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How to compute the ratio of the insection point in each grid on the boundary of a real structure

Asked by Jiali on 9 Sep 2019
Latest activity Edited by Bruno Luong
on 16 Sep 2019 at 7:08
In the Cartesian grid, the real structures are pixelized into the staircase approximation. To reduce the staircase error, the effective permittivity is computed, which is weighted by the fraction of the mesh step inside the material 1 or 2 (See the below figure). For instance, the structure is sphere. How I can compute the fraction ratio of four edges in each grid on the boundary of the sphere? Could you please give me some suggestions?
% build Cartesian grid
[Y1,X1]=meshgrid(y_1grid,x_1grid);
% build structure (sphere)
UCC= ((X1+dx/2).^2 + (Y1+dy/2).^2) <= (diam/2.0)^2;
eps_cc=eps(2)*(UCC==1)+eps(1)*(UCC==0);
% find the boundary
[Fx,Fy]=gradient(eps_cc);
boundary=(Fx~=0)|(Fy~=0);
% compute the ratio of intersection point
??
Big Thanks to you

  2 Comments

Yes, I want to compute fraction for each pixel. That's why I try to subgrid the current one pixel into 10 smaller grids and determine whether each point is inside the circel or not.

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2 Answers

Answer by Bruno Luong
on 9 Sep 2019
Edited by Bruno Luong
on 9 Sep 2019
 Accepted Answer

Assuming the boundary has one connexed piece
x=-3:12;
y=-3:10;
[X,Y]=meshgrid(x,y);
cx = 4;
cy = 3;
r = 5;
Z = (X-cx).^2 + (Y-cy).^2 - r^2;
C = contourc(x,y,Z,[0 0]);
Cx = C(1,2:end);
Cy = C(2,2:end);
[Cx; Cy] % fractional crossing is grouped here
close all
hold on
plot(X,Y,'-b');
plot(X',Y','-b');
plot(Cx,Cy,'-r');
plot(cx,cy,'ko');
axis equal

  2 Comments

Your method is very smart. However, the key is to know the fraction of each pixel. Thus, based on the crossing points calculated by your method, I write the below code to compute the fraction for each pixel. Could you please give me some suggestions for the following code?
ratio_x=ones(length(x),length(y));
ratio_y=ratio_x;
for k=1:length(Cx)
for ii=1:length(x)
for jj=1:length(y)
if Cx(k)==x(ii) && (Cy(k)<=y(jj)+dy && Cy(k)>=y(jj))
if (x(ii)-cx)^2+(y(jj)-cy)^2-r^2<=0
ratio_y(ii,jj)=abs(y(jj)+dy-Cy(k))/dy;
else
ratio_y(ii,jj)=abs(y(jj)-Cy(k))/dy;
end
end
if Cy(k)==y(jj) && (Cx(k)>=x(ii) && Cx(k)<=x(ii)+dx)
if (x(ii)-cx)^2+(y(jj)-cy)^2-r^2<=0
ratio_x(ii,jj)=abs(x(ii)+dx-Cx(k))/dx;
else
ratio_x(ii,jj)=abs(x(ii)-Cx(k))/dx;
end
end
end
end
end
The pixels coordinates are
ii=interp1(x,1:length(x),Cx,'previous')
jj=interp1(y,1:length(y),Cy,'previous')
And the corresponding fractional are
ratio_x=interp1(x,1:length(x),Cx)-ii
ratio_y=interp1(y,1:length(y),Cy)-jj
Just loop on the list [ii; jj; ratio_x; ratio_y] and do whatever you need to do.
for pixel = [ii; jj; ratio_x; ratio_y]
i = pixel(1);
j = pixel(2);
rx = pixel(3);
ry = pixel(4):
% do something with i,j,rx,ry ...
end
Alternatively, to avoid using INTERP1 you can call contour with pixel coordinates
C = contourc(1:size(Z,2),1:size(Z,1),Z,[0 0]);
Cx = C(1,2:end);
Cy = C(2,2:end);
% these replace the INTERP1 stuffs
ii = floor(Cx);
jj = floor(Cy);
ratio_x = Cx-ii;
ratio_y = Cy-jj;
The process ii,jj,ratio_x,ratio_y

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Answer by darova
on 11 Sep 2019

'That's why I try to subgrid the current one pixel into 10 smaller grids and determine whether each point is inside the circel or not.'
THAT IS GREAT IDEA
See attached script

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