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Step Function

  1. Dec 27, 2005 #1
    Cable System - Step Function

    Hi, I want to do an experiment using two dissimilar ropes in my basement and see if I can produce some results using a rope analysis. Here’s the set up. I want to tie a light rope to a heavy rope, and be able to characterize the curvature the system takes on. I understand the basic equations involved that I must use to come up with a solution; however, they are based on a uniform weight distribution. Since I am tying a light rope to a very heavy rope, I have to use a step function, and I'm thinking of going the Laplace route to integrate. I would appreciate the help you could provide me with. I don't know if it will work or not, but I figure what the hell why not at least try. The laplace will make it very messy, Unfortunately (I think).
     
    Last edited: Dec 27, 2005
  2. jcsd
  3. Dec 27, 2005 #2
    The first function I know is that: [tex] dy/dx = \frac {1}{F_H} \int w(s)ds [/tex]
    Where, w(s), is the weight per unit length.
    and I will state that:
    [tex] w(s) = \varrho_1 [/tex] for [tex] 0 \leq x \leq \alpha [/tex]
    and
    [tex] w(s) = \varrho_2 [/tex] for [tex] \alpha \leq x \leq \beta [/tex]

    where [tex] \alpha [/tex] is where the two ropes are tied, and [tex] \beta [/tex] is the end of the rope.
     
    Last edited: Dec 27, 2005
  4. Dec 27, 2005 #3

    HallsofIvy

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    "I want to tie a light rope to a heavy rope, and be able to characterize the curvature the system takes on."

    Why should there be any "curvature" at all? Just lay the ropes in a line on the floor!

    If you are talking about wave motion of the two ropes then please say that.
     
  5. Dec 27, 2005 #4
    No, this is not wave motion. Its a composite catenary system. Putting them on the floor would make no sense.
     
  6. Dec 27, 2005 #5
    I don't think using laplace transform will do me a bit of good, because I don't have a differential equation im working with. My integral would just be equivalent to having an expression for the total weight of the rope. Im thinking it would look like this: [tex] \int w(s)ds = w_1s + w_2u_{\alpha}(s) + C_1[/tex] Im just calling u-sub alpha the step function, or if you guys think thats bad notation, I can use < > and make it a singularity function, that turns on after you pass point beta along the rope. Im saying the two dissimilar ropes are tied together at alpha, and the weight of the rope suddenly 'jumps' much heavier (or lighter), after that point.

    So now I have to integrate this mess:

    [tex] x = \int \frac {ds} { [1+ \frac{1}{F^2_H} ( \int w(s)ds)^2 ]^{ \frac{1}{2}}}[/tex]

    Can someone please help me?

    Idealy, what I REALLY want is one smooth function of s that I can write that has a built in step to it and gives me the same results, so the second integration is easier. Is that possible?
     
    Last edited: Dec 27, 2005
  7. Dec 27, 2005 #6
  8. Dec 28, 2005 #7

    benorin

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    Catenary

    You are (it seems) looking to describe the shape or curve that a flexible rope (cord, chain, telephone wire, whatever) would assume when both ends are suspended and the length of rope between them is acted on by gravity, (viz. catenary, and also here).

    Good stuff.
     
  9. Dec 28, 2005 #8
    Yeah, but those links are for a uniform weight distribution, not a step change, like two ropes tied together.
     
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