# How can i calculate the length of curve?

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Volcano on 26 Aug 2022
Edited: Stephen23 on 9 Mar 2023 at 15:25
Hi,
I have a curve which includes X (meter) and Y (meter) data. Is there any way to obtain the length of curve easily?
Thanks a lot,
X=[96.0741000000000,97.1940000000000,98.3139000000000,99.4338000000000,100.553700000000,101.673600000000,102.793500000000,103.913400000000,105.033300000000,106.153200000000,107.273100000000,108.393000000000,109.512900000000,110.632800000000,111.752700000000,112.872600000000,113.992500000000,115.112400000000,116.232300000000]
X = 1×19
96.0741 97.1940 98.3139 99.4338 100.5537 101.6736 102.7935 103.9134 105.0333 106.1532 107.2731 108.3930 109.5129 110.6328 111.7527 112.8726 113.9925 115.1124 116.2323
Y=[-4.13836296940031,-4.10455468315876,-4.05645426203322,-3.99617782198545,-3.92344322326347,-3.83385191481492,-3.73582865974161,-3.61740402741020,-3.49399064332423,-3.35231953224592,-3.20552503148528,-3.04892626846560,-2.88658570885772,-2.72091440408539,-2.55226630046971,-2.38425597793465,-2.21787687713447,-2.05656258174384,-1.89889800700337]
Y = 1×19
-4.1384 -4.1046 -4.0565 -3.9962 -3.9234 -3.8339 -3.7358 -3.6174 -3.4940 -3.3523 -3.2055 -3.0489 -2.8866 -2.7209 -2.5523 -2.3843 -2.2179 -2.0566 -1.8989
Volcano on 26 Aug 2022
Moved: Stephen23 on 30 Jan 2023

Ankit on 26 Aug 2022
Edited: Ankit on 26 Aug 2022
X=[96.0741000000000,97.1940000000000,98.3139000000000,99.4338000000000,100.553700000000,101.673600000000,102.793500000000,103.913400000000,105.033300000000,106.153200000000,107.273100000000,108.393000000000,109.512900000000,110.632800000000,111.752700000000,112.872600000000,113.992500000000,115.112400000000,116.232300000000];
Y=[-4.13836296940031,-4.10455468315876,-4.05645426203322,-3.99617782198545,-3.92344322326347,-3.83385191481492,-3.73582865974161,-3.61740402741020,-3.49399064332423,-3.35231953224592,-3.20552503148528,-3.04892626846560,-2.88658570885772,-2.72091440408539,-2.55226630046971,-2.38425597793465,-2.21787687713447,-2.05656258174384,-1.89889800700337];
len_curve = sum(vecnorm(diff( [X(:),Y(:)] ),2,2));
% the 2-norm along the rows of a matrix: vecnorm(A,2,2) , where A is a
% vector
% diff: Difference and approximate derivative.
##### 2 CommentsShowHide 1 older comment
Guntz Romain on 9 Mar 2023 at 14:57
le boss

Star Strider on 26 Aug 2022
Possibly —
X=[96.0741000000000,97.1940000000000,98.3139000000000,99.4338000000000,100.553700000000,101.673600000000,102.793500000000,103.913400000000,105.033300000000,106.153200000000,107.273100000000,108.393000000000,109.512900000000,110.632800000000,111.752700000000,112.872600000000,113.992500000000,115.112400000000,116.232300000000];
Y=[-4.13836296940031,-4.10455468315876,-4.05645426203322,-3.99617782198545,-3.92344322326347,-3.83385191481492,-3.73582865974161,-3.61740402741020,-3.49399064332423,-3.35231953224592,-3.20552503148528,-3.04892626846560,-2.88658570885772,-2.72091440408539,-2.55226630046971,-2.38425597793465,-2.21787687713447,-2.05656258174384,-1.89889800700337]
Y = 1×19
-4.1384 -4.1046 -4.0565 -3.9962 -3.9234 -3.8339 -3.7358 -3.6174 -3.4940 -3.3523 -3.2055 -3.0489 -2.8866 -2.7209 -2.5523 -2.3843 -2.2179 -2.0566 -1.8989
dX = gradient(X); % Numerical Derivative
dY = gradient(Y); % Numerical Derivative
Len = cumtrapz(X,hypot(dX,dY)) % Integrate The Hypotenuse Of The Numerical Derivatives Of The Segments
Len = 1×19
0 1.2549 2.5102 3.7662 5.0231 6.2812 7.5405 8.8012 10.0634 11.3271 12.5922 13.8584 15.1256 16.3934 17.6616 18.9298 20.1976 21.4648 22.7315
figure
plot(X, Y, '.-')
hold on
plot(X, Len, '.-')
hold off
grid
.
Star Strider on 29 Jan 2023
My code calculates the trapezoidal integral of the gradients (numerical derivatives) of ‘X’ and ‘Y’.

Torsten on 26 Aug 2022
Edited: Torsten on 26 Aug 2022
I'd say Ankit's solution is the more intuitive.
But Star Strider's solution should be second-order accurate while Ankit's is only first-order accurate.
X=[96.0741000000000,97.1940000000000,98.3139000000000,99.4338000000000,100.553700000000,101.673600000000,102.793500000000,103.913400000000,105.033300000000,106.153200000000,107.273100000000,108.393000000000,109.512900000000,110.632800000000,111.752700000000,112.872600000000,113.992500000000,115.112400000000,116.232300000000];
Y=[-4.13836296940031,-4.10455468315876,-4.05645426203322,-3.99617782198545,-3.92344322326347,-3.83385191481492,-3.73582865974161,-3.61740402741020,-3.49399064332423,-3.35231953224592,-3.20552503148528,-3.04892626846560,-2.88658570885772,-2.72091440408539,-2.55226630046971,-2.38425597793465,-2.21787687713447,-2.05656258174384,-1.89889800700337];
length = 0;
for i = 1:numel(X)-1
length = length + sqrt((X(i+1)-X(i))^2 + (Y(i+1)-Y(i))^2);
end
length
length = 20.2980

Tamas Rozsa on 29 Jan 2023
Edited: Tamas Rozsa on 30 Jan 2023
Based on @Star Strider's answer, but with correct result:
% option 1
Len = cumsum(hypot(dX,dY)) % if sublengths of all segments also needed
% option 2
Len = sum(hypot(dX,dY)) % if only total length needed
As @Star Strider also highlighted in comment, gradient() may be substituted with diff(), but gradient() may give more satisfactory (i.e., smoother) result in most cases. [UPDATE: in some cases, and depending on the actual use-case]
Unlike @Star Strider's original answer, this code gives correct result even in case of arbitrary spacing in the input data as well as in case of vertical line segments.
Paul on 30 Jan 2023
I think the example I showed reinforces your concerns.

Stephen23 on 30 Jan 2023
Edited: Stephen23 on 30 Jan 2023
A very simple approach is to download John D'Errico's excellent ARCLENGTH function:
X = -1:.01:1;
Y = sqrt(1-X.^2);
L = arclength(X,Y,'spline')
L = 3.1416
L-pi
ans = 5.0768e-07
For the sample curve, this gives a more accurate solution.
Stephen23 on 9 Mar 2023 at 15:24
Edited: Stephen23 on 9 Mar 2023 at 15:25
Tested on the cases given here:
Arbitrary spacing (L=pi/2):
X = logspace(-10,0,200);
Y = sqrt(1-X.^2);
L = arclength(X,Y,'spline')
L = 1.5709
L - pi/2
ans = 8.9196e-05
Vertical lines (L=10):
X = ones(1,200);
Y = logspace(-10,1,200);
L = arclength(X,Y,'spline')
L = 10.0000
L - 10
ans = -1.0000e-10