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I'm trying to get a chain physics model with segments. I'm feeling rather stupid, but how can I calculate which is the new position of each dot when the chain moves its ends?

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expressed in another way and considering just three segments of the same length: which is the way for automatically calculating the different P1, P2. when an end moves?

enter image description here

Qmechanic
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galtor
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    Create a free body diagram for each segment? This should give you a set of simultaneous equations for the $x$ and $y$ components of the forces acting on each dot. – Gert Mar 18 '20 at 11:08
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    You may find my answer here helpful: https://physics.stackexchange.com/a/421965/123208 – PM 2Ring Mar 18 '20 at 11:26
  • @PM2Ring I did see it, however i cannot understand how to get the equilibrium in the formulas of your answer. But that's exactly what I want. – galtor Mar 18 '20 at 11:32
  • Sorry, but I don't know how to explain it better than what I said in that answer. But if you'd like to know how to calculate the points on a chain given its length and the coordinates of its end points see the comments at the start of my old POV-Ray code here Also see here. – PM 2Ring Mar 18 '20 at 11:59
  • Thanks for the reply, @PM2Ring. What I cannot understand is how you fix the two ends. In your drawing, P0 is the lowest point, and the rest of coordinates come from it, isnt? – galtor Mar 18 '20 at 12:47
  • That's right. Working from the lowest point makes things relatively easy. If you don't know the lowest point, but you know the coords of the end points & the length of the chain then you can use the technique that my POV-Ray code uses. There's also good info on https://en.wikipedia.org/wiki/Catenary – PM 2Ring Mar 18 '20 at 13:07
  • @PM2Ring is your model suitable for ropes hanging freely from one point? – galtor Mar 18 '20 at 13:09
  • No, it assumes there are 2 fixed endpoints. A rope hanging freely from 1 point should just hang in a vertical line. – PM 2Ring Mar 18 '20 at 13:11

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If your chain has more than just a few links, it can be modeled with a catenary curve. (See https://en.wikipedia.org/wiki/Catenary . Scroll down for derivations.) In the formula, y = a cosh(x/a), x= 0 at the bottom of the loop, and y = a at x=0. Calculate a = (s^2 – h^2)/(2h) where, s, is half the length of your chain, and, h, is the vertical distance from the bottom of the loop to the point of support. For any other, s, along the arc from the bottom, x = a arcsinh(s/a) .

R.W. Bird
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