# Yet another airplane question

The point of discussion here is the pressure distribution across an airfoil.

In order to simplify the question, I'd like to consider an airfoil which looks like a triangle wedge with the blunt face of the wedge facing into the wind (i.e. the wing is traveling westward, to the left) <- direction of travel

This would be an incredibly draggy airfoil, but should create lift even with zero angle of attack due to the low-pressure region above the wing (note there is not a significant high-pressure region region below the wing).

There are a couple different ways, I understand, to think about why this would create lift. One way is the following: we can imagine as the airfoil moves leftward into the wind, that the region behind the airfoil is displaced. That is, like a piston sucking air, as the airfoil moves through the wind, it displaces air behind it and thus creates a suction force. The air around the airfoil then gets pulled into the displaced region by the airfoil as it moves. This low-pressure region of displaced air, because of the shape of the airfoil, is above the hypotenuse of the triangle. Therefore, the airfoil experiences lift.

However, there are two problems here that I'm having a hard time reconciling.

1. If we look at a standard pressure distribution of a NACA airfoil, we find that the lowest region of pressure (i.e. the region of highest lift) is towards the tip of the airfoil, rather than in the area the air is displaced by the moving airfoil. Wouldn't the lowest pressure region be precisely the region of air that is displaced since that displacement is what causes the air to be sucked in (and thus also lower pressure)?

1. If we imagine the scenario of this inefficient wedge-shaped airfoil in a wind tunnel, we see that the displaced-air way of thinking about the lift doesn't seem to work. Reason being, in a wind tunnel, the airfoil itself is stationary and thus is not actually displacing any air as it moves. So in this situation we can't use the same logic for why the air seems to be "sucked" into the lower pressure region of displaced air. Yet windtunnnels are often used to measure the pressure distribution of an airfoil. I suppose the question here is - is there any difference between an airfoil's pressure distribution measured by moving air past the airfoil (i.e. in a windtunnel) vs moving the airfoil through air (i.e. in actual flight).
• We can imagine even car or house flying in the air stream with sufficient velocity. But have you any data concerning drag and lift force acting on this airfoil? Mar 14 '21 at 13:07

That is, like a piston sucking air, as the airfoil moves through the wind, it displaces air behind it and thus creates a suction force. The air around the airfoil then gets pulled into the displaced region by the airfoil as it moves.

Molecules and atoms do not "hold onto" or "grab" each other when in stable states, so there is no such thing as "sucking" or "suction force." These are really just pressure gradients acting on some part of the system.

1. If we look at a standard pressure distribution of a NACA airfoil, we find that the lowest region of pressure (i.e. the region of highest lift) is towards the tip of the airfoil, rather than in the area the air is displaced by the moving airfoil. Wouldn't the lowest pressure region be precisely the region of air that is displaced since that displacement is what causes the air to be sucked in (and thus also lower pressure)?

First, the reason planes fly is clearly stated in many of the answers to the following question What really allows airplanes to fly?.

Second, the exact location of the lowest or highest region of pressure is not really critical for flight. Well, I suppose the airfoil is likely intentionally designed to have the region with the largest pressure gradient being the most structurally sound. However, my point is that the airfoil could literally be a sheet of plywood, as stated in the following answer. The angle of attack is the critical design feature while the shape of the airfoil is often designed for specific purposes (e.g., reduce drag or provide lift regardless of fuselage orientation relative to the horizon, i.e., both rightside up and upside down).

If we imagine the scenario of this inefficient wedge-shaped airfoil in a wind tunnel, we see that the displaced-air way of thinking about the lift doesn't seem to work.

Why not? The wedge you drew would be a terrible airfoil, not because it is incapable of lift if given a proper angle of attack but because the drag it would cause would be enormous. As long as there is an angle of attack and you push hard enough, there should be a finite amount of lift. Whether the lift is sufficient to keep the object from falling is another question.

Reason being, in a wind tunnel, the airfoil itself is stationary and thus is not actually displacing any air as it moves.

