The roundness ratio of exact circle is less than 1

hello everyone,

i am new to matlab and i am trying to understand it by myself. I am plotting an exact circle. however, when i try to find the roundness ratio, matlab gives me a ratio of less than 1.

Answer 1

[Image Analyst]

How are you finding it? Normally it's the perimeter squared divided by (4*pi*area), or the inverse of that. What formula are you using?

[Mahmoud]

I am using the same formula you mentioned. I used Autocad to draw an exact circle a certain radius to check if Matlab can give the ratio to be 1. however, I got it 0.92. I want any approach that can help me differentiate ovality from an exact circle. Please, can you provide any help? I appreciate your response :)

[Image Analyst]

Attach your m-file and image.

Answer 2

[John D'Errico]

Show how you are computing that number. Without knowing how you did it, one can only surmise what you did.

In fact, I can imagine any number of ways one might do so, and I would expect to see SOME number that is not exactly 1. The fact is, no matter what you do, you will not get the exact answer, UNLESS you use the circle itself. Perhaps this quote applies:

"The best material model of a cat is another, or preferably the same, cat."
   Philosophy of Science (1945) (with A. Rosenblueth)

The point is, any approximation is only that, an approximation.

For example, perhaps you computed the roundness-ratio as the ratio of areas, comparing the area of a polygonal set of segments that will approximate a circle (using points on that circle) to the known area of that circle.

It matters not that you used many points in that approximation. (In fact, that was one way that some early mathematicians were able to approximate the value of pi.) But no matter how many points taken, that ratio of areas will always be less than 1, and only approach 1 as a limit.

[Mahmoud]

I am using the ratio of 4*pi*area/perimeter^2 I drew a circle using AutoCAD to make sure it is an exact circle, but I got a ratio of 0.92 in Matlab

[John D'Errico]

Why do you assume it is "exact"? In fact, it will be only an approximation, especially if it results in an image.

[Image Analyst]

Khalid, what is the aspect ratio of your image? Have you checked/verified the circle width and height in another program such as Photoshop? And why are you making us work blind? Would you consider showing us your image so we can help you? Or is it so secret and proprietary that you can't upload it?

[Mahmoud]

I have drawn this on AutoCAD to make sure it is a perfect circle. Still, I dont get a ratio of 1

[Image Analyst]

How are you getting values for the area and perimeter from that picture?

[John D'Errico]

A picture of a circle is not a circle, any more than a picture of me is actually me.

In fact, even autocad does not create a mathematically exact circle, only an approximation to one. Then if you base your computation on a picture thereof, you have yet another approximation.

Need I repeat my quote about a model of a cat?

[Christovão Abrahão]

I understand that the analog world can not be represented exactly by a digital model (discrete). But a circle drawn by Autocad should have greater roundness higher the resolution or the absolute size of the picture, right?

[John D'Errico]

It depends entirely upon how the image is created, something about which I can have no knowledge. Start with a low res image, then resize it, and you will gain nothing. Anyway, be careful about the measured length of a curve, when that curve is a city block one, as one might find in an image.

p1 = [0 1;1 1;1 0];
p2 = [0 1;.5 1; .5 .5; 1 .5;1 0];
p3 = [0 1;.25 1;.25 .90;.5 .90; .5 .75; .75 .75;.75 .5;.90 .5;.90 .25;1 .25;1 0];

plot(p1(:,1),p1(:,2),'r-',p2(:,1),p2(:,2),'g-',p3(:,1),p3(:,2),'b-')

See that each of those curves have the same length, though the third would seem to be a bit better approximation to a circle. In fact, ANY curve that proceeds in a city block fashion from the point (0,1) to the point (1,0) has the same length (2 units) as long as it moves monotonically between the two end points.

Answer 3

[Image Analyst]

Mahmoud, I know I asked you before and you didn't respond with your code, so perhaps you realized your mistake. But I thought I'd show the circularity with my code and your image.

clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
% clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 20;

%===============================================================================
% Read in a gray scale demo image.
folder = pwd;
baseFileName = 'trial.png';
% Get the full filename, with path prepended.
fullFileName = fullfile(folder, baseFileName);
% Check if file exists.
if ~exist(fullFileName, 'file')
  % File doesn't exist -- didn't find it there.  Check the search path for it.
  fullFileNameOnSearchPath = baseFileName; % No path this time.
  if ~exist(fullFileNameOnSearchPath, 'file')
    % Still didn't find it.  Alert user.
    errorMessage = sprintf('Error: %s does not exist in the search path folders.', fullFileName);
    uiwait(warndlg(errorMessage));
    return;
  end
end
grayImage = imread(fullFileName);
% Get the dimensions of the image.  
% numberOfColorBands should be = 1.
[rows, columns, numberOfColorChannels] = size(grayImage);
if numberOfColorChannels > 1
  % It's not really gray scale like we expected - it's color.
  % Convert it to gray scale by taking only the green channel.
  grayImage = grayImage(:, :, 2); % Take green channel.
end
% Display the image.
subplot(2, 2, 1);
imshow(grayImage, []);
axis on;
title('Original Grayscale Image', 'FontSize', fontSize, 'Interpreter', 'None');

% Set up figure properties:
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
% Get rid of tool bar and pulldown menus that are along top of figure.
set(gcf, 'Toolbar', 'none', 'Menu', 'none');
% Give a name to the title bar.
set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off') 

% Unfortunately the image was saved as color so we need to get the circle only.
% Threshold the image at 192 to binarize it.
binaryImage = grayImage == 192;
% Display the image.
subplot(2, 2, 2);
imshow(binaryImage, []);
axis on;
title('Binary Image', 'FontSize', fontSize, 'Interpreter', 'None');

% Now fill it so we can get the area and perimeter
binaryImage = imfill(binaryImage, 'holes');
% Display the image.
subplot(2, 2, 3);
imshow(binaryImage, []);
axis on;
title('Filled Binary Image', 'FontSize', fontSize, 'Interpreter', 'None');

labeledImage = bwlabel(binaryImage);
measurements = regionprops(labeledImage, 'Perimeter', 'Area');
circularity = measurements.Perimeter^2 ./ (4*pi*measurements.Area)

% Tell user we're done
message = sprintf('The circularity = %f', circularity);
uiwait(helpdlg(message));

So I get a circularity = 0.987191321413265 which is reasonably close to 1, especially considering the jaggy nature of the low resolution image you uploaded. If you're still interested in learning why your code gave a wrong value of 0.92, then you need to attach your code.