comair25 From Germany, joined Sep 2006, 192 posts, RR: 0 Reply 1, posted (3 months 4 weeks 1 day 4 hours ago) and read 7297 times:
I haven't even watched the video, but I have actually never heard of the reasoning for lift because of the speed of air flowing over the wing. That's a new one to me.
frmrCapCadet From United States of America, joined May 2008, 1379 posts, RR: 1 Reply 2, posted (3 months 4 weeks 1 day 3 hours ago) and read 7153 times:
Any number of commonly understood physical phenomena are actually a lot more complicated than what we were taught in high school.
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bellancacf From United States of America, joined May 2011, 51 posts, RR: 0 Reply 3, posted (3 months 4 weeks 17 hours ago) and read 6433 times:
I'd feel a lot more enlightened if they hadn't shown a symmetrical airfoil approaching a stall. That lower surface was doing a fair imitation of a barn door.
Gingersnap From United Kingdom, joined Aug 2010, 687 posts, RR: 6 Reply 4, posted (3 months 4 weeks 17 hours ago) and read 6353 times:
Quoting bellancacf (Reply 3):
I'd feel a lot more enlightened if they hadn't shown a symmetrical airfoil approaching a stall. That lower surface was doing a fair imitation of a barn door.
Agreed. There's no way that video shows anything conclusive.
It does give a nice visual as it relates to the bernoulli effect, and the effects of induced drag with the attitude of the wing. But being so close to a stall position, I'm going to pass on this professors explanation.
jfritz From United States of America, joined Jun 2011, 38 posts, RR: 0 Reply 6, posted (3 months 4 weeks 16 hours ago) and read 5385 times:
Well it is about the air going faster over the top, what is not discussed is the second half of that, Because that air is moving faster, the air pressure decreases while the bottom air is at a higher pressure. The pressure differential causes lift in that the higher pressure on bottom pushing toward the low air pressure on the top.
Fly2HMO From , joined Dec 1969, posts, RR: Reply 7, posted (3 months 4 weeks 16 hours ago) and read 5383 times:
This is a piece of poor journalism on a piece of a poorly over-simplified concept. Those of us who actually took aerodynamics courses knew this already.
And I don't see how this makes the understanding of how lift is produced any more clear to your average John Doe
cschleic From United States of America, joined Feb 2002, 879 posts, RR: 0 Reply 8, posted (3 months 4 weeks 16 hours ago) and read 5300 times:
It simply seemed to be saying that the air traveling over the top of the airfoil reaches the back of the wing at a different time than the same air traveling over the bottom side. That seemed to be true in this case. But it didn't say anything about how that relates to lower pressure over the top of the wing vs the bottom, thus generating lift.
tdscanuck From Canada, joined Jan 2006, 11032 posts, RR: 72 Reply 9, posted (3 months 4 weeks 16 hours ago) and read 5271 times:
Quoting parapente (Thread starter):
You thought it was about air speeding up over the upper surface.
It is (partly) about the air speeding up over the lower surface...this video is addressing a frequent but incorrect explanation for *why* the air speeds up over the upper surface. It's still true that the air does speed up over the upper surface.
Quoting comair25 (Reply 1): I have actually never heard of the reasoning for lift because of the speed of air flowing over the wing. That's a new one to me.
Speeding up and lowering pressure are one and the same thing; the two explanations are equivalent. This video is addressing the "equal transit time" explanation for lift, which is wrong because it's based on the incorrect assumption that molecules that arrive at the leading edge at the same time but go above and below the wing must end up at the trailing edge at the same time.
Quoting bellancacf (Reply 3): I'd feel a lot more enlightened if they hadn't shown a symmetrical airfoil approaching a stall.
The same thing happens even at low AoA...the air over the top does not meet up with it's "partner" on the lower side. The effect is more pronounced at high AoA, which is presumably why they did the video at high AoA. Being symmetric has nothing to do with it (the path from the leading edge stagnation point to the trailing edge over the upper surface is longer on a symmetric airfoil with non-zero AoA)
Quoting Gingersnap (Reply 4): There's no way that video shows anything conclusive.
