# Finding closest circle to a general circle

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Mirzobek Malikov am 26 Dez. 2022
Bearbeitet: Matt J am 27 Dez. 2022
Hi,
For my research, I have to calculate three circular orbits. One original orbit and two other orbits. I have to find the most closest circle to the original circle. From visualization, it can be seen that the red circle is the closest one to original (green) cirle. But I am looking for analytical way to make conclusions on circles. Because sometimes circles become to closer to original circle, where visualization works not fine, in this case I have to decide by looking at analytical result. But I am in a doubt how to calculate analytically. I have used some methods which gives different unrelated results.
Would be nice if someone helps me to solve it:))
Green - original (3rd circle)
Red - 1st circle
Blue - 2nd circle
(2D view)
From pictures may look like ellipses but orginally they are inclined circles! Figures are provided from different angles. 2D view also provided
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KSSV am 26 Dez. 2022
How did you plot the circles?
Mirzobek Malikov am 26 Dez. 2022
long history. but let's think that i have each circle's datas. each circle is drawn by using discretization rule, obtaining set of (x, y, z) coordinates for each

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### Akzeptierte Antwort

Matt J am 27 Dez. 2022
Bearbeitet: Matt J am 27 Dez. 2022
Here's another possible method using pdist2. Assume we have Nx3 matrices XYZ1, XYZ2 whose rows are the points (x,y,z) on orbit 1 and orbit 2 respectively. Then,
difference=max( pdist2(XYZ1,XYZ2,'euc','Smallest',1) )
measures the largest gap between the 2 orbits. Example:
t=(0:0.5:359.5)';
difference=max( pdist2(XYZ1,XYZ2,'euc','Smallest',1) )
difference = 0.2000
plot(XYZ1(:,1), XYZ1(:,2),...
XYZ2(:,1), XYZ2(:,2)); axis equal
##### 2 KommentareKeine anzeigenKeine ausblenden
Mirzobek Malikov am 27 Dez. 2022
@Matt J thansk but can you explain it more briefly
Matt J am 27 Dez. 2022
Bearbeitet: Matt J am 27 Dez. 2022
I don't think I can get more brief than the 3 lines above. What isn't clear?

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### Weitere Antworten (1)

Karim am 26 Dez. 2022
You could try to fit a plane trought each circle, comput the normal of that plane. And then determine the angle between the normal of each circle and the reference circle. See below for a demonstration of the concept.
First we need a bit of code to generate a circle in 3D:
% create a circle
tht = linspace(0,2*pi,100);
Cirle_Grid = [cos(tht); sin(tht); 0*tht];
% define some rotation functions so that we can orient our circle in 3D
% space
RotX = @(t) [1 0 0; 0 cos(t) -sin(t); 0 sin(t) cos(t)];
RotY = @(t) [cos(t) 0 sin(t); 0 1 0;-sin(t) 0 cos(t)];
RotZ = @(t) [cos(t) -sin(t) 0; sin(t) cos(t) 0; 0 0 1];
% create some random oriented circles, we choose the data such that circle
% 2 will be the closest and we have a third circle which is fully random
C1x = rand;
C1y = rand;
C1z = rand;
C2x = C1x + 0.1*rand;
C2y = C1y + 0.1*rand;
C2z = C1z + 0.1*rand;
Circ1 = RotZ(C1x) * RotY(C1y) * RotX(C1z) * Cirle_Grid;
Circ2 = RotZ(C2x) * RotY(C2y) * RotX(C2z) * Cirle_Grid;
Circ3 = RotZ(rand) * RotY(rand) * RotX(rand) * Cirle_Grid;
% make a plot to see if evrything is working
figure
hold on
plot3(Circ1(1,:),Circ1(2,:),Circ1(3,:),'b')
plot3(Circ2(1,:),Circ2(2,:),Circ2(3,:),'r')
plot3(Circ3(1,:),Circ3(2,:),Circ3(3,:),'g')
hold off
grid on
view(3)
Now lets look for the circle closest to a reference circle, here let's take Circ1 as the reference circle.
% first find the normals to the circles, i included the function below
[N1,C1] = CircleNormal(Circ1);
[N2,C2] = CircleNormal(Circ2);
[N3,C3] = CircleNormal(Circ3);
% plot the circles and the normal's
figure
hold on
plot3(Circ1(1,:),Circ1(2,:),Circ1(3,:),'b')
plot3(Circ2(1,:),Circ2(2,:),Circ2(3,:),'r')
plot3(Circ3(1,:),Circ3(2,:),Circ3(3,:),'g')
quiver3(C1(1),C1(2),C1(3),N1(1),N1(2),N1(3),'b')
quiver3(C2(1),C2(2),C2(3),N2(1),N2(2),N2(3),'r')
quiver3(C3(1),C3(2),C3(3),N3(1),N3(2),N3(3),'g')
hold off
grid on
view(3)
% now determine the angles between the normals
Ang_21 = atan2(norm(cross(N2,N1)), dot(N2,N1))
Ang_21 = 0.0883
Ang_31 = atan2(norm(cross(N3,N1)), dot(N3,N1))
Ang_31 = 0.4256
And indeed, as we can see Ang_21 is smaller en hence is the circel which is "closest" to our reference circle :)
function [N,Center] = CircleNormal(Grid)
Center = mean(Grid,2);
Grid = Grid - repmat(Center, 1, size(Grid,2));
[U,S,V] = svd(Grid',0);
N = -1/V(end,end)*V(:,end);
end
##### 6 Kommentare4 ältere Kommentare anzeigen4 ältere Kommentare ausblenden
Mirzobek Malikov am 26 Dez. 2022
sorry for not providing enough data. @Matt J yes, I know each circle's radii, center and plane.
Mirzobek Malikov am 26 Dez. 2022
@Karim it looks good, but i want to compare this method with other methods if other methods available. because i cannot conclude by comparing angle of normal vectors.

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