Nothing to do with intelligent design, as such, but Sal Cordova brought up the issue of how wings produce lift in this thread.
hmmm.
[Edited to add xkcd, which my son just reminded me of….]
Like most people, I bought the airfoil myth for many years until I actually started to think about designing a kite (it was a fun first-year architecture project). And then I didn’t think much more about it until an interminable thread about a wind-powered cart that could travel downwind faster than the wind arrived at Talk Rational, in the course of which I discovered something I hadn’t appreciated about air, which is that it isn’t readily compressible unless in a confined space (I’d always imagined propellers in water behaving rather differently to propellers in air, but it turns out they don’t). Anyway, thought it might make a change from ID, and maybe Gil Dodgen might like to weigh in on Something Completely Different :). I’ll invite him.
People need to get the idea out of their minds that the “Bernoulli effect” and the “flow turning/redirection effect” are independent phenomena that make varying contributions to the lift of a wing.
The “Bernoulli effect” is simply a proxy for saying that there is a differential pressure between the top and bottom surfaces of the wing, hence generating a net force up or down. The fact that there is a pressure difference is explained by Bernoulli’s principle given that there is a differential in speed of airflow above and below.
Pressure is how we account for air molecules striking the wing surface. That is the only mechanism by which air can have any effect at all on a wing. If there is no pressure differential, there is no lift force. Period.
If there is a pressure differential, it will be due to a velocity differential. No Bernoulli effect, no lift. (The only “exception” is thrust vectoring. But in that case the vertical force is not lift, per se, it is vertical thrust, similar to a rocket.)
But if there is a pressure differential on the wing, and hence a net vertical force on the wing due to the air, then Newton’s third law says that the wing exerts an equal and opposite force on the air. Hence air is pushed in a vertical direction (typically downward for upward lift). Thus whenever there is a pressure differential (due to Bernoulli’s effect), there HAS TO BE a corresponding redirection of the air flow. The two are inseparable. They are two sides of the same coin. You can’t have one without the other. Bernoulli’s effect always produces the same amount of lift as flow turning — but they are not two separate contributions to lift, they are the same thing.
Pat Peebles’s Fan Wing uses a rotating cylinder to generating lift — a lot! Similar in concept is this prototype UVA from Austrian Innovative Aeronautical Technology (IAT21) .
I don’t think this is correct. For example look at this NASA link on the F-14 tomcat. In subsonic flight the Lift to Drag ratio is 14
http://www.hq.nasa.gov/pao/History/SP-468/ch11-6.htm
That means for every 15 pounds of weight you only need 1 pound of thrust. This can’t be achieved by simple deflection of air or vertical thrust.
High performance air-to-air missles or surface-to-air don’t need much bernoulli lift, but a fighter needs it so it doesn’t run out of fuel when trying to get to its destination.
The bernoulli effect is very necessary for energy efficient flight. Even fighter planes need some energy efficiency.
The F-14 will have a lower lift-to-drag ratio when its wings are swept back for high speed flight. The space shuttle has a lift to drag ratio of about 4. A glider can reach ratios of 70. That means for every 70 pounds of weight it only needs 1 pound of thrust to keep it airborne (assuming it isn’t using thermal lift)
The lift force depends on airspeed, wing surface area, wing shape (particularly the camber — which represents the curvature of the wing), and angle of attack.
When a plane is rolling on the runway at takeoff Bernoulli’s effect (and flow turning, since the go hand-in-hand) is creating lift just like in steady flight. But the airspeed on the runway is much lower than in steady flight. You would need to be going way to fast on the ground to lift off without rotating. Rotation serves the purpose of increasing the angle of attack of the wing, which raises the lift considerably and facilitates liftoff. There are other things done to increase lift at low speeds during takeoff. Flaps are extended. These serve two purposes. They increase the wing area and they increase camber. Both of these things increase lift for a given air speed. Once altitude is achieved the flaps are withdrawn and the aircraft levels out because at higher speeds the increase in camber, surface area, and angel attack all increase drag. But you no longer need those lift boosters because the higher speed accommodates the necessary lift. But at no time is Bernoulli’s effect (i.e., the effect of a pressure differential on the wing) negligible. Without a pressure differential you can’t fly.
I proposed two experiments, and I even gave preliminary results. Here are more details from my 2 minute experiment.
With one sheet of paper hanging, I blew air parallel to the right of paper and the bernoulli effect was in evidence as the paper moved TOWARD the airstream (to the right). If I turned about 20 degrees into the paper (so the airstream is mostly parallel but now with an angle of incidence), the paper moved AWAY from the airstream (to the left).
Clearly according to theory, the bernoulli effect should be in play in both airstream orientations, since the airstream on the right side is faster than that of the left side (zero). So why does the paper move away in the second case? I argue that another mechanism of force is in play, namely dynamic pressure. The bernoulli effect is still in play, but it is overcome by the deflection (or flow turning) effect.
In the case of this experiment, we can get the bernoulli effect to work in opposition to the flow turning effect, whereas in an airplane, the two mechanisms can work to gether (albeit possibly at the expense of each other).
