Need Help to find the error
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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 Commenti
Risposta accettata
Azzi Abdelmalek
il 21 Ott 2012
Modificato: Azzi Abdelmalek
il 21 Ott 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 Commenti
Azzi Abdelmalek
il 21 Ott 2012
Modificato: Azzi Abdelmalek
il 21 Ott 2012
do you want the distance between two consecutive points?
then look at Star's answer
Più risposte (1)
Star Strider
il 21 Ott 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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