赛车道路况分析问题精品文档.docx
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赛车道路况分析问题精品文档.docx
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赛车道路况分析问题精品文档
赛车道路况分析问题
数学10-03班
王玉刚10104477
吴曦10104478
徐晓10104479
1、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):
由A→B各点的位置坐标(单位:
km)
横坐标x
0.34.566.459.7113.1716.2318.3620.5323.1526.49
纵坐标y
6.565.284.685.192.346.945.559.865.283.87
横坐标x
28.2329.130.6530.9231.6733.0334.3535.0137.5
纵坐标y
3.042.883.682.382.062.582.161.456
由D→C→B各点的位置坐标(单位:
km)
横坐标x
1.84.906.519.7313.1816.2018.9220.5023.2325.56
纵坐标y
19.8924.5234.8240.5437.6741.3830.0019.6814.5618.86
横坐标x
28.3129.4530.0030.9231.6733.3134.2335.8137.5
纵坐标y
18.5522.6618.2815.0613.4211.867.689.456
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) holdon; 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; fori=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 4.赛道路面情况,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=[0.30,1.80,4.90,6.51,9.73,13.18,16.20,18.92,20.50,23.23,25.56,28.31,29.45,30.00,30.92,31.67,33.31,34.23,35.81,37.50]; 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([-540-545]); grid; fori=1: length(x1)-1 l=0; t1=x1(i): 0.01: x1(i+1); d1=spline(x1,y1,t1); forii=1: length(d1)-1 l=l+sqrt((0.01)^2+(d1(ii+1)-d1(ii))^2); end v1(i)=l*4; ifv1(i)<=10 holdon; plot(t1,d1,'k','linewidth',3); elseifv1(i)>30 holdon; plot(t1,d1,'m','linewidth',4); else holdon; plot(t1,d1,'r','linewidth',5); end end forj=1: length(x2)-1 ll=0; t2=x2(j): 0.01: x2(j+1); d2=spline(x2,y2,t2); forjj=1: length(d2)-1 ll=ll+sqrt((0.01)^2+(d2(jj+1)-d2(jj))^2); end v2(j)=ll*4; ifv2(j)<=10 holdon; plot(t2,d2,'k','linewidth',2); elseifv2(j)>30 holdon; plot(t2,d2,'g','linewidth',3); else holdon; plot(t2,d2,'r','linewidth',4); end End 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
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