赛车道路况分析问题.docx

1、赛车道路况分析问题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AB各点的位置坐标（单位：km）横坐标x0.3 4.56 6.45 9.71 13.17 16.23 18.36 20.53 23.15 26.49纵坐标y6.56 5.28 4.68 5.19 2.34 6.94

2、 5.55 9.86 5.28 3.87 横坐标x28.23 29.1 30.65 30.92 31.67 33.03 34.35 35.01 37.5纵坐标y3.04 2.88 3.68 2.38 2.06 2.58 2.16 1.45 6由DCB各点的位置坐标（单位：km）横坐标x1.8 4.90 6.51 9.73 13.18 16.20 18.92 20.50 23.23 25.56纵坐标y19.89 24.52 34.82 40.54 37.67 41.38 30.00 19.68 14.56 18.86横坐标x28.31 29.45 30.00 30.92 31.67 33.31

3、34.23 35.81 37.5纵坐标y18.55 22.66 18.28 15.06 13.42 11.86 7.68 9.45 6y=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

4、,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=0.3,4.56,6.45,9.71,1

5、3.17,16.23,18.36,20.53,23.15,26.49,28.23,29.1,30.65,30.92,31.67,33.03,34.35,35.01,37.5,35.81,34.23,33.31,31.67,30.92,29.65,29.8,28.31,26.56,23.23,20.50,18.32,16.20,13.18,9.73,6.51,4.90,1.8,0.3;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.

6、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

7、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=0.00,0.00,6.56,5.28,4.68,5.19,2.34,6.94,5.55,9.86,5.2

8、8,3.87,3.04,2.88,3.68,2.38,2.06,2.58,2.16,1.45,6,9.45,7.68,11.86,12.42,14.06,17.28,20.66,17.55,19.86,14.56,18.68,35.24,42.38,38.67,41.54,35.82,24.52,19.89,6.56;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=1:97

9、5 L=L+vv(i)*0.01;endL所以，L180.4573.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.

10、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

11、-s1所以，S 750.20034.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

12、.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(-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:leng

13、th(d1)-1 l=l+sqrt(0.01)2+(d1(ii+1)-d1(ii)2);endv1(i)=l*4;if v1(i)30 hold on; plot(t1,d1,m,linewidth,4);else hold on; plot(t1,d1,r,linewidth,5);endend 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);endv2(j)=ll*4;if v2(j

14、)30 hold on; plot(t2,d2,g,linewidth,3);else hold on; plot(t2,d2,r,linewidth,4);endEnd 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=0.3,4.56,6.45,9.71,13.17,16.23,18.36,20.53,23.15,26.49,28.23,29.1,30.65,30.92,31.67,33.03,34.35,35.01,37.5,35.81,34.23,33.31,31.67,30.92,29.65,29.8,28.31,26.56,23.23

15、,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);

16、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&vv(i)30 plot(xx(i),yy(i),r+,markersize,2); hold on;else plot(xx(i),yy(i),ko,markersize,2); hold on;endEnd 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

17、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。