Is there a solution to use "circcirc" with continuous variable as radius?
5 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Miro Mitev
el 10 de Ag. de 2016
Comentada: Miro Mitev
el 10 de Ag. de 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);
0 comentarios
Respuesta aceptada
Pawel Ladosz
el 10 de Ag. de 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 comentarios
Pawel Ladosz
el 10 de Ag. de 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
Más respuestas (0)
Ver también
Categorías
Más información sobre Software Development Tools en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!