How to obtain union of three shapes given the coordinates

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Oluwaseyi Ogun
Oluwaseyi Ogun 2021 年 12 月 13 日
コメント済み: Oluwaseyi Ogun 2021 年 12 月 13 日
I am trying to obtain the union of shapes comprising of 2 rectangles and a circle. Please does anyone know how to get the union coordinates? .Below is the matlab code, and the corresponding figure.
cy_l=0.3; length of rectangle outside the circle
r=0.5; % radius of circle
a=0.3; % width of rectangle
b=2*r+2*cy_l; % length of rectangle
centre=[0.5 0.3]; %circle centre
theta=0:2*pi/360:2*pi;
circ=[r*cos(theta')+centre(1) r*sin(theta')+centre(2)]; % circle coordinates
%% rectangle
xy=[centre(1)-(r+cy_l), centre(2)+a/2]; % top left coordinates of rectangle1
xy1=[centre(1)-a/2, centre(2)+(r+cy_l)]; %top left coordinate of rectangle2
R1=[xy(1), xy(1), xy(1)+b, xy(1)+b, xy(1);
xy(2)-a, xy(2), xy(2), xy(2)-a, xy(2)-a]; % coordinates of rectangle 1
R2= [ xy1(1), xy1(1), xy1(1)+a, xy1(1)+a, xy1(1);
xy1(2)-b, xy1(2), xy1(2), xy1(2)-b, xy1(2)-b]; %coordinates of rectangle 2
plot(circ(:,1), circ(:,2), 'k', R1(1,:), R1(2,:),'k', R2(1,:), R2(2,:), 'k', 'LineWidth', 2)
grid on
axis equal

採用された回答

Steven Lord
Steven Lord 2021 年 12 月 13 日
I'd use polyshape.
cy_l=0.3; % length of rectangle outside the circle
r=0.5; % radius of circle
a=0.3; % width of rectangle
b=2*r+2*cy_l; % length of rectangle
centre=[0.5 0.3]; %circle centre
theta=0:2*pi/360:2*pi;
circ=polyshape(r*cos(theta')+centre(1), r*sin(theta')+centre(2)); % Circle
Warning: Polyshape has duplicate vertices, intersections, or other inconsistencies that may produce inaccurate or unexpected results. Input data has been modified to create a well-defined polyshape.
%% rectangle
xy=[centre(1)-(r+cy_l), centre(2)+a/2]; % top left coordinates of rectangle1
xy1=[centre(1)-a/2, centre(2)+(r+cy_l)]; %top left coordinate of rectangle2
R1=polyshape([xy(1), xy(1), xy(1)+b, xy(1)+b, xy(1)], ...
[xy(2)-a, xy(2), xy(2), xy(2)-a, xy(2)-a]); % rectangle 1
R2= polyshape([xy1(1), xy1(1), xy1(1)+a, xy1(1)+a, xy1(1)], ...
[xy1(2)-b, xy1(2), xy1(2), xy1(2)-b, xy1(2)-b]); % rectangle 2
plot([circ, R1, R2])
axis equal
% Show the union in a separate figure for comparison
figure
plot(union([circ, R1, R2]), 'FaceColor', 'g')
axis equal
  1 件のコメント
Oluwaseyi Ogun
Oluwaseyi Ogun 2021 年 12 月 13 日
yes, thanks. This is fine

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その他の回答 (2 件)

Alex Alex
Alex Alex 2021 年 12 月 13 日
編集済み: Alex Alex 2021 年 12 月 13 日
may be command "polyxpoly" help you
[xi1,yi1] = polyxpoly(circ(:,1), circ(:,2), R1(1,:), R1(2,:))
[xi2,yi2] = polyxpoly(circ(:,1), circ(:,2), R2(1,:), R2(2,:))
[xi3,yi3] = polyxpoly(R1(1,:), R1(2,:), R2(1,:), R2(2,:))
plot(xi1,yi1, 'o')
plot(xi2,yi2, 'o')
plot(xi3,yi3, 'o')
  1 件のコメント
Oluwaseyi Ogun
Oluwaseyi Ogun 2021 年 12 月 13 日
ok. i tried that but what i got is not the union of the figure. I got an 'N' shape

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John D'Errico
John D'Errico 2021 年 12 月 13 日
Tivial.
  1. generate the three objects as polyshapes.
  2. Compute the union.
For example...
cy_l=0.3; % length of rectangle outside the circle
r=0.5; % radius of circle
a=0.3; % width of rectangle
b=2*r+2*cy_l; % length of rectangle
centre=[0.5 0.3]; %circle centre
theta=0:2*pi/360:2*pi;
circ=[r*cos(theta')+centre(1) r*sin(theta')+centre(2)]; % circle coordinates
%% rectangle
xy=[centre(1)-(r+cy_l), centre(2)+a/2]; % top left coordinates of rectangle1
xy1=[centre(1)-a/2, centre(2)+(r+cy_l)]; %top left coordinate of rectangle2
R1=[xy(1), xy(1), xy(1)+b, xy(1)+b, xy(1);
xy(2)-a, xy(2), xy(2), xy(2)-a, xy(2)-a]; % coordinates of rectangle 1
R2= [ xy1(1), xy1(1), xy1(1)+a, xy1(1)+a, xy1(1);
xy1(2)-b, xy1(2), xy1(2), xy1(2)-b, xy1(2)-b]; %coordinates of rectangle 2
PSc = polyshape(circ(:,1),circ(:,2));
Warning: Polyshape has duplicate vertices, intersections, or other inconsistencies that may produce inaccurate or unexpected results. Input data has been modified to create a well-defined polyshape.
PSr1 = polyshape(R1(1,:),R1(2,:));
PSr2 = polyshape(R2(1,:),R2(2,:));
PSu = union(union(PSc,PSr1),PSr2);
plot(PSu)

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