It conclusively shows the equal transit time theory isn't true at that flight condition, which was the narrator's point. It happens to be true for lower AoA's too.
Quoting Gingersnap (Reply 4): It does give a nice visual as it relates to the bernoulli effect
Bernoulli relates pressure to velocity. Since there is no pressure measurement in the video it actually doesn't show Bernoulli at all.
Quoting Gingersnap (Reply 4): and the effects of induced drag with the attitude of the wing
This is a 2D airfoil; it doesn't have any induced drag. I think you mean form drag.
ordjoe From United States of America, joined Aug 2010, 459 posts, RR: 0 Reply 11, posted (3 months 4 weeks 15 hours ago) and read 4253 times:
I am only a chemist, so any aerospace engineers will know beter, but always thought due to the shape of the wing air on top is faster and exerts less pressure per bernoulis equation, thus lift
WingedMigrator From United States of America, joined Oct 2005, 1944 posts, RR: 57 Reply 12, posted (3 months 4 weeks 14 hours ago) and read 4155 times:
Quoting Fly2HMO (Reply 7): Those of us who actually took aerodynamics courses knew this already.
You'd be surprised how many of those who've taken aerodynamics courses still don't believe or understand that airplanes fly quite simply by shoving air downwards!
aircellist From Canada, joined Oct 2004, 1302 posts, RR: 9 Reply 13, posted (3 months 4 weeks 14 hours ago) and read 4015 times:
Quoting WingedMigrator (Reply 12): You'd be surprised how many of those who've taken aerodynamics courses still don't believe or understand that airplanes fly quite simply by shoving air downwards!
And the plane is lifted in a case of reaction of equal force in opposite direction, is it? Funny, I thought so, and have been told no no no...
"Mort aux cons!" (anonyme) "Vaste programme..." (Charles de Gaulle)
UALWN From Andorra, joined Jun 2009, 1753 posts, RR: 1 Reply 14, posted (3 months 4 weeks 14 hours ago) and read 3808 times:
Quoting tdscanuck (Reply 9): This video is addressing the "equal transit time" explanation for lift, which is wrong because it's based on the incorrect assumption that molecules that arrive at the leading edge at the same time but go above and below the wing must end up at the trailing edge at the same time.
I always thought the "equal transit time" explanation to be correct, at least in first approximation. If we place ourselves in the air molecules' rest frame, they see the wing coming over, splitting them in two sets (one above and one below the wing), and then leaving them where they were before (neglecting drag). If we shift now to the wing rest frame, what has happened can only be described as "equal transit time." Therefore, since the path over the wing is longer, the air moving over the wing has had to move faster, and, as Bernouilli's principle tells us, has exerted less pressure, hence the lift.
tdscanuck From Canada, joined Jan 2006, 11032 posts, RR: 72 Reply 15, posted (3 months 4 weeks 14 hours ago) and read 3510 times:
Quoting ordjoe (Reply 11): always thought due to the shape of the wing air on top is faster and exerts less pressure per bernoulis equation, thus lift
Bernoulli requires incompressibility, which isn't valid for modern airliner wings, but the general statement that the shape of the wing causes the airflow to go faster over the top, which causes lower pressure over the top, which causes lift, is true.
This is identically equivalent to the idea that the wing shoves air downwards...it's exactly the same physics, the only difference is whether you look at pressure or momentum (two sides of the equals sign in the same equation).
Quoting aircellist (Reply 13): And the plane is lifted in a case of reaction of equal force in opposite direction, is it?
Yes.
Quoting aircellist (Reply 13): Funny, I thought so, and have been told no no no...
It's still true. The fight usually comes between those explaining lift via pressure and those explaining it via momentum. You have to get into some fairly mathmatically-based aerodynamics (specifically, control volumes) to see really concretely that those are really the same explanation but with the control volume drawn in two different places.
Quoting UALWN (Reply 14): I always thought the "equal transit time" explanation to be correct, at least in first approximation. If we place ourselves in the air molecules' rest frame, they see the wing coming over, splitting them in two sets (one above and one below the wing), and then leaving them where they were before (neglecting drag).
