how can I sweep a triangle area using two loops

조회 수: 2 (최근 30일)
mohammad mortezaie
mohammad mortezaie 2021년 6월 20일
편집: Scott MacKenzie 2021년 6월 20일
Hello,
I want to sweep all points inside a triangle by coordinates (0,0), (2*pi/(3*a),0) and (0,4*pi/(3*a)) and by deviding this into two triangles and calculating line equations i wrote this code
for j=1:101
x=2*(j-1)*pi/(100*sqrt(3)*a);
for l=1:101
y=-(l-1)*x/(100*sqrt(3))+2*pi*(l-1)/(3*a*100);
for k=1:101
x=2*(k-1)*pi/(100*sqrt(3)*a);
for m=1:101
y=-(m-1)*x/(100*sqrt(3))+2*pi*(m-1)/(100*3*a)+2*pi/(3*a);
But this code doesn't cover some y points inside the triangle. What can i do for solving this problem?
  댓글 수: 3
mohammad mortezaie
mohammad mortezaie 2021년 6월 20일
This is code:
acc=1.42;
a=sqrt(3)*acc;
c=3.3;
gama0=3.16;
gama1=0.39;
gama2=-0.020;
gama3=0.315;
gama4=0.044;
gama5=-0.04;
V=0.;
iform=complex(0.0,1.0);
iden= eye([6 6]);
delta=0.0015;
eta=0.05;
for i=1:201
e=-0.5+4*(i-1)*delta;
Density=0;
for j=1:101
k_x=2*(j-1)*pi/(100*sqrt(3)*a);
for l=1:101
k_y=-(l-1)*k_x/(100*sqrt(3))+2*pi*(l-1)/(3*a*100);
f0=exp(-iform*acc*k_x)+2*cos(sqrt(3)*k_y*acc)*exp(iform*k_x*acc);
f1=exp(iform*k_x*c);
f3=f0*exp(iform*c*k_x);
H6=[-V gama0*f0 gama4*f3 gama3*f3 gama2*f1 0;
gama0*conj(f0) -V gama1*f1 gama4*f3 0 gama5*f1;
gama4*conj(f3) gama1*conj(f1) 0 gama0*f0 gama4*conj(f3) gama1*f1;
o vv gama3*conj(f3) gama4*conj(f3) gama0*conj(f0) 0 gama3*conj(f3) gama4*conj(f3);
gama2*conj(f1) 0 gama4*f3 gama3*f3 V gama0*f0 ;
0 gama5*conj(f1) gama1*conj(f1) gama4*f3 gama0*conj(f0) V];
G=(e*iden+eta*iform*iden)-eig(H6).*iden;
Green=inv(G);
Den=trace(Green);
Density=Density-imag(Den)/pi;
end
end
for k=1:101
k_x=2*(k-1)*pi/(100*sqrt(3)*a);
for m=1:101
k_y=-(m-1)*k_x/(100*sqrt(3))+2*pi*(m-1)/(100*3*a)+2*pi/(3*a);
f0=exp(-iform*acc*k_x)+2*cos(sqrt(3)*k_y*acc)*exp(iform*k_x*acc);
f1=exp(iform*k_x*c);
f3=f0*exp(iform*c*k_x);
H6=[-V gama0*f0 gama4*f3 gama3*f3 gama2*f1 0;
gama0*conj(f0) -V gama1*f1 gama4*f3 0 gama5*f1;
gama4*conj(f3) gama1*conj(f1) 0 gama0*f0 gama4*conj(f3) gama1*f1;
gama3*conj(f3) gama4*conj(f3) gama0*conj(f0) 0 gama3*conj(f3) gama4*conj(f3);
gama2*conj(f1) 0 gama4*f3 gama3*f3 V gama0*f0 ;
0 gama5*conj(f1) gama1*conj(f1) gama4*f3 gama0*conj(f0) V];
G=(e*iden+eta*iform*iden)-eig(H6).*iden;
Green=inv(G);
Den=trace(Green);
Density=Density-imag(Den)/pi;
end
end
i
T(i,1)=e;
T(i,2)=Density/8000;
end
plot(T(:,1),T(:,2))
Yes from points i mean (x,y) coordinate.
mohammad mortezaie
mohammad mortezaie 2021년 6월 20일
this is triangle that I mean. Strip area is needed to be sweep or broom.

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답변 (1개)

Scott MacKenzie
Scott MacKenzie 2021년 6월 20일
편집: Scott MacKenzie 2021년 6월 20일
No need for loops. The tests are built in to the inpolygon function when you provide matrices as input. Assuming you want to find points inside the triangle, given a "sweep" grid of x-y test points (integers or reals), then I think this achieves what you are after:
% your triangle points
a = 0.1;
xt = [0, 2*pi / (sqrt(3)*a), 0, 0];
yt = [0, 2*pi/(3*a), 4*pi/(3*a), 0];
% plot triangle with margin around edges
margin = 5;
x1 = round(min(xt)- margin);
x2 = round(max(xt)+ margin);
y1 = round(min(yt)- margin);
y2 = round(max(yt)+ margin);
axis([x1 x2 y1 y2]);
hold on;
plot(xt,yt);
% make a grid of points to find points inside and outside triangle
delta = 1; % granularity of grid (adjust as desired)
[X,Y] = ndgrid(x1:delta:x2,y1:delta:y2);
% test the points (logical matrix 'in' identifies points in/out)
in = inpolygon(X,Y,xt,yt);
% output number of points inside triangle
sum(sum(in))
% plot the points inside (blue) and outside (light blue)
plot(X(in),Y(in),'.b') % points inside
plot(X(~in),Y(~in),'.', 'color', [.7 .7 1]) % points outside
  댓글 수: 4
Scott MacKenzie
Scott MacKenzie 2021년 6월 20일
편집: Scott MacKenzie 2021년 6월 20일
I'm not sure what you mean by "all points" or "all appropriate y coordinates". The number of points inside the triangle is infinite. I included a variable delta for setting up ndgrid. The lower the value of delta the greater the number of points inside the triangle. Some stats here are...
Delta Number of Points inside triangle
1 781
0.1 76186
0.01 7599720
mohammad mortezaie
mohammad mortezaie 2021년 6월 20일
Thank you so much

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