Aerodynamic Center

Finding the Aerodynamic Center

So far we’ve defined the geometry of the planform and found the mean aerodynamic chord. But we need to know where the wing’s aerodynamic center is so that we can determine the tail length for stability calculations. The task here is to determine the spanwise position of the MAC, and to determine the x-coordinate of its quarter-chord point.

Finding these two dimensions, and , is similar to the way we found in the last post, and you could use any of the three methods already presented.

The two equations we’ll use are:

$$\scriptsize \text{(1) spanwise position of MAC:}\:\:\bar{y} = \frac{2}{S_{ref}}\int_{0}^{\frac{b}{2}} c \:y \,dy \ \\ \scriptsize \text{and}$$

$$\scriptsize \text{(2) x-coordinate of a.c.:}\:\: x_{a.c.} = \frac{2}{S_{ref}}\int_{0}^{\frac{b}{2}} x_{c/4} \: c \,dy \ \text{,} \\ \scriptsize \text{where:}$$

$$\scriptsize \text{chord length:}\:\:c = c_s\sqrt {1-\left( \frac{y}{b/2}\right)^2}$$

$$\scriptsize \text{wing axis x-coordinate:}\:\:x_{c/4} = 0.6c_s\left(1-\sqrt {1-\left( \frac{y}{b/2}\right)^2}\right)$$

$$\scriptsize \text{span:}\:\:b = 400\text{ in}$$

$$\scriptsize \text{root chord:}\:\:c_s = 60.18\text{ in}$$

I chose to use a spreadsheet to perform the numerical integration. This sacrifices a tiny bit of accuracy but has the advantage that all of the calculations are contained in the spreadsheet. Here are the results:

$$\scriptsize\bar{y} = 84.72\text{ in}$$

$$\scriptsize x_{a.c.} = 5.45\text{ in}$$

An Alternate Method

It’s a good idea to check the results using another method. We’ll check using definite integration, but to save a little time, we’ll use a graphing calculator that can do the manual labor for us. While we could use this as our primary calculation method, we’d have to go back and forth between the spreadsheet and the graphing calculator every time a change is made, and there’s opportunity that something doesn’t get entered consistently in both applications.

You can click the graph below to visit the graphing calculator and see how the equations are entered.

Results:

$$\scriptsize\bar{y} = 84.88\text{ in}$$

$$\scriptsize x_{a.c.} = 5.46\text{ in}$$

Here again, we find that numerical integration got us within two-tenths of one percent of the exact answer. (In the case of , that’s within about 3/16″–definitely good enough for our purposes.)