Even neglecting drag, that's not what happens. There is no physical reason that the air over the top should end up at the trailing edge at the same time as the air over the bottom (the upper air has no idea what the lower air is doing so it can't reliably "get there" at the same time)...in fact, not only does that not happen, it can't happen and still provide lift. This is easier to see when you use the momentum explanation than the pressure explanation but it's a subtle point.
Basic wind tunnel tests (like the video but any angle of attack other than 0), the equivalent CFD runs (with or without viscosity), or the painful but mathematically accurate method of integrating a particle over the velocity field very quickly show that the equal transit time theory is wrong. The air over the top does speed up but it gets to the back of the trailing edge *before* its "mate" on the underside...it speeds up "too much" for the equal transit time theory.
Mir From United States of America, joined Jan 2004, 17876 posts, RR: 59 Reply 17, posted (3 months 4 weeks 13 hours ago) and read 3296 times:
Quoting ordjoe (Reply 11): I am only a chemist, so any aerospace engineers will know beter, but always thought due to the shape of the wing air on top is faster and exerts less pressure per bernoulis equation, thus lift
This is absolutely true. But it's the question of why the wing air on the top is faster that creates the confusion.
Quoting UALWN (Reply 14): I always thought the "equal transit time" explanation to be correct, at least in first approximation. If we place ourselves in the air molecules' rest frame, they see the wing coming over, splitting them in two sets (one above and one below the wing), and then leaving them where they were before (neglecting drag).
But they're not where they were before - one is further ahead of the other.
-Mir
7 billion, one nation, imagination...it's a beautiful day
UALWN From Andorra, joined Jun 2009, 1753 posts, RR: 1 Reply 19, posted (3 months 4 weeks 13 hours ago) and read 3194 times:
Quoting tdscanuck (Reply 15): There is no physical reason that the air over the top should end up at the trailing edge at the same time as the air over the bottom
If you look at the problem from the reference frame in which the air is at rest and the wing is coming towards it, you see two air molecules sitting one just atop the other, then the wing comes in and splits them apart, but after the wing has passed, the two molecules are still where they used to be, one just atop the other (they have never moved). If we look at this now from the reference frame in which the wing is at rest, we see the two molecules travelling one over the wing and one below, they start together and they end together, hence the equal transit time. Where's the catch?
Quoting UALWN (Reply 19): but after the wing has passed, the two molecules are still where they used to be, one just atop the other (they have never moved).
They are not one atop the other after the passage of the wing. They will have moved.
-Mir
7 billion, one nation, imagination...it's a beautiful day
BMI727 From United States of America, joined Feb 2009, 11547 posts, RR: 27 Reply 21, posted (3 months 4 weeks 13 hours ago) and read 3040 times:
Quoting tdscanuck (Reply 15): Bernoulli requires incompressibility, which isn't valid for modern airliner wings, but the general statement that the shape of the wing causes the airflow to go faster over the top, which causes lower pressure over the top, which causes lift, is true.
It's worth noting that the Bernoulli equation is basically a special case of the Euler equation, which is beyond the Discovery Channel level of detail, so Bernoulli usually has to suffice.
Quoting UALWN (Reply 19): the two molecules are still where they used to be, one just atop the other (they have never moved).
Why should that have to be the case? If I understand your statement correctly, the particle on the top should have moved faster due to the bound vortex created about the airfoil.
Why do Aerospace Engineering students have to turn things in on time?
ikramerica From United States of America, joined May 2005, 20630 posts, RR: 62 Reply 22, posted (3 months 4 weeks 13 hours ago) and read 2904 times:
Read "Stop Abusing bernoulli" as it explains this and why the simplified explanation we give children and even teach at the air and space museum is a load of crap. The lift due to the pressure change alone based on equal transit is insufficient for high speed flight with heavy aircraft with small surface area wings, yet these aircraft fly every day.
Because its simply not true. Nor is there any physical reason it should be true. That's the basis of the "lie" we are taught for some reason. Fluids don't work like that. Water molecules entering a pipe at pressure will exist faster when in the center of the stream and slower and later the closer to the walls. The molecules directly touching the walls of the pipe barely move at all. Why wouldn't air around a wing act similarly?
