# How can i calculate the length of curve?

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Volcano il 26 Ago 2022
Commentato: Stephen23 il 25 Gen 2024
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
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Volcano il 26 Ago 2022
Spostato: Stephen23 il 30 Gen 2023

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### Risposte (6)

Stephen23 il 30 Gen 2023
Modificato: Stephen23 il 30 Gen 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.
##### 5 CommentiMostra 3 commenti meno recentiNascondi 3 commenti meno recenti
Torsten il 24 Gen 2024
Modificato: Torsten il 24 Gen 2024
For the case of a curve in the plane:
arclength = integral_{s=a}^{s=b} sqrt(x'(s)^2+y'(s)^2) ds
Example: Half circle
syms s
a = 0;
b = 1;
x = cos(pi*s);
y = sin(pi*s);
arclength = int(sqrt(diff(x,s)^2+diff(y,s)^2),s,a,b)
arclength =
π
pi
ans = 3.1416
Stephen23 il 25 Gen 2024
Some more comparisons with ARCLENGTH:
a = linspace(0, 2*pi, 1E+3);
X = cos(a);
Y = sin(a);
Len = cumtrapz(hypot(dX,dY));
Len(end) - 2*pi
ans = -4.1393e-05
sum(hypot(dX,dY)) - 2*pi
ans = 0.0062
arclength(X,Y,'spline') - 2*pi
ans = -1.3272e-11
Also with unequal spacing:
X = logspace(-10,0,200);
Y = sqrt(1-X.^2);
Len = cumtrapz(hypot(dX,dY));
Len(end) - pi/2
ans = -0.0093
sum(hypot(dX,dY)) - pi/2
ans = 0.2245
arclength(X,Y,'spline') - pi/2
ans = 8.9196e-05
And with vertical lines:
X = ones(1,200);
Y = logspace(-10,1,200);
Len = cumtrapz(hypot(dX,dY));
Len(end) - 10
ans = -1.0000e-10
sum(hypot(dX,dY)) - 10
ans = 0.5976
arclength(X,Y,'spline') - 10
ans = -1.0000e-10

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Ankit il 26 Ago 2022
Modificato: Ankit il 26 Ago 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 CommentiMostra NessunoNascondi Nessuno
Volcano il 26 Ago 2022
Modificato: Volcano il 26 Ago 2022
Thanks a lot, but there is small difference between your answer and other one.
Guntz Romain il 9 Mar 2023
le boss

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Star Strider il 26 Ago 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
.
##### 7 CommentiMostra 5 commenti meno recentiNascondi 5 commenti meno recenti
Star Strider il 29 Gen 2023
My code calculates the trapezoidal integral of the gradients (numerical derivatives) of ‘X’ and ‘Y’.
Star Strider il 25 Gen 2024
Following up —
In light of @Torsten’s Comment, using cumtrapz is correct, however including the independent variable as the first argument to it in this instance may not be —
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];
dX = gradient(X); % Numerical Derivative
dY = gradient(Y); % Numerical Derivative
Len = cumtrapz(hypot(dX,dY)); % Integrate The Hypotenuse Of The Numerical Derivatives Of The Segments
Len_end = Len(end)
Len_end = 20.2978
figure
plot(X, Y, '.-', 'DisplayName','Data')
hold on
plot(X, Len, '.-', 'DisplayName','Length')
hold off
grid
title('Provided Data')
legend('Location','best')
axis('equal')
a = linspace(0, 2*pi, 1E+3);
X = cos(a);
Y = sin(a);
dX = gradient(X); % Numerical Derivative
dY = gradient(Y); % Numerical Derivative
Len = cumtrapz(hypot(dX,dY)); % Integrate The Hypotenuse Of The Numerical Derivatives Of The Segments
Len_end = Len(end)
Len_end = 6.2831
figure
plot(X, Y, '.-', 'DisplayName','Data')
hold on
plot(X, Len, '.-', 'DisplayName','Length')
hold off
grid
legend('Location','best')
axis('equal')
My code is unchanged, however it now includes a second data set (the circle) whose result can be checked.
.

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Torsten il 26 Ago 2022
Modificato: Torsten il 26 Ago 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
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Tamas Rozsa il 29 Gen 2023
Modificato: Tamas Rozsa il 30 Gen 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.
##### 3 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
Tamas Rozsa il 30 Gen 2023
Because it also depends on the input data.
My main message was not really about the accuracy, but to point out that @Star Strider's solution is buggy and shall never be used by anyone in that way.
It accidentally gave kinda fair result for the OP input data, but fails heavily e.g. in the two scenarios I've mentioned:
• arbitrary spacing, e.g.:
X = logspace(-10,0,200);
Y = sqrt(1-X.^2);
(result is 0.11 instead of ca. 1.57 (pi/2))
• vertical line segments, e.g.:
X = ones(1,200);
Y = logspace(-10,1,200);
(result is 0 instead of ca. 10).
The accuracy is another topic; diff() is indeed better here than gradient().
Paul il 30 Gen 2023
I think the example I showed reinforces your concerns.

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Walter Roberson il 21 Lug 2023
No, you cannot really get the length of a curve defined by a finite list of points. A finite list of 2D points does not define a curve: a finite list of 2D points defines a polygon at best (possibly a self-interesecting one.)
In order to get a curve length, you either have to be given a curve equation, or else you have to be willing to approximate the true curve length by using a model. The model might over-estimate or under-estimate the true curve length.
Mathematically it is impossible to be given a finite set of points that are finitely expressed, and use them to come up with "the" defining curve. Mathematically given any finite set of points that are finitely expressed, there are an uncountable infinity of curves that go through the given points. (Allowing for the possibility that there is noise or round-off or truncation error in the list of coordinates does not increase the number of possible curves, as uncountable infinity is the largest infinity already until you get into abstractions such as Aleph-One )
The answers posted by the other participants are either finding total segment lengths (treating the point list as a polygon whose perimeter is to be found), or else are using different models of how to interpolate the points into a curve. They produce different results because they use different interpolation methods. That does not make any of them "wrong", just different. Unless you know the form of the original function, you just have to accept that the problem is under-specified.
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