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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 comentarios
Azzi Abdelmalek
el 21 de Oct. de 2012
which distance do you want, maybe there is a better way to do it
Respuesta aceptada
Azzi Abdelmalek
el 21 de Oct. de 2012
Editada: Azzi Abdelmalek
el 21 de Oct. de 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 comentarios
Azzi Abdelmalek
el 21 de Oct. de 2012
Editada: Azzi Abdelmalek
el 21 de Oct. de 2012
do you want the distance between two consecutive points?
then look at Star's answer
Más respuestas (1)
Star Strider
el 21 de Oct. de 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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