making a connectivity array for an ellipse based of rectangles and traingles

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Saim Ehtesham 2022-12-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1868028-making-a-connectivity-array-for-an-ellipse-based-of-rectangles-and-traingles

评论： Saim Ehtesham 2022-12-5

采纳的回答： Matt J

在 MATLAB Online 中打开

I have the following code:

clear all

close all

clc

n = 3; % number of divisions on the boundary

x1 = linspace(0,2,n);

y1_x1 = sqrt(1-( (x1.^2)/4 ) );

y2_x1 = -sqrt(1-( (x1.^2)/4 ) );

x2 = linspace(0,-2,n);

y1_x2 = sqrt(1-( (x2.^2)/4 ) );

y2_x2 = -sqrt(1-( (x2.^2)/4 ) );

X1 = repelem(x1,[n:-1:1]);

X2 = repelem(x2,[n:-1:1]);

Y1_X1 = [];

for i = 1:n

Y1_X1 = [Y1_X1, y1_x1(i:end)]; % concatenate

end

Y2_X1 = [];

for i = 1:n

Y2_X1 = [Y2_X1, y2_x1(i:end)]; % concatenate

end

Y1_X2 = [];

for i = 1:n

Y1_X2 = [Y1_X2, y1_x2(i:end)]; % concatenate

end

Y2_X2 = [];

for i = 1:n

Y2_X2 = [Y2_X2, y2_x2(i:end)]; % concatenate

end

x = [X1 X1 X2 X2];

y = [Y1_X1 Y2_X1 Y1_X2 Y2_X2];

figure()

scatter(x,y) %shows all points of mesh

Now, I want to assign a number to each point in this ellipse such that a 4xN matrix is formed that has either 4 values (for square elements that are on the inside) and 3 values (for all traingles formed from 3 points on the outside line) with one NaN. For this example, at n = 3, the output connected matrix should look like this:

[1 3 2 NaN
4 3 NaN
6 5 NaN
3 7 6
4 8 7
9 8 NaN
6 10 NaN
7 11 10
8 12 11
9 12 NaN
11 13 NaN
12 13 NaN]

Here, each row represents a shape, and the numbers in the shape represent the number assigned to its coordinates.

The logic behind numbering the coordinate points is it starts at the lowest point then goes above, but from left end to right end.

And the assembly of the array is such that it starts from the left bottom shape, goes to the right and takes one step upwards (but starts at the left everytime), and the assignment of numebrs within the array is counterclockwise for that shape. The assignment code should be such that it works for any value of n (for example more points would be generated if n is increased from 3 to any larger number in line 4 of initial code).

Can this be done by a loop?

I am attaching a sketch where I show the shapes with encircled numbers and numbers assigned to each point without the encircling.

Hope this helps.

Thanks a lot in advance

Feel free to ask questions.

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Matt J 2022-12-1

Wouldn't a cell array be better?

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采纳的回答

Matt J 2022-12-1

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1868028-making-a-connectivity-array-for-an-ellipse-based-of-rectangles-and-traingles#answer_1116903

编辑：Matt J 2022-12-1

在 MATLAB Online 中打开

xymesh.mat

Using spatialgraph2D from this FEX download,

https://www.mathworks.com/matlabcentral/fileexchange/73630-spatialgraph2d

load xymesh

xy=sortrows( unique([x(:),y(:)],'rows'),[2,1] );

x=xy(:,1); y=xy(:,2);

DT=delaunayTriangulation(x,y);

E=DT.edges;

Eb=DT.freeBoundary;

Eint=setdiff(E,Eb,'rows');

fn=@(z) abs(diff(z(Eint),1,2));

keep=xor(fn(x)<1e-6,fn(y)<1e-6);

E=[Eb;Eint(keep,:)];

G=graph(table(E,'Var',{'EndNodes'}));

obj=spatialgraph2D(G,x,y);

mosaic(obj);

[p,IDs]=obj.polyshape;

[~,is]=sortrows( cell2mat( arrayfun(@(z) {max(z.Vertices)},p(:)) ) ,[2,1]);

p=p(is); IDs=IDs(is);

N=numel(p);

out=nan(N,4);

for n=1:N

id=IDs{n}; m=length(id);

out(n,1:m)=id;

end

out=sort(out,2)

out = 12×4

1 2 3 NaN 1 3 4 NaN 2 5 6 NaN 2 3 6 7 3 4 7 8 4 8 9 NaN 5 6 10 NaN 6 7 10 11 7 8 11 12 8 9 12 NaN

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Matt J 2022-12-2

编辑：Matt J 2022-12-2

在 MATLAB Online 中打开

You can get a handle to the label text with findobj. Then, you can modify it or delete it as you wish, e.g.,

mosaic(obj);

ht=findobj(gca,'type','text');

set(ht,'FontSize',12);

Saim Ehtesham 2022-12-5

Thanks a lot for all the help Matt. As you asked, the purpose of that specific arrangement is to do Finite element analysis on the elliptic area. My partner and I are making the connectivity matrix to multiply the shape functions from a rectangular or square master element but the shape functions are stored counter clockwise, starting from lower left corner to keep a homogenous approach throughout the solution.

Again, thanks for all the help.

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