赛车道路况分析问题Word下载.docx
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由D→C→B各点的位置坐标（单位：
1.84.906.519.7313.1816.2018.9220.5023.2325.56
19.8924.5234.8240.5437.6741.3830.0019.6814.5618.86
28.3129.4530.0030.9231.6733.3134.2335.8137.5
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<
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）;
图象：
（2）插值法：
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:
xx=spline（t,x,tt）;
yy=spline（t,y,tt）;
--ms'
xx,yy,'
k'
'
LineWidth'
1,'
MarkerEdgeColor'
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）;
9.5;
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,'
yy2=interp1（x2,y2,xx,'
plot（xx,yy1,'
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;
length（x1）-1
l=0;
t1=x1（i）:
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）;
v1（i）=l*4;
ifv1（i）<
=10
holdon;
plot（t1,d1,'
linewidth'
3）;
elseifv1（i）>
30
4）;
else
5）;
end
forj=1:
length（x2）-1
ll=0;
t2=x2（j）:
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）;
v2（j）=ll*4;
ifv2（j）<
plot（t2,d2,'
2）;
elseifv2（j）>
plot（t2,d2,'
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
x=[0.3,4.56,6.45,9.71,13.17,16.23,
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