CODEWITHSUNDEEP
matlabmatlab-figure
XJ.C
XJ.C

Reputation: 215

To calculate the length of a curve on the image

Given the data of the curve and the size of the image, how to calculate the length of the curve ?

curve = [9.93,4;9.87,4.10;9.80,4.20;9.74,4.30;9.68,4.40;9.63,4.50;9.59,4.60;9.55,4.70;9.53,4.80;9.51,4.90;9.50,5;9.50,5.10;9.51,5.20;9.48,5.30;9.55,5.40;9.45,5.47;9.55,5.52;9.45,5.59;9.55,5.65;9.45,5.72;9.55,5.77;9.45,5.84;9.55,5.90];
backgroud = ones(20,20);
imshow(backgroud) 
hold on, plot(curve(2,:),curve(1,:),'r');

Upvotes: 0

Views: 886

Answers (1)

jodag
jodag

Reputation: 22304

To expand on the comment, you can simply sum the lengths of the line segments that make up the curve. This could be accomplished using the following code.

curve = [9.93,4;9.87,4.10;9.80,4.20;9.74,4.30;9.68,4.40;9.63,4.50;9.59,4.60;9.55,4.70;9.53,4.80;9.51,4.90;9.50,5;9.50,5.10;9.51,5.20;9.48,5.30;9.55,5.40;9.45,5.47;9.55,5.52;9.45,5.59;9.55,5.65;9.45,5.72;9.55,5.77;9.45,5.84;9.55,5.90];

len = sum(sqrt(sum(diff(curve).^2,2)))

result

len =

    2.4757

Edit: As beaker pointed out, the indices are backwards in your answer. Take a look at the values of curve_x1, curve_x2, etc... and you will see that they are just a single value. If you reverse the indices we get the same result. Also, as a sanity check take a look at the plot of the curve, it spans about 2 units in the first dimension and about 0.4 units in the second dimension so a number near 2.5 seems reasonable, a number like around 8 is much too large.

curve_x1 = num2cell(curve(1:end-1,1));
curve_y1 = num2cell(curve(1:end-1,2));
curve_x2 = num2cell(curve(2:end,1));
curve_y2 = num2cell(curve(2:end,2));

instance_length = cellfun(@(x1,y1,x2,y2) sqrt((x2-x1)^2+(y2-y1)^2), curve_x1,curve_y1,curve_x2,curve_y2);
distance = sum(instance_length)

Upvotes: 3

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