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Aerodynamics Question

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Aerodynamics Question

Old 13th Mar 2013, 06:52
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Aerodynamics Question

Hey all,

First of all the diagrams of the things I'm trying to explain are as follows





I have a question, I'm in the Blackhawk course for the Army. In our aerodynamics manual it states that the AoA (angle of attack) is the same for IGE and OGE conditions "...Induced flow velocity is increased causing a decrease in AOA. A higher blade pitch angle is required to maintain the same AOA as in IGE hover."

I have a problem fully accepting this notion because the as the AoA increases this will shift the Total aerodynamic force (TAF) to the rear and if the lift vector were to remain the same, the lift in the positive Y direction would be reduced. The only thing I can think of that would counteract this would be an increased lift from the increased overall velocity of the resultant relative wind caused by the increased induced flow.

I know this may be hard to read but any insight is really appreciated. Or if you know a reference that confirms or denies this that is really what I'm looking for. I looked through several aerodynamics manuals of other services and the FAA and none specifically mention this specific. Basically I'm looking for a source that doesn't reference the 1-203 Fundamentals of flight manual

Thanks for your time!!
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Old 13th Mar 2013, 09:34
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Shifty - try a search on PPRuNe for urban myths debunked - a thread by Nick Lappos. One of the issues covered was that of ground effect and the myth of the 'bubble' of high pressure air under the disc reducing the IF when close to the ground.

The actual explanation isn't really any clearer since it involves the statement that the proximity of the ground affects the angle of the IF above the disc in such a way as to affect the power required.

AFAIK there is no explanation that doesn't involve pages of maths - perhaps that is why the 'bubble' explanation was so popular - it explained what happened in an easily understood way. Now it is just a question of 'it happens - but we're really not sure why'.

Your manuals seem to obfuscate the issue with such statements as AoA is the same in an IGE and OGE hover which may or may not be strictly true.

However, with reference to your diagrams, if the pitch angle stays the same but the AoA is increased (by reducing IF somehow) then the relative airflow is now coming from a greater angle to the blade which will tilt the lift vector forwards and increase it. Then the drag will increase appropriately and the total reaction will show more lift in the vertical axis and less drag (so less Tq required) in the horizontal axis.

We know we require less power in the IGE hover - that can be demonstrated easily - the mystery is to explain exactly why if there is no pressure increase under the disk.

Last edited by [email protected]; 13th Mar 2013 at 09:37.
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Old 13th Mar 2013, 11:14
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A simple explanation.

Picture an airfoil section with the leading edge to the left, trailing edge to the right, and inclined at a given angle from the horizontal.

Got it?

OK, now picture relative wind coming horizontally from the left to meet the leading edge. Of course, the angle of inclination that you have pictured in your mind represents the angle of attack.

Still with me?

OK, now imagine the air flowing over the airfoil and departing the trailing edge. It is deflected downward, right? So the airflow is now no longer horizontal, but at an angle downward from the horizontal, right?

Draw all this out.

Got the picture?

OK, now here we go: If you make a vector diagram of the relative wind meeting the leading edge and the airflow departing the trailing edge...that is, join the tails of the two vectors you've pictured, and then picture the resultant...you will see that the resultant is angled down from the horizontal to a degree.

That angling down of the vector resultant represents a reduction in the angle of attack and therefore a reduction in lift.

This is exactly the situation in OGE hover.

Think this over.

Now let's look at what happens in IGE hover.

Start again with the same picture in your mind. Now, as the air moved away from the trailing edge, it's downward deflection becomes less because of the proximity of the ground...that is, the solid ground is a boundary that prevents the pressure field from bending downward so much.

Again, make a vector diagram: Join the tail of the relative wind with the tail of the less-downwardly-deflected air leaving the trailing edge in this IGE situation.

You should be able to see that less downward deflection IGE means an increased angle of attack, even though the blade angle from the horizontal has not changed.

...and therefore, more lift is produced IGE, meaning that less power is required to hover IGE, a fact which is well known.

