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Catenary - find rope length Maple must be used

  1. Jun 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Utilities companies usually use a separation between poles about 340 feet. These poles are 34 feet tall and due to restrictions due to maximum height of trucks using interstates and state routes, the minimum clearance is 20 feet (and also considering the increase of sag of the cable due to expansion in the summer time). Compute the required length of the power line between poles.

    Note 1: From Eq. (1) for y(170)=34 and y(0)=20, it can be solved (numerically) the parameter a from a single nonlinear equation. Use Newton method or any "Find root" instruction to solve this nonlinear equation. The starting guess value a0 for these techniques can be obtained graphically (show this graph). Once the value of parameter a has been found (4 decimal places accuracy), use it in Eq. (4) to find the length of the cable.

    Note 2: Derive Eq. (4) from the computation of the integral of the arc length of y(x) in Eq. (1). Compute symbolically this integral to verify the result of Eq. (4).

    2. Relevant equations

    (1) y(x) = (a)(cosh(x/a)) + C
    (2) a = H/w
    (3) cosh(x) = ((e^x) + (e^-x))/2
    (4) L = (2a)(sinh(M/a))

    3. The attempt at a solution

    I attempted using Maple's "solve" command to solve for L, a, and C. Needless to say it did not work.

    The answer yielded without evalf() was as follows for a:

    a = 28*exp(RootOf(14*_Z*exp(_Z)-85*(exp(_Z))^2+170*exp(_Z)-85))/(exp(RootOf(14*_Z*exp(_Z)-85*(exp(_Z))^2+170*exp(_Z)-85))-1)^2

    With the evalf() i yielded:

    a = Float(infinity), L = Float(undefined), C = Float(-infinity)

    I also substituted as suggested using y(170)=34 and y(0)=20. I solved for "C" in y(170) and yielded:

    C= -a*cosh(170/a)+34

    I substituted this into y(0) and yielded:

    a-a*cosh(170/a)+14 = 0

    Now that i have an equation equal to zero with a single variable I can now use Newton's method. So I assigned the new equation to f(a):

    f(a) = a-a*cosh(170/a)+14

    I next assigned g(a) to the derivative of f(a):

    g(a) = 1-cosh((170)/(a))+(170 sinh((170)/(a)))/(a)

    I assigned h(a) to Newton's Method. Now the issue is that no matter which value I choose it doesn't seem to work. I graphed the function f(x) and g(x) and it does not seem to intersect with the x-axis whatsoever so I am not quiet sure if I derived the correct equation by my substitution. I would really appreciate some expert advice of which the members on this site have an abundance. Thank for your time. I realize this is a long question, but I don't think the question is simple...in my opinion.
     
  2. jcsd
  3. Jun 17, 2010 #2
    I have solved the problem, but it makes me feel good to see that a bunch of people viewed the thread and no replies. I assume it was a brain buster for the people who viewed it. Fact remains I couldn't find an exact value, but i achieved the desired decimal place accuracy.
     
  4. Nov 11, 2010 #3
    Hi,

    I have the exact same problem and information with which to work out the length of a catenary. I don't have maple at my disposal!

    Can you please advise on how I might solve this problem??

    Your help would be much appreciated!
     
  5. Nov 11, 2010 #4
    buy it, check your college library or computer lab, or download it using torrents
     
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