Downwash wake
Photographer Paul Bowen, courtesy of Cessna Aircraft, Co.

I remember being on a neighbour’s boat when i was nine or ten and the owner of the boat showed me a boomerang he had sculpted. I had never seen one of these things so he satisfied my curiosity by explaining what it was for and how it worked. He then went on to explain how lift is produced by an airfoil; the air gets split into two parts, the part that goes over the top of the wing and the part that passes below. Since the top is curved (and therefore longer than the bottom) and the two flows need to meet back up again at the same time at the back the top flow needs to go faster. Further, since the faster the stream the lower the pressure, the pressure over the top of the wing decreases, thus “sucking” the top of the foil up.

Unfortunately, this often repeated explanation for lift is rather misleading, so i’ll set out here to give a more thorough explanation.

Basic aerodynamics 101;

Lift is defined as being the total force experienced by an object in a fluid flow orthogonal to the far field flow direction. In other words, the total force at ninety degrees to the relative motion between the undisturbed fluid and the object.

In order to have a force one must have a counter force (Newton’s third law; action and reaction) so the air that is pushing the foil up, means that the foil is also pushing the air down.
Furthermore, we have by Newton’s second law,

$F = \dfrac{d(mv)}{dt}$
Force is equal to the time rate of change of momentum.
This means that as the foil moves through the fluid it is constantly deflecting the flow downwards, imparting new momentum to it continually and it is this that creates the lift force.
So whatever happened to the longer path and the pressure drop?

Upper and lower flow past lifting foil;
the fluid particles never meet up again

Well, this is also happening, however it is entirely false to assume that the fluid particles have to meet up again behind the wing. They never do, unless there is no lift and no viscocity and no turbulence. In fact the flow over the top of the wing is much faster than necessary to make up the difference in upper and lower path length.
This actually makes sense when you think that the passage of the wing has deflected air downwards behind it, thus pulling back over the top of the wing the required extra air and similarly, slowing down the air that is passing underneath.

Now we also have Bernoulli’s equation for incompressible flow;

$P + \frac{1}{2}\rho{U^2} = const$

where $\rho$ is fluid density, U is flow velocity and P is pressure.

This is a conservation of energy equation since pressure is a form of potential energy and the second term is the so called “dynamic pressure”, which is the kinetic energy. Therefore, it is indeed true that there is a lowering of pressure over the top of the wing, although there is also a raise in pressure over most of the underside as well.

Typical pressure distribution upper and lower surface

In the above image the fractional distance along the chord (line connecting leading and trailing edges) is on the x axis and the y axis is negative coefficient of pressure. I generated a Naca 64209 foil and ran Javafoil with the foil at ten degrees to obtain this graph.

Of course the total force experienced by any object in a fluid flow is the integral of the pressure (normal) and shear (tangential) forces upon the wetted skin.

——–*———

How it all works in practice.

We need to introduce the concept of circulation. Circulation is just what it sounds like; the air circulates around, in this case the foil. It is a key concept to understand. Imagine a vortex spinning around the foil such that the axis of rotation of the vortex runs from wingtip to wingtip. Now imagine adding, superimposing, the straight line flow to this vortex; what happens is that the flow is accelerated on one side, decellerated on the other, and deflected one way ahead of the foil, the other way behind.

Mathematically this is;

$\Gamma = -\oint_C V\cdot{ds}$

Which is the closed line integral (enclosing the foil), V is the velocity of the fluid element and ds is the differential length along the loop.

from Marchaj “Aero/HydroDynamics of Sailing”

One may object that this seems to imply that the upwash (air being sucked upwards towards the low pressure zone above the wing) cancels out the downwash behind the wing thus leaving no overall momentum change to the fluid stream, but in fact the downwash is always twice as great as the upwash.

To understand why this is so, one must go back to the beginning of the path of motion of the foil.

The following images show the initial conditions when the wing commences its path;

from Marchaj “Aero/HydroDynamics of Sailing”

What happens is that as the foil begins to move forward at an angle of attack, the air tries to make up the short path difference by curling around the trailing edge, initiating a vortex which quickly gets left behind but which is the exact reciprocal of the wing’s bounding vortex and which forces the establishment of the bounding vortex itself. Only like so can the overall circulation in the total airmass remain zero, as it must be according to the law of conservation of angular momentum.

Therefore, the downwash is actually due to the sum of the two vortices, whereas the upwash ahead of the wing is proper only to the wing’s bounding vortex. And when tacking a sailboat, the circulation decays to zero, only to re-establish itself in the opposite direction by shedding a new starting vortex behind.

——*——

So, to recap;

The lift created by a foil or other lifting body always involves imparting by means of an effective angle of attack of the lifting body, new momentum to the fluid flow, in  a direction orthogonal to the stream and opposite to the direction of the lift force.

The mechanism for achieving this invariably involves circulation.

This circulation imposes a pressure distribution on the foil; low pressure above and high pressure below.

—-*—-

Some astute readers may object to me using an equation for incompressible flows (Bernoulli’s eqn) when air is in fact compressible. This is a very good point, but the reason that it is valid is because for small mach numbers (the ratio of flow speed to the speed of sound) , less than ~0.3, the amount of compression or expansion air undergoes is so small that it can be taken out of the equations with negligible loss of accuracy, greatly simplifying the mathematics.