You need to draw this out on paper until you see it. It would be easier if I could do it for you, unfortunately this explanation is the best I can do.
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Old 13th Mar 2013, 11:22
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Any textbook that refers to 'wind' in this context needs approaching with some scepticism!

Last edited by 212man; 13th Mar 2013 at 12:41. Reason: joined text to book!
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Old 13th Mar 2013, 11:30
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I have just read this, I don't know if it will help in a general context?

As the rotor blades rotate they generate what is called rotational relative wind. This airflow is characterised as flowing parallel and opposite the rotor’s plane of rotation and striking perpendicular to the rotor blades leading edge. This rotational relative wind is used to generate lift. As rotor blades produce lift, air is accelerated over the foil and projected downward. Anytime a helicopter is producing lift, it moves large masses of air downward through the rotor system. This downwash, referred to as induced flow can significantly change the efficiency of the rotor system. Rotational relative wind combines with induced flow to form the resultant relative wind. As induced flow increases when transitioning to forward flight, resultant relative wind becomes less horizontal. Since angle of attack is determined by measuring the difference between the chord line and the resultant relative wind, as the resultant relative wind becomes less horizontal, angle of attack decreases.

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Old 13th Mar 2013, 12:12
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Your weight is the same whether you are IGE or OGE. So, the total reaction vertical component, balancing weight, is the same.

What changes is the angle that the induced flow comes in - OGE it is pretty much vertical, but IGE it is restricted by the proximity of the ground, and its angle changes.

The end result is that in OGE, the total reaction has tilted back a bit, making the Induced Drag bigger and requiring more Power (collective pitch, leading to more N1 and fuel) to hold the same weight up there.

The only difference between IGE and OGE is the Induced Drag and the power required to overcome it. As Nick has said, there is no pressure bubble, it is simply a convenient way to describe the effects.
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Old 13th Mar 2013, 15:47
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Shifty, the AOA is not the same IGE and OGE. What is the same is the Vertical component of Total Reaction( ie Thrust = Weight) and this is achieved at a lower blade pitch angle and a slightly reduced AOA.
What you need is "Principles of Flight" by Wagtendonk p60-62( ISBN 1-56027-217-1) which explains all.
Hope this helps.
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Old 13th Mar 2013, 16:21
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If the induced flow (or inflow) changes, the angle of attack cannot stay the same within the same blade pitch angle.

Phil

Last edited by paco; 13th Mar 2013 at 16:21.
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Old 13th Mar 2013, 18:58
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L = Cl * A * .5 * r * V^2
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Old 13th Mar 2013, 22:51
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POFMAN,
The angle of attack will be the same, to produce the same amount of force to hold the weight up in the air. The "half rho V squared S" is constant, the CL is the same.

What HAS changed is the pitch angle, because you need less power to overcome the induced drag.
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Old 13th Mar 2013, 23:26
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Ascend Charlie

I agree with your first statement
The angle of attack will be the same, to produce the same amount of force to hold the weight up in the air. The "half rho V squared S" is constant, the CL is the same.
But struggle with the second because you are contradicting yourself

What HAS changed is the pitch angle, because you need less power to overcome the induced drag.
Yes the pitch angle has change but the amount of lift generated must still be the same as you point out in the first statement as the aircraft still weighs the same. As I remember "induced drag" is short for "lift induced drag" so if the lift generated is the same then surely the drag is still the same. Maybe the answer is in the increased "form drag" of the blade OGE?

Its all just made to make the gray matter hurt lol

Last edited by snakepit; 13th Mar 2013 at 23:28. Reason: Drink = poor spelling
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Old 13th Mar 2013, 23:59
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Look at your vector diagrams - angle of attack is between the Pitch angle and the relative air flow. RAF is a combination of rotational airflow and induced airflow. Change the induced AF, you change the AoA (for a constant pitch angle)

If the RAF comes in from a different position, but the AoA is the same, then the pitch angle has changed, because of a different induced airflow.