But I see another dynamic at play in this discussion. If we have a hard time arguing over slightly controversial topics, how much more will we argue of evolution.
I mentioned earlier an old text book I had in aerospace class. It was by John Anderson. Anderson wrote:
I’ll go with the bernouilli effect over explanations that resort to deflecting air downward. Further, I’ll argue a typical kite doesn’t work like an airfoil that uses the bernoulli principle, but a simple deflection device. I provided experiments that might help interested readers to decide for themselves whether typical kites fly on different principles than airiplanes.
To be fair the right brothers first planes were glorified kites, but where they excelled was they built kites that used the bernoulli principle versus crude deflection.
Airplanes stall (lose lift) when the wings exceed a certain angle of attack (which depends on speed). At cruising speed the stall angle might be in the range of 15 degrees, give or take.
Without Bernoulli’s effect (that is, assuming it is something separate from the flow turning/redirection effect) stall cannot be explained. Stall happens because the air flow over the wing separates from the upper wing surface. Before separation, the air over the top of the wing is at high speed, and hence low pressure. After separation, the speed of the air near the upper surface of the wing drops precipitously (similar to a back water or eddy in a flowing river), hence the pressure jumps up, killing the lift. If you only think of the lift as being due to the flow hitting the underside of the wing and being redirected (like a fire hose directed at an inclined surface), then there is no reason for the lift to disappear at an angle above about 15 degrees. To the contrary, in the fire hose case the lift would continue to grow well past that angle.
This example is irrelevant. There is no fluid on the other side of the door. The door is not immersed in a fluid that flows around it.
What you are doing in effect, Sal, is arguing that momentum conservation beats Newton’s second law. 🙂
This fallacy is dealt with here.
Leviathan,
I used to buy simple rubber powered balsa wood models with flat wings that flew well.
Again, if you were right, when flying upside down, a plane should be sucked down to the ground by its “Bernoulli effect”, but…., that doesn’t happen.
This has turned out to be a fascinating post.
If I put a parachute on my back, and then tie a rope between myself and a boat, could I fly?
Suppose I replaced the boat and rope, with a motor and propeller?
Does the parachute allow me to fly because of the “Bernoulli effect”?
Bernoulli’s effect works for a flat plate just fine. At zero angle of attack a flat plate produces no lift because of symmetry (no difference top or bottom. At a positive angle of attack, the flat plate produces lift. The flow over the top is faster than that underneath EVEN THOUGH the length of the plate is the same on top as on bottom. The flow accelerates over the top because of what’s known as the Kutta condition — the physical requirement that flow leave the trailing edge parallel to that edge both top and bottom. This physical reasons for this requirement, and its effect on velocity above and below, is not simply explained by just the distance traveled top and bottom.
Bernoulli’s effect in no way prevents a plane from flying upside down (in principle). Most wing airfoil shapes are asymmetric (they have a camber that is concave down) which means they generate lift even at zero angle of attack (plane flying perfectly level). If the plane were to fly upside down perfectly level, it would drop like a rock. That is why, if you think about the stunt planes you’ve seen flying upside down, the always fly tilted way up in front. That is because they have to achieve a high upside down angle of attack in order to overcome the camber and begin generating lift in the opposite direction.
Every asymmetrical (cambered) airfoil has a zero-lift angle of attack which is negative. That is, at zero angle of attack (perfectly level) the airfoil will produce lift. Tilted up (at positive angle of attack) it will produce more lift (until stall). Tilted down, the lift will decrease till at some negative angle the lift will go to zero. Tilt it further down to a greater negative angle and you will start to generate negative lift. If you turn it over, the negative lift will become positive lift.
Flying upside down doesn’t change anything, not one whit, about the physics of flight. The only thing that changes is the angle of attack you have to have to generate the lift you want. You can’t generate lift flying upside down at zero angle of attack. You have to tilt way up
Leviathan,
I agree, which is why I disagree with the notion that wings create lift because of the “Bernoulli effect”.
There’s a way to prove that.
Take a wing, bolt it to a shaft at a five degree angle of attack, blow wind “only” over the top and measure the torque.
Then repeat the experiment by blowing wind “only” under the wing.
I predict the air being blown under the 5 degree wing, will produce magnitudes more lift.
On the other hand, if the “Bernoulli effect” is what is keeping aircraft in the air, the torque readings should be close to equal.
Toronto,
You may be correct about the results of your hypothetical experiment, but your experiment doesn’t have much to do with flight. The physics of a wing (airfoil) depend on the wing being completely immersed in a moving fluid. In your hypothetical experiment you are simply blowing a jet of air over either the top or bottom, but not both. The effects in your experiment don’t translate to the real case because in the real case the flow over the top is intimately coupled to the flow over the bottom.
Toronto,
Here is a computational fluid dynamics (CFD) calculation of the flow over a NACA 0012 airfoil at 5 degree angle of attack.
