re. finding perimeter of a 2D shape

Question

Nithin 2018-6-14

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评论： Nithin 2018-6-15

Profgraph.xls

Hi, I have coordinates of a 2D shape (x,y) and I am trying to determine the perimeter of the shape. I had initially tried the below code on circles etc, using the alpha shape function and perimeter with a given alpha radius of 2.5 and it worked fine. But my current shape I cant seem to use the alpha radius, as with alpha radius the perimeter is coming as zero. But without it works but I dont know if the result if correct as while plotting the figure seems a bit incomplete ie, the area within the contour is not completely filled. I am attaching the shape file, Thank you, Nithin

the code:

    [Data.num] = xlsread('Profgraph.xls', 'sheet2'); 
    nx50 = Data.num(1:end,1 );  ny50 = Data.num(1:end,2);
   shp = alphaShape(nx50,ny50);
   shp.Alpha=2.5;
   plot(shp)
   totalperim = perimeter(shp)

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Answer 1

Anton Semechko 2018-6-14

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% Contour coordinates
A=xlsread('Profgraph.xls','sheet2');
% Compute perimeter
N=size(A,1);
D=A(:,2:N)-A(:,1:(N-1));
D=sqrt(sum(D.^2,2)); % edge lenghts
P=sum(D); % perimeter
disp(P)

Answer comes out to 386.19

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Nithin 2018-6-14

编辑：Nithin 2018-6-14

Thank you for the alternative solution. Will look into this. :) Nithin There is some indexing error in the second line d = y-x. But hope to fix it.

Sorry, When I run the code with D=A(:,2:end)-A(:,1:(end-1));

I get around 585.71, Dont know what mistake I am doing.

Anton Semechko 2018-6-14

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My bad:

% Compute perimeter
N=size(A,1);
D=A(2:N,:)-A(1:(N-1),:);
D=sqrt(sum(D.^2,2)); % edge lenghts
P=sum(D); % perimeter
disp(P)

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Answer 2

Steven Lord 2018-6-14

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Is an alpha value of 2.5 going to be appropriate for all data sets? Probably not.

You probably want to use the criticalAlpha and/or alphaSpectrum function to find an appropriate alpha value for your data set. Using the example from the alphaSpectrum page:

% Generate sample data and plot it
th = (pi/12:pi/12:2*pi)';
x1 = [reshape(cos(th)*(1:5), numel(cos(th)*(1:5)),1); 0];
y1 = [reshape(sin(th)*(1:5), numel(sin(th)*(1:5)),1); 0];
x = [x1; x1+15;];
y = [y1; y1];
plot(x,y,'.')
axis equal
hold on
% Build an alphaShape using the default alpha value
shp = alphaShape(x,y);
% Determine alpha values that produce unique shapes
alphaspec = alphaSpectrum(shp);
% Iterate through the alpha values. Show each one in turn
% with a 0.1 second pause between each shape
for k = 1:length(alphaspec)
    shp.Alpha = alphaspec(k);
    h = plot(shp);
    title(sprintf('Alpha value of %.16f', alphaspec(k)));
    pause(0.1)
    delete(h)
end