Need Help to find the error
6 ビュー (過去 30 日間)
古いコメントを表示
Hello, I need from an array consisting of 10,000 items and are the coordinates (X and Y) of the ellipse to determine the minimum and maximum distance between two points. The problem is that the origin of 10000 should have 9999 range, the mine, the program calculates only 9998. you can verify using the command size(dist).
clear all
clc
A=[2/3 -2/3; 2/3 7/3];
[eigenvektor, eigenvalue]=eig(A);
n = 10000;
theta = linspace(0,2*pi,n);
r = 1;
x =r.*cos(theta);
y =r.*sin(theta);
plot(x,y,'.')
v = [x;y];
w=A*v;
xcoor=w(1,:);
ycoor=w(2,:);
i=0;
while i<n
i=i+1;
dxx2=xcoor(n-i:length(xcoor)-i);
dxx1=xcoor(n-(i+1):length(xcoor)-(i+1));
dyy2=xcoor(n-i:length(ycoor)-i);
dyy1=xcoor(n-(i+1):length(ycoor)-(i+1));
dist(i)=sqrt( (dxx2-dxx1)^2 + (dyy2-dyy1)^2);
end
2 件のコメント
採用された回答
Azzi Abdelmalek
2012 年 10 月 21 日
編集済み: Azzi Abdelmalek
2012 年 10 月 21 日
r = 1;
x =r.*cos(theta);
y =r.*sin(theta);
plot(x,y,'.')
v = [x;y];
w=A*v;
xcoor=w(1,:);
ycoor=w(2,:);
figure,plot(xcoor,ycoor)
dist=sqrt(xcoor.^2+ycoor.^2)*2
min(dist)
max(dist)
If the origin of your ellipse is not (0,0) we can make translation
Also you don't need 10000 points, 5000 points are enough
dist=sqrt(xcoor(1:5000).^2+ycoor(1:5000).^2)
3 件のコメント
Azzi Abdelmalek
2012 年 10 月 21 日
編集済み: Azzi Abdelmalek
2012 年 10 月 21 日
do you want the distance between two consecutive points?
then look at Star's answer
その他の回答 (1 件)
Star Strider
2012 年 10 月 21 日
Instead of your while loop, I suggest:
dxx = diff(xcoor);
dyy = diff(ycoor);
dist = hypot( dxx, dyy);
Does this do what you want?
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Creating and Concatenating Matrices についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!