Of all the things to worry about... the Wookie has no pants.
tdscanuck From Canada, joined Jan 2006, 11032 posts, RR: 72 Reply 23, posted (3 months 4 weeks 12 hours ago) and read 2901 times:
Quoting UALWN (Reply 19): If you look at the problem from the reference frame in which the air is at rest and the wing is coming towards it, you see two air molecules sitting one just atop the other, then the wing comes in and splits them apart, but after the wing has passed, the two molecules are still where they used to be, one just atop the other (they have never moved).
This would only be correct if the particles experienced no fore/aft force...i.e. that there was no pressure gradient along the length of the airfoil. However, this isn't true...there is a front-to-back pressure gradient along the forward upper surface of the airfoil (this is what speeds the air up) and a back-to-front pressure gradient along the aft upper surface of the airfoil (this is what slows the air back down). In general, these gradients aren't symmetric so the particle accelerations aren't the same, so the particle final positions aren't the same.
Quoting UALWN (Reply 19): If we look at this now from the reference frame in which the wing is at rest, we see the two molecules travelling one over the wing and one below, they start together and they end together, hence the equal transit time. Where's the catch?
They don't end together. The particles don't just move up and down, they move forward and backwards. Except in special cases, the particles don't end up in the same position after the airfoil has passed.
Quoting BMI727 (Reply 21): It's worth noting that the Bernoulli equation is basically a special case of the Euler equation, which is beyond the Discovery Channel level of detail, so Bernoulli usually has to suffice.
Very true. And Euler is a special case of Navier-Stokes...the simplifications to Bernoulli break down at transsonic speeds so that Bernoulli is quantitatively incorrect but, since it's really just an expression of conservation of energy, it remains qualitatively true in most real-world cases and suffices for general explanations unless you're trying to actually derive values.
Quoting BMI727 (Reply 21): If I understand your statement correctly, the particle on the top should have moved faster due to the bound vortex created about the airfoil.
Exactly. The fact that the upper air goes faster, that it gets to the trailing edge first, that the upper pressure is lower, that there's circulation about the airfoil, and that the air gets net downward motion are all reflections of exactly the same physics.
UALWN From Andorra, joined Jun 2009, 1753 posts, RR: 1 Reply 25, posted (3 months 4 weeks 12 hours ago) and read 3027 times:
Quoting tdscanuck (Reply 23): This would only be correct if the particles experienced no fore/aft force...i.e. that there was no pressure gradient along the length of the airfoil.
You are again looking at this from the wing-at-rest reference frame. Which is fine, of course. How do you account for the aft/fore force from the air-a-rest reference frame?
Mir From United States of America, joined Jan 2004, 17876 posts, RR: 59 Reply 27, posted (3 months 4 weeks 12 hours ago) and read 3061 times:
Quoting tdscanuck (Reply 23): In general, these gradients aren't symmetric so the particle accelerations aren't the same, so the particle final positions aren't the same.
Which is a good thing, since if the pressure gradients were symmetric there would be no lift at all.
Quoting UALWN (Reply 24): Why should they move at all if they are not dragged on by the passing wing?
Some of them are dragged on. That's what the boundary layer is all about.
Quoting UALWN (Reply 25): How do you account for the aft/fore force from the air-a-rest reference frame?
Pressure gradients. Air is pulled toward the low pressure in the middle of the wing (not really in the exact middle - it's further forward than that), and then as it passes through that pressure, it will slow down again somewhat due to the tendency of a fluid not to move away from low pressure.
If you take your time to go through it you will learn a lot, it is very nicely explained. For those in more hurry here is the index and you can dive directly to the subject of interest:
Starlionblue From Greenland, joined Feb 2004, 15102 posts, RR: 69 Reply 29, posted (3 months 4 weeks 11 hours ago) and read 3040 times:
An old Howstuffworks.com article which I unfortunately has been dumbed down since then described the whole thing pretty nicely without getting into maths. Basically they wrote that Bernouilli and Newtonian lift are both somewhat correct but also both "incomplete" which I think is a good way of putting it. It also says that to accurately describe lift, you need fairly complex math which is beyond the scope of laymen.
