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In many textbooks, such as Anderson (Fundamentals of Aerodynamics), Bertin (Aerodynamics for Engineers), and Houghton (Aerodynamics for Engineering Students), the authors present the idea that there is a starting vortex which is formed. This starting vortex and the bound vortex that form (due to Kelvin's Theorem) co-exist with lift (as Mclean puts it in Understanding Aerodynamics). However, I am unsure on why the starting vortex forms on the upper surface. Why does it not form on the lower surface and have clockwise rotation and lead to counter-clockwise bounded circulation at the airfoil?

While I know that this is not experimentally observed nor does it make intuitive sense, I would like to know why the starting vortex does not form on the bottom.

To shed more light, here are some instances of where authors state that the starting vortex starts on the upper surface, but never explain why it doesn't start on the lower one: (taken from Bertin): "At the instant of starting, the flow is a potential flow without circulation, and the streamlines are as shown in Figs. 6.3 a, with a stagnation point occurring on the rear upper surface"

enter image description here

(taken from Understanding Aerodynamics) "It can be shown, based on starting the flow from rest, that in the absence of viscosity, the nonlifting flow pattern of Figure 7.1.3a is the one that would occur... "

enter image description here

So how can it be shown (mathematically) or using strong rigorous theoretical arguments that the stagnation point cannot be at the lower surface before circulation?

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The illustrations under the time-invariant, inviscid, irrotational and incompressible (potential) description is a bit misleading, in my opinion. In the real world, assuming the airfoil starts from rest (no flow field) and you start accelerating it to some airspeed, viscosity should start generating vorticity as soon as airspeed becomes non-zero. Therefore, the non-lifting flow field would never exist in the first place.

The non-lifting flow field serves better as an illustration of the flow solution using uniform + doublet flows. In this case, there is a unique solution for a uniform flow past a cylinder with radius $R$:

  • Radial speed: $V_r=V_\infty cos\theta(1-\frac{R^2}{r^2})$
  • Tangential speed: $V_\theta=-V_\infty sin\theta(1+\frac{R^2}{r^2})$

Since we can conformally transform most airfoil geometries into a cylinder, flow solutions past a cylinder is a generalization to those past airfoils. The above result assumes zero angle of attack. But since cylinder is symmetrical, a non-zero AOA simply rotates the result by that AOA, and that's why you have the rear stagnation point on the upper airfoil surface.

For a lifting flow, we would need to model with uniform + doublet + vortex flows. In this case, a set of solutions can be obtained for a cylinder:

  • Radial speed: $V_r=V_\infty cos\theta(1-\frac{R^2}{r^2})$
  • Tangential speed: $V_\theta=-V_\infty sin\theta(1+\frac{R^2}{r^2})-\frac{\Gamma}{2\pi r}$

Notice that there is, at this point, no imposition on what the vortex strength ($\Gamma$) has to be. You can have an infinite number of lifting solutions with an arbitrary value of vortex strength (at the same AOA). I think that's what (b) and (c) of your second set of figures are trying to illustrate. We get around this by specifying the Kutta-Condition and impose the stagnation point at the trailing edge to get a unique solution.

Since the total flow field must have zero circulation, the starting vortex must have a counter-clockwise circulation.

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  • $\begingroup$ +1 and it aligns with the observed flowfield for an impulsively started airfoil $\endgroup$ – AEhere supports Monica Oct 18 at 23:27
  • $\begingroup$ @AEhere that is an amazing photo. I had imagined the downward deflection of the airflow wing bottom (when it was pitched up) contributes to lower pressure behind the wing, but had no idea it would form a second vortex, helping draw air over the top of the wing. Do you think the bottom vortex contributes to an explanation of ground effect? $\endgroup$ – Robert DiGiovanni Oct 20 at 19:43
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    $\begingroup$ @RobertDiGiovanni the image I linked is of a very contrived situation showcasing a transient aerodynamic effect. It has little to nothing to do with ground effect, and everything to do with circulation. $\endgroup$ – AEhere supports Monica Oct 21 at 14:08
  • $\begingroup$ @Jimmy You state in your reply that "...Therefore, the non-lifting flow field would never exist in the first place." But, if there is always a "lifting" flow field, how will the starting vortex ever have been formed? And moreover, under what mechanism does the starting vortex ever form? $\endgroup$ – Nick Hill Oct 22 at 15:56
  • $\begingroup$ @NickHill The circulation is started by viscosity (i.e. the airfoil boundary wall), and in large part due to the no-slip condition at the wall. You can take a look at this answer and see if it helps your understanding: physics.stackexchange.com/questions/135707/… $\endgroup$ – JZYL Oct 22 at 19:39
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Nick presents a very interesting way of looking at changes in airfoil circulation, not by changing AOA, but by changing speed. In a review of polar diagrams, it is becoming apparent that lift creation from circulation (the "Bernoulli bump" in the lift vs airfoil AOA curve), where much lift can be had with very little increase in drag, requires a Reynolds number of sufficient value. For the same chord (of the same airfoil), this means sufficient speed to bring the stagnation point off the top of the wing to behind the wing in order to accelerate air over the top of the wing.

This is difficult to present in diagrams without picturing the motion. As a wing, pitched to a positive AOA, speeds up, what happens is analogous to watching the transom of a boat as it accelerates to the point where the boiling wake separates from the boat and forms the "vee" behind the boat .

With a wing, once the "wake" is behind the wing, air is pulled across the top of the wing (and from all directions) to fill in the lower pressure area. But as the wing is continuously moving forward, this is a constantly repeating process as long as AOA and velocity are maintained.

Slow the boat down, and the boiling wake returns to the transom. It is in this light that both sets of diagrams appear to be non-contradictory.

Since the wing is pitched up, we would expect the stagnation point to form on the upper surface first.

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  • $\begingroup$ @Nick Hill it might be easiest to model a fully symmetrical airfoil. I suspect the stagnation point would be always behind the TE unless it was pitched. Then the progression of it's movement with increasing speed could be observed (in a wind tunnel) $\endgroup$ – Robert DiGiovanni Oct 20 at 4:49
  • $\begingroup$ The rear stagnation point of a fully attached airfoil is always at the trailing edge. $\endgroup$ – JZYL Oct 20 at 14:11

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