No, there is no difference between an airfoil in a wind tunnel and one on a plane in flight (from the airfoil's perspective). That is, the wind tunnel is just like imagining you are in the plane's rest frame so that the air moves over/under the wing.

So in this situation we can't use the same logic for why the air seems to be "sucked" into the lower pressure region of displaced air. Yet windtunnnels are often used to measure the pressure distribution of an airfoil.

Again, air does not get "sucked" into different regions, but the molecules can be pushed by other molecules. The macroscopic view of this would be a pressure gradient.

Regardless, the ability to measure pressure does not depend on whether the airfoil is stationary in a lab with air moving past it or if it is on an airplane moving relative to the ground. In both scenarios, there is an airspeed – the speed of the airfoil relative to the stationary air – and a ground speed – speed of airfoil relative to stationary ground.

I suppose the question here is - is there any difference between an airfoil's pressure distribution measured by moving air past the airfoil (i.e. in a windtunnel) vs moving the airfoil through air (i.e. in actual flight).

Yes, but it's for subtle and often negligible reasons. That is, the air moving in a wind tunnel is often intentionally laminar and almost always parallel to what would be the plane fuselage. When airplanes actually fly, they are often moving through mildly turbulent air that has a bulk flow speed (wind) relative to the ground and that is not aligned with the fuselage.

However, for practical purposes there is no significant difference.

Update in response to edited post

In the new post with the sideways wedge-like airfoil (triangular cross-section), the airfoil will generate lift due to viscosity and geometry inducing circulation according to the Kutta–Joukowski theorem. However, it's important to note that while there may be some lift that does not equate to a net upwards force in the limit of a zero angle of attack. That is, while there may be some lift, it may not overcome gravity and the object attempting to fly using such an airfoil may fall.

As I said before, the answers linked to above go into much more detail on this topic, specifically this excellent answer, which explains the Kutta–Joukowski theorem in gory detail.

• Thanks for the answer Vivere. This does make sense in describing the lift generated by the high-pressure region beneath a wing caused by a non-zero angle of attack. But I'm looking to understand the pressure distribution and cause of lift generated for zero-camber wings with zero-angle of attack, such as the example drawn in the question. Mar 5 '21 at 22:00
• I think you may be misunderstanding some things. If you wish to think in terms of pressure, that is fine, but make sure you recognize what actually causes it. That is, the faster moving air over the top of the wing is not at lower pressure because it moves faster. In fact, there is no reason for two stream lines that separated at the leading edge to meet up at the trailing edge. A flat piece of plywood will only generate lift with a finite angle of attack. In your bad triangle example, rotate it clockwise slightly and you will have lift. Mar 5 '21 at 22:05
• You may still get lift due to circulation according to the Kutta–Joukowski theorem without an angle of attack due to asymmetric flow (i.e., viscosity and geometry induce circulation). But as I said in my answer (and others in the links I provided), that does not mean the object will fly. In your example, the drag would completely dominate and the lift would be negligible. Mar 5 '21 at 22:08
• Hi Vivere - we may be talking past each other, but let me give this a go. 1. I agree - a flat piece of plywood will only generate lift with a finite angle of attack. 2. A piece of plwyood, flat on the bottom, but shaved on the top so as to come to a point on the trailing edge, will also generate lift with a zero angle of attack 3. My question is to understand a) the shape of the low-pressure distribution at the top of the asymmetrical wedge-shaped airfoil given. b) whether this low-pressure region would show up at all if the same airfoil were measured in a wind tunnel. Mar 5 '21 at 22:16
• Yes, but the shaved plywood's lift results from what I said, viscosity and geometry induce circulation from the Kutta-Joukowski theorem. That circulation can generate some lift, yes. But again, your reference to the pressure distribution makes me worry you are inappropriately applying an idea to this scenario. If you are thinking of the full pressure tensor and not some scalar value at different locations on the wing, then I might not be so trepidatious. Mar 5 '21 at 22:51

I have tested with FEM model incompressible flow around this airfoil for Reynolds numbers range from 250 up to 1000. For every test the lift force is a positive one. Hence, there is effect of lift in the case of viscous incompressible flow. Lift force decreases with Reynolds number increasing. The velocity streamlines (left) and pressure distribution (right) for $$Re=250, 1000$$ are shown in Figure 1 - upper line and second line consequently. For higher Reynolds number (about 1000) the stationary flow around this airfoil is unstable one (in numerical model).

