comair25 From Germany, joined Sep 2006, 192 posts, RR: 0 Reply 1, posted (3 months 4 weeks 1 day 3 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.
Buffet: the airline business...has eaten up capital...like..no other (business)
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.
25 UALWN: 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
26 UALWN: Non-viscous perfect fluids do. That's why I always thought this was at least a good first approximation to the explanation.
27 Mir: Which is a good thing, since if the pressure gradients were symmetric there would be no lift at all. Some of them are dragged on. That's what the bou
28 ferpe: NASA has excellent explanations of this all at this web site: http://www.grc.nasa.gov/WWW/K-12/airplane/guided.htm If you take your time to go through
29 Starlionblue: An old Howstuffworks.com article which I unfortunately has been dumbed down since then described the whole thing pretty nicely without getting into ma
30 BMI727: 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
31 tdscanuck: Because pressure isn't the same thing as viscous drag. You have pressure gradients even if you assume an inviscid fluid. Pressure is a scalar...the p
32 UALWN: 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 conceptual
33 David L: 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
34 tdscanuck: 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 e
35 UALWN: I feel much better now! Thanks again for the additional explanation.
36 Gingersnap: 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 fo
37 tdscanuck: 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'
38 bellancacf: What an excellent thread this turned out to be!
39 faro: 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 wi
40 titanmiller: 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
41 Starglider: Except when "Compression Lift" is included in the design of the aircraft. Applicable to supersonic aircraft it reduces induced drag (e.g. increases l
42 Vzlet: Smoke puffs in the tunnel stream provide a way to debunk the "meeting molecules" view:
43 bellancacf: So is the displacement of the upper sheet relative to the lower directly proportional to the lift generated?
Sponsor Message: 
From United Kingdom, joined Mar 2006, 1209 posts, 
From Germany, joined Sep 2006, 192 posts, 
From United States of America, joined May 2008, 1379 posts, 
From Canada, joined Jan 2006, 11032 posts, 
From Andorra, joined Jun 2009, 1753 posts, 
Sponsor Message: 
Printer friendly format
