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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):
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
6.56 5.28 4.68 5.19 2.34 6.94 5.55 9.86 5.28 3.87
y
横坐标
28.23 29.1 30.65 30.92 31.67 33.03 34.35 35.01 37.5
x
纵坐标
3.04 2.88 3.68 2.38 2.06 2.58 2.16 1.45 6
y
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 横坐标
x
纵坐标
19.89 24.52 34.82 40.54 37.67 41.38 30.00 19.68 14.56 18.86
y
28.31 29.45 30.00 30.92 31.67 33.31 34.23 35.81 37.5
横坐标
x
纵坐标
18.55 22.66 18.28 15.06 13.42 11.86 7.68 9.45 6
y
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=[0.3,1.8,4.90,6.51,9.73,13.18,16.20,18.92,20.50,23.23,25.56,28.31,29.45,3 0.00,30.92,31.67,33.31,34.23,35.81,37.5]; y=[6.56,19.89,24.52,34.82,40.54,37.67,41.38,30.00,19.68,14.56,18.86,18.55,2 2.66,18.28,15.06,13.42,11.86,7.68,9.45,6]; [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. 06,2.58,2.16,1.45,6,9.45,7.68,11.86,12.42,14.06,17.28,20.66,17.55,19.86,14.56,1 8.68,35.24,42.38,38.67,41.54,35.82,24.52,19.89,6.56]; 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','MarkerFaceCo lor','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=[0.00,0.00,0.3,4.56,6.45,9.71,13.17,16.23,18.36,20.53,23.15,26.49,28.23,2 9.1,30.65,30.92,31.67,33.03,34.35,35.01,37.5,35.81,34.23,33.31,31.67,30.92,29.6 5,29.8,28.31,26.56,23.23,20.50,18.32,16.20,13.18,9.73,6.51,4.90,1.80,0.30]; 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 4.赛道路面情况,以及对选手的建议。 方法一 clear; clc; x1=[0.30,4.56,6.45,9.71,13.17,16.23,18.36,20.53,23.15,26.49,28.23,29.10,30.65,3 0.92,31.67,33.03,34.35,35.01,37.50]; 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=[6.56,19.89,24.52,34.82,40.54,37.67,41.38,30.00,19.68,14.56,18.86,18.55,22.6 6,18.28,15.06,13.42,11.86,7.68,9.45,6.00]; axis([-5 40 -5 45]); grid; for i=1:length(x1)-1 l=0; t1=x1(i):0.01:x1(i+1); d1=spline(x1,y1,t1); for ii=1:length(d1)-1 l=l+sqrt((0.01)^2+(d1(ii+1)-d1(ii))^2); end v1(i)=l*4; if v1(i)<=10 hold on; plot(t1,d1,'k','linewidth',3); elseif v1(i)>30 hold on; plot(t1,d1,'m','linewidth',4); else hold on; plot(t1,d1,'r','linewidth',5); end end for j=1:length(x2)-1 ll=0; t2=x2(j):0.01:x2(j+1); d2=spline(x2,y2,t2); for jj=1:length(d2)-1 ll=ll+sqrt((0.01)^2+(d2(jj+1)-d2(jj))^2); end v2(j)=ll*4; if v2(j)<=10 hold on; plot(t2,d2,'k','linewidth',2); elseif v2(j)>30 hold on; plot(t2,d2,'g','linewidth',3); else hold on; 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=[0.3,4.56,6.45,9.71,13.17,16.23,18.36,20.53,23.15,26.49,28.23,29.1,30.65,30.9 2,31.67,33.03,34.35,35.01,37.5,35.81,34.23,33.31,31.67,30.92,29.65,29.8,28.31,2 6.56,23.23,20.50,18.32,16.20,13.18,9.73,6.51,4.90,1.8,0.3]; 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); dx=diff(xx)./0.01; dy=diff(yy)./0.01; vv=(dx.^2+dy.^2).^(1/2); for i=1:1:925 if vv(i)>0&vv(i)<10 plot(xx(i),yy(i),'g*','markersize',5); hold on; elseif vv(i)>10&vv(i)<30 plot(xx(i),yy(i),'r+','markersize',2); hold on; else plot(xx(i),yy(i),'ko','markersize',2); hold on; end End 因此,选手要想取得优异的成绩,必须在不同路段选择不同的速度。也就是要在不用的时间段选择不同的速度,只要这样才可以顺利的完成比赛,不至于造成危险。 课题总结: 在本次实验中我们可以明显看到,这是一道综合性的题目,但仔细分析就可以发现,这其实是我们最近几次实验内容的有机组合。因此,在解决问题的时候,我们可以应用最近几次实验的方法,将复杂问题分解为一个个简单问题,一点一点来解决。由此,我们也深刻体会到了平时训练的重要性,只有做好了平时每一次训练,才可以解决难题;只有掌握了每一种方法,并将其融会贯通,才可以熟练的应用这些方法去解决综合性的问题。另外,我们应该注意培养自己分析问题的能力,这是至关重要的,这也是解决问题的关键。至于MATLAB 的一些基本操作,随着学习的深入进行,我们已经基本上掌握了。所以,我们必须将分析问题能力的培养放在第一位,这才是我们继续学习数学实验课的目标和我们学习这门课程的初衷—分析和解决实际问题。通过这次实验,我们也再一次认识到了团结合作精神的重要性,它是我们顺利解决问题的保证,也是我们当代大学生所应当具备的基本素质,更是我们将来走向社会后顺利开展工作的保证,是我们所应该重点培养的。 总之,虽然经历了许多困难,但是我们最后终于将问题解决。这次实验,使得我们的能力得到了进一步的提高,也收获了很多课本以外的东西,更重要的是,它为我们以后更好的开展这门课的学习打下了坚实的基础。