How can I write the coordinates of this geometry?

Hello there,

I have to input the coordinates of this shape in a matlab code as an array of one row, to be used later in solving a BEM problem. I have managed to get the points of the curved parts, but I am not sure how to compile the points all togther (the 4 corner points and the curved points in a counter clockwise direction).

last value before the last value

(0,h) (h,h)

(0,0) (h,0)

first value second value

This is the code I tried to formulate,

h= 10;
xc=zeros;
yc=zeros;
%the left side 
th = linspace( pi/2, -pi/2, 100);
R = (h/4);  %or whatever radius you want
x1 = R*cos(th) ;
y1=(h/4); 
y1 = y1+ R*sin(th) + (h/4) ;
% plot(x1,y1); axis equal;

%the right side 
x2 = - R*cos(th)+ h;
y2=(h/4); 
y2= y2+ R*sin(th) + (h/4);
% plot(x2,y2); axis equal;

j=0;
k=0;
for i=1:204
    if i==1
        yc(i)= 0;
        xc(i)= 0;
    elseif i==2
        yc(i)= 0;
        xc(i)= h;
    elseif i> 2 && i<104
  
        xc(i) = xc(i) + x2(j);
        yc(i) = yc(i) + y2(j);
        j= j+1
    elseif i ==104
        yc(i)= h;
        xc(i)= h;
    elseif i==105
        yc(i)= h;
        xc(i)= 0;
    else
        xc(i) = xc(i) + x1(k);
        yc(i) = yc(i) + y1(k);
        k= k+1
    end
end

I am getting this error,

Index exceeds the number of array elements (2).

Error in checkingGeometry (line 29)
        xc(i) = xc(i) + x2(j);
 

Answer 1

[Les Beckham]

You don't need a loop.

h = 10;
th = linspace(pi/2, -pi/2, 100);
R = (h/4);  %or whatever radius you want
x1 = R*cos(th) ;
y1=(h/4); 
y1 = y1+ R*sin(th) + (h/4) ;
% plot(x1,y1); axis equal;

%the right side 
x2 = - R*cos(th)+ h;
y2=(h/4); 
y2= y2+ R*sin(th) + (h/4);
% plot(x2,y2); axis equal;

% add the corners
y = [0 0 flip(y2) h h y1 0];
x = [0 h flip(x2) h 0 x1 0];

plot(x,y)
axis equal 
grid on
xlim([-1 11])
ylim([-1 11])

[Maram]

Thank you!

[Les Beckham]

You are quite welcome. Glad I could help.

Answer 2

[John D'Errico]

Simplest? Use a polyshape.

H = 10;
Ps0 = polyshape([0 0;0 H;H H;H 0])

Ps0 = 
  polyshape with properties:

      Vertices: [4×2 double]
    NumRegions: 1
      NumHoles: 0

plot(Ps0)

Next, create a pair of circles.

t = linspace(0,2*pi,250)';
t(end) = [];
p = 2; % radius of the semi-circular cutout
C = p*[cos(t),sin(t)];
PsC1 = polyshape(C + [0,H/2]); % semi-circle, centered along each edge
PsC2 = polyshape(C + [H,H/2]);
Psfinal = subtract(subtract(Ps0,PsC1),PsC2);
plot(Psfinal)
axis equal

You can extract the points trivially.

XYpoly = Psfinal.Vertices

XYpoly = 257×2

0 0 0.0000 3.0001 0.0126 3.0000 0.0631 3.0010 0.1135 3.0032 0.1638 3.0067 0.2141 3.0115 0.2642 3.0175 0.3141 3.0248 0.3639 3.0334

Easy peasy. No loops. Just few calls to polyshape to do all the hard work.

[Maram]

Oh I didn't know that, wonderful!

Thanks a lot.