CFD computation of flow over airfoil at 5 degree angle of attack
The interesting thing about the NACA 0012 is that it is symmetric (i.e., no camber), and relatively thin, so it behaves somewhat similar to a flat plate. So at 5 degree angle of attack, this calculation is in the ballpark of your hypothetical example. The colors in this picture represent pressure magnitudes. Green is the ambient pressure (the pressure in the air without the airfoil). Yellow, orange, and red are increasing pressure (over ambient). Light to dark blue is decreasing pressure (below ambient).
As you can see, there is a high pressure spot near the lower side of the leading edge. This is called the stagnation point, where the flow of air is brought to a standstill. Air above the stagnation point goes over the wing, air below goes under. There is also a stagnation point on the trailing edge (where it has to be because of the Kutta condition).
Okay, now is where it gets interesting. If you look at the lower surface, the pressure is light blue, which means that the pressure has actually dropped below ambient, which, if you ignore everything else, means that there is actually in some sense a downward “suction” from the bottom surface, rather than the increase in upward pressure on the bottom that you might imagine based on your intuition.
But if you look at the top surface, you can see that there is a much larger pressure drop along the top. Therefore, despite the fact that the pressure dropped along the bottom surface, the pressure on the bottom is still much larger than the pressure along the top, creating a net upward force; i.e., lift. But as counter-intuitive as it might seem, the bulk of the lift comes, in a very real sense, more from a drop in pressure along the top than from any increase in pressure along the bottom. And the reason the pressure drops so much along the top is because the flow velocity is much greater there.
Leviathan,
But for our intents and purposes, air is NOT a fluid since it is compressible.
Secondly, my experiment shows airflow = 0 at one time for the top, and another time airflow = 0 for the bottom.
There is nothing wrong with that as we have a delta between top and bottom.
For instance, a top of 15 and a bottom of 10 should be equivalent to a top of 5 and a bottom of 0 for purposes of calculating pressure differentials.
That’s a great plot you supplied by the way.
What do the numbers actually represent?
What I’m asking is are they absolute values of pressure that we can compare?
Toronto,
Air is always a fluid – fluids can be either liquids or gases – but I understand your point that air, being a gas, is compressible. The plot I linked is for Mach 0.3, which is low enough speed that air flow can be accurately modeled as incompressible. But even at speeds approaching Mach 1.0, for which compressibility can’t be ignored, the basic physical explanations don’t change even though the calculations being more complicated.
I did not generate the plot, but I believe that the numbers on the color scale are absolute pressures in Pascals (Newtons per meter squared).
I’m going to rescind what I said earlier about you probably being right about the results of your hypothetical experiment. On second thought, I think you’d be surprised by the results. Watch this video of water flowing only over the top edge of an inclined flat plate: Flow over top of plate. Notice that water, which is only going over the “top” side of the plate, is redirected parallel to the angled plate, just as it would be if is was flowing over the “bottom” side of the plate! What this means is that if the flow is turned to the left in the video, the plate must be exerting a force on the water to the left. Therefore, by Newton’s third law, the water is exerting a force on the plate to the right, which is a lift force. This force is of similar magnitude regardless of whether the water is flowing down the “top” side or “bottom” (right or left) of the plate. As counter-intuitive as it might seem to you, the fluid flowing over the “top” of the plate creates significant lift.
Similarly in your experiment, the torque would be about the same regardless of whether the air flows over the top or bottom.
Now look at this video and you can see actual smoke streamlines of air flowing over an airfoil as the angle of attack changes. They do a pulsed smoke that allows you to see the acceleration over the top compared to the bottom (and hence a differential pressure per Bernoulli): Smoke streamlines.
The other thing my butter “experiment” taught me is that shear is a better way of thinking about what is going on than compression.
Pressure, sure, but not compression.
As Leviathan said, fluids are compressible to various degrees. Whether density changes are sufficiently large to merit consideration a quantitative question in each particular situation. Earlier in this thread, I provided an estimate for a pressure difference for wings of a Boeing 777. It is a few percent of the atmospheric pressure. That translates into density variations of a few percent, a minor effect. You don’t need to worry about air compressibility unless your aircraft flies at a speed comparable to the speed of sound.
That was an illustration, but the same complication is more in evidence with the paper airstream experiment. The “no fluid” objection can’t be used to criticize the obvious result of flow turning working in opposition to the bernoulli effect. The problem with flow turning arguments is one of semantics. Anderson’s book for example never refers to flow turning as an explanation, and for good reason.
Regarding Flow turning from the wiki entry on LIFT:
Total pressure is defined as static pressure plus dynamic pressure. Static pressure is the pressure we typically think of when the weatherman gives us a barometric reading. Dynamic pressure is what one experiences when they put their hand out of the windown in a moving car.
Pressure (both dynamic, static, and total) is generated by a change of momentum. If one is speaking about the ‘flow” it usually is in reference to the net motion of a body of fluid.
Of course pressure is generated by “deflection” at the molecular level. Even with no net motion in a the air one experiences pressure because the air molecules deflect (bounce) off surfaces and creates a reaction force. But that is hardly considered “flow” as in the notions of fluid mechanics. The notion of ‘flow’ is more consistent with the net velocity of a body of fluid, not the individual molecules in isolation.