XKCD puts it well as usual:
Quoting KC10Guy (Reply 16):
I always thought money (lots and lots of it) is what made airplanes fly?
That and paperwork.
Tact Is For People Who Aren't Witty Enough To Be Sarcastic
BMI727 From United States of America, joined Feb 2009, 11547 posts, RR: 27 Reply 30, posted (3 months 4 weeks 11 hours ago) and read 3037 times:
Quoting UALWN (Reply 24): Why should they move at all if they are not dragged on by the passing wing?
Because it is being acted on by the bound vortex of the wing.
It's important to understand how vortices and circulation play into creating lift. Take a basic cylinder in a uniform flow. It will have two stagnation points: one right at the front and one right at the rear and the streamlines split perfectly above and below the cylinder. (Just assume that there's no flow separation) Now if we add a vortex flow, as if adding backspin to the cylinder, and the stagnation points move towards the bottom of the cylinder. Looking at the associated streamlines, the circulation results in them becoming closer together on the top of the cylinder and further apart on the bottom. If we assume incompressible, we can directly associate the space between streamlines with velocity since the mass flow between streamlines must be equal. The closer together the streamlines become, the faster the air is moving, so clearly the air on top is moving faster. Apply the Bernoulli equation, and there's the lift.
That, in a nutshell, is the Kutta-Joukowski Theorem which relates the lift (per unit span) of a cylinder to the circulation around it, fluid density, and uniform flow velocity.
The crucial point in making this useful is the Kutta condition, which states that an object with a sharp trailing edge (like an airfoil) will hold the rear stagnation point at that sharp trailing edge. Like the cylinder, with no circulation the stagnation points will be directly at the front and rear so the airfoil must create a circulation about itself to satisfy the condition. With the circulation comes lift.
Now here's where it gets cool. We can go back and forth between a cylinder and an airfoil using a Joukowsky Transform. It's a mathematical way to switch from one coordinate system to another and altering the characteristics of the cylinder to change things like the thickness and camber of the airfoil. Anyway, the upshot of all that is that early aerodynamicists could make their analyses easier by using the cylinder with equivalent circulation rather than the airfoil.
Anyway, in order for the Kutta condition to be satisfied, there must be a circulation about the wing to keep the stagnation point at the rear which is referred to as the bound vortex. If you picture a wing flying right to left, the bound vortex will be moving clockwise, creating upwash ahead of the wing, downwash behind it, pushing the air rearward on top of the wing, and pushing it forward below it.
It's worth noting that the lift depends directly on the strength of the bound vortex. It makes sense intuitively, since if you picture a wing at a high angle of attack and a wing at a lower angle of attack, the wing at the higher angle of attack will have to move its stagnation point more than the wing at the lower angle of attack. That means the first wing requires a stronger bound vortex and therefore creates more lift. It also means that anytime the wing changes lift, it has to shed a new starting vortex equal and opposite to the change in the bound vortex.
Why do Aerospace Engineering students have to turn things in on time?
tdscanuck From Canada, joined Jan 2006, 11032 posts, RR: 72 Reply 31, posted (3 months 4 weeks 10 hours ago) and read 3023 times:
Quoting UALWN (Reply 24): Why should they move at all if they are not dragged on by the passing wing?
Because pressure isn't the same thing as viscous drag. You have pressure gradients even if you assume an inviscid fluid.
Quoting UALWN (Reply 25):
Quoting tdscanuck (Reply 23):
This would only be correct if the particles experienced no fore/aft force...i.e. that there was no pressure gradient along the length of the airfoil.
You are again looking at this from the wing-at-rest reference frame. Which is fine, of course. How do you account for the aft/fore force from the air-a-rest reference frame?
Pressure is a scalar...the pressure field is the same whether you take the wing-at-rest frame (intuitively equivalent to a wind tunnel) or the air-at-rest frame (intuitively equivalent to an actual airplane). You have fore/aft pressure gradients in either frame. The results in terms of lift, drag, and airflow are exactly the same.
