赛车道路况分析问题

1、赛车道路况分析问题 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CD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）：4540353025201510500510152025303540-5由AfB各点的位置坐标（单位：km）横 0.3 4.56 6.45 9.71 13.17 16.23坐 18.36 20.53 23.15 26.49标X6.

2、56 5.28 4.68 5.19 2.34 6.945.55 9.86 5.28 3.8728.23 29.1 30.65 30.92 31.67 33.0334.35 35.01 37.5纵 3.042.883.682.382.06坐 2.58 2.16 1.45 6由CfB各点的位置坐标（单位：km）1.84.906.519.7313.1816.20 18.92 20.50 23.23 25.5619.89 24.52 34.82 40.54 37.67 41.3830.00 19.68 14.56 18.8628.31 29.45 30.00 30.92 31.67 33.3134.2

3、3 35.81 37.518.55 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<v<30)、松软泥泞路(v<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

4、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);x

5、x=0:0.001:38.1;yy=polyval(a,xx);plot(x,y,'o:m',xx,yy, 'LineWidth ' ,2) hold on;x=0.3,1.8,4.90,6.51,9.73,13.18,16.20,1 8.92,20.50,23.23,25.56,28.31,29.45,30.00, 30.92,31.67,33.31,34.23,35.81,37.5;y=6.56,19.89,24.52,34.82,40.54,37.67,4 1.38,30.00,19.68,14.56,18.86,18.55,22.66, 18.28,15

6、.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: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 ,31.67,33.03,34.35,3 5.01,37.5,35.81,34.23 ,33.31,31.67,30.92,29.65,29.8, 28.

7、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=0:0.01 925;xx

8、=spli ne(t,x,tt);yy=spli ne(t,y,tt);plot(x,y/-msxx,yy/kVL ineWidth1,1 /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=0.00,0.00,0.3,4.56,6.45,9.71,13.17,16.23,18.36,20.53,23.15,26.49,28.23,29.1,3

9、0.65,30.92,31.67,33.03,34.35,35.01,37.5,3 5.81,34.23,33.31,31.67,30.92,29.65,29.8,2 8.31,26.56,23.23,20.50,18.32,16.20,13.18, 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.

10、55,19.86,14.56 ,18.68,35.24,42.38,38.67,41.54,35.82,245 2,19.89,6.56;dx=diff(x)./0.25;dy=diff(y)./0.25;v=(dx.A2+dy.A2).A(1/2);t=0:0.25:9.5;tt=0:0.01:9.75;vv=interp1 (t,v,tt/cubic)；plot(3v',t3vv,'r')L=0;for i=1:975L=L+vv(i)*0.01; end所以，L= 180.4571401201006040203 .所围面积x1=0.3,4.56,6.45,9.71

11、,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. 86,5.28,3.87,3.04,2.88,3.68,2.38,2.06,2.5 8,2.16,1.45,

12、6;y2=19.89,24.52,35.82,41.54,38.67,42.38,3 5.24,18.68,14.56,19.86,17.55,20.66,17.28, 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= 750.20034 .赛道

13、路面情况，以及对选手的建议。方法一clear;clc;x1=0.30,4.56,6.45,9.71,13.17,16.23,18.36 ,20.53,23.15,26.49,28.23,29.10,30.65,30.92,31.67,33.03,34.35,35.01,37.50;y1=6.56,5.28,4.68,5.19,2.34,6.94,5.55,9.86,5.28,3.87,3.04,2.88,3.68,2.38,2.06,2.58,2.16,1.45,6.00;x2=0.30,1.80,4.90,6.51,9.73,13.18,16.20,18.92,20.50,23.23,25.

14、56,28.31,29.45,30.00 ,30.92,31.67,33.31,34.23,35.81,37.50;y2=6.56,19.89,24.52,34.82,40.54,37.67,41.38,30.00,19.68,14.56,18.86,18.55,22.66,18.28,15.06,13.42,11.86,7.68,9.45,6.00;axis(-5 40 -5 45);grid;for i=1:length(x1)-11=0;t1=x1 (i):0.01:x1 (i+1);d1=spline(x1,y1,t1);for ii=1 :length(d1 )-1l=l+sqrt(

15、0.01 )八 2+(d1(ii+1 )-d1 (ii)八 2);endv1(i)=l*4;if v1(i)<=10hold on;plot(t1 ,d1 ,'k','linewidth',3);elseif v1 (i)>30hold on;plot(t1 ,d1 ,'m',linewidth',4);elsehold on;plot(t1 ,d1 ,'r','linewidth',5);endend for j=1 :length(x2)-111=0;t2=x2(j):0.01 :x2(j+

16、1); d2=spline(x2,y2,t2);for jj=1:length(d2)-1ll=ll+sqrt(0.01 )八 2+(d2(jj+1 )-d2(jj)八 2); end v2(j)=ll*4;if v2(j)<=10hold on; plot(t2,d2,'k',linewidth1,2);elseif v2(j)>30hold on; plot(t2,d2,'g',linewidth',3);elsehold on; plot(t2,d2,'r',linewidth1,4);endEnd40-35-30-25

17、-20-15-10-5-0-可以看出，这个程序比较复杂。于是，我们又用了另外一种方法，程序如下：方法二clear;clc;x=0.3,4.56,6.45,9.71,13.17,16.23,18.36,2 0.53,23.15,26.49,28.23,29.1,30.65,30.92,3 1.67,33.03,34.35,35.01,37.5,35.81,34.23,3 3.31,31.67,30.92,29.65,29.8,28.31,26.56,2 3.23,20.50,18.32,16.20,13.18,9.73,6.51,4. 90,1.8,0.3;y=6.56,5.28,4.68,5.

18、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);dx=diff(xx)./0.01;dy=diff(yy)./0.01;vv=(dx.A2

19、+dy.A2).A(1 /2);for 1=1:1:925if vv(i)>0&vv(i)<10plot(xx(i),yy(i),'g*','markersize',5);hold on;elseif w(i)>10&vv(i)<30plot(xx ,yy ,'r+','markersize',2);hold on;elseplot(xx(i),yy(i)/ko7markersize2); hold on;endEndFigure 1OJFile Edit View Insert Tools desktop AVindAw Help35302520151051