# Downwash created by horseshoe vortex at zero travelling speed?

This is a simple aerodynamics question about the horse shoe vortex. As it seems, the trailing edges on the side cause the downwash on the wing of the plane. However, since the circulation is constant, the downwash is constant as well. This intuitively doesnt make sense to me, since if the airplane is not moving, there would still be downwash. How? Or am I terribly misunderstanding something?!

the horseshow vortex model

You are misunderstanding a number of things. If the airplane is not moving, then there is no horseshoe vortex. If the wing is moving (or we have a wing generating lift in uniform flow), then the downwash is not constant. As a matter of fact, for the simple Prandtl lifting line model the downwash will become singular at the wingtips, which makes the model unphysical, see the Wikipedia article on the subject. To obtain a model that can at least approximate the flow in a physically meaningful way, you need to use lifting-line theory which assumes an infinite superposition of infinitesimal horseshoe vortices. The resulting circulation distribution will depend on the geometry of the wing, and can never be constant for a finite wing.

The magnitude of the circulation $\Gamma$ at any spanwise position along the wing follows from a kinematic condition (the Kutta condition), and will be proportional to the free-stream velocity.

• I agree with what you say, but the superposition of these horse shoe vortices are constant by themselves. Meaning that there is downwash even without travelling speed, no? – user3604362 Dec 28 '16 at 19:36
• It even says it in the wikipedia article, that downwash is only a function of circulation. – user3604362 Dec 28 '16 at 19:38
• No. What do you mean by "constant by themselves"? Constant with respect to what? – Pirx Dec 28 '16 at 19:38
• The circulation around a wing is a function of the speed of the uniform flow past it. At constant angle of attack, circulation is proportional to the flow speed, approximately. – Pirx Dec 28 '16 at 19:40
• If you have L= rhoVGamma, then Lift can be determined only if we have set Gamma, the circulation. – user3604362 Dec 28 '16 at 20:35

In the early years of flight, electricity was new and exciting, and it just happened that the equations which could calculate the strength of an electromagnetic field worked equally well when calculating the local flow change effected by a wing. What is the electrical current in a wire became the vorticity in a vortex, and the strength and orientation of the induced magnetic field were equivalent to the induced flow changes. So the vocabulary of electricity was copied over to aerodynamics, just like brain research used vocabulary from computer science when that was a hot topic.

Now we are left with abstract concepts like circulation or induced drag. It would be so much more descriptive to use proper concepts or names, but the authors of technical books learned it that way and are much too lazy to explain aerodynamics any better.

And now let me give you my advice: If you do not want to operate or to write a potential flow code, do yourself a favor and forget all that. It is much better to interpret lift as the consequence of a pressure field around a wing which accelerates the air flowing around this wing downwards. When a wing approaches at subsonic speed, the low pressure area over its upper surface will suck in air ahead of it. The air will rise and accelerate towards the wing and be sucked into that low pressure area. Once there, it will "see" that the wing below it curves away from its path of travel, and if that path would remain unchanged, a vacuum between the wing and the air would form. Reluctantly, the air will change course and follow the wing's contour. This requires even lower pressure to make the molecules change their direction. This fast-flowing, low-pressure air will in turn suck in new air ahead and below of it, will go on to decelerate and regain its old pressure over the rear half of the wing, and will flow off with its new flow direction.

Note that lift can only happen if the upper contour of the wing will slope downwards and away from the initial path of the air flowing around the wing's leading edge. This could either be camber or angle of attack - both will have the same effect. Since camber allows for a gradual change of the contour, it is more efficient than angle of attack.

If there is no movement of air (either caused by the wing's movement or by the rotation of a Flettner rotor), no suction field will develop and no downward acceleration of air over a cambered or inclined wing. Lift will be zero, as will be downwash and, yes, circulation.

And if you insist on an explanation within the theory of potential flow: When the wing decelerates at a constant angle of attack, the vortex strength diminishes. When the wing has come to a complete stop, the bound vortex will have disappeared, too.

• @Pirx "a number of errors"? Name one! And you imply that potential theory is the One Truth, all other explanations are just "interpretations". If you read the question carefully it becomes clear that the doubts are coming from the very unintuitive nature of potential theory, and instead of explaining that further, it helps more to look at what really happens. And please explain how "the vortex will have disappeared" does not address the question. – Peter Kämpf Dec 30 '16 at 23:53

In the comments above you seem to have answered you own question without knowing it. Take for example, the Kutta-Joukowski theorem, which describes the lift around a two-dimensional object. The lift per unit span is given as, $$L = -\rho_{\infty} V_{\infty} \Gamma$$ where $\rho_{\infty}$ is the freesteam density, $V_{\infty}$ is the freestream velocity, and $\Gamma$ is the circulation around the object. The first thing you missed was that if there is no freestream velocity (i.e. $V_{\infty} = 0$), then there is no lift. That part is obvious, but the lift is zero regardless of the value of circulation. However, the definition of circulation is given by the line integral, $$\Gamma = \oint_C V \cdot ds$$ where $C$ is a closed contour around the two-dimensional object, $V$ is the velocity along the contour, and $ds$ is an infinitesimal distance along the contour. Obviously, in the case of an airfoil where there is no freestream velocity, then $\Gamma = 0$. Hence, in the event $V_{\infty} = 0$, we obtain zero lift from the first expression, but also zero circulation from the second expression.

• Careful here: $\Gamma=0$ does not immediately follow from the definition of circulation and the fact that $V_\infty$ vanishes. $\Gamma=0$ for $V_\infty=0$ is a consequence of the Kutta condition for airfoils. If you consider a circular cylinder, say, instead, you can have $\Gamma\ne0$ for $V_\infty=0$ – Pirx Dec 28 '16 at 23:13
• You are correct regarding the Kutta condition at the trailing edge of airfoils. I guess technically speaking, it is true that even with $V_{\infty} = 0$, there are cases where $\Gamma \neq 0$. A common flow is the rotating cylinder or circle in a stationary fluid. – TRF Dec 29 '16 at 0:14