Quoting UALWN (Reply 26): Quoting ikramerica (Reply 22):
Fluids don't work like that.
Non-viscous perfect fluids do. That's why I always thought this was at least a good first approximation to the explanation.
Non-viscous perfect fluids do do that *if* you don't apply the Kutta condition...but if you don't apply the Kutta condition you get completely incorrect flows. As BMI727 captured:
Quoting BMI727 (Reply 30): The crucial point in making this useful is the Kutta condition, which states that an object with a sharp trailing edge (like an airfoil) will hold the rear stagnation point at that sharp trailing edge.
The Kutta condition covers up the effects you lost by dropping viscosity. If you drop viscosity and compressibility and don't enforce the Kutta condition you get zero drag but also zero lift since there's nothing to prevent the fluid from navigating arbitrarily sharp corners like the trailing edge. In order to restore reality to the situation you either need viscosity or the Kutta condition...either one will cause the equal transit time theory to fall apart.
Thanks a lot to all of you for your detailed explanations. I really thought that equal transit was a good way to understand lift, at least conceptually, for a perfect fluid. I knew reality is way more complicated, but still....
The funny (or sad) thing is that I"m a physicist, although the last time I was exposed to non-perfect fluid dynamics was just over 30 years ago (and I hated it back then!). I feel (moderately) stupid... Anyway, thanks a lot. I learned quite a bit today!
David L From United Kingdom, joined May 1999, 8994 posts, RR: 44 Reply 33, posted (3 months 4 weeks 6 hours ago) and read 2914 times:
Quoting UALWN (Reply 32): The funny (or sad) thing is that I"m a physicist
I suffered the same "handicap", with the equal transit theory being taught to me, even at university. To be fair to them, it was only a small part of one lecture and the object of the excercise was to understand Bernouilli's principle rather than the principle of flight. It wasn't until I took part in discussions here that I realised how incomplete my understanding was.
tdscanuck From Canada, joined Jan 2006, 11032 posts, RR: 72 Reply 34, posted (3 months 4 weeks ago) and read 2726 times:
Quoting UALWN (Reply 32): I really thought that equal transit was a good way to understand lift, at least conceptually, for a perfect fluid....The funny (or sad) thing is that I"m a physicist
In all fairness, from a physicist point of view what you were taught may have been correct. If you have a perfect inviscid incompressible fluid the equal transit time is actually true as long as your body doesn't have sharp edges (like a trailing edge). Mathematically it's true even with a sharp edge but then you get a singularity (infinite acceleration) at the edge and the solution isn't physically realizable. This is a totally valid method for a wide class of fluid problems...it just happens that aerodynamic lift isn't one of them. If you're just working on the mathematics of potential flows and smooth bodies like cylinders, spheres, Rankine ogives, or you're working with super low Reynolds number flows, it's a completely fine view of the fluid world.
The problem is that viscosity is absolutely essential to lift in the real world and if you drop it by invoking an inviscid condition to clean up your mathematics, you have to replace it with something else (typically the Kutta condition) that mimics the effect.
UALWN From Andorra, joined Jun 2009, 1753 posts, RR: 1 Reply 35, posted (3 months 3 weeks 6 days 23 hours ago) and read 2715 times:
Quoting tdscanuck (Reply 34): In all fairness, from a physicist point of view what you were taught may have been correct. If you have a perfect inviscid incompressible fluid the equal transit time is actually true as long as your body doesn't have sharp edges (like a trailing edge).
Gingersnap From United Kingdom, joined Aug 2010, 687 posts, RR: 6 Reply 36, posted (3 months 3 weeks 6 days 22 hours ago) and read 2648 times:
Quoting tdscanuck (Reply 9): This is a 2D airfoil; it doesn't have any induced drag. I think you mean form drag.
I assumed it was a 3D model side on my apologies.
It isn't clear how fast the airflow is over the airfoil, but I was initially led to believe that form drag was more prevalent at higher airspeeds. Whereas induced drag was more prevalent at lower airspeeds, especially when a higher AOA is a factor. At least that's how I understood it from various drag curve diagrams.
