赛车道路况分析问题.docx
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赛车道路况分析问题.docx
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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
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=[0.3,4.56,6.45,9.71,13.17,16.23,18.36,20.53,23.15,26.49,28.23,29.1,30.65,30.92,31.67,33.03,34.35,35.01,37.5,35.81,34.23,33.31,31.67,30.92,29.65,29.8,28.31,26.56,23.23,20.50,18.32,16.20,13.18,9.73,6.51,4.90,1.8,0.3];
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);
dx=diff(xx)./0.01;
dy=diff(yy)./0.01;
vv=(dx.^2+dy.^2).^(1/2);
fori=1:
1:
925
ifvv(i)>0&vv(i)<10
plot(xx(i),yy(i),'g*','markersize',5);
holdon;
elseifvv(i)>10&vv(i)<30
plot(xx(i),yy(i),'r+','markersize',2);
holdon;
else
plot(xx(i),yy(i),'ko','markersize',2);
holdon;
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。
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