But since one can argue that even in the static case, “deflection” of individual molecules takes place when it bounces off a surface, the “flow turning” explanation becomes confused because it doesn’t make a distinction between momentum change from the net velocity change (associated with dynamic pressure) and momentum change from velocity change of a single molecule (associated with total pressure). Hence no new insight is gained, only confusion.
If one wants to integrate the “effects of flow” which means changes in pressure, that’s fine, but that adds really no new insight and adds a confusion factor. The lack of rigor to the “flow turning” explanation is well-placed.
The simple airstream and paper experiment highlights the confusion that “flow turning” explanations are prone to give. It adds little new insight, and only confuses the matter more, imho.
Balloons employing archemedes principle don’t resort to flow explanations. One could argue that deflection at the molecular level and thus change of momentum is involved. There is no need to invoke the momentum of the flow as an explanation (since there is no flow momentum to begin with). This is “static lift”, but I point this out to highlight that maybe the momentum of the flow has less to do with aerodynamic lift than is supposed by the NASA weblink. This ought to be very much in evidence by the fact we have planes with lift-to-drag ratios of 40 or more.
Differing velocities around the wing result in differing pressures resulting from change in momentum. But trying to explain the pressure changes in terms of changes of momentum of the flow is forced at best.
In the airstream experiment with the paper being moved toward the airstream because of the bernoulli effect, it is hard to see that the momentum of the airstream is really being changed. And in the case where the momentum of the airsteam is obviously changed (by blowing 20 degrees into the paper), the paper moves in opposition to the force created by the bernoulli effect. Thus it seems “flow turning” explanations for lift adds little new insight, maybe only more confusion.
Leviathan,
Those were great links but the one I like is the smoke one showing a stall.
I can now see where the “wing” stalls but it is way before the turbulence over the top and the video claims that it stalls.
The video shows the angle increasing much beyond 15 degrees before it claims a stall, more likely 40 degrees or so.
What I noticed when you look at the top of the leading edge as the angle increases, is that there is a pronounced downward force on the wing right at that leading edge.
Before we even get to turbulence, there is already a force fighting lift pushing down on the leading edge of the wing.
If the “Bernoulli effect” doing anything for lift were valid, a water skier would need water flowing over the top of his ski to “fly” but the inclined plane of the ski is more than enough.
I’m not strong enough in math to do proper service to Mr. Bernoulli but everything I have ever read about it seems to suggest that what he is talking about is closer to Ohm’s Law than small objects moving through a large gas.
In Ohm’s Law, 1 A * 2 V = 2 W but 2 A * 1 V = 2W also.
It’s the same with hooking up a 1/2″ inch garden hose nozzle to a 1″ inch hose.
The water will speed up to try and compensate for a smaller hole.
None of this really has much to do with free air.
What I see in the video confirms to me that lift primarily, maybe 95 %, is due to the planing effect not the pressure differential caused by the result of air flowing over the top and bottom of the wing.
When the top of the wing becomes turbulent, your control surfaces such as your ailerons, will no longer be functional, but if the bottom of the wing can generate a larger force than the force on top, your wing may not be controllable but you will have lift.
olegt,
That’s true which is why I say that the “Bernoulli effect” claimed as lift, doesn’t really contribute much to lift.
It would seem to me that you could simply measure the pressure differential and that would give you an upper limit to its contribution.
petrushka,
Yes, measuring it would be a good experiment but I don’t know a fair cheap way to do that.
By running air at the same time over the top and bottom, a simple torque reading would give us net torque, not separate readings for top and bottom.
Running only one surface at a time would seem unfair to those who claim both sides need airflow for a fair test even though I think it would be fair.
I think I have a fair test!
Maybe.
We make a model wing and slice it into top and bottom pieces.
Use a very light glue to hold the top and bottom together again.
Run air over the wing.
If the “Bournelli effect” is what contributes mostly to lift, the wing should tear itself apart by the top section leaving the bottom as the top negative pressure should be contributing as much force as the positive planing pressure from the bottom.
If its mainly planing pressure from the bottom, the top of the wing should stay where it is.
Anybody here make balsa wood models?
There are simpler ways to try this out. Some are very elegant. There used to be a collection of methods in Scientific American’s Amateur Scientist series. One was how to make a smoke chamber. Another was to use water flowing over permanganate crystals embedded in a strip at intervals, this produces streams of dyed water and one could watch the flow of the streams around objects placed in the wider stream.
Anyhow… the usual method is to make an airfoil and to drill tiny holes in the surface. Each hole is connected to an inclined manometer. As air flows over (and under) the airfoil it produces a direct physical image of the pressure at various points over the wing. I saw such a setup decades ago being operated in the wind tunnel at MIT.
There is an interesting discussion of this at the hyperphysics website, already linked but it bears repeating:
http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html#airf
Which basically says… it is both, Bernoulli and Newton. Depending on the angle of attack one or the other becomes more dominant.
