Relative Airflow? Help With an Explanation!

[twcpilot]

I have read that relative airflow is always parallel and opposite to the aircraft's path. I also know that the angle of attack is the is degrees between the wing chord and the realative airflow. But here what i dont understand! If you increses your angle of attack (which means hitting the relative airflow with the wing chord at a greater angle) would the relative airflow not change? If the relative airflow is always parallel to the path and you incresed the angle of attack with a constant speed your path would change. This then changes the realtive airflow and then you would loose the angle of attack (angle between wing chord and relative airflow)! I dont quite understand relative airflow and angle of attcka and how they realte. And explantion would be greatly appreciated!

[Bushav8er]

A simple way to look at it...take a plane that is moving fast (low angle of attack) at 4000 feet - what does its direction and relative airflow look like? Now slow the plane down, increase the angle of attack and maintain the 4000 feet - what is its direction and relative airflow now? See the difference?

[joco]

If you trim up, the nose goes up, the angle of attack increases, indicated airspeed decreases but the altitude stays the same, and the path of the a/c stays the same.
Remember that you change altitude with power. Having constant power, you maintain altitude therefore your a/c path vs realtive airflow stays the same.

Does it make sense?

[iflyforpie]

This then changes the realtive airflow and then you would loose the angle of attack (angle between wing chord and relative airflow)!

What you are describing is Aerodynamic Damping and it happens just as you say. The wings always want to balance the load they are carrying and they do this by trying to stay at the same angle of attack for a given airspeed.

The plane is flying along straight and level. Relative airflow is from straight ahead and the angle of attack is positive at around 3 or 4 degrees. We want to climb (let's just make it a cruise climb to keep things simple)... We increase the angle of attack which increases lift which overcomes weight.

The aircraft's flight path--and therefore, the relative airflow--is now positive. If we set a specific climb attitude (we do usually, don't we), the increased angle of the relative airflow will reduce the angle of attack to close to what it was before. Now the lift is again roughly equal to weight

Our aircraft climbs because it has excess power from the engine beyond what is required to maintain airspeed, not because lift is greater than weight.

We only need to unbalance the forces (lift and weight) to change our flight path; the rest of the time they are nearly equal as a result of Aerodynamic Damping.

[loopa]

So how do you explain it?

Relative Airflow - Student, remember when you drive in a car and stick your hands out? What do you feel? "student says... Oh i feel the wind blowing against my hand."

Good Work ! That's relative airflow. Now try changing the angle of your hand, exposing more of your hand to this "relative airflow." What do you feel? "student says, my hand gets pushed up." GOOD !

So what is relative airflow? Sure in the book it says it's "blah blah... reference it" but in the most simplified manner, airflow is the movement of the air in relation to your hand. In an airplane, you hand represents the wing.

Relative airflow is the movement of the air with respect of the wing. So if the wing is moving FORWARD, the relationship is that the AIR IS MOVING BACKWARDS.

Then you take that DEFINITION, and talk about Lift and Drag.

Ok... so let's look at lift

it's all yours

Now to ANGLE OF ATTACK. Remember, it's the angle between the chord line and the relative airflow. But how do we get RELATIVE AIRFLOW? It's as a result of the FLIGHT PATH of the aircraft. Remember, that there are are primarily two sources of lift in an aerofoil. One is ANGLE OF ATTACK and one is DYNAMIC PRESSURE (your TRUE AIRSPEED, or the speed of the air flowing over the wings). It's the combination of these two that gives you LIFT.

For Angle of Attack, you could be in LEVEL FLIGHT, but have a 12 degree angle of attack. What's your flight path though? it's still going forward. THUS, the RELATIVE AIRFLOW is going OPPOSITE TO YOUR FLIGHT PATH. If your flight path is STRAIGHT (as in not going up or down), then the relative airflow is also STRAIGHT, but going opposite. So if you have more of the wing exposed to this straight airflow, your angle of attack will increase because the angle between the CHORD LINE and the RELATIVE AIRFLOW has increased.

Makes more sense now doesn't it ?

Remember, flying is a 3 dimensional science. You can be nose up, nose down, wings banked, and so forth, but STILL SUSTAIN STRAIGHT AND LEVEL FLIGHT. So it's VERY IMPORTANT to understand that the ATTITUDE of the airplane has ABSOLUTELY NOTHING to do with the airplanes angle of attack.

