賽車道路況分析問題

1、賽車道路況分析問題 賽車道路況分析問題數學10-03班王 玉剛10104477吳 曦10104478徐曉10104479一、題目?，F要舉行一場山地自行車賽，為了了解環(huán)行 賽道的路況，現對一選手比賽情況進行監(jiān)測，該選 手從A地出發(fā)向東到B,再經CD回到A地（如下 圖）?，F從選手出發(fā)開始計時，每隔15min觀測其 位置，所得相應各點坐標如下表（假設其體力是均 衡分配的）：4540353025201510500510152025303540-5由AfB各點的位置坐標（單位：km）橫 0.3 4.56 6.45 9.71 13.17 16.23坐 18.36 20.53 23.15 26.49標X6.

2、56 5.28 4.68 5.19 2.34 6.945.55 9.86 5.28 3.8728.23 29.1 30.65 30.92 31.67 33.0334.35 35.01 37.5縱 3.042.883.682.382.06坐 2.58 2.16 1.45 6由CfB各點的位置坐標（單位：km）1.84.906.519.7313.1816.20 18.92 20.50 23.23 25.5619.89 24.52 34.82 40.54 37.67 41.3830.00 19.68 14.56 18.8628.31 29.45 30.00 30.92 31.67 33.3134.2

3、3 35.81 37.518.55 22.66 18.28 15.06 13.42 11.86 7.68 9.45 6 假設：1 .車道幾乎是在平原上，但有三種路況(根 據平均速度v (km/h)大致區(qū)分)：平整沙土路(v>30)、坑洼碎 石路(10<v<30)、松軟泥濘路(v<10);2 .車道是一條連續(xù)的可以用光滑曲線來近似的 閉合路線；3 .選手的速度是連續(xù)變化的.求解：1.模擬比賽車道的曲線和選手的速度曲線；2 .估計車道的長度和所圍區(qū)域的面積；3 .分析車道上相關路段的路面狀況(在車道上用不同顏色標記出來)；4 .對參加比賽選手提出合理建議.二、問題分 析以及

4、求解。1.賽道：根據圖可知直接求解出y=f (x)比較困 難，故可采用參數函數的形式。這里使用了插值法 以及多項式擬合法。(1 )多項式擬合：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);x

5、x=0:0.001:38.1;yy=polyval(a,xx);plot(x,y,'o:m',xx,yy, 'LineWidth ' ,2) hold on;x=0.3,1.8,4.90,6.51,9.73,13.18,16.20,1 8.92,20.50,23.23,25.56,28.31,29.45,30.00, 30.92,31.67,33.31,34.23,35.81,37.5;y=6.56,19.89,24.52,34.82,40.54,37.67,4 1.38,30.00,19.68,14.56,18.86,18.55,22.66, 18.28,15

6、.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)圖象: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 ,31.67,33.03,34.35,3 5.01,37.5,35.81,34.23 ,33.31,31.67,30.92,29.65,29.8, 28.

7、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=0:0.01 925;xx

8、=spli ne(t,x,tt);yy=spli ne(t,y,tt);plot(x,y/-msxx,yy/kVL ineWidth1,1 /MarkerEdgeColor'/k'/MarkerFaceColor', 'g')圖像：由以上兩種方法的對比可以看出，插值法的效果明 顯好于多項式擬合。2.速度曲線，賽道長度。根據相鄰兩點求出直線斜率，及該段內的平均 速度，利用自動插值可求出速度變化曲線。x=0.00,0.00,0.3,4.56,6.45,9.71,13.17,16.23,18.36,20.53,23.15,26.49,28.23,29.1,3

9、0.65,30.92,31.67,33.03,34.35,35.01,37.5,3 5.81,34.23,33.31,31.67,30.92,29.65,29.8,2 8.31,26.56,23.23,20.50,18.32,16.20,13.18, 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.

10、55,19.86,14.56 ,18.68,35.24,42.38,38.67,41.54,35.82,245 2,19.89,6.56;dx=diff(x)./0.25;dy=diff(y)./0.25;v=(dx.A2+dy.A2).A(1/2);t=0:0.25:9.5;tt=0:0.01:9.75;vv=interp1 (t,v,tt/cubic)；plot(3v',t3vv,'r')L=0;for i=1:975L=L+vv(i)*0.01; end所以，L= 180.4571401201006040203 .所圍面積x1=0.3,4.56,6.45,9.71

11、,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. 86,5.28,3.87,3.04,2.88,3.68,2.38,2.06,2.5 8,2.16,1.45,

12、6;y2=19.89,24.52,35.82,41.54,38.67,42.38,3 5.24,18.68,14.56,19.86,17.55,20.66,17.28, 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= 750.20034 .賽道

13、路面情況，以及對選手的建議。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.

14、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)-11=0;t1=x1 (i):0.01:x1 (i+1);d1=spline(x1,y1,t1);for ii=1 :length(d1 )-1l=l+sqrt(

15、0.01 )八 2+(d1(ii+1 )-d1 (ii)八 2);endv1(i)=l*4;if v1(i)<=10hold on;plot(t1 ,d1 ,'k','linewidth',3);elseif v1 (i)>30hold on;plot(t1 ,d1 ,'m',linewidth',4);elsehold on;plot(t1 ,d1 ,'r','linewidth',5);endend for j=1 :length(x2)-111=0;t2=x2(j):0.01 :x2(j+

16、1); d2=spline(x2,y2,t2);for jj=1:length(d2)-1ll=ll+sqrt(0.01 )八 2+(d2(jj+1 )-d2(jj)八 2); end v2(j)=ll*4;if v2(j)<=10hold on; plot(t2,d2,'k',linewidth1,2);elseif v2(j)>30hold on; plot(t2,d2,'g',linewidth',3);elsehold on; plot(t2,d2,'r',linewidth1,4);endEnd40-35-30-25

17、-20-15-10-5-0-可以看出，這個程序比較復雜。于是，我們又用了另外一種方法，程序如下：方法二clear;clc;x=0.3,4.56,6.45,9.71,13.17,16.23,18.36,2 0.53,23.15,26.49,28.23,29.1,30.65,30.92,3 1.67,33.03,34.35,35.01,37.5,35.81,34.23,3 3.31,31.67,30.92,29.65,29.8,28.31,26.56,2 3.23,20.50,18.32,16.20,13.18,9.73,6.51,4. 90,1.8,0.3;y=6.56,5.28,4.68,5.

18、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.A2

19、+dy.A2).A(1 /2);for 1=1:1:925if vv(i)>0&vv(i)<10plot(xx(i),yy(i),'g*','markersize',5);hold on;elseif w(i)>10&vv(i)<30plot(xx ,yy ,'r+','markersize',2);hold on;elseplot(xx(i),yy(i)/ko7markersize2); hold on;endEndFigure 1OJFile Edit View Insert Tools desktop AVindAw Help35302520151051