📄 measuringangleofintersection.m
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%Measuring Angle of Intersection
%********************************
%A common task in machine vision applications is hands-free measurement
%using image acquisition and image processing techniques.
%Your goal is to measure the angle and point of intersection between two beams
%using bwtraceboundary, which is a boundary tracing routine.
%********************************
%Step 1: Load image Read in gantrycrane.jpg and draw arrows pointing to two beams of
%interest. It is an image of a gantry crane used to assemble a bridge.
RGB = imread('gantrycrane.png');
figure,imshow(RGB);
text(size(RGB,2),size(RGB,1)+15,'Image courtesy of Jeff Mather',...
'FontSize',7,'HorizontalAlignment','right');
line([300 328],[85 103],'color',[1 1 0]);
line([268 255],[85 140],'color',[1 1 0]);
text(150,72,'Measure the angle between these beams','Color','y',...
'FontWeight', 'bold');
%Step 2: Extract the region of interest
%Crop the image to obtain only the beams of the gantry crane chosen earlier.
%This step will make it easier to extract the edges of the two metal beams.
% you can obtain the coordinates of the rectangular region using
% pixel information displayed by imview
start_row = 34;
start_col = 208;
cropRGB = RGB(start_row:163, start_col:400, :);
figure,imshow(cropRGB)
% Store (X,Y) offsets for later use; subtract 1 so that each offset will
% correspond to the last pixel before the region of interest
offsetX = start_col-1;
offsetY = start_row-1;
%Step 3: Threshold the image
%Convert the image to black and white for subsequent extraction of the edge coordinates
%using bwtraceboundary routine.
I = rgb2gray(cropRGB);
threshold = graythresh(I);
BW = im2bw(I,threshold);
BW = ~BW; % complement the image (objects of interest must be white)
figure,imshow(BW)
%Step 4: Find initial point on each boundary
%The bwtraceboundary routine requires that you specify a single point on a boundary.
%This point is used as the starting location for the boundary tracing process.
%To extract the edge of the lower beam, pick a column in the image and inspect it
%until a transition from a background pixel to the object pixel occurs.
%Store this location for later use in bwtraceboundary routine.
%Repeat this procedure for the other beam, but this time tracing horizontally.
dim = size(BW);
% horizontal beam
col1 = 4;
row1 = min(find(BW(:,col1)));
% angled beam
row2 = 12;
col2 = min(find(BW(row2,:)));
%Step 5: Trace the boundaries
%The bwtraceboundary routine is used to extract (X, Y) locations of the boundary points.
%In order to maximize the accuracy of the angle and point of intersection calculations,
%it is important to extract as many points belonging to the beam edges as possible.
%You should determine the number of points experimentally. Since the initial point for
%the horizontal bar was obtained by scanning from north to south, it is safest to set
%the initial search step to point towards the outside of the object, i.e. 'North'.
boundary1 = bwtraceboundary(BW, [row1, col1], 'N', 8, 70);
% set the search direction to counterclockwise, in order to trace downward.
boundary2 = bwtraceboundary(BW, [row2, col2], 'E', 8, 90,'counter');
figure,imshow(RGB); hold on;
% apply offsets in order to draw in the original image
plot(offsetX+boundary1(:,2),offsetY+boundary1(:,1),'g','LineWidth',2);
plot(offsetX+boundary2(:,2),offsetY+boundary2(:,1),'g','LineWidth',2);
%Step 6: Fit lines to the boundaries Although (X,Y) coordinates pairs were obtained
%in the previous step, not all of the points lie exactly on a line. Which ones should
%be used to compute the angle and point of intersection? Assuming that all of the
%acquired points are equally important, fit lines to the boundary pixel locations.
%The equation for a line is y = [x 1]*[a; b]. You can solve for parameters 'a' and 'b'
%in the least-squares sense by using polyfit.
ab1 = polyfit(boundary1(:,2), boundary1(:,1), 1);
ab2 = polyfit(boundary2(:,2), boundary2(:,1), 1);
%Step 7: Find the angle of intersection
%Use the dot product to find the angle.
vect1 = [1 ab1(1)]; % create a vector based on the line equation
vect2 = [1 ab2(1)];
dp = dot(vect1, vect2);
% compute vector lengths
length1 = sqrt(sum(vect1.^2));
length2 = sqrt(sum(vect2.^2));
% obtain the larger angle of intersection in degrees
angle = 180-acos(dp/(length1*length2))*180/pi
%Step 8: Find the point of intersection
%Solve the system of two equations in order to obtain (X,Y) coordinates of the
%intersection point.
intersection = [1 ,-ab1(1); 1, -ab2(1)] \ [ab1(2); ab2(2)];
% apply offsets in order to compute the location in the original,
% i.e. not cropped, image.
intersection = intersection + [offsetY; offsetX]
%Step 9: Plot the results.
inter_x = intersection(2);
inter_y = intersection(1);
% draw an "X" at the point of intersection
plot(inter_x,inter_y,'yx','LineWidth',2);
text(inter_x-60, inter_y-30, [sprintf('%1.3f',angle),'{\circ}'],...
'Color','y','FontSize',14,'FontWeight','bold');
interString = sprintf('(%2.1f,%2.1f)', inter_x, inter_y);
text(inter_x-10, inter_y+20, interString,...
'Color','y','FontSize',14,'FontWeight','bold');
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