To summarize, relative airflow is the air that moves in relation to the wing. If your flight path is resulting in LEVEL FLIGHT, the relative airflow is travelling exactly opposite to your flight path. Now if your wing is meeting this "straight" flight path (relative airflow) at a higher angle, then you have a HIGHER ANGLE OF ATTACK.

UNDER WHAT CONDITIONS WOULD YOU ENCOUNTER THIS?

Well I said it's combination of the ANGLE OF ATTACK and DYNAMIC PRESSURE (TAS) that give you LIFT. If you were to decrease DYNAMIC PRESSURE, then you have to increase the ANGLE OF ATTACK to maintain the same amount of lift.

This is not a formula but just a way to think of it.

30 ANGLE OF ATTACK + 70 DYNAMIC PRESSURE = 100 units of lift

IF YOU DECREASE TRUE AIRSPEED (DYNAMIC PRESSURE)

such as 30 ANGLE OF ATTACK + 60 DYNAMIC PRESSURE = 90 units of lift ................ then the airplane won't maintain level flight

As a RESULT you have to increase angle of attack by 10 UNITS to COMPENSATE for the loss in TAS

FINAL

40 ANGLE OF ATTACK + 60 DYNAMIC PRESSURE = 100 UNITS OF LIFT

The lift formula in reality is 1/2pv^2 times CL X S

1/2 is a constant, P = AIR DENSITY, V is your TAS, CL is coefficient of lift (ANGLE OF ATTACK) and S is your SURFACE AREA.


It's important to understand, you can hit the CRITICAL ANGLE OF ATTACK at ANY AIRSPEED. Because as we just examined, it's the angle between the CHORD LINE and the FLIGHT PATH (relative airflow) that gives you ANGLE OF ATTACK. And you can reach that anywhere. So never MISTAKENLY think that a nose down ATTITUDE can't result in a stall.



Hope that helps !

[twcpilot]

Wow! thanks everyone for the prompt replies! I understand it now, it has alo to do with difrerring speeds.

Bushav8er: Now I understand, it has to travel quicker becasue there's less airflow over the wing generating lift

Joco: thanks for the reminder that when you trim an aircraft it will not affect attitude only airspeed, to a point i guess right?

Iflyforpie: aerodynamic dampening, a new term i learned, thank you thats were i headed on the wrong track

Loopa: Thank you very much, great expalnation, i now understand simply how it works! Maybe you should be the first to write flight theory for Dummies!!!

[loopa]

Wow! thanks everyone for the prompt replies! I understand it now, it has alo to do with difrerring speeds.

Bushav8er: Now I understand, it has to travel quicker becasue there's less airflow over the wing generating lift

Joco: thanks for the reminder that when you trim an aircraft it will not affect attitude only airspeed, to a point i guess right?

Iflyforpie: aerodynamic dampening, a new term i learned, thank you thats were i headed on the wrong track

Loopa: Thank you very much, great expalnation, i now understand simply how it works! Maybe you should be the first to write flight theory for Dummies!!!

Haha thank you !

Just to dig deeper into this angle of attack concept (since you seem to have such a good understanding of it now).

The stall speed, is just a REFERENCE. What it is REFERing to is the point at which you will hit the critical angle of attack. Because the CRITICAL ANGLE OF ATTACK is the point where the wing is no longer producing SUFFICIENT LIFT to sustain the airplane in level flight. But the reason I say it's a reference is because, in a very steep turn, you will stall the plane at a faster speed. The reason for that is that when you do very steep turns, the nose drops, and as a result you INCREASE THE ANGLE OF ATTACK to maintain level flight. Because of this, you are now closer to the critical angle of attack and will infact stall the plane at a speed HIGHER than the normal stall speed.

When you go onto doing POWER ON STALLS or also known as Departure stalls, some airplanes will actually stall at a LOWER SPEED because the of the propeller. The propeller is simply a pair of wings attached to a rotor. They are mounted in such a way where they produced THRUST. In reality, this "thrust" can also be viewed as "horizontal lift" as it really is the lift produced by the propeller that pushes the plane through the sky. But in a power on stall, when you pitch the airplane up, more and more of this "horizontal thrust" is being shifted over to a VERTICAL component. Thus it's producing a VERTICAL COMPONENT OF LIFT. Thereby, allowing the plane to maintain level flight at a LOWER AIRSPEED. Makes sense?