The interesting thing (to me anyhow) is that it IS a bit more complicated AND it should be looked in to further so that it is understood. XKCD gets it right.
There is also the Kutta-Jukowski theroem which appears to contradict the Bernoulli effect (but doesn’t really)
http://hyperphysics.phy-astr.gsu.edu/hbase/fluids/kutta.html#c1
What it does is illustrate how things are not always easy to simplify and explain with an over-simplified application of the theory.
The first time I saw the rotating cylinder experiment I got the predicted behavior and the explanation wrong. Experiment trumps theory (or misunderstanding of theory) every. single. time.
Once more unto the breach…
As I’ve previously stated, there is no “opposition” between pressure-based explanations (e.g., Bernoulli) and momentum-based explanations (e.g., flow turning).
Let’s examine your examples more closely to see why they don’t illustrate any conflict.
When you hang two sheets separated by a small gap, then blow through the gap, the sheets move towards each other. The velocity of the moving air between, presumably (due to Bernoulli) results in a lower pressure in the gap than in the static air in contact with the back sides of the sheets, creating net forces to push the sheets inward. As you have observed, and as symmetry would dictate, the air blown between the sheets will continue straight beyond the sheets without turning, thus seemingly rendering the momentum (flow turning) idea moot.
But it is not moot. Because you have a symmetric pair of sheets, identical lift forces are generated on the two sheets in opposite directions (both forced towards the center). Newton’s 3rd law requires the sheets to exert equal and opposite forces on the air, hence the air is subject to two forces of equal magnitude butin opposite directions. In other words there is no net lateral force acting on the air and it therefore continues straight. The principles of momentum are at work in this case just as much as any other, but in this case lateral momentum happens to be conserved.
Now, before I look at your other case, in which you blow onto one sheet at 20 degrees, I have to address another issue, and I hope that I can explain my point clearly.
In your examples (as well as in Toronto’s experiment of measuring the torque on a wing mounted on a shaft), air is only blown over one surface of the paper/wing, which creates a fundamental difference between those cases and the case of an actual wing immersed in moving air. It doesn’t mean the basic principles (Bernoulli & momentum) don’t apply, just that one has to be very careful in how one interprets what’s going on. It is easy to think of it way to simplistically and run into the weeds.
I’ll address this difference in the next post.
Look at the video I linked previously of the smoke streamlines , which is representative of a wing moving through air (or air moving over the wing, if you prefer). The streamlines entering the left of the picture are all parallel and traveling with the same velocity (called the “freestream” velocity). The presence of the wing obviously disrupts these streamlines.
Bernoulli’s principle says that total pressure, which is the sum of static pressure and dynamic pressure, must be constant along a streamline. Thus, if a given streamline speeds up (going over the top of the wing, say), then its dynamic pressure will increase and its static pressure will correspondingly decrease in order to keep the total pressure the same. But it is important to note a couple of things. First, when calculating the pressure forces on the object, we use the static pressure since that is the actual physical pressure being exerted.
Second, because the incoming streamlines are all identical, they all have the same total pressure. Thus, for a airfoil in a uniform flow field, the total pressure is always the same at every point. Therefore, if the air goes faster over the top than under the bottom, the dynamic pressure above is greater than the dynamic pressure below, and hence the static pressure below must be greater than the static pressure above, and hence there must be a net upward lift force.
In the next post I will explain why this is a significant departure from the thought experiments that have been proposed using flow only over one surface.
In the experiments y’all have devised for which air is blown over one side only of an object, there is not a uniform freestream condition. That is, air on one side is static while the air coming in on the other is moving. An the source of the moving air, whether blowing with your mouth or using a fan or whatnot, is adding energy to the air that the static atmospheric air on the other side does not have. The consequence of this is the the air being blown on one side is at a higher total pressure than the static air on the other side.
As we said, total pressure is the sum of static and dynamic pressures, or pt = ps + q.
Total pressure still must be preserved along a streamline, so Bernoulli’s principle still applies, but unlike the actual wing case, now we can’t simply determine which side will have the highest static pressure (and hence which way the lift force will go) because the total pressure is different on the two sides.
So here is what is happening in the two paper experiments. When we blow through the gap between two sheets, and we think it is a clear demonstration of Bernoulli, we are not being completely accurate. The total pressure between the sheets is greater than the total pressure outside because we have added energy with our mouth. So we can’t just say that because the air is going faster between, there must be a differential pressure, because that conclusion is based on naively assuming the total pressure must be the same everywhere, which it isn’t!
However, we still get the result we expect because, even if for the wrong reason, because the static pressure in the air moving between the sheets is still lower than the atmospheric pressure on the outsides. But since the total pressures are different, this doesn’t have to be the case in principle, it just happens to be the case for this circumstance.
I will finish up with the paper blown at 20 degree angle in the next post.
That’s mostly drag if I am not mistaken.
Leviathan,
But by what amount do they differ?
If we’re in the range of a 5% differential, is that enough to lift the wing and more importantly, the aircraft it’s attached to?
I think “olegt” came up with something like 4% pressure differential on a 747 wing.
