Location of Mean Chord on complex wing shapes

[Steve Helmick]

Found this on the shop wall, during renovations, and thought it might 'splain the process for finding the MAC relative to the LE at the root. Once you have the MAC, then you can accurately locate the CG at the theoretical safe starting point (i.e., "Ted's Law" of 18% for a design with 18% tail area, etc.). I hope somebody will find this useful, even if you don't build double tapered wings like the example. If you paid attention in Algebra class, it should be pretty easy to work it out. If you didn't, so sorry...

This was given to me by Dan Tracy. Dan is a retired Boeing engineer, past Jr. National Champion, Sr. National Champion, and previously a member of the US F1B (Wakefield) team. His latest build is an RV-7 that he flies from his hanger at Auburn Municipal. Sometimes he stops by and watches us fly stunt at the 2nd best stunt site in the NW. He can fly CL, so there's hope that he'll get the bug.  Steve

[Don Hutchinson AMA5402]

For a complex shape such as the T-Bird, cut out a smaller cardboard copy of the wing. suspend it on the point of a pin until it balances, that is the midpoint of the MAC. The rest is easy from here.
Don

[Dan Bregar]

 Steve

  Ask Dan Tracy what ever happened to his deHavilland Comet.  

[Randy Powell]

Or just ask Gary Letsinger. He did a whole series of articles in Stunt News on this awhile back.

[Howard Rush]

For something so basic, there's a heap of misinformation about.  No, it's not the average chord.  http://en.wikipedia.org/wiki/Chord_%28aircraft%29 has the definition.  Better for CL models is to put something inside the integral to account for the inside wing tip going slower than the outside wing tip.  Multiply that c by (r/R)^2, where r is the distance from the control handle to y, and R is the distance from the control handle to the center of the airplane.  That has been kinda useful for making funny shaped combat planes come out so they don't need weight in either tip.  Mean aerodynamic chord is kinda rule-of-thumby, so one wouldn't want to split hairs.

[Larry Cunningham]

Suppose you knew the mean aerodynamic chord. Precisely.

What would you do with it? I'm not asking what Bill Netzband would do with it, I'm asking the average CL Stunt jock.

To be useful, it would need to be something more complex than the mean chord of the perimeter shape. You might want to consider the centroid of the volume of the wing. But that wouldn't be quite useful enough, without considering other features (like a big "flap" on its trailing edge) and perhaps even the difference in length of the flight path, as mentioned by Howard. If you're serious, you need to break it into billions of tiny little triangular pieces and use finite element analysis.

But what are you actually doing with the MAC value? Are you trying to find a center of aerodynamic lift? Or just a CG for balancing purposes? Or are you trying to figure out where the CG needs to be for that crazy canard stunter you're building with forward swept wings and tip chord and thickness greater than for the root?

I'm really getting old, I can tell. I think the MAC data is clearly a hog looking at a wrist watch for about 99% of CL Stunt fans. We'll have to justify it simply as satisfying our interest, which is enough. ;->

L.

"The meta-Turing test counts a thing as intelligent if it seeks to devise and apply
Turing tests to objects of its own creation." -Lew Mammel, Jr.

[Steve Helmick]

IMO, the geometric mean chord as calculated using this "simple" method is plenty good enough to find a starting point for the C of G, from which we will all adjust our models to suit our preferences, right or wrong. Largely because most of us "balance" our models on our fingertips and say "that's good enough to try it."  Worrying about the different airspeed of the inboard wing is simply not worth a fart in a hurricane. I heard that Boeing has something like 900 lbs. of tailweight in one model of the 747. Obviously not Howard's way, because they got the airplane finished and flying, and then fixed it. Falls into the "trimming" catagory, I reckon.   Steve

[Serge_Krauss]

This is just an addition.

I'm not going to get into arguments about whether anyone should care or how accurate or useful any of this stuff is. If you don't like it, just buzz on off.

