MAC Length

Mean Aerodynamic Chord

The length of the MAC is an important figure that will be used later in stability calculations, so we might as well find it now that we’ve defined the geometry of the wing.

Assuming the wing has no twist and consists of a single airfoil (from root to tip) the MAC will be the same as the mean geometric chord (MGC). There is a simple method for determining the MAC graphically for straight-tapered wings which you may have seen in aircraft design textbooks, but it doesn’t apply an elliptical planform. Instead, the length of the MAC, , is found with an integral:

$$\scriptsize \bar{c} = \frac{2}{S_{ref}}\int_{0}^{\frac{b}{2}} c^2 \,dy \ $$

Wait, nobody said there was going to be calculus involved here! Don’t worry, it’s not so complicated. I’ll show you three ways we can do it. In plain language, the expression above is just a set of instructions that says:

• Divide up the wing (from buttline 0 to b/2, that is from root to tip) into a bunch of chordwise strips with a spanwise width of dy. Ideally, dimension dy would be infinitesimally small to get a really accurate result, but you can choose whatever dimension you like.
• Find the area of each strip (chord times strip width) and multiply each area by the chord.
• Add up the sum of all of these area-chord products.
• Divide the sum by the whole area.

Method 1: Numerical Integration – Graphic

You can do this easily with pencil and paper (an engineer’s scale and a calculator would be handy too). The strips can be fairly wide where there isn’t much change in chord from one strip to the next. Out near the wingtip where the chord changes rapidly, narrower strips are used for a more accurate result.

It’s also possible to use trapezoids instead of rectangular strips, but we’ll keep things simple with rectangles–the result will be surprisingly close to the exact value we calculate later in Method #3.

Method 2: Numerical Integration – Spreadsheet

Now we’ll perform the same operation, but instead of measuring the chord lengths from the drawing, we’ll let the spreadsheet calculate it for us. This is very handy because we don’t have to make dozens of measurements every time we adjust the span or aspect ratio. Automation is a good thing!

In the last post, we found an equation for the wing chord as a function of span. You plug in a buttline dimension for y and the function spits out the chord at that station along the wing. Or you can enter some fraction of the semispan in place of . We’ll enter this equation into each cell in our chord column. Instead of c_s and 2y / b, we’ll enter the cell address for root chord and percent semispan.

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

The result is within about two-tenths of one percent of the answer we found earlier. If for some reason we needed even more accuracy, we could simply divide up the wing into more strips.

Method 3: Definite Integration

If we want to know the exact answer, all we have to do is evaluate the definite integral, and there’s no need to bother with measuring drawings and making spreadsheets.

$$\scriptsize \bar{c} = \frac{2}{S_{ref}}\int_{0}^{\frac{b}{2}} c^2 \,dy \ $$

The other values we’re working with are:

$$\scriptsize \text{reference area:}\:\:S_{ref}=131.29\text{ ft}^2=18,906\text{ in}^2$$

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

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

The c in the expression above is equal to:

$$\scriptsize c_s\sqrt {1-\left( \frac{y}{b/2}\right)^2}$$

So plugging this in, we have:

$$\scriptsize \bar{c} = \frac{2}{S_{ref}}\int_{0}^{\frac{b}{2}} \left( c_s \sqrt {1-\left( \frac{y}{b/2}\right)^2} \right) ^2 \,dy \ $$

$$\scriptsize \bar{c} = \frac{2c_s^2}{S_{ref}} \int_{0}^{\frac{b}{2}} 1-\frac{4}{b^2}y^2 \,dy \ $$

$$\scriptsize \bar{c} =\frac{2c_s^2}{S_{ref}} \left( y-\frac{4}{3b^2}y^3 \right) ,\:y=\frac{b}{2}$$

$$\scriptsize \bar{c} =\frac{2(60.18\text{ in})^2}{18,906\text{ in}^2} \left[\frac{400\text{ in}}{2}-\frac{4}{3(400\text{ in})^2}\left(\frac{400\text{ in}}{2}\right)^3 \right] $$

So, now with the exact answer in hand, we can see that the other two methods are more than acceptably accurate as well–each within a sixteenth of an inch. Next we’ll look at how to locate the aerodynamic center of the wing.