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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?

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    $\begingroup$ 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. $\endgroup$ – Gert Mar 18 at 11:08
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    $\begingroup$ You may find my answer here helpful: physics.stackexchange.com/a/421965/123208 $\endgroup$ – PM 2Ring Mar 18 at 11:26
  • $\begingroup$ @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. $\endgroup$ – galtor Mar 18 at 11:32
  • $\begingroup$ 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. $\endgroup$ – PM 2Ring Mar 18 at 11:59
  • $\begingroup$ 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? $\endgroup$ – galtor Mar 18 at 12:47
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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) .

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  • $\begingroup$ Yes, that's the case for two hanging points. But my project was just for one $\endgroup$ – galtor Mar 19 at 20:50
  • $\begingroup$ If there is only one point of support, the chain will hang straight down. $\endgroup$ – R.W. Bird Mar 19 at 21:11
  • $\begingroup$ Yes, but that's the quasistatic state. If I drive the rope, I would like to see the dynamics. $\endgroup$ – galtor Mar 19 at 21:57
  • $\begingroup$ How do you plan to drive the rope? $\endgroup$ – R.W. Bird Mar 20 at 14:19
  • $\begingroup$ I've started to model it like a single pendulum for a start. The pivot is moved in a horizontal movement with a sudden stop. $\endgroup$ – galtor Mar 20 at 14:34

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