Aspect Ratio Understanding

[flusterfly]

O.K. here goes... Why does a high aspect ratio wing have less induced drag? I have a pretty limited aerodynamic understanding. Doesn't reducing the cord decrease the reynolds numbers resulting in more drag? Can some one help me get my head around this one?

Adam

[Sparky Paul]

Science stuff.... :)
http://www.allstar.fiu.edu/aero/flight45.htm

[Ollie]

A wing has drag and two parts, airfoil drag and induced drag. Dw= Do + Di

The induced drag coefficient is about equal to lift coefficient squared divided by pi times aspect ratio (depending on elliptcal lift distribution).
CDi ~ Cl^2 / ( pi * AR )

See:
http://aero.stanford.edu/WingCalc.html
Play around with the program. The wing platform doesn't say anything about airfoil drag.

[vintage1]

That's the how of it, the why of it is allegedly that wingtip vortices are the biggest contributor to induced drag, and the bigger the tip chord, the bigger the drag..there is a spanwise flow from the root to the tip that 'spills over' the tip and creates trailing spiralling vortices..these have energy that is wasted.

[Tom Harper]

flusterfly,

You are correct - reducing the Reynolds number does increase the drag of the wing and that's what happens when you increase AR. It's a matter of how much.

Consider a wing 'A' with an AR of 3:1 (30" span, 10" chord) and a wing 'B' with an AR of 6:1 (42" span, 7" chord). They will both have the same lift. Wing 'A' will have a little more profile drag than wing 'B' and wing 'B' will have considerably less induced drag than wing 'A'. It's a matter of degree. You benefit more from higher aspect ratio than you lose in Reynolds number.

The equation for induced drag is:

CDi ~ (Cl^2 / ( pi * AR ))* e

The 'e' term is the wing efficiency factor. It takes planform and wing twist into account. Reynolds number is accounted for in Cl.

So, while your observation is astute, there are other factors with greater impact than the decrease in Reynolds number.

[flusterfly]

How can I find the balance between AR and airfoil drag?

Adam

[Tom Harper]

Adam,

Look at the plot of characteristics of an airfoil. There will be a L/D ratio line. Where that line peaks (usually around Cl of .4 or .5) it will be optimum. That is the design point of the section.

[pmackenzie]

That's the how of it, the why of it is allegedly that wingtip vortices are the biggest contributor to induced drag, and the bigger the tip chord, the bigger the drag..there is a spanwise flow from the root to the tip that 'spills over' the tip and creates trailing spiralling vortices..these have energy that is wasted.

The way the theory works is that the trailing vortex sheet interacts with the incoming airflow to cause the local angle of attack to lean back a bit.
This causes the lift vector to lean back so a portion of the lift becomes the lift induced drag.

By having high aspect ratios the vortex sheet is shed in smaller amounts, so the tilting of the inflow is less.

If you work out all of the theory and math you get the classical result of an elliptical lift distribution as having minimum induced drag.
In effect the tilt is the same for the entire span so the entire wing is working together.

Pat MacKenzie

[Ollie]

Adam's question:
"How can I find the balance between AR and airfoil drag?"

It depends on the type of aircraft. See:
http://www.charlesriverrc.org/articles/design/jeffreid_aspectratiovsdrag.htm

I think your question is about DS type aircraft. Correct me if I'm wrong.

DS design depends on maximum L/D for high speed. See:
http://www.charlesriverrc.org/articles/flying/markdrela_ds.htm

My point of view: "The basic equation derived by Wurts and Drela states that the ultimate DS speed is some constant times the lift to drag ratio of the plane. The constant expresses the ratio of energy gained to energy lost in a circuit of the course. The greater the wind shear the greater the energy gained. The bigger the circle the more energy is lost. The trade is between the ability of the pilot to control the plane in a small circle versus the ability of the plane to maintain its shape and stay in one piece under high G forces.

So, the better the L/D the faster in given conditions and the smaller the constant can be for a given speed. I am supposing that the larger the circle for a given speed the easier it is to pilot the plane accurately. If the circle gets too small, pilot skill may become a limiting factor. The less the G forces the less strain on the structure and the bigger the circle the less the strain on the structure."