But as it isn't clear what speed the air is travelling over the airfoil, so I'll accept what you're thinking.
Also sorry for my confusion about the Bernoulli effect. As it states that there is an increase of speed with a reduction of pressure, I must have got my wires crossed. I do take into account that the principle is based on fluid with no viscosity.
Thanks for taking the time to ensure I was on the right track however. I am only a first year Aeronautical student so I have much to learn. Thanks again.
tdscanuck From Canada, joined Jan 2006, 11032 posts, RR: 72 Reply 37, posted (3 months 3 weeks 6 days 17 hours ago) and read 2492 times:
Quoting Gingersnap (Reply 36): I assumed it was a 3D model side on my apologies.
If it's a 3D they've done some video processing to remove the stinger/balance and the smoke trails running between the wingtips and the camera. That's certainly possible but I think it's a lot more likely that they just used a 2D tunnel. You get better raw airfoil data that way anyway.
Quoting Gingersnap (Reply 36): It isn't clear how fast the airflow is over the airfoil, but I was initially led to believe that form drag was more prevalent at higher airspeeds. Whereas induced drag was more prevalent at lower airspeeds, especially when a higher AOA is a factor.
It's not exactly about airspeed, although there are correlations, and you need to be careful about the difference between force and coefficient. Unless you change Mach number significantly, the form drag coefficient is relatively constant at fixed AoA. That means the form drag *force* varies with the square of the airspeed (assuming all other things equal) so it is indeed more significant at high airspeeds.
Induced drag coefficient is proportional to the square of the lift coefficient (not lift force). For equal lift, at high airspeed, you need a much lower lift coefficient (lower AoA) so the induced drag coefficient and induced drag force drop off significantly at high airspeed in practical conditions. Lift coefficient varies linearly with AoA (to first order).
However, when you're talking about pure airfoil properties, at fixed AoA you have a constant lift coefficient (hence constant induced drag coefficient) and a constant form drag coefficient. Since the coefficients are airspeed insensitive if we ignore Mach effects, the ratios of all the coefficients and the forces (i.e. how significant each one is) is the same at any airspeed as long as you don't change AoA.
bellancacf From United States of America, joined May 2011, 51 posts, RR: 0 Reply 38, posted (3 months 3 weeks 6 days 14 hours ago) and read 2420 times:
faro From Egypt, joined Aug 2007, 1327 posts, RR: 0 Reply 39, posted (3 months 3 weeks 6 days 5 hours ago) and read 2300 times:
and let's not forget the AoA of the wing and fuselage. In the cruise these are typically at 1°-3° nose up and produce (a little) lift even if the wing were a flat plank of wood.
titanmiller From United States of America, joined exactly 6 years ago today! , 71 posts, RR: 0 Reply 40, posted (3 months 3 weeks 6 days 3 hours ago) and read 2269 times:
Quoting aircellist (Reply 13): You'd be surprised how many of those who've taken aerodynamics courses still don't believe or understand that airplanes fly quite simply by shoving air downwards!
I've never found it hard to understand, especially if you consider a propeller to be a rotating wing...it pushes A LOT of air backward, so why wouldn't a wing push air downward?
Starglider From Netherlands, joined Sep 2006, 642 posts, RR: 44 Reply 41, posted (3 months 3 weeks 5 days 12 hours ago) and read 1991 times:
Quoting faro (Reply 39): and let's not forget the AoA of the wing and fuselage. In the cruise these are typically at 1°-3° nose up and produce (a little) lift even if the wing were a flat plank of wood.
Except when "Compression Lift" is included in the design of the aircraft. Applicable to supersonic aircraft it reduces induced drag (e.g. increases lift) by 30 percent relative to a conventional design.
It is a good example to illustrate the air flowing over the top of a relatively flat wing surface reaching the trailing edge sooner than the air flowing below the wing which is slowed down and compressed due to deceleration in the supersonic shock waves.
bellancacf From United States of America, joined May 2011, 51 posts, RR: 0 Reply 43, posted (3 months 3 weeks 4 days 15 hours ago) and read 1624 times:
So is the displacement of the upper sheet relative to the lower directly proportional to the lift generated?
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