How to find the distance between two points along a curve?

I have a set of generated X and Y axis data, which have given a curved line. Now, I need to find the distance between two specified points along the curve, not the straight shortest distance between two points but along the curve path. Can someone help me finding this numerically?

[Dyuman Joshi]

See - https://in.mathworks.com/matlabcentral/fileexchange/34871-arclength

[Sam Chak]

Can you help the users by providing the math formula to find the distance between two specified points along the curve? The rest is just typing out the formula in MATLAB:

%% formula:
xdata    = ...;
ydata    = ...;
distance = ...;

[Sathiya]

The coordinates

th = linspace(-pi/2, pi/2, 100);
R = 200;
X = R * sin(th) ;  % X-coordinates
Y = R * cos(th) ;  % Y-coordinates

The plot

plot(X, Y, 'r*');

To find the distance between first point and second point along the curve path, I need a robust function.

The dr here gives the euclidian distance only,

for i = 1:1:numel(X)-1
dr = sqrt((X(i+1) - X(i))^2 +  (Y(i+1) - Y(i))^2);
end

But, I need the distance along the curve like this between the specified points (index of value in that array).

[Sathiya]

@Sam Chak, is it understandable now?

[Sam Chak]

I tested and it returned this result.

n   = 101;
th  = linspace(-pi/2, pi/2, n);
R   = 200;
X   = R * sin(th) ;  % X-coordinates
Y   = R * cos(th) ;  % Y-coordinates

dr  = zeros(1, numel(th)-1);
for i = 1:1:numel(X)-1
    dr(i) = sqrt((X(i+1) - X(i))^2 +  (Y(i+1) - Y(i))^2);
end
plot(dr, '.'), grid on

[Mathieu NOE]

dr represents only the increment of arc length (quite constant in this example)

now you need to do a sum of them - see my answer

[Sathiya]

@Sam Chak, dr here gives the shortest distance (displacement) between two points. For eg. lets say, I need the distance along the curve from 1st point and 50th point, this formula finds the straight distance between data points, not along the curve.

Answer 1

[Mathieu NOE]

hello

try this

th = linspace(-pi/2, pi/2, 100);
R = 200;
X = R * sin(th) ;  % X-coordinates
Y = R * cos(th) ;  % Y-coordinates

dx = diff(X);
dy = diff(Y);
ds = sqrt(dx.^2+dy.^2); % increment of arc length
s = cumsum(ds); % integration => total arc length 
s = [0 s]; % add first point : arc length = 0

% compute arc length between two points (defined by index position)
k1 = 25;
k2 = 59;
d = s(k2) - s(k1)

d = 215.7771

plot(X, Y, 'r.-',X(k1:k2), Y(k1:k2), 'db');
axis square

[Sathiya]

@Mathieu NOE, thanks for the comment. But I am not sure that the dx and dy are measured along the curve or taking the shortest path betweem two points.

[Mathieu NOE]

hello

so we have discretized a curve and we have points (X and Y coordinates)

let's call ds the distance between two successive points, and we assume that this distance is small enough vs the curvature of the curve, so we approximate the curve as a succession of small rectilinear segments; this approximation gets better if you increase the resolution (the number of points per unit length)

now ds is a vector of coordinates (dx,dy) where dx = diff(X) is the difference of coordinates of two successives points, same for dy = diff(Y)

then computing a arc length between two points is just a matter of summing all ds numbers - as demonstrated in the code

in short , ds represent a increment of arc length , dx and dy it's coresponding coordinates (projections on X and Y axes)

[Torsten]

If you have an explicit equation of your curve, you can use the usual formula for arclength:

R = 200;
fun = @(t)sqrt((R*cos(t)).^2+(R*(-sin(t))).^2)

fun = function_handle with value:
    @(t)sqrt((R*cos(t)).^2+(R*(-sin(t))).^2)

length_of_curve = integral(fun,-pi/2,pi/2)

length_of_curve = 628.3185

2*pi*R/2

ans = 628.3185

[Sathiya]

Yes, but I donot have a curve fitted equation for my curve.

Answer 2

[Walter Roberson]

Use https://www.mathworks.com/matlabcentral/fileexchange/34871-arclength

[Sathiya]

@Mathieu NOE, understood. Now it is perfect to use then. Thanks Mathieu and all others who spent their time for my problem.

[Mathieu NOE]

my pleasure !!