Later when you get into more complex planes, you can actually change the "angle of attack" of the propeller using a PROP LEVER - it's blue and is positioned to the right hand side of the throttle. Again, the idea of this is to set up for the ideal propeller efficiency for the appropriate phase of flight be it take off, climb, cruise, descent, landing and so forth These type of propellers are called "CONSTANT SPEED PROPELLERS."

Another IMPORTANT concept to dig into is that the greater the angle of attack, the greater drag you are now producing.

We have two different kind of categories when it comes to drag. INDUCED drag and PARASITE drag.

Induced drag refers to the exposure of the wing's surface area to the airflow. So it would make sense that when you have a higher angle of attack, your INDUCED drag would increase.

Parasite drag would be the kind of drag that increases with the TAS of the airplane. The faster the plane flies through the air, the more parasite drag it is to encounter.

Induced drag can be considered as the FUNCTION of FUEL FLOW. This is CRUICAL when it comes to long cross countries.

You can SAVE FUEL by reducing the amount of induced drag. How can you do this? One significant way is in the way you position the CENTRE OF GRAVITY of the airplane.

Remember, the MORE AFT the C of G is, the SHORTER THE LEVER between the HORIZONTAL STABILIZER and the C of G. As a result, the HORIZONTAL STAB no longer needs to provide as much TAIL DOWN FORCE; THUS, you can reduce the angle of attack (which reduces induced drag), and as a result, reduce POWER and maintain similar cruise settings at a LOWER FUEL FLOW.


Take this material of info and digest it and learn it really well. Angle of attack is one of those things are gets asked by the interview committee at the airlines, so you can see how it's an important concept !

That's enough for today haha

[Strega]

My Gawd...

What does fuel flow have to do with angle of attack?

I think alot of you need to go back to basics, and read FTGU.

Angle of attack of the wing is the same (+- some constant) to the relative airflow over the plane, regardless of power setting, fuel flow, airspeed, heading yada yada yada.

Tell me again why its not mandatory for people to have some REAL education in this biz?

[mcrit]

Tell me again why its not mandatory for people to have some REAL education in this biz?

....cuz then you'd need to get a real job and you wouldn't have all sorts of spare time to post pithy (or pissy) stuff on avcanada.

[loopa]

My Gawd...

What does fuel flow have to do with angle of attack?

I think alot of you need to go back to basics, and read FTGU.

Angle of attack of the wing is the same (+- some constant) to the relative airflow over the plane, regardless of power setting, fuel flow, airspeed, heading yada yada yada.

Tell me again why its not mandatory for people to have some REAL education in this biz?

I'm doing pretty well in a job thank you, mind correcting me ? Maybe my post was so long you lost track. My bad.

I was taught that the distance from C of G to the horizontal stab is a function of you being able to reduce angle of attack. So the shorter the arm between the two, the less tail down force, therefore you can reduce angle of attack. Thus you have less induced drag, so to maintain the same airspeed, you can reduce POWER ... reduced fuel flow.

Where am I going wrong with this?

mcrit, you posting something that's irrelevant to the thread is equally as pissy... go and get your self a job just being a devils advocate

[Strega]

I was taught that the distance from C of G to the horizontal stab is a function of you being able to reduce angle of attack.

I would advise you go and find the person that taught you this, and kick them in the balls.....

The tail of an aeroplane is there to counteract the pitching moment of the wing, thats it, no more, nadda......

Think about that for a bit..

[loopa]

I was taught that the distance from C of G to the horizontal stab is a function of you being able to reduce angle of attack.

I would advise you go and find the person that taught you this, and kick them in the balls.....

The tail of an aeroplane is there to counteract the pitching moment of the wing, thats it, no more, nadda......

Think about that for a bit..

Yea i know it's there to counteract the pitching moment. But why would the plane pitch down to begin with? it would pitch down because the "weight" (c of g) of the airplane being in front of the wing's lift force, Centre of Pressure. Therefore weight is an unbalanced force causing the plane to pitch down. Between weight (c of g) and the horizontal stab, is a wing.. producing lift.