I don’t think that’s enough.
Don’t helicopter blades have a zero degree angle of attack when running up the engine?
If a pressure differential due to the “Bournelli effect” was at work, the helicopter would get lighter when running up the engine despite the blade angle.
We could try this with an RC helicopter and a bathroom scale.
That’s the pressure differential required to lift the aircraft. Aircraft weight divided by the wing area. So it’s enough by definition. 🙂
olegt,
I see I misunderstood as your figure is for what’s “required” to lift that particular aircraft, given its wing area.
Thanks!
I never knew there was contention on how planes fly. I read about this on wiki.
They can’t even get that figured out but do know how bacteria became buffalos.
Its a clue that evolution has been a ongoing error because its more slippery to test or prove then this flight stuff.
Even flying which is right in front of us has enough options to put conclusions about it in doubt.
Think about that the next time you get on a plane. Or drive a car. Or type on your computer. Scientists don’t know anything. You can really only trust preachers to tell you the truth.
Robert Byers,
Are you suggesting that planes fly because the “Intelligent Designer” picks up them up with his tiny invisible fluffy mittens?
Invisible pink fluffy mittens.
Today we learned faith is easy and science is hard.
Okay, lets do a calculation using some real values. Here is a website that provides data for a light sport aircraft called “One Design”. Figures 4 & 5 toward the bottom of the page show pressure coefficients for the upper and lower wing surfaces for this plane’s airfoil and a similar NACA airfoil. The lift on the wing is related to the difference in pressure coefficient between top and bottom. That difference varies from the front to the back of the wing — larger toward the front and tapering to zero toward the end. This is a rough guesstimate from the graphs, but the average difference in coefficient of pressure from front to back is something in the ballpark of 0.5 give or take.
The actual pressure differential between top and bottom is this pressure coefficient differential times the freestream dynamic pressure (0.5*density*velocity squared). Using the air density and flight speed values given on the page, the dynamic pressure can be calculated as:
q = 0.5 * 0.0023769 slug/ft3 * (176 ft/sec)^2 = 36.8 lb/ft2 = 0.26 psi
The average pressure differential is therefore approximately
delta p = 0.5 q = 18.4 lb/ft2 = 0.13 psi
The wing surface area is given as A = 75.55 ft2, so the total lift is
L = delta p * A = 18.4 lb/ft2 * 75.55 ft2 = 1390 lb
The aircraft weight is given as 1000 lb (on another website I found a value for maximum weight of 1140 lb).
Therefore a pressure differential between top and bottom wing surfaces of only 0.13 psi (which is less than 1% of atmospheric pressure at sea level) is more than sufficient to lift the aircraft.
In the case of bigger planes traveling at much higher speeds, it would not be unusual to see a pressure differential in the range of 1 psi, which for the 5500 ft2 wing area of a 747 translates into a lift force of 79,200 lbs, which is about the same as the max gross take-off weight of that airplane.
Leviathan,
Thanks for all the work you put into that comment.
I didn’t realize that such a small percentage of difference was involved.
It looks like Sal’s “Bournelli effect” could actually be quite meaningful in providing lift with such a small difference involved.
Okay, to finish my final installment from earlier….
If you hang a piece of paper and blow onto the near side at an angle of 20 degrees you’ll find that the paper moves away from you (as intuition would likely suggest). The air on the far side of the paper is not moving. The air on the near side clearly is moving. The simple understanding of Bernoulli’s principle says that moving air should have a lower pressure than non-moving air, which suggests that the pressure should be higher on the back side of the paper causing the paper to move toward you. This doesn’t happen, so it leads to the natural conclusion that for some reason Bernoulli doesn’t work here or that some other competing phenomenon is opposing Bernoulli’s principle (namely that the direct impingement of the air on the near surface is pushing the paper in the way that is similar to directing a fire hose at a surface.
But the explanation above is incorrect because it is based on a misunderstanding/misapplication of Bernoulli’s principle. As I discussed in a previous post, Bernoulli’s principle says that total pressure (static + dynamic) is conserved along streamlines. But in the case of blowing against one side of the paper at a 20 degree angle, the air on the near side has been given energy (by your mouth) not possessed by the air on the back side. Therefore the total pressure on the near side is higher than that on the backside (which is than different for a wing immersed in flow for which the total pressure is the same everywhere. Bernoulli’s principle is an expression of the conservation of energy principle, and total pressure is an expression of total energy in the air.
The air on the back side is not moving, thus has no dynamic pressure, hence its total pressure is the same as its static pressure and will be equal to atmospheric pressure.The air on the near side, having been given additional energy by your mouth, has a total pressure higher than atmospheric pressure, part of which is dynamic because the air is moving and part of which is static. Initially, the static part might be lower than atmospheric (and lower than that on the back side), but doesn’t have to be depending on how much energy have been put into the flow by your mouth. As the moving air strikes the paper at the 20 degree angle and is deflected (turned) by the paper, the velocity of that air is reduced somewhat by the interference of the paper. This causes the dynamic pressure to drop (proportional to the square of the change in velocity). But Bernoulli’s principle still applies as long as we are just talking about one side of the paper, so the static pressure on the near side will go up by the same amount that the dynamic pressure went down. When it goes up, it exceeds atmospheric pressure (if it didn’t already start out that way) and thus achieves a higher pressure than that on the back side, creating a net lift force to push the paper away from you.
