How to get the distance references of geometry and curvature of the object?

Dear members, I would be truly grateful if you could assist me in extracting automatically the distances indicated in the image and obtaining the curvature of its extreme points on the right and left. My goal now is to obtain the curvature of the object as it flows.

Could you please guide me on the correct approach for achieving this?

I have tried to obtain the distances but it was manual.

%%

clc

clear all

close all

% Load the image

I = imread('20000002002.bmp');

% Binarize

BW = imbinarize(I);

% Extract the ROI

BW = ~BW;

se = strel("disk", 5);

BW = imopen(BW, se);

BW = imclearborder(BW);

BW = bwareafilt(BW, 1);

BW = imfill(BW, "holes");

% Calculate the centroid

s = regionprops(BW, "Centroid");

% Calculate the extreme points

idx = any(BW);

pt1 = find(idx, 1);

pt2 = find(idx, 1, "last");

% Visualize the result

imshow(I)

hold on

% Adiciona uma linha vertical constante em x = 860

xline(860, "g", "X = 860", "FontSize", 18);

% Calcula a distância entre pt1 e 860

distancia_pt1_860 = abs(860 - pt1);

% Exibe as linhas e pontos

xline(pt1, "r", sprintf("X = %d", pt1), "FontSize", 18)

xline(pt2, "r", sprintf("X = %d", pt2), "FontSize", 18)

h = scatter(s.Centroid(1), s.Centroid(2), "r", "filled");

% Adiciona linhas horizontais com o centroide no meio delas

distancia_entre_linhas = 183;

linha_superior_y = s.Centroid(2) + distancia_entre_linhas/2;

linha_inferior_y = s.Centroid(2) - distancia_entre_linhas/2;

% Adiciona a linha superior

plot([pt1, pt2], [linha_superior_y, linha_superior_y], 'b', 'LineWidth', 1);

% Adiciona a linha inferior

plot([pt1, pt2], [linha_inferior_y, linha_inferior_y], 'b', 'LineWidth', 1);

% Adiciona os pontos extremos e o centroide

scatter([pt1, pt2, s.Centroid(1)], [linha_superior_y, linha_superior_y, s.Centroid(2)], "b", "filled");

% Adiciona os pontos extremos e o centroide

scatter([pt1, pt2, s.Centroid(1)], [linha_inferior_y, linha_inferior_y, s.Centroid(2)], "b", "filled");

% Exibe a distância entre pt1 e 860

disp(['Distância entre pt1 e x=860: ', num2str(distancia_pt1_860)]);

[Matt J]

Your attached image has all kids of junk on the left and right. Do you really have to deal with that automatically, or can you crop the image to the object of interest?

[Rhandrey Maestri]

Hi, these junks are not relevant for me. I agree with croping the image to only the object of interest.

Can it be done automatically as well?

[Matt J]

Why is there a chequerboard pattern in your image background? Can the image be retaken without it? Most of the effort in your code is devoted towards getting rid of it.

[Rhandrey Maestri]

I do not know. I have not paid attention to that before. Now it would be difficult to retake these pictures.

Answer 1

[Matt J]

I was able to do it with the help of some FEX downloads:

https://www.mathworks.com/matlabcentral/fileexchange/90762-further-tools-for-analyzing-objects-in-n-dimensional-images

https://www.mathworks.com/matlabcentral/fileexchange/87584-object-oriented-tools-for-fitting-conics-and-quadrics

I'm not sure if it's a coincidence that the two circles intersect at the minimum dip point. You would know better than me, I'd imagine.

BW=~imbinarize(imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/1553107/20000002002.bmp'));

BW=BW(round(end/4):round(2*end/3),round(end/3):round(2*end/3));

BW=imfill(BW,'holes');

BW=imclose(imerode(BW,[1 0;0 1]),ones(2));

BW=bwareaopen(BW,30);

bullet=bwareafilt(BW,1);

bg=bwareafilt(bwlalphaclose(BW&~bullet,5),4);

sbullet=boundaryXY(bullet);
sborder=boundaryXY(bwmorph(bg,'skel',inf));

ydip=max(sborder(2).y);
xdip=mean(sborder(2).x(sborder(2).y==ydip));

[xdip,ydip] %Minimum Dip Point

ans = 1×2

220 92

distance=abs(median(sborder(1).y)-median(sborder(4).y))  %distance ebtween top and bottom lines

distance = 180

XY=[sbullet.x, sbullet.y]';

xmin=min(XY(1,:),[],2);
xmax=max(XY(1,:),[],2);


XY1=XY(:,XY(1,:)<=xmin+4);
XY2=XY(:,XY(1,:)>=xmax-4);

circ1=circularFit(XY1)

circ1 = 
  circularFit with properties:

    center: [161.2045 153.6687]
    radius: 86.6585

circ2=circularFit(XY2)

circ2 = 
  circularFit with properties:

    center: [259.8320 153.7851]
    radius: 72.6716

imshow(BW);hold on

showfit(circ1);
showfit(circ2);
scatter(xdip,ydip,'c','filled','SizeData',80);  
hold off

function s=boundaryXY(bw)

 B=bwboundaries(bw);

 for i=1:numel(B)
  s(i).x=B{i}(:,2);
  s(i).y=B{i}(:,1);
 end
 
end

[Rhandrey Maestri]

Thanks for your answer!

I had some errors with some images, Is there any parameter that I can change to fit longer objects that I am measuring these radius?

[Matt J]

Of course! You can change anything in the code that you want.

[Matt J]

If I had to guess (I do have to because I don't have the error-generating image), the image cropping step is not compatible with the longer object,

BW=BW(round(end/4):round(2*end/3),round(end/3):round(2*end/3));

but all I can really do is guess.

