Is there a solution to use "circcirc" with continuous variable as radius?

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Miro Mitev
Miro Mitev le 10 Août 2016
Commenté : Miro Mitev le 10 Août 2016
Hello, I am trying to implement trilateration algorithm and I am using "circcirc" command to find intersections of the circles.
My problem is: I need to find intersections of circles, which changes their radius every step but they have the same centers. Is it possible without re-writing the command or with another command?
This is my code:
Center1=[0.75,5];
Center2=[6.98,4.31];
Center3=[3.46,1.48];
radii1=[1; 3; 4];
radii2=[2.5; 3; 2.3];
radii3=[3; 1; 5];
[x_intersection1_2,y_intersection1_2] = circcirc(Center1(:,1),Center1(:,2),radii1,Center2(:,1),Center2(:,2),radii2);
[x_intersection1_3,y_intersection1_3] = circcirc(Center1(:,1),Center1(:,2),radii1,Center3(:,1),Center3(:,2),radii3);
[x_intersection2_3,y_intersection2_3] = circcirc(Center2(:,1),Center2(:,2),radii2,Center3(:,1),Center3(:,2),radii3);

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Pawel Ladosz
Pawel Ladosz le 10 Août 2016
Hi Miro,
You may want to use for or while loops. Please find description here. The exact code would vary depending on which radius you want to change at what time.
  6 commentaires
Pawel Ladosz
Pawel Ladosz le 10 Août 2016
yes, I made a mistake regarding the number of outputs of function. x and y are now 2x3 matrix where each row is a different coordinate and each column different radius. Let me know whether it works now.
Center1=[0.75,5];
Center2=[6.98,4.31];
Center3=[3.46,1.48];
radii1=[1, 3, 4];
radii2=[2.5, 3, 2.3];
radii3=[3, 1, 5];
%initialize them as 2 by 3 rather than 1 by 4
x1_2=zeros(2,3);
y1_2=zeros(2,3);
x1_3=zeros(2,3);
y1_3=zeros(2,3);
x2_3=zeros(2,3);
y2_3=zeros(2,3);
for n = 1:3
[x1_2(:,n),y1_2(:,n)] = circcirc(Center1(:,1),Center1(:,2),radii1(n),Center2(:,1),Center2(:,2),radii2(n));
[x1_3(:,n),y1_3(:,n)] = circcirc(Center1(:,1),Center1(:,2),radii1(n),Center3(:,1),Center3(:,2),radii3(n));
[x2_3(:,n),y2_3(:,n)] = circcirc(Center2(:,1),Center2(:,2),radii2(n),Center3(:,1),Center3(:,2),radii3(n));
end
Miro Mitev
Miro Mitev le 10 Août 2016
Thanks a lot. Yes it is working this time perfectly.
I have fixed it by the way bellow but your approach is much more simple.
[c,c1] = circcirc(Center2(:,1),Center2(:,2),radii2(n),Center3(:,1),Center3(:,2),radii3(n));
x2_3(1,n)=c(1);
y2_3(1,n)=c1(1);
x2_3(2,n)=c(2);
y2_3(2,n)=c1(2);

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