-4
$\begingroup$

I recently learned that the pressure on the bottom of the wing is increased, and air moves faster on the bottom part of the wing. The speed of the air on top of the wing is not increased. So I am confused about what is the point of the slope on top of the airfoil. The Wright brothers' airfoil was flat, but still had the slight slope to it, and now modern aircraft have a huge slope to them. So what is the reason for the slope?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Hi Ethan. You should know by now that this is not how wings work. You have taken part in enough Q&As on here to know that, for example, this one. $\endgroup$
    – Simon
    Mar 3, 2016 at 7:06
  • $\begingroup$ To understand how a wing maintains the aircraft airborne, in addition of the multiple answers you have got on this site, you may read the full explanation here. In particular you should first understand this image: The same wing, on the top it doesn't produce lift, on the bottom it produces lift because air is deflected downward by the wing. Read it until you understand most of it, as other details are then explained from the assumption you understand this important basis. $\endgroup$
    – mins
    Mar 3, 2016 at 9:11

3 Answers 3

Reset to default
-1
$\begingroup$

First of all, wherever you learned that,

"...the pressure on the bottom of the wing is increased, and air moves faster on the bottom part of the wing. The speed of the air on top of the wing is not increased."
It is wrong, and citations would be useful, if only to educate the owner of that website.

A wing provides lift by imparting momentum to the air through which it passes equal to the airplane's weight. If the wing is cambered, it imparts more acceleration to the air through which it passes. (i.e. the air coming off the trailing edge has a faster downward velocity than compared to a non-cambered airfoil of equal area), thus making the wing more efficient for its area. It also affects things like stall speed.

Improve this answer
$\endgroup$
1
  • $\begingroup$ " imparting momentum to the air through which it passes equal to the airplane's weight" That's not possible. The units of momentum are mass-times-velocity and the units of weight are mass-times-acceleration. $\endgroup$
    – David Richerby
    Mar 3, 2016 at 9:11
-2
$\begingroup$

The slope (known as the camber) is for better stalling characteristics so that the turbulent flow separates from the airfoil slowly instead of suddenly because the curve allows the air more surface to stick on at the AoA increases to a stall.

This stall diagram shows how the cambered wing performs in a stall: enter image description here

Image Source

Also, according to the gas laws, the air in higher pressure moves slower than air in low pressure. (P1*V1 = P2*V2) The air on the upper surface does move faster at higher AoAs due to the lower pressure created by the AoA.

Improve this answer
$\endgroup$
-2
$\begingroup$

You are wrong about the air under the wing travelling faster. It is the air travelling over the top of the wing that is faster.

The "slope" is actually called Camber. Camber makes the air flