# How can i calculate the length of curve?

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Volcano on 26 Aug 2022
Answered: Volcano on 26 Aug 2022
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

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 26 Aug 2022
I use the gradient function to calculate the derivatives., It produces a different (and in my opinion more accurate) estimate of the derivative than diff (that by definition also results in a vector that is one element shorter than the original), while the length of the gradient result is the same as the original.

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.
Volcano on 26 Aug 2022
Edited: Volcano on 26 Aug 2022
Thanks a lot, but there is small difference between your answer and other one.

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

Volcano on 26 Aug 2022

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