So here’s the summary. The ONLY WAY to get lift is to have a pressure differential between the sides of a wing (or paper or whatever). Bernoulli’s principle always applies. In the cases where air is only blown on one side of the wing/paper, the reason Bernoulli sometimes doesn’t seem to apply is because of the error of unconsciously importing the incorrect assumption that the total energy is the same on both sides (as it is for an actual wing). But when you account for the fact the air on one side is being energized in a way that the air on the other side is not, then there is no paradox. You are adding energy to one side of the object and not the other.
But with an actual wing, the total pressure IS the same on top and bottom, which means that the total energy is the same top and bottom. Because of this, the ONLY way to get lift is to get the air to move faster over the top than the bottom.
The importance of what’s happening on top of a wing, i.e., that it is not just the air pushing from the bottom, is illustrated by the use of spoilers on airplane wings. The spoilers are simply hinged plates on the top side of he wing that pop up. These are often used on airliners when beginning a descent from altitude and also after touchdown. The purpose of raising the spoilers on the top side of the wing is to cause flow separation on the top, which kills the flow speed next to the wing surface and raises the pressure. If all that mattered for lift was the air pushing on the bottom of the wing, there would be no reason for a spoiler to kill lift since it doesn’t really affect the flow striking the bottom surface.
Exactly! Your position rests entirely on faith as it doesn’t have any science to support it.
Well I don’t know what a scientist is but researchers or teachers should know something or everything about the thing they put their minds too and get paid for.
anyways this flying stuff shouldn’t be a mystery by this time.
If these things are contentious then surely evolutionary biology is more so and if not then because its not open to proper investigation techniques.
Origin subjects are not repeatable or testable.
Robert Byers,
If two creationists argue about what type of wood the Ark was made of, can I claim the Ark never existed because they don’t agree on that detail?
Yeah, you’d think that before building supersonic and even hypersonic aircraft, and fly-by-wire passenger airliners, that they would know something.
I guess they’re held aloft by the prayers of creationists.
Seriously though, you might consider, just for a moment, that when you don’t understand a scientific explanation, it might be because a journalist is trying to reduce a scientific explanation to metaphors that can be understood by the mathematically illiterate.
Macaroni and cheese, and kidney pie.
I don’t think that is correct. You’re such a gentleman, so I will simply disagree rather than say NO NO NO. 🙂
The actual physical pressure being meausred is the perpendicular component of the dynamic pressure plus the static pressure. Strictly speaking the physical pressure measured can be:
1. the total pressure
2. the static pressure
3. somewhere in between
Look at the architecture of the Pitot tube to get insight.
http://en.wikipedia.org/wiki/Pitot_tube
Dynamic pressure is not experienced unless the flow has a perpendicular component to the surface it is going against. Since in the case of the pitot tube, the freestream velocity is perpendicular to the cross section of the opening, the physical pressure (or Pitot pressure) in this case is equal to the total pressure.
But that is not always the case that the force experienced on a surface is the “total pressure” of the streamline. The experiment I provided shows in
case 1:
the paper being moved because of decrease in static pressure inside the the airstream, the total pressure on the stream line is different than the static pressure that is moving the paper to the right. The physical pressure in this case is ONLY the static pressure.
case 2:
the paper is being moved because a perpendicular component of the dynamic pressure is creating a left-pointing force on the paper that is sufficiently higher than the static pressure in the air stream creating right-pointing force. The physical pressure is a mix of the static pressure plus dynamic pressure, but it is not quite the total pressure (staic plus dynamic).
I deliberately created this experiment to show that the analysis of the forces is more nuanced than just invoking bernoulli lift. In case 2 we have the forces from dynamic pressure actually working in opposition to forces associated with static pressure. And to top it off, the net force in case 2 cannot be described exactly by total pressure (static + dynamic) since the streamline airflow comes at an angle of 20 degrees.
The reason I gave this example is to illustrate that static and dynamic pressure can be made to work against each other, and thus highlight the confusion factor that appeals to flow turning can make. Of course flow turining could be argued to be correct but it is idiosynratic, and offers not a lot of insight above other means of analysis.
See also:
http://www.dot.state.mn.us/aero/aved/pdf/2004%20TEA%20Academy/Liftofanairplanewing.pdf
So again, appeals to flow turning don’t explain anything more than what would already be known via other means.
Pressure is sensing the average change of momentum in the molecules, but that change of momentum isn’t the same as the change of momentum associated with the flow, unless one uses an idiosyncratic notion of the momentum of flow being other than mass times mean velociy of the body of air.
