45 6 由D→C→B各点的位置坐标（单位：km） 1.8 4.90 6.51 9.73 13.18 16.20 18.92 20.50 23.23 25.56 19.89 24.52 34.82 40.54 37.67 41.38 30.00 19.68 14.56 18.86 28.31 29.45 30.00 30.92 31.67 33.31 34.23 35.81 37.5 18.55 22.66 18.28 15.06 13.42 11.86 7.68 9.45 6 假设：1. 车道几乎是在平原上，但有三种路况（根据平均速度v(km/h)大致区分）：
平整沙土路（v>30）、坑洼碎石路（10 2. 车道是一条连续的可以用光滑曲线来近似的闭合路线； 3．选手的速度是连续变化的. 求解：1. 模拟比赛车道的曲线和选手的速度曲线； 2．估计车道的长度和所围区域的面积； 3．分析车道上相关路段的路面状况（在车道上用不同颜色标记出来）； 4．对参加比赛选手提出合理建议. 二、问题分析以及求解。 1.赛道：根据图可知直接求解出y=f(x)比较困难，故可采用参数函数的形式。这里使用了插值法以及多项式拟合法。 （1）多项式拟合： 程序： x=[0.3,4.56,6.45,9.71,13.17,16.23,18.36,20.53,23.15,26.49,28.23,29.1,30.65,30.92,31.67,33.03,34.35,35.01,37.5]; y=[6.56,5.28,4.68,5.19,2.34,6.94,5.55,9.86,5.28,3.87,3.04,2.88,3.68,2.38,2.06,2.58,2.16,1.45,6]; [a,s]=polyfit(x,y,9); xx=0:0.001:38.1; yy=polyval(a,xx); plot(x,y,'o:m',xx,yy,’LineWidth’,2) hold on; x=[0.3,1.8,4.90,6.51,9.73,13.18,16.20,18.92,20.50,23.23,25.56,28.31,29.45,30.00,30.92,31.67,33.31,34.23,35.81,37.5]; y=[6.56,19.89,24.52,34.82,40.54,37.67,41.38,30.00,19.68,14.56,18.86,18.55,22.66,18.28,15.06,13.42,11.86,7.68,9.45,6]; [a,s]=polyfit(x,y,11); xx=0:0.001:38.1; yy=polyval(a,xx); plot(x,y,'o:m',xx,yy,’LineWidth’,2) 图象： （2）插值法： 程序： x=[0.3,4.56,6.45,9.71,13.17,16.23,18.36,20.53,23.15,26.49,28.23,29.1,30.65,30.92,31.67,33.03,34.35,35.01,37.5,35.81,34.23,33.31,31.67,30.92,29.65,29.8,28.31,26.56,23.23,20.50,18.32,16.20,13.18,9.73,6.51,4.90,1.8,0.3]; y=[6.56,5.28,4.68,5.19,2.34,6.94,5.55,9.86,5.28,3.87,3.04,2.88,3.68,2.38,2.06,2.58,2.16,1.45,6,9.45,7.68,11.86,12.42,14.06,17.28,20.66,17.55,19.86,14.56,18.68,35.24,42.38,38.67,41.54,35.82,24.52,19.89,6.56]; t=0:0.25:9.25; tt=0:0.01:9.25; xx=spline(t,x,tt); yy=spline(t,y,tt); plot(x,y,'--ms',xx,yy,'k','LineWidth',1,'MarkerEdgeColor','k','MarkerFaceColor','g') 图像： 由以上两种方法的对比可以看出，插值法的效果明显好于多项式拟合。 2.速度曲线，赛道长度。 根据相邻两点求出直线斜率，及该段内的平均速度，利用自动插值可求出速度变化曲线。 x=[0.00,0.00,0.3,4.56,6.45,9.71,13.17,16.23,18.36,20.53,23.15,26.49,28.23,29.1,30.65,30.92,31.67,33.03,34.35,35.01,37.5,35.81,34.23,33.31,31.67,30.92,29.65,29.8,28.31,26.56,23.23,20.50,18.32,16.20,13.18,9.73,6.51,4.90,1.80,0.30]; y=[0.00,0.00,6.56,5.28,4.68,5.19,2.34,6.94,5.55,9.86,5.28,3.87,3.04,2.88,3.68,2.38,2.06,2.58,2.16,1.45,6,9.45,7.68,11.86,12.42,14.06,17.28,20.66,17.55,19.86,14.56,18.68,35.24,42.38,38.67,41.54,35.82,24.52,19.89,6.56]; dx=diff(x)./0.25; dy=diff(y)./0.25; v=(dx.^2+dy.^2).^(1/2); t=0:0.25:9.5; tt=0:0.01:9.75; vv=interp1(t,v,tt,'cubic'); plot(t,v,'*',tt,vv,'r') L=0; for i=1:975 L=L+vv(i)*0.01; end L 所以，L＝180.457 3.所围面积 x1=[0.3,4.56,6.45,9.71,13.17,16.23,18.36,20.53,23.15,26.49,28.23,29.1,30.65,30.92,31.67,33.03,34.35,35.01, 37.5]; x2=[0.3,4.90,6.51,9.73,13.18,16.20,18.32,20.50,23.23,26.56,28.31,29.8,29.65,30.92,31.67,33.31,34.23,35.81, 37.5]; y1=[6.56,5.28,4.68,5.19,2.34,6.94,5.55,9.86,5.28,3.87,3.04,2.88,3.68,2.38,2.06,2.58,2.16,1.45,6]; y2=[19.89,24.52,35.82,41.54,38.67,42.38,35.24,18.68,14.56,19.86,17.55,20.66,17.28,14.06,12.42,11.86,7.68, 9.45,6]; xx=0.2:0.1:37.5; yy1=interp1(x1,y1,xx,'cubic'); yy2=interp1(x2,y2,xx,'cubic'); plot(xx,yy1,'r',xx,yy2,'b') s1=trapz(xx,yy1); s2=trapz(xx,yy2); s=s2-s1 所以，S＝ 750.2003