So we have weight causing a down force in front of the centre of pressure... and we have aerodynamic weight (lift produced by the stab) producing a force aft of the centre of pressure causing a teeter totter motion. And in a teeter totter, the perpendicular distance to it's arm is referred to as torque. Thus, the aerodynamic weight produced by the horizontal stabilizer is torque.

Are we on the same page so far?

Now the significance of the tail down force is directly proportional to the C of G or in other words, the WEIGHT of the plane. Cause we have established that these are the two forces that are producing this teeter totter motion.


The position of the C of G is crucial to the amount of tail down force needed by the horizontal stabilizer. Because if the plane weighs 2000 lbs on the ground, it's going to weigh maybe 2200 lbs in the air cause of that extra weight added by the horizontal stab. If the C of G is even further away, a stronger AERODYNAMIC WEIGHT is required to prevent the plane from pitching down. So a plane that normally weighs 2200 in the air, now weighs 2400 lbs in the air. So as a result it needs to produce 200 lbs more lift. If the wing shape, and dynamic pressure is remained the same, the only factor that can increase the LIFT is Coefficient of Lift. Which is going to be our angle of attack.
This brings us back to 1/2pv^2 times CL times S.

So you can see how the relationship of the C of G with respect to the horizontal stabilizer affects the angle of attack at which an airplane needs to be operated. Thereby, if we have a higher angle of attack, we have more induced drag, and thus we need more thrust to over come this. This thrust comes from our engine, and since we need more work from our engine we need to burn more fuel.

On the other hand, if the arm between c of G and the tail is smaller, you don't have to exert as much aerodynamic weight... and as a result, the 2000 lb plane on the ground would maybe weigh 2100 lbs in the air and as a result not need as much lift to be produced as the 2400 lb plane; thus, you can reduce the angle of attack, expose the plane to a lesser degree of induced drag and either A) attain a better TAS or B) reduce power and save fuel.

So as a result, the position of the centre of gravity is in direct relationship to the tail down force. And the combination of those are a function of the Coefficient of Lift (Angle of Attack) required to maintain level flight.

I've seen this be retaught through my PPL, CPL, and I even was asked this question in an interview for a pilot position and what's exactly what I told them... they didn't object me there either.




Am I wrong Strega?

[Strega]

Yea i know it's there to counteract the pitching moment. But why would the plane pitch down to begin with?

Because it is trying to create lift.

Yes a plane will go faster as you move the c of g rearward to some point, most modern jetliners do this by moving fuel to the tailplane so as it will fly at zero degrees aoa, but you seem to be missing the whole point of things and how they work,,

What happens in a flying wing?

and as for the "interviewer" saying your answer was ok, well thats not saying much, as most people involved in hiring in aviation these days dont know shit themselves..

[loopa]

I beg to differ. I know my stuff pretty well dude. I'm enjoying a really good life style out of the flying I'm doing and I wouldn't be here if it wasn't for a big part of my career which involved in instruction. The fact that you go and throw remarks like "you don't know how things work" is uncalled for strega.

It's not about who knows his shit or not, please tell me this is not why you're on a forum? to gain personal satisfaction.

If you read through my post thoroughly when it's not midnight it will probably make sense.

I'm not questioning your credibility on knowing your stuff theory, but my ground school teacher's back in the day for both my PPL and CPL were taught by experienced pilot's. Not some class 4. And what I posted above comes right from what I was taught. And I've seen that proof be used a bunch of times.

As for the "interviewer" it consisted of a chief pilot and 1 management hr lady too.


Have a good night.

[Strega]

the 2000 lb plane on the ground would maybe weigh 2100 lbs in the air

nuff said..

[loopa]

it would.

A plane at 2000 lbs on the ground is not flying through the air. Thus there is no relative airflow over the horizontal stab. The whole purpose of the stab is to produce a force that opposes plane from pitching down. It's an inverted wing, so it produces lift downwards.

If it's producing lift towards mother earth, it's like saying the plane is producing weight downwards. The lift vector adds to the total weight of the airplane. Thus the dynamic weight of the plane (when the plane is in motion) is greater than when the plane is on the ground.

Not personally attacking you here, but just getting back at you for telling me that I don't know my stuff; it really seems like you lack understanding in this area.

I started questioning that maybe you were right, but I was just talking to a few airline captains who told me the same thing I just wrote to you man. I even read it in Advanced Pilot Flight Manual. Look at the theory of flight part and you'll see more pictures and details on what I'm talking about.


