# Real-world airfoil problem

1. Apr 6, 2013

### Quivil

Hello all,

I'm in the process of building an ultralight glider. (to be piloted by me)

As part of the construction process, I need to bend some aluminum tubing into an airfoil shape. This tubing will form the top of the wing. The bottom of the wing is flat.

The shape of this particular airfoil is such that the forward section is a 1/4 ellipse, and the rear section is a large-radius circular section.

When bending aluminum, the material exhibits a property called bounce-back. If you bend it around a circular wooden form, it will bounce back to a larger diameter circular shape.

I'd like to focus on the ellipse portion of the airfoil. I wish to cut a wooden form such that I can bend my aluminum tubing around the form and it will bounce back to the elliptical shape I desire.

The shape I create will need to have the same length as the final ellipse section. It will also need greater curvature at every point so that my tubing will bounce back to the shape I desire.

In bending an aluminum bar, (at a single point) I found that the bounce-back was about 5 degrees. I see no reason not to use that as a starting point.

How is curvature defined for an object like an ellipse? I need to know the curvature at each point so that I can increase it by 5 degrees at each point.

Can anyone suggest how I might go about creating the shape of the wooden form I need?

Thanks!

2. Apr 6, 2013

### SteamKing

Staff Emeritus
If your wing skin is built out of aluminum tubing, you are going to make an ultraheavy glider. If you look at some examples of UL gliders on the web, the wings seem to be made from some sort of fabric stretched over a frame.

I would also research a proper airfoil shape.

3. Apr 6, 2013

### jackmell

Yeah, start simple and first get curvature down then move to the airfoil. Get the Calculus book out. I'm doing that now and bam!

How about we just do curvature for the ellipse section in the first quadrant of the equation:
$$R(t)=3\cos(t)i+2\sin(t)j$$
That's an elliplse right? Plot it parametrically and see. Then
$$\frac{dR}{dt}=-3\sin(t)i+2\cos(t)j$$
so that the tangent vector is:
$$T(t)=\frac{\frac{dR}{dt}}{\left|\frac{dR}{dT}\right|}$$
And then the curvature vector is:
$$K(t)=\frac{\frac{dT}{dt}}{\left|\frac{dR}{dT}\right|}$$
so that
$$\text{curvature}=C(t)=\left|K(t)\right|$$

Ok, bingo-bango for a first start:

$$\text{my wooden form=C(t)+5}$$
from t=0 to $\pi/2$. I'm gettin' the saw out now . . .

4. Apr 6, 2013

### Quivil

Thanks Jackmell...trying to digest this.

I need something I can plot on a piece of plywood and then cut out!

If I understand your last equation correctly, you've given the magnitude of the curvature at any point "t". (then added my additional 5 degrees)

It seems to me that I could plot 0 to pi/2 on the "x" axis of my plywood (scaled to the dimension I need) but then I don't see how the magnitude of the curvature helps me.

I follow the steps of what you did fairly well (I think):

So the first derivative you took was to find the slope at any point "t", then to get the tangent vector you divided by the magnitude of that vector, correct? (as in Pythagorean) That's what the vertical lines mean, right?

Then you took the derivative again to show how fast the tangent is changing at any point "t" (leading to the magnitude of the curvature at any point)

I'm lost, however, as to how this helps me plot points that I can then connect and cut!

SteamKing, these pieces I'm making ARE the frame upon which the fabric will be stretched. This is an airfoil of convenience. It's not flying at any speed that will make the shape of the airfoil all that critical. My design is based on the "Goat" of Mike Sandlin, if you're interested.

5. Apr 6, 2013

### jackmell

Ok, I left out some things. Let me rephrase the question independent of flying: given a function, we seek to find a second function such that the second function has a curvature that is the curvature of the first function plus some constant. Let's start with my ellipse: $R(t)=3 \cos(t)i+2\sin(t)j$. Now, the expression for the curvature vector is complicated and I just did it all in Mathematica. I eventually come out with the curvature function $c(t)$ which I plotted below in red. Now, suppose we want to find another function such that it has a curvature function that is $c(t)+0.1$ indicated by the dashed black line. Now, what function, any function cus' we're not proud, would produce a curvature function which would track the black dashed line? Let's just try to guess that it's a smaller ellipse $R_2(t)=h \cos(t)i+k \sin(t)j$. Now, how do we find h and k? Well, for starters now, just guess at them. In Mathematica, we can set up a Manipulate function that can vary h and k interractively and I can do that until I get my new function to line up close to the dashed lines. I did that which is shown in green:

$$R_2(t)=1.732 \cos(t) i+1.014 \sin(t)j$$

It's not perfect but when you're starting, it's encouragement. Also, the curvature is in radian measures I think so if you want it to be 5 degrees more, then I suppose we need to have $c(t)+\frac{5\pi}{180}$

You know how to transfer $R_2(t)$ to a piece of wood? Not an insulting question right? Get the ellipse you want in terms of $R_2(t)$, plot the points on a large transparent grid, connect the points, move it to the wood, cut it out. I'm sure you'll make sure everything is $10^{-6}$ accurate before you fly it. :)

#### Attached Files:

• ###### curvature plot.jpg
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Last edited: Apr 6, 2013
6. Apr 6, 2013

### Quivil

That's interesting, that a smaller ellipse would be the solution as far as curvature goes, but it misses an important constraint--the curve length needs to be the same--I have an airfoil of a certain length, and can't change that.

Any easy solution to that? It may turn out that it's not even a function--it may double-back on itself.

I've plotted a curve for the airfoil I'm trying to achieve in Maya, and using Mel, Maya's embedded programming language, I've been able to calculate the radius of curvature at any point along the length of the airfoil curve.

The problem is, I haven't figured out how I can use that information.

7. Apr 8, 2013

### Quivil

I managed to code this in Maya in a sort of iterative fashion. Depending upon the value of bounceback used, it will curve more or less, even into a spiral, and the length of the curve stays constant. Thanks for your help on this, Jackmell, very kind of you.