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Math Help - Solving the constants of a catenary

  1. #1
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    Solving the constants of a catenary

    Hi all.
    I have managed to derive the equation of a catenary from an initial set up. Using calculus of variations and lagrange multipliers. However I have become stuck on finding some remaining constants.

    The equation I have at the moment is:
    y=(lambda)+Acosh(x/A)

    I need to find lambda and A in terms of the length of the chain, L, and possibly the end points of the chain. (-d,0) and (d,0)

    I have tried using the integral form of an arc length equal to L, but I can't see how to solve for A, with it being in the argument of cosh, and multiplying it.

    Many thanks for any help you can give
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  2. #2
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    I think you're going to need two equations. Your y is symmetric, so you're only going to get one equation out of stipulating that y(d) = 0. I mean, the equation y(-d) = 0 is superfluous. The other equation you get from, as you hinted, using the arc length formula. That is, set

    \displaystyle\int_{-d}^{d}\sqrt{1+(y'(x))^{2}}\,dx=L.

    What do you get?
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  3. #3
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    The end point gives: (lambda) + Acosh(d/A) = 0
    And the integral for the length constraint gives: 2Asinh(d/A)=L

    Would the next step to be to get A = -(lambda)/cosh(d/A) from the first equation, and substitute it into the second?
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  4. #4
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    That substitution won't eliminate anything for you. I would try using some hyperbolic trig identities in order to eliminate the hyperbolic trig functions. Try setting

    \cosh^{2}(d/A)-\sinh^{2}(d/A)=1.

    That is, solve each of the two equations you have for the hyperbolic trig functions, and then substitute into this identity. This will give you another equation, still containing both \lambda and A, but more easily allowing you to solve for one of them. What does that do for you?
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  5. #5
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    Rearranging my two equations gives:
    cosh(d/A) = -(lambda)/A and sinh(d/A) = L/2A
    Using the identity you suggest leads me to A = { (lambda)^2 - (L/2)^2 }^1/2
    A has to be positive or it would invert the catenary.

    Is this the right path? How would I then find a value for lambda?
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  6. #6
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    Looks great. I agree with your physical reason for ruling out the negative solution for A. To find \lambda, I would plug this solution for A into the endpoint condition. The resulting equation for \lambda is only solvable numerically, I would think. You'd have to have d and L specified. If you were to do it the other way, where you solve for \lambda first, you'd still get a nasty enough equation for A that you'd have to solve it numerically.

    This is about as far as you can go analytically, I think.
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  7. #7
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    Thank you for your help, unfortunately I am trying to solve for arbitrary L and d.
    My intuition tells me that \lambda should be negative, as it has to bring the end points to y=0
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  8. #8
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    Is this a problem in a textbook? If so, are you fairly sure that you can get an analytical solution? I'm not sure I see how.

    I would agree that lambda is negative.
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  9. #9
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    No it was a question set by a lecturer. Who said that the constants should be determined in terms of one or all of:
    mass per unit length \rho
    strength of gravitational field g
    the chain length L
    and the horizontal separation of the two end points 2d < L

    From the question I am assuming an analytical solution is possible.
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  10. #10
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    Hmm. Well, like I said, I don't know how to solve for the two remaining parameters analytically, if those questions are correct. I am not sure but what you can't. Maybe someone else like CaptainBlack or mr fantastic or Opalg could do it. Maybe I'm missing something.
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  11. #11
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    My lecturer has said that it can't be solved analytically! So you aren't missing anything any tricks.
    He said the length constraint gives an implicit equation for A and thats as good as it gets.
    Thank you again for all your help
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  12. #12
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    You're very welcome! Have a good one.
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