Azzera filtri
Azzera filtri

How to split a polygon.

23 visualizzazioni (ultimi 30 giorni)
Carlos Zúñiga
Carlos Zúñiga il 31 Ago 2020
Commentato: Bruno Luong il 31 Ago 2020
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 Commenti
the cyclist
the cyclist il 31 Ago 2020
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 il 31 Ago 2020
Modificato: Carlos Zúñiga il 31 Ago 2020
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!

Accedi per commentare.

Risposta accettata

Bruno Luong
Bruno Luong il 31 Ago 2020
Modificato: Bruno Luong il 31 Ago 2020
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 Commenti
Carlos Zúñiga
Carlos Zúñiga il 31 Ago 2020
Actually I already answer myself!
Thank you so much Mr. Bruno!
Bruno Luong
Bruno Luong il 31 Ago 2020
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

Accedi per commentare.

Più risposte (0)

Categorie

Scopri di più su Elementary Polygons in Help Center e File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by