フィルターのクリア

How to split a polygon.

23 ビュー (過去 30 日間)
Carlos Zúñiga
Carlos Zúñiga 2020 年 8 月 31 日
コメント済み: Bruno Luong 2020 年 8 月 31 日
Hello everyone.
If I have a polygon with the following coordinates:
x=[0 4 7 5 1]; %Polygon x-coordinates
y=[0 -2 0 10 8]; %Polygon y-coordinates
How can I split the polygon formed by the coordinates shown bellow in for example six parts which area is equal to each other?
  2 件のコメント
the cyclist
the cyclist 2020 年 8 月 31 日
Two questions before anyone spends time thinking about this:
  • Is this a homework assignment?
  • Is the only requirement that the six parts have equal area? I'm wary of other assumptions you may be neglecting to mention. For example, would it be ok to just make vertical slices? Or do you need to find a single point in the interior, such that lines to the vertices separate the area equally?
Carlos Zúñiga
Carlos Zúñiga 2020 年 8 月 31 日
編集済み: Carlos Zúñiga 2020 年 8 月 31 日
Hello, thank you for your answer.
Actually it is not a homework. It is a problem that I couldn't achieve.
Yes, just as I said, all the areas must be equal. About the vertical slices, yes, we can use vertical slices because I'm traying to do the same but rotating the coordinets according to a slope with a rotation matrix.
Greetings and thank you so much for your time!

サインインしてコメントする。

採用された回答

Bruno Luong
Bruno Luong 2020 年 8 月 31 日
編集済み: Bruno Luong 2020 年 8 月 31 日
Each slice has area of 9.5
x=[0 4 7 5 1]; %Polygon x-coordinates
y=[0 -2 0 10 8]; %Polygon y-coordinates
n = 6;
P = polyshape(x,y);
A = P.area/n;
xmin = min(x); xmax = max(x);
ymin = min(y); ymax = max(y);
x0 = xmin+0.01;
b = zeros(1,n-1);
Q = cell(1,n);
Qk = polyshape(); % empty
for k=1:n-1
x0 = fzero(@(x) areafun(P, xmin, x, ymin, ymax)-k*A, x0);
b(k) = x0;
Qp = Qk;
[s, Qk] = areafun(P, xmin, b(k) , ymin, ymax);
Q{k} = subtract(Qk, Qp);
end
Q{n} = subtract(P, Qk);
close all;
figure
hold on
for k=1:n
Q{k}.area
plot(Q{k});
end
axis equal
function [s, Q] = areafun(P, xmin, xmax, ymin, ymax)
R = polyshape([xmin xmax xmax xmin],[ymin ymin ymax ymax]);
Q = intersect(P,R);
s = Q.area;
end
  6 件のコメント
Carlos Zúñiga
Carlos Zúñiga 2020 年 8 月 31 日
Actually I already answer myself!
Thank you so much Mr. Bruno!
Bruno Luong
Bruno Luong 2020 年 8 月 31 日
Star-like partitioning
x=[0 4 7 5 1]; %Polygon x-coordinates
y=[0 -2 0 10 8]; %Polygon y-coordinates
n = 6;
P = polyshape(x,y);
A = P.area/n;
xmin = min(x); xmax = max(x);
ymin = min(y); ymax = max(y);
b = zeros(1,n-1);
Q = cell(1,n);
[xc,yc] = P.centroid;
r = sqrt(max((x-xc).^2+(y-yc).^2))*1.1;
Qk = polyshape(); % empty
x0 = 2*pi/n;
for k=1:n-1
x0 = fzero(@(tt) areafun(P, xc, yc, tt, r)-k*A, x0);
b(k) = x0;
Qp = Qk;
[s, Qk] = areafun(P, xc, yc, x0, r);
Q{k} = subtract(Qk, Qp);
end
Q{n} = subtract(P, Qk);
close all;
figure
hold on
for k=1:n
Q{k}.area
plot(Q{k});
end
axis equal
function [s, Q] = areafun(P, xc, yc, tt, r)
ntt = max(ceil(abs(tt)*128),2);
phi = linspace(0,tt,ntt);
Q = polyshape([xc xc+r*cos(phi)],[yc yc+r*sin(phi)]);
Q = intersect(P,Q);
s = sign(tt)*Q.area;
end

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeElementary Polygons についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by