Gnight.

[Strega]

Loopa,

you really don't understand this.

An aircrafts weight has nothing to do if it is moving, flying, being turned. it is its WEIGHT!!!! Weight=(MASS)(Gravity)

you are trying to oversimplify things, and when the statement is made that "an aircrafts weight is higher in the air than on the ground" I can clearly see this.

For the Record, I have a degree in Aeronautical engineering. Trust me I understand how an aircraft flies.

Again I'll say the only reason you have a the horizontal stab on the plane is to counteract the piitching moment from the wing, when you move the cg rearward as you describe, you increase the moment arm the from the c of g to the aerodynamic center of pressure, but in doing this, you add a big negative the the overall stability equation, yes there is some penalty from the induced drag the tailplane is making, and from the wing to counteract the forces from the tailplane, but this is critical to the longitudinal stability of the aeroplane.

Have you ever been on a 737NG at 410? ever noticed the AOA of the plane? Im not checked out in one, but I would be somewhat surprised if the tailplane is flying with zero AOA.



S

[mcrit]

Loopa, I got a real education and that's what has helped me avoid the need for a real job all these years! (I don't consider getting paid to do something you like work....so I haven't had a real job in about 20 years)

BTW....you are very much right about a rearward C of G being better for fuel.

[mcrit]

Loopa,

you really don't understand this.

An aircrafts weight has nothing to do if it is moving, flying, being turned. it is its WEIGHT!!!! Weight=(MASS)(Gravity

......really? That runs counter to what us physicist use. Mass is constant regardless of motion (assuming non-relativistic velocity). Weight is the force being applied to a body due to it's motion or the local gravitational field. Mass = constant, weight = context dependent.

[Stan Darsh]

These threads always degrade into demonstrations of intellectual superiority. The OP's question was answered sufficiently 10 times over.....SHUT UP! Actually no, keep going, it's funny to read.

[mcrit]

These threads always degrade into demonstrations of intellectual superiority. The OP's question was answered sufficiently 10 times over.....SHUT UP! Actually no, keep going, it's funny to read.

Conversations only really get interesting when they head off on a tangent.

[hairdo]

the 2000 lb plane on the ground would maybe weigh 2100 lbs in the air

nuff said..

Strega, if you were in a properly co-ordinated 60degree bank turn, how much would a 2000lb airplane (on the ground), now weigh? (Remember, weight, NOT mass.)

[Strega]

Loopa,

you really don't understand this.

An aircrafts weight has nothing to do if it is moving, flying, being turned. it is its WEIGHT!!!! Weight=(MASS)(Gravity

......really? That runs counter to what us physicist use. Mass is constant regardless of motion (assuming non-relativistic velocity). Weight is the force being applied to a body due to it's motion or the local gravitational field. Mass = constant, weight = context dependent.

Mcrit, I agree,

But when we are talking about an aircraft that is flying close to the surface of the earth, G is constant.

[Strega]

the 2000 lb plane on the ground would maybe weigh 2100 lbs in the air

nuff said..

Strega, if you were in a properly co-ordinated 60degree bank turn, how much would a 2000lb airplane (on the ground), now weigh? (Remember, weight, NOT mass.)

It would still weigh 2000 lbs. or the vertical component (normal to the surface of the earth) of the lift generated is still 2000#

[iflyforpie]

Mass: the amount of matter in an object. Always the same.

Force: Mass times acceleration.

Weight: A specific force where mass is multiplied by acceleration due to gravity.

The only way weight changes with constant mass is if the acceleration due to gravity changes. Considering the space shuttle feels 90% of the acceleration due to gravity; in aircraft it isn't going to change that much.


So if you have forward C of G, a conventional plane will not weigh any more. The horizontal stab will produce more down-force (and more drag), so the wings will have to lift more; creating more drag.

Similarity, the aircraft has the same weight during a turn.

Also, be cautious about the requirement for a horizontal stabilizer to produce down-force. Canard aircraft, flying wings, and delta aircraft can all operate with no down-force whatsoever.

I've also heard that if you load a 172 up to max aft C of G, the tail will actually produce positive lift. This is as a result of the thrust-drag couple and the wing pitch-up moment overcoming the CL/CG couple.