Solve Arc Length Problem: Find Circumference of Wing Section

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Homework Help Overview

The discussion revolves around computing the circumference of a wing section by using cubic splines to define the airfoil's arc pieces. The original poster presents a specific function for an arc and attempts to apply the arc length formula to find its length, raising questions about the integration process.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the derivative of the function and the challenges in expressing it for integration. There are considerations of integrating by parts and expressing the curve in terms of parameters. Some participants question the feasibility of finding an exact integral for the square-root of a quartic expression.

Discussion Status

Participants are exploring various methods to approach the problem, including the potential use of approximation methods. There is a suggestion to simplify the problem by using second degree polynomials for the airfoil sections, which may lead to a more manageable integration process.

Contextual Notes

There are constraints regarding the use of mathematical notation due to technical issues with the forum's LaTeX support. Participants are also navigating the complexity of integrating functions derived from cubic splines and the implications of using different polynomial degrees.

Gordon Arnaut
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Homework Statement



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

Homework Equations



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


L = ∫ab √(1+[dy/dx])^2dx


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:
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Gordon Arnaut said:
Sorry about the lack of math characters. I tried to use latex but it would not work in my browser.

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:)

/b
L = | sqrt(1+[dy/dx]^2)dx
/a

I started by getting the derivative of the function:

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

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:
 
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.
 
Hi Gordon! :smile:

(you could have missed out the ^s also :wink:)
Gordon Arnaut said:
Why do you say there is not an exact way of integrating this expression? Can you explain?

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).
 
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?
 
Gordon Arnaut said:
I hate to give up so quickly. What about integrating by parts? That way you could have a quadratic function in each part?

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:
 
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
 
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.
 
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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