Pitch is connected to Power via the governors, and it changes because the power required to overcome induced drag has changed.
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Old 14th Mar 2013, 05:01
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I think we all need to remember that we are trying to simplify a concept which is actually quite a lot more complex than the basics we teach and learn at flight school. For example the vector diagram (in its various formats depending which book you read) is only considering one particular point along the span of a MR blade.
That said, IMHO, (not being an ADY expert but tending towards the simple (KISS) explanation if possible), there seems to be a fair bit of confusion in some of the responses so far.
Who cares what happens to the A of A in an IGE vs OGE scenario?
In my view, the opinion offered by pofman is closest to the money. If the weight of the helicopter is the same in both scenarios, then the vertical component of the Total Reaction must remain the same. To achieve this situation OGE the Pitch Angle must be increased (due to the increased "Induced Flow") thereby increasing Rotor Drag. The Rotor Drag obviously has to increase (OGE) due to the change in position of the TR. The vertical component of the TR will be the same as it was in the IGE situation but the Rotor Drag component(ie the component of the TR acting in the plane of rotation) will be increased causing an increase in the power required to hover OGE vs that required IGE.
It's easy to see it that way rather than introducing the Lift Formula and its different elements, including A of A.
What does your engine do for you Grasshopper?
Why it overcomes Rotor Drag, Sir.
Good answer son!
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Old 14th Mar 2013, 10:13
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Aerodynamics Question

Old rotor head
Good explanation. Makes more sense to me that way round
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Old 14th Mar 2013, 18:28
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my method

here goes:

You load up a bunch of guys (or stuff) in the back of a blackhawk (or any other helicopter). You pull enough pitch to get the thing up to about a 5 foot hover. You look at your performance planning and if you are at least 10 percent under the calculated max for the day you are good to go. Now, push ever so slighly forward on the cyclic. The thing will gently move forward and go through this thing called ETL (for short). If you do nothing else it might settle a little but soon it will begin a sweet climb, as sweet as mom's apple pie. I can't explain why, but it works every time. You might get nervous and add a little power to overcome that settling feeling and thats okay. Now then, when you feel like you are getting to an altitude that satisfies you and your boss, gently push down on the collective until the dial that is marked VSI is pointing at "0". (you don't even need to know what VSI means, just make the pointer point at "0")For complicated aerodynamic reasons that you don't need to know about, the aircraft will essentially stay at that altitude with slight variations which you can overcome with small corrections. Trust me, it will, you don't even have to know why, it just does. Now then when ready to land, begin by lowering the collective and gently pulling back on the cyclic. Aerodynamically speaking, this will cause the machine to slow down and eventually begin a gentle descent toward the earth. As you approach your intended landing area, continue to adjust the collective and cyclic as necessary (oh yeah and the pedals, they have some aerodynamic forces as well) to bring the machine to about a five foot hover. Gently set it down on the ground. With good practice and experience, if you use this method or slight variations, you will bring all the vortex ring states, slugs of drag, variations in density altitude, the driving, driven and stall regions and all other variables of the lift/drag ratios into perfect alignment and become a damn good helicopter pilot. Using this tried and true method, used by millions of rotor pilots over the years, will free up brain cells to think about more important things. I don't charge for my expertise.

Cheers

Last edited by grumpytroll; 14th Mar 2013 at 18:31.
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Old 14th Mar 2013, 18:43
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Ascend Charlie.
The simple version is that the Angle of Attack is the same but strictly :-
IGE the IF is reduced so the RAF lies closer to the Plane of Rotation. Therefore both the Lift(90degrees to the RAF) and Total Reaction vectors are moved closer to the axis of rotation so the component felt along the axis of rotation is larger. For a given AOA the TR is the same magnitude but at a lower BPA the inclination is towards the AOR so there is a greater vertical component. OGE and IGE the Vertical component needs to be the same so a very small reduction in AOA is needed to achieve this.
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Old 14th Mar 2013, 19:45
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Raise the lever.
Science happens.
Up you go.
Easy.
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Old 14th Mar 2013, 21:25
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Good place to start Helicopter Urban Myths
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