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Arc Length Problem

  1. Apr 22, 2009 #1
    1. The problem statement, all variables and given/known data

    I'm trying to compute the circumference of a wing section. I have broken up the airfoil circumference into arc pieces and used cubic splines to come up with an equation for each piece.

    For example, the arc nearest the leading edge of the wing is the function:

    y = 0.0290X^3-0.3334x^2+1.8645x+1.4155

    To find the length of this arc, I use the Arc Length Formula:


    L = ∫ab √(1+[f'(x)])^2dx


    The integration limits are: a = 0.75, b = 5

    2. Relevant equations

    Alternate notation for the arc length formula gives the derivative as dy/dx:


    L = ∫ab √(1+[dy/dx])^2dx


    3. The attempt at a solution

    I started by getting the derivative of the function:

    f'(x) = dy/dx = 3(0.0290X^2)-2(0.3334x)+1.8645

    I then substituted u for the derivative term:

    u = 3(0.0290X^2)-2(0.3334x)+1.8645

    Here's where I get bogged down. How do I get du in terms of dx? I tried differentiating u but that gives negative numbers so something is wrong.

    I'm wondering if I should try integrating in parts? Or maybe I should first try to factor that polynomial?

    Sorry about the lack of math characters. I tried to use latex but it would not work in my browser.
     
    Last edited: Apr 23, 2009
  2. jcsd
  3. Apr 23, 2009 #2

    tiny-tim

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    Hi Gordon! :smile:

    (have a square-root: √ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)

    It's not your browser … reality is at fault! :biggrin:

    (There are server problems, and the LaTeX facility is off. :redface:)

    hmm … you seem to be trying to integrate

    ∫ √(1 + (ax2 + bx + c)2) dx

    I don't think there's any exact way of doing that. :frown:
     
  4. Apr 23, 2009 #3
    Thanks Tiny-Tim.

    I went back and put in those characters.

    Why do you say there is not an exact way of integrating this expression? Can you explain?

    Regards,

    Gordon.
     
  5. Apr 23, 2009 #4

    tiny-tim

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    Hi Gordon! :smile:

    (you could have missed out the ^s also :wink:)
    Because it's the square-root of a quartic expression, and I just don't think there is (most square-roots don't have an "exact" integral).
     
  6. Apr 23, 2009 #5
    Thanks again Tim.

    I put in those missing superscripts. It's looking better now.

    I hate to give up so quickly. What about integrating by parts? That way you could have a quadratic function in each part?
     
  7. Apr 23, 2009 #6

    tiny-tim

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    If you're going to give up, then giving up quickly is best!

    Integrating by parts would give you the square-root of a quadratic function in each part, which gives you a trig thing times another square-root :frown:
     
  8. Apr 23, 2009 #7
    What about expressing the curve in terms of parameters?

    Then we can use the formula for parametric equations:

    L = ∫ba √(dx/dt)^2+(dy/dt)^2 dt
     
  9. Apr 23, 2009 #8

    tiny-tim

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    Try it if you like,

    but my guess is you'll end up with the same problem, and have to use approximation methods and a computer.
     
  10. Apr 23, 2009 #9
    Thanks, Tim.

    I'm glad I asked. I actually have a solution to the problem and that is to use curves with second degree polynomials.

    I can just cut the airfoil into smaller pieces and use curves defined by three points instead of four.

    The result is a function that looks like this:

    y = -0.2x^2+1.69x+1.48

    This should integrate nicely in the arc formula.

    In fact these curves are much nicer to work with (parabolic, so they pass through the x-axis twice and therefore have two roots each).
     
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