Controlwing

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Jeremy

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I've recently been re-reading the various articles by the George Spratt's on the controlwing. I've also read a couple of papers about it on the NASA tech server.

As far as I can see, it seems a simple principle, quite well suited to a microlight (ultralight) or low speed light aircraft. The advantages in terms of control surface simplicity alone make it an apparently good option for such a very light aircraft as a weight-limited microlight.

I understand the speed range issue (essentially it doesn't have a speed range without a AoA "bias loading" system or variable wing pivot point geometry), but apart from this disadvantage, it seems have a lot of positive points. I see that a French company seems to be developing a microlight version (see http://spratt.103.free.fr/spratt103_english/welcome.htm ), but as far as I know, no one else is working on a man-carrying controlwing (Freewing seem to be exclusively UAV now).

Is it just the odd configuration that makes this principle unpopular, or is there some unspoken major flaw with the whole idea?

Jeremy
 

Dana

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It makes sense if you're trying to market a "safety" plane like the Ercoupe, but many (if not most) sport pilots enjoy the required skills of controlling their aircraft, so anything that makes it too easy (or less maneuverable) is kinda boring.

-Dana

The difference between genius and stupidity is that genius has limits.
 

Jeremy

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Feb 23, 2003
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Salisbury, England
I guess you may well be right when it comes to a light aircraft, but perhaps this is less of a drawback with a microlight/ultralight, where I'm sure the "easy to fly" aspect could be a real plus to some people.

My concern is that, like a gyroplane, the controlwing may have some undiscovered handling attribute that makes it seem deceptively safe. I don't know about the rest of the world, but here in the UK gyroplanes were originally "marketed" as being very safe and "stall proof". Whilst true in many respects, this encouraged a lot of people to build them many years ago in the belief that they only needed minimal training to fly, with the inevitable consequence that the accident rate rocketed, earning them a poor reputation very quickly.

If I get some spare time over the coming winter I may have a go at making a controlwing model to see for myself whether or not the design lives up to it's reputation.

Jeremy
 

rtfm

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Hi Jeremy,
I know this is an old thread, but I'm doing due diligence in reading everything I can find on the freewing design.

I'm currently building a freewing-based single seater in my workshop. Early days yet - but the fuse is taking shape quite nicely. However, I have one very basic and very simple question: without augmenting the AoA of the wing by adding manual override, what AoA will the wing make with the wind? In other words, left to its own devices, what sort of AoA can I expect? If I knew that, I could work out the dimensions of the wing.

Failing that, if I could work out the pivot point at which the wing will be in trim at my target cruising speed, I could adjust my geometry accordingly - but I can't seem to find this information either.

I'll be using the NACA747a315 airfoil, because it has even lower pitching moment than the "traditional" NACA23012, while it has greater lift. I use Profili software to calculate lift, drag, pitch moments etc - but I need to know the "natural" AoA of the wing.

Any help you can provide would be appreciated.

Regards,
Duncan
 
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Jeremy

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Salisbury, England
Hi Duncan,

I think this is pretty easy to estimate. The wing AoA will be dependent on the lift it has to generate, so will vary with airspeed. The lift requirement is fairly straightforward to work out; in straight and level flight for the control wing (no tail down force assumed) it will equal the aircraft weight.

Using the wing area and the anticipated trim airspeed will allow you to work out the wing lift coefficient (not the aerofoil section coefficient). Looking back at the change in lift coefficient with AoA plot for the wing will allow you to see what the AoA will be.

For the purpose of rough calculation you can shortcut this process and assume that the wing lift coefficient changes at about 0.1 per degree of AoA change. This rule-of-thumb is good enough to get some ball park figures.

Required lift coefficient, Cl, can be worked out from:

Cl = weight / (0.5 x 1.225 x velocity^2)

Where 1.225 = density of air at sea level in kg/cu m

So, if you had a 200kg aircraft, with a 10 sq m wing, flying at 25 m/S, then it would need a lift coefficient of about 0.52 for straight and level flight. Using the 0.1 per degree approximation gives a AoA of about 5 degrees.

For a 60 degree banked turn, the lift required doubles (as this is a 2g turn). This pushes the lift coefficient required up to 1.04, assuming that the airspeed stays the same, which takes the AoA up to a rather steep 10 degrees or so for this example. Increasing the speed a bit brings it back down to a safer level, as lift, and hence AoA relate to the square of airspeed..

Hope this helps. It's a bit rough and ready, but should give you some idea of where to start.

Jeremy
 

rtfm

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Jeremy,
You're a star, mate. I've had a number of other answers as well - but this is really quite simple. I'm off to play with some numbers and see what I come up with...

Cheers,
Duncan
 

mfc

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Apr 23, 2009
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Coventry, UK
Minor correction/clarification to the formula above -- I believe it should read:

Required lift coefficient, Cl, can be worked out from:

Cl = weight / (0.5 x 1.225 x velocity^2 x A)

Where
weight = mass * g = mass (in kg) * 9.8 (in m/s^2)
1.225 = density of air at sea level (in kg/m^3)
velocity = airspeed (in m/s)
A = wing area (in m^2)

So, if you had a 200kg aircraft, with a 10 sq m wing, flying at 25 m/S, then it would need a lift coefficient of about 0.51 for straight and level flight. Using the 0.1 per degree approximation gives a AoA of about 5 degrees.

Mike
 
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lr27

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Nov 3, 2007
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If I'm following this correctly, the above tells you what cl to try for. However, the wing will trim itself so that the pitching moment times dynamic pressure times MAC (in other words, the physical pitching moment) is equal to the weight times the distance the c.g. is ahead of 25% MAC. (approximately).

Let's say the average chord is 4 feet and the pitching moment is .04 (positive, i.e. nose up) If the c.g. is at 21 percent MAC, or 10 inches behind the L.E. , then the trimmed CL will be 1. If the c.g. was at 17 percent MAC, then the trimmed CL would be 0.5. The pivot point is the c.g. It trims just like any flying wing. However, I would imagine it would adjust angle of attack to gusts much faster.

I would also think, in the absence of a trim mechanism, that the aircraft would speed up in a banked turn until it was enough faster that the trimmed Cl provided enough lift.
 
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