[Rhandrey Maestri]

Look where it is getting the values for this image. Can you please tell me how to fix?

Thanks!

[Rhandrey Maestri]

In this image the code does no run. It has an error.

Error using horzcat

Dimensions of arrays being concatenated are not consistent.

Error in CODEnew (line 61)

XY=[sbullet.x, sbullet.y]';

[Rhandrey Maestri]

% Binarize

BW = imbinarize(I);

% Extract the ROI

BW = ~BW;

se = strel("disk", 5);

BW = imopen(BW, se);

BW = imclearborder(BW);

BW = bwareafilt(BW, 1);

BW = imfill(BW, "holes");

% Calculate the centroid

s = regionprops(BW, "Centroid");

% Armazena a posição do centróide no vetor

posicoes_centroid(i, :) = s.Centroid;

% Calculate the extreme points

idx = any(BW);

pt1 = find(idx, 1);

pt2 = find(idx, 1, "last");

% Armazena a posição do pt1 e pt2 no vetor

posicoes_pt1(i, :) = [pt1, 0];

posicoes_pt2(i, :) = [pt2, 0];

%

% ... restante do código sem plotar

figure;

imshow(I)

This code works well to get the centroid and last borders. MAybe we can work on it to get the curvatures ?

[Matt J]

This version seems to work more robustly:

BW=~imbinarize(imread('1bubble-exported0001437.bmp'));

processImage(BW)

ans = 1×2

219 91

distance = 177

circ1 = 
  circularFit with properties:

    center: [204.1143 151.8610]
    radius: 85.5361

circ2 = 
  circularFit with properties:

    center: [310.0993 152.5000]
    radius: 41.6114

BW=~imbinarize(imread('1bubble-exported0000200.bmp'));

processImage(BW)

ans = 1×2

218 91

distance = 179

circ1 = 
  circularFit with properties:

    center: [313.4939 153.7820]
    radius: 92.1933

circ2 = 
  circularFit with properties:

    center: [402.5134 155.1992]
    radius: 41.5158

BW=~imbinarize(imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/1553107/20000002002.bmp'));

processImage(BW)

ans = 1×2

219 91

distance = 177

circ1 = 
  circularFit with properties:

    center: [153.3680 154.0047]
    radius: 78.7668

circ2 = 
  circularFit with properties:

    center: [272.3128 153.5000]
    radius: 60.0612

function processImage(BW)

BW=BW(round(end/4):round(2*end/3),round(end/3):round(2*end/3));
BW=imfill(BW,'holes');
BW=imclose(imerode(BW,[1 0;0 1]),ones(2));
BW=bwareaopen(BW,30);

bullet=bwareafilt(~bwlalphaclose(~BW,15),1);

bg=BW&~bullet;
bg=medfilt2(bg,[1,11]);
bg([1,end],:)=0; bg(:,[1,end])=0;
bg=bwlalphaclose(bg,5);
bg=bwareafilt(imerode(~bg,strel('disk',3)),2,'smallest');
bg=bwmorph(~imdilate(bg,strel('disk',3)),'remove');
bg([1,end],:)=0; bg(:,[1,end])=0;

sbullet=boundaryXY(bullet);
sborder=boundaryXY(bg);

ydip=max(sborder(2).y);
xdip=mean(sborder(2).x(sborder(2).y==ydip));

[xdip,ydip] %Minimum Dip Point

distance=abs(median(sborder(1).y)-median(sborder(4).y))  %distance ebtween top and bottom lines



XY=[sbullet.x, sbullet.y]';

xmin=min(XY(1,:),[],2);
xmax=max(XY(1,:),[],2);


XY1=XY(:,XY(1,:)<=xmin+4);
XY2=XY(:,XY(1,:)>=xmax-4);

circ1=circularFit(XY1)
circ2=circularFit(XY2)


imshow(BW);hold on

showfit(circ1);
showfit(circ2);
scatter(xdip,ydip,'c','filled','SizeData',80);  
hold off
shg

end

function s=boundaryXY(bw)

 B=bwboundaries(bw);

 for i=1:numel(B)
  s(i).x=B{i}(:,2);
  s(i).y=B{i}(:,1);
 end
 
end

[Image Analyst]

What is showfit()? I've heard of the built-in viscircles but not showfit.

help viscircles

 viscircles - Create circle
    This MATLAB function draws circles with specified centers and radii onto
    the current axes.

    Syntax
      viscircles(centers,radii)
      viscircles(ax,centers,radii)
      viscircles(___,Name=Value)
      h = viscircles(___)

    Input Arguments
      centers - Coordinates of circle centers
        two-column numeric matrix
      radii - Circle radii
        positive number | column vector
      ax - Axes in which to draw circles
        handle

    Name-Value Arguments
      EnhanceVisibility - Augment drawn circles with contrasting features to
      improve visibility
        true or 1 (default) | false or 0
      Color - Color of boundary
        "red" (default) | RGB triplet | hexadecimal color code |
        color name | short color name
      LineStyle - Line style of circle edge
        "-" (default) | "--" | ":" | "-." | "none"
      LineWidth - Width of circle edge
        positive number | 2 (default)

    Output Arguments
      h - Circles drawn
        hggroup object

    Examples
      openExample('images/DrawLinesAroundBrightAndDarkCirclesInImageExample')
      openExample('images/ClearAxesBeforePlottingCirclesExample')

    See also Image Viewer, visboundaries, imfindcircles, imdistline,
      drawcircle

    Introduced in Image Processing Toolbox in R2012a
    Documentation for viscircles
       doc viscircles

[Matt J]

It is a method defined for the circularFit class in this FEX fitting library,

https://www.mathworks.com/matlabcentral/fileexchange/87584-object-oriented-tools-for-fitting-conics-and-quadrics