The equations for straight-tapered wings are easy to use with a pocket calculator, but a bit long. It's easier just to go to a site like the Palos R/C site, put in your wing measurements, and read off the answer. If you juggle the c.g. so that it comes out at 24%-25% of the MAC, then you have the root position of the wing's neutral point. There are a couple sites that approximate the n.p. for the entire plane, probably not too accurately.

However, as I've posted elsewhere, elliptical wings are easier to compute, and if the wing actually has any elliptical chord distribution, the computations are easier than for straight tapered wings and faster on a pocket calculator than making a cardboard model, although that is a quite valid method, which I have used for complex shapes. MAC and d are simple multiplications of the root chord or span by one number each. If you have the MAC, root chord, and position of the alignment (necessary to have built the model), then the root neutral point position is easy. Three of several ways are shown below.

I derived equations for finding the MAC , N.P., and their locations, and drew sample wings, as shown in the image below, posting and publishing them elsewhere previously. There is erroneous stuff on the internet, but these are well checked and valid. In them are the following variables:

b = half span.

Cr = root chord

MAC = length of the mean aerodynamic chord. (please no arguments about teminology here; even the NASA higher-ups disagree on it).

d = distance of MAC from root chord (this is often wrong on the internet; actual local chord and mean chord don't coincide)

p = tip sweep distance (usually zero; the distance the tip is behind the sweep line's intersection with the root; p = 0 for all but the far right wing).

F = the fractional positions where all local chords are aligned along a straight line. For instance, if all chords are aligned at their quarter-chord points, then F = 1/4. That "Spitfire"-like wing is illustrated in the third picture in the top row. A negative value makes an apparent, curved forward sweep. The far right picture is a wing with straight sweep, with the 30% points of the chords aligned and the tip 30% of the root chord behind the sweep line's root position. If F>1.0, then the tip trails the root trailing edge. (Edited to correct inequality limit to read 'If F>1.0...')

x' = the distance that the neutral point or aerodynamic center is behind the root leading edge.

The diagram was used to derive all equations, many versions of which do not appear here. Anyone interested in the derivations of anything else can e-mail me off forum. I'll check to see whether there's a simple diagram to explane the final results.

[john e. holliday]

I am so glad and thankful I have you guys around to do all the math and designing of the model planes.  Really flunked algebra but the teach let me pass so he would not have me in his class again.   

[Serge_Krauss]

Aw, Doc. He probably liked you!

[Jim Thomerson]

I've been under the undocumented impression that MAC was the chord where half the area was inboard and half was outboard.  If this is the case, then the cut out a piece of cardboard and find the balance point would would be accurate. Is this so wrong that it wont vanish into my measuring errors?   

[Serge_Krauss]

Jim-

The MAC takes into account the distances the tiny elements of area that make up the whole wing are from the root, so that lift or weight moments about the MAC are equal. It's not sufficient in tapered, rounded, or other wings more complex than rectangles to just have half the area on each side. Those elements of area (lift or weight) that are furthest away exert more lifting moment (or in the case of the cardboard model, downward - weight - leverage) about a given point. For instance in the extreme case of straight taper, the delta (pointed) wing, the MAC is 2/3 of the root chord and is located 1/3 of the way out along the half span. Computing areas on each side of it, you find that the inner area is 5bc/18 and the outer area is 2bc/9 or only 4bc/18, where "b" is the half span and "c" is the root chord. So the inboard area is 25% greater.

For a more extreme example, think of a wing with only two significant parts: a 10.0" x 1.00" part at the root and centered .5" out.and a 1.0" x 1.0" centered 30 " further out at the tip, separated by an ultra thin section of .001 in² (negligable) in area. The MAC would be centered very close to (1/11)(.5 x 10 + 30 x 1) = 35/11" = 3.2" outboard, where the actual local chord is negligable, and 91% of the area is inboard of the MAC. The "theory", really an assumption known to be only approximate, is that each element of area provides the same lift and that the MAC is located at the point where outboard and inboard lifts balance. The MAC for this example would be 7.3" in length (not at all equal to the local chord).

Edited for grammar, red type, and another order of ten to make an area truly negligable in the second example.

SK