how to generate multiple curves with the same track length

I want to know how to generate multiple curves with the same track length. In other words, how can we get the curve of a rope with a certain length (such as 1m) after its random bending.

[J. Alex Lee]

It seems like the harder part of this question is how to make the rules for randomly bending...after that, measuring the track length (arc length) of a curve is relatively easy, just us pythagoreas on each segment making up the curve and add it all up.

Answer 1

[Mathieu NOE]

hello

try this

% rope at rest (straight) would have length = 1 (= max(t))
n = 1000;
t = (0:n-1)./n;
max_length = 0.7; % we remove x , y points from curve s that exceed  max_length (should be < 1)

figure(1)
hold on
for ck = 1:10
    w = 2*pi*rand(1);
    x = t;
    y = rand(1).*t+cos(w.*t);
    
    % distance formula : ds² = dx² + dy² , then s = cumsum of ds
    dx = diff(x);
    dy = diff(y);
    ds = sqrt(dx.^2 + dy.^2);
    s = [0 cumsum(ds)];
    id = (s>max_length);
    x(id) = [];
    y(id) = [];
    % recompute s after points removal (to check it works);
    dx = diff(x);
    dy = diff(y);
    ds = sqrt(dx.^2 + dy.^2);
    s = sum(ds) % should remains below or equal to max_length
    
    plot(x,y);
end
hold off

[lei kong]

Thanks for your help! The way you calculate the length of the curve is very useful to me.

[Mathieu NOE]

My pleasure !

if my suggestion has fullfilled your expectation, don't hesitate to accept it (or someone else answer if it better deserves your project)

Answer 2

[Matt J]

Using interparc from the File Exchange

https://www.mathworks.com/matlabcentral/fileexchange/34874-interparc?s_tid=srchtitle

format long
for i=1:4
    
    [Px,Py,TrackLength]=makeRope(2);
    
    TrackLength
    
    plot(Px,Py);hold on
    
end

TrackLength = 
   1.999999999963130

TrackLength = 
   1.999999999957910

TrackLength = 
   1.999999999541550

TrackLength = 
   1.999999999896211

function [Px,Py,Lfinal]=makeRope(targetLength)

    L=targetLength;   %target length
    
    N=100; %number of knots
    
    px=sqrt(L)*rand(1,N); py=sqrt(L)*rand(1,N);
    
    t=linspace(1,N,1e5);
    
    Px=interp1(px,t,'spline');
    Py=interp1(py,t,'spline');
    
    Lc=cumsum(vecnorm(diff([Px;Py],1,2),2,1));
    
    tc=find(L<Lc,1);
    
    Px=Px(1:tc+1); Py=Py(1:tc+1);
    
    Ltot=sum(vecnorm(diff([Px;Py],1,2),2,1));
    
    pt=interparc(L/Ltot, Px,Py)';
    
    P=[Px;Py];
    
    [~,k]=mink( vecnorm(P-pt,2,1) ,2);
    k=min(k)-1;
    
    Px=[Px(1:k),pt(1)]-Px(1); %final points
    Py=[Py(1:k),pt(2)]-Py(1); 

    Lfinal=sum(vecnorm(diff([Px;Py],1,2),2,1)); %confirm the length
end

[J. Alex Lee]

neat!

[lei kong]

Thank you for your help. The matlab functions you used are strange to me. I need a little time to learn, and the results look good! thank you!