The reason for the lift force is the structure of the flow as I discussed this before in my answer here.

Just imagine a fan blade in this wedged shape, assembled in a way that the blunt side cut the air. It is clear that the (weak) wind generated by this fan goes from straight to wedged side (bottom to top of your drawing). I see no reason to be the other way.

If the air flows in that direction, the reaction force on the blades is in the opposite direction. In the case of a plane, it is a downward force. It doesn't help to lift it.

• This is a good way of thinking about it, but I disagree that the wind fan would push air from bottom-to-top of the fan blade (when circulating such that the blunt side is the leading edge). Clearly when the wedged side is the leading edge, the fan would push air from bottom to top (bottom being the straight-horizontal side of the blade). But when running the fan in the opposite direction, with the wedged side as the trailing edge, I would expect the fan to push air from top to bottom due to the small vacuum in the displaced region of the fan as the blades of the fan move through the air. Mar 6 '21 at 17:07
• @mcFreid By the same logic, an L-shaped wing should have a lift. That doesn't seem plausible, tbh. Mar 12 '21 at 23:33

First of all, the figure showing the pressure distribution across the airfoil is not a measurement but a theoretical calculation based on questionable concepts like Bernoulli's law ( https://www.mh-aerotools.de/airfoils/velocitydistributions.htm ). Secondly, the pressure (momentum transfer) on the wing surface only depends on the density and vertical speed component of molecules relative to the surface. So it should not matter for the pressure distribution whether the airfoil moves relative to the air or the air relative to the airfoil.

With your own schematic model, one can actually get quite a good estimate of the aerodynamic lift of e.g. a Boeing 747:

consider first a plate of a size $$1 m^2$$ moving head-on with a velocity of 250 m/s in air; air has a density of $$3*10^{25}$$ molecules/$$m^3$$, so in 1 sec the plate will be hit by $$3*10^{25}*250 = 7.5*10^{27}$$ molecules. If you assume that each molecule has a weight of $$4.5*10^{-26} kg$$, this means that the force on the plate is $$7.5*10^{27} *4.5*10^{-26} *250$$ = $$8.4*10^4 N = 20,000 lb$$ (rounded). Of course, the wing surface is not directly facing into the airstream but only at a very shallow angle. Let's assume that this angle (the average slope of the upper wing surface) is about 5 deg; this means that the force calculated above has to be multiplied by a factor $$sin(5)*cos(5)$$ to obtain the lift and by a factor $$sin^2(5)$$ to obtain the drag force, which results in about $$1,700 lb$$ and $$150 lb$$ respectively. Now this would be for a wing surface of $$1m^2$$; however the total wing area of the Boeing 747 is $$541 m^2$$ ( http://www.airliners.net/info/stats.main?id=100 ), so the forces become about $$920,000 lb$$ for the lift and $$81,000 lb$$ for the drag. The former corresponds practically exactly to the weight, and the latter is about equivalent to the thrust of one of the engines (so one engine compensates for about the drag by the wings, the other 3 apparently have to overcome the drag by the airplane's fuselage).

This calculation is based on the assumption of an inviscid gas, so effects due to the viscosity of the air (i.e. collisions between air molecules) are neglected. The airflow around the wing (which the figure you reproduced above are based on) is indeed only a secondary effect caused by the viscosity of the air. It does not cause the aerodynamic lift, which, as demonstrated by my calculation would exist for an inviscid gas as well. I have explained this in more detail on my page https://www.physicsmyths.org.uk/bernoulli.htm

• You can do the computational fluid dynamics simulation for free on simscale.com I use this sometimes but you need to provide your own CAD drawing
– user291777
Mar 13 '21 at 17:42