# What does centre of lift depend on?

I've read in many places that centre of lift is about quarter chord of the wing and that post-stall lift (the part developed on lower surface) has centre midchord. The later makes sense; the pressure is distributed more or less uniformly (or at least symmetrically) on the lower surface. But what determines the centre of pressure on the upper surface?

The streamlines normally look like this:

and the pressure field like this:

But I've never seen explanation why the pressure should be lowest in the forward part (though it's the trailing edge that actually drives the circulation). Or can it only be explained by numeric calculation of the pressure and velocity field?

Also is it possible to describe (at least approximately) how this depends on the shape of the airfoil (like just flat plate or supercritical airfoil with the thickest point further aft) or is it again only possible by numeric calculation?

• I wonder if the image shared in the recent question physics.stackexchange.com/q/129502/26969 would be helpful. It shows the large forward component of force on the upper surface. Acceleration of air over the airfoil? – Floris Aug 5 '14 at 8:13
• @Floris: The image shows that the pressure field is indeed such that the centre of lift is in forward quarter of length. It does not explain why it is the case. – Jan Hudec Aug 5 '14 at 11:38
• It's a good question. I like this source. The program he uses to calculate flow about an airfoil should give the answer. – Mike Dunlavey Aug 5 '14 at 12:00
• @MikeDunlavey: I've read that source some time already. It shows the pressure field and that, but then it just glosses over that it depends on the angle of attack and shape and does not really explain it much. – Jan Hudec Aug 5 '14 at 12:26
• @Jan: That part doesn't bother me - off the cuff, it's where there's the greatest curvature of flow. Anyway, as I said, I think it's a good question. – Mike Dunlavey Aug 5 '14 at 13:08

In the inviscid approximation, assuming potential flow, the surface pressure distribution on the airfoil ultimately follows from the solution of an integral equation. Without going into the details this means that the value of the pressure coefficient at a point on the surface of the airfoil will depend on the airfoil shape everywhere along the airfoil chord. A corollary remark is that intuitive notions of pressure being affected by the local curvature of the airfoil, while fundamentally still valid, of course, will not be sufficient to understand the pressure distribution overall. I will note that the typical, classical pressure distribution on an airfoil, like this:

certainly does not hold for all airfoils, and most certainly not for modern high-performance airfoils. An example of an airfoil shape and pressure distribution for a so-called laminar airfoil is this:

You can see that in this case we have a completely different pressure distribution, with a center of pressure that is roughly at mid-chord, not at $c/4$.

Finally, we find that, to a good approximation at least for practical airfoils, we can use thin-airfoil theory which tells us that the location of the center of pressure is mostly determined by the shape of the airfoil's camber line (roughly the line in the middle between the upper and lower surface), and only mildly affected by its thickness distribution. See the Wikipedia article on airfoils for some more details.

I think it's a good question.

Without delving into numerical calculation, I assume the lowest pressure above the wing is in the area where the curvature of flow is greatest.

As far as the upwash, I'm not sure, but the leading air may be pulled upward by the reduced pressure above.

ADDED: This is essentially what @Floris meant in his comment, referring to acceleration.

• This was basically what I meant when I said "greatest acceleration" in my first comment - the air has to change direction and a force is needed to allow that to happen. The force is provided by the air pressure further out so it gets "used up" and the top of the air foil is "protected" from the pressure further out. – Floris Aug 5 '14 at 15:29