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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 Comments
Accepted Answer
Azzi Abdelmalek
on 21 Oct 2012
Edited: Azzi Abdelmalek
on 21 Oct 2012
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 Comments
Azzi Abdelmalek
on 21 Oct 2012
Edited: Azzi Abdelmalek
on 21 Oct 2012
do you want the distance between two consecutive points?
then look at Star's answer
More Answers (1)
Star Strider
on 21 Oct 2012
Instead of your while loop, I suggest:
dxx = diff(xcoor);
dyy = diff(ycoor);
dist = hypot( dxx, dyy);
Does this do what you want?
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