I pointed out that in the case of purely static lift (like that of a balloon), one is generating lift without even appealing to momentum of the flow, since no flow exists! So how is momentum change accomplished in the case of purely static lift? There is momentum change at the molecular level since there is molecular motion in the air molecules. This momentum change can be related to the slighltly lower static pressure on top of the ballon versus the slightly higher pressure under the ballloon. It is still a momentum change explanation, though not obviously so.
In like manner, the bernoulli principle is a momentum change explanation also, but that is NOT the same as the change in momentum of the flow.
Sal – is there any meainstream science you DO agree with? It might help us to know.
Thanks for not screaming NO at me. 🙂
Every discipline has its jargon, and sometimes that jargon can arise for reasons other than precision in description. Sometimes jargon, rather than having literal meaning, has metaphorical meaning, or at times has meaning that derives from analogy.
Unfortunately this can sometimes cause confusion when people, quite reasonably, are prone to interpret words literally. This is a case in point.
At any given point in a fluid there exists a pressure. It is a single thing. It has physical effects. But why then, we might ask, does Bernoulli’s equation talk about Total Pressure, Static Pressure, and Dynamic Pressure, as if pressure were made up of different parts? It does so because those terms are jargon that has proven useful as a way for physicists/engineers to think about Bernoulli’s equation, but which don’t literally represent different types of pressures, or even pressures at all in the case of total and dynamic pressures.
Bernouilli’s equation is a conservation of energy equation. If we write the equation in the simple form that applies to what we’ve been talking about, it says that along a given streamline:
Total Pressure = Static Pressure + Dynamic Pressure = Constant
or
pt = ps + q = constant
or
pt = ps + 0.5*RHO*V^2 = constant
Where the dynamic pressure q is calculated from the fluid density RHO and the fluid velocity V.
But the terms in this equation are not really pressures. They look like pressures, they have units of pressure, and they can be usefully thought of as if they were pressures, but they are not pressures. The terms in the equation are energy densities, or energies per unit volume. The precise way to state Bernoulli’s equation is:
Total Energy per unit volume = Potential Energy per unit volume + Kinetic Energy per unit volume = constant
It just so happens that the Potential Energy per unit volume of a fluid is the same as the pressure in the fluid, so that term came to be called “Static Pressure” since (a) it is a pressure and (b) it represents potential energy, which is that portion of the energy not associated with motion. Since the Potential Energy per unit volume term is for all practical purposes simply the pressure, and since the Total Energy per unit volume and Kinetic Energy per unit volume terms also have the same units as pressure, those latter two terms have come to be known as Total Pressure and Dynamic Pressure.
Despite this terminology, there ARE NOT real pressures, they are energy densities. Only the Static Pressure is a real, physical pressure. If you want to assess the forces exerted on something by the pressure in a fluid, it is the Static Pressure that matters and nothing else. I will demonstrate that with the example of a Pitot-Static tube, which you mentioned.
If we rearrange the Bernoulli eqn, then we can write:
q = .5 * RHO * V^2 = pt – ps
or
V = square root ( 2 * (pt – ps) / RHO)
So if we know the total and static pressures of a flow, and we know the air density, then we can calculate the flow velocity. More precisely, all we need to know is the difference between the total and static pressures since that is what appears in the equation.
This is the relationship used to calculate the airspeed or an airplane in flight. The aircraft will have a Pitot-static tube projecting, typically from the nose in order to sample the freestream air before it encounters the airplane. The tube has two pressure ports (holes), one at the end and one on the side. The one at the end senses the Total Pressure while the one on the side senses the Static Pressure. But despite that terminology, both ports are sensing a static pressure. How is that?
The end port hole is facing straight into the flow. Air impinging directly into that hole will be brought to a standstill, hence the pressure there is also known as the stagnation pressure. But since the velocity goes to zero, the Dynamic Pressure at that point will be zero and, consequently, the Total Pressure is equal to the Static Pressure. Let me express that again. We say that the end port senses the Total Pressure. But pressure ports can only ever sense static pressures since static pressures are the only kind of real pressures that actually exist. The reason we can say we are sensing the Total Pressure is because by pointing the tube directly into the flow we have eliminated the Dynamic Pressure term and made the Total Pressure equal to the Static Pressure. We had to do this to get the Total Pressure since static pressure is the only thing we are capable of sensing with a pressure gauge!
Now, because the Pitot-static tube is small and straight and smooth, we can make the assumption that the air flowing past the pressure port on the side of the tube is traveling at a speed that is approximately equal to the freestream speed, which is also the airspeed. Therefore, we can assume that the pressure we sense there is the Static Pressure that exists when the dynamic pressure is equal to the dynamic pressure of the freestream. Thus, when we then use a pressure gauge to measure the difference in the pressures at the two ports, and then use the equation above to calculate V, we can assume the V we calculate is the airspeed.
You can put pressure measuring ports all over an airplane or airplane wing, as is done for wind tunnel models. It doesn’t matter which direction the air is coming at the port, whether perpendicular, parallel, or at any other angle whatsoever, nor does it matter what speed the air is going. The pressure ports only ever measure static pressure, which of course varies from point to point, because static pressure is the only kind of pressure there really is.