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

2016 8 10
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更新时间：2016 8 10
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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 2016-8-10

1 个投票

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.

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Miro Mitev
Miro Mitev 2016-8-10
编辑：Miro Mitev 2016-8-10
Hello Pawel,
I want to change all the radii at each next iteration. Now I tried this, but it gives me mistake:
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];
x1_2=zeros(1,4);
y1_2=zeros(1,4);
x1_3=zeros(1,4);
y1_3=zeros(1,4);
x2_3=zeros(1,4);
y2_3=zeros(1,4);
for n = 2:3
[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
Pawel Ladosz
Pawel Ladosz 2016-8-10
Hi Miro,
I am not sure why you getting this error. As a solution I propose assigning the output of function to temporary variables a,b,c and then assigning it to actual variables, please find code below.
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];
x1_2=zeros(1,4);
y1_2=zeros(1,4);
x1_3=zeros(1,4);
y1_3=zeros(1,4);
x2_3=zeros(1,4);
y2_3=zeros(1,4);
for n = 2:3
a = circcirc(Center1(:,1),Center1(:,2),radii1(n),Center2(:,1),Center2(:,2),radii2(n));
x1_2(n)=a(1);
y1_2(n)=a(2);
b = circcirc(Center1(:,1),Center1(:,2),radii1(n),Center3(:,1),Center3(:,2),radii3(n));
x1_3(n)=b(1);
y1_3(n)=b(2);
c = circcirc(Center2(:,1),Center2(:,2),radii2(n),Center3(:,1),Center3(:,2),radii3(n));
x2_3(n)=c(1);
y2_3(n)=c(2);
end
Miro Mitev
Miro Mitev 2016-8-10
Thanks a lot for your help. That is perfect! :)
Miro Mitev
Miro Mitev 2016-8-10
No actually it is not, now i see. It gives me just 1 coordinate for X and 1 for Y. But the circles have 2 intersections. Still something missing :/
Pawel Ladosz
Pawel Ladosz 2016-8-10
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 2016-8-10
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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