Can you tell me how to calculate the mean aerodynamic chord of the wing on a canard aircraft?

...over the past few years I have gotten the bug to design canard configured aircraft. I have read all your comments on your site and having said that I am still unclear on a couple of things. They are: Can you tell me how to calculate the mean aerodynamic chord of the wing?

From : Don Stackhouse

Assuming the wing is reasonably straight (if the sweep along the wing is not constant, it gets more complicated), the MAC is the chord at which half the wing area is inboard of that point, and half is outboard.

For a straight taper (i.e.: trapezoidal) wing, take the root chord and draw a line the same length forward from the leading edge of the tip. Draw a line the same length as the tip chord aft from the trailing edge of the root. Connect the far ends of the two lines to each other with a diagonal line. Now, draw another line from the middle (i.e.: 50% chord location) of the tip chord to the middle of the root chord. The point where this line crosses the diagonal line is the location of the MAC.

If the tip isn't squared-off, such as elliptical or rounded, etc., you can "fudge" a squared-off tip that has about the same area (i.e.: the part of the rounded tip that's outside of the squared-off tip is about equal to the extra area enclosed inside the corners of the squared-off tip shape) and still be reasonably close.

If the wing has multiple tapers, is elliptical, or is otherwise so far from being a constant taper that you simply can't get away with "fudging" an equivalent constant-taper wing, then try breaking it up into a series of trapezoids that are a reasonable approximation of the shape. Figure the MAC's of each trapezoid, then find their combined MAC by using a weighted average.

For example, let's assume a double-tapered wing, 10" per panel in span each, with a 7" root chord, 6" at the panel break and 4" at the tip. Using our graphical method above, we find that the MAC of the first panel is 4.872" outboard from the root, and the MAC of the outer panel is 4.667" outboard of the panel break.

The area of the inner panel is 65 in^2, and the area of the outer is 50 in^2. This means that the total area is 115 in^2.

Now we're ready to find the MAC of the total wing. Let's kill 2 birds with one stone and find the aerodynamic center ("AC") of the whole wing in the same operation. Draw a line from the 25% chord point of the MAC of the inboard panel (its AC) to the 25% chord point of the outer panel's MAC. The resulting (slightly swept) line happens to be 9.855" long. The AC of the entire wing will lie somewhere along this line, at a distance proportional to the ratio of the areas of the inboard and outboard panels.

The area of the outboard panel is 50 in^2, the total area is 115 in^2, so the AC will be 50/115, or 43.5% of the way along the line, starting from the INBOARD panel's AC. If you have trouble keeping track of which end of the line to start from, just remember that the AC of the total will be closest to the panel that has the most area (in this case, the inboard). 43.5% of 9.855" is 4.285". Using the AC of the inboard panel as the center, strike an arc across our AC line at 4.285". This intersection point is the aerodynamic center of the whole wing panel, and the spanwise location of the wing's MAC, which we can now measure as 9.131" from the root.

We can use a similar weighted average method to find the chord length of the wing's MAC. Note, it will NOT be the same as the chord length of the wing at that location, that's only the case for a SINGLE tapered panel.

The chord length of the inboard panel's MAC is 6.513" (this sort of stuff is where a CAD system is REALLY helpful!), and the outboard panel's MAC is 5.067" long. Multiply each chord length by the area of its panel, add them together, then divide by the total area:

6.513 * 65 = 423.3 5.067 * 50 = 253.4 423.3 + 253.4 = 676.7 676.7 / 115 = 5.884

The length of the wing's MAC is 5.884". The AC of that MAC is at the location we determined in the previous step, so the leading edge of the wing's MAC is 1.471" (that's 25% of 5.884") ahead of the wing's AC.

Yes, it's tedious, but it works, and it works for any planform. If you have more than two segments, find the area, MAC length and AC location of the first two, then find the area, MAC length and AC location of those two plus the third segment, and so on. This will even work for planforms with irregular tapers, non-constant sweeps, and also for multiple wings such as biplanes and triplanes.

Can you tell me how to find the aerodynamic center of the aircraft?

I'll bet you've already guessed that. Just find the AC and MAC of the wing and of the tail or canard, and use the weighted averages method above to find the AC of the combination of the two.

Based on my limited information, I believe that the C/G will be ahead of the leading edge. If this is true, will a foward sweep wing cure my problem of leadout placement?

Certainly. Just put in enough forward sweep to put the tip where you want your control line leadout wires to be, while keeping the MAC in the same place that it was to begin with. However, if this requires more than about 5 degrees or so of forward sweep, be sure to read the articles in AJ&D on the aerodynamic and structural effects of forward sweep. While you're probably not going to have trouble with structural divergence at the mild sweep angles you'll probably need, even small amounts of forward sweep definitely could have an effect on yaw stability. You may need to add some more fin area.

Don Stackhouse
DJ Aerotech