Read the whole thing:
http://www.rcgroups.com/forums/showthread.php?t=101242

[Ollie]

There's more.
An aerobatic plane uses low aspect ratio wing because it doesn't care about CDi and thrust is much larger than drag.

A long range plane or a DS plane design cares about high aspect ratio because of minimized CDi and maximum L/D. See:
http://en.wikipedia.org/wiki/Lift-induced_drag

A duration plane cares about (L^1.5)/D so it has a very, very high aspect ratio and low speed.

A pylon racer has a comprised aspect ratio between tuning and straight away.

A wing with high aspect ratio has a structure problem compared to low aspect ratio wings because of strength and stiffness.

See:
http://www.flyforfun.net/BP-wings.html

[ThinnedDope]

Guys,

I’m late to the party on this one, but this is one of those areas I have spent a lot of time working on. While the various equations and vectors allow you to calculate the induced drag, this doesn’t really answer the question of why a higher aspect ratio wing has less induced drag.

Here’s my attempt at a short answer. By the way, I’m an engineer, so my definition of a short answer is probably long by most standards. Here goes…

The bottom line is wings generate lift be accelerating air down. Regardless of the various mechanics involved in accelerating the air downward, this explains ALL the lift generated. Yes there are pressure differences, but this is just another aspect of the problem. In the end, lift, just like thrust, is created by accelerating a mass in the opposite direction from the force being generated. If you doubt that a wing moves air down to generate lift, just stand under a helicopter rotor as it lifts off. The rotor is just a rotating wing. Nuff said.

Now, there are two components to generating this force, how much air (mass) you accelerate and what speed you ultimately accelerate it to. It’s a momentum thing. You can generate the same amount of lift by moving a little bit of air very fast or a lot of air very slow. Either way can generate the same lift. However, both approaches require very different amounts of energy. Note: This speed is the speed imparted to the air downward. This is NOT the airspeed of the plane.

Prior to the wing passing through it, the air was not moving. But after the wings flies through the air is actually moving down. It has gained a speed. Therefore this mass of air has to have gained kinetic energy. This energy came from the wing (and therefore from the plane). And in fact, it is this energy, imparted to the air, that represents the induced drag.

Unlike the lift force, the energy is not linearly proportional to the speed of the air, but rather it is proportional to the speed squared (speed X speed). So for the same lift, a wing that moves a lot of air mass, at a low speed will impart less energy into the air than a wing moving a little air very fast. This is at the root of why a high aspect ratio wing has lower induced drag. High AR wings move more air, but at a lower speed and therefore less energy goes into generating the lift.

A wing acts like a big air scoop which moves air downwards. It turns out that the effective area of this scoop is the same as a circle with a diameter equal to the wing span. So a wider wing span acts like a bigger scoop and therefore, all things being equal (same air speed), it will act on more mass. As a result, it will generate the same lift (as a lower AR wing) , but will impart less energy into the air (less induced drag) to generate lift.

The amount of air moved is actually very much higher than common sense would seem to indicate. It’s actually very easy to estimate. Calculate the “scoop” area as a circle with a diameter equal to the wing span. Multiply this by the airspeed and you get a volume per unit time. Something like meters-cubed per second. Then multiply this by the density of the air (about 1.2 kg/m^3 at sea level) and you will have a pretty good idea how much air the plane is moving.

Here’s a few examples. A hand launch glider wing has a wing span of 1.5 meters and cruises at roughly 5 m/s. The 1.5 m span equates to a scoop area of about 1.77 m^2. Multiply these by the air density of 1.2 kg per cubic meter and you have 10.6 kg/second. Yes, 10.6 kg or over 23 pounds of air per second. Sounds crazy, but it’s the very reason it flies. It is much more efficient to move this rather large amount of air at a very low speed than to move a smaller amount of air at the speeds need to generate the lift.

For a real eye opener, a Boeing 747 has a wing span of about 60 m and it’s cruise speed is about 800 km/h = 220 m/s. At its cruising altitude of 10,000 m the air density is about 0.4 kg/m^3. Therefore the mass flow rate is a whopping 250,000 kg/second = 275 tons per second or almost 1 million tons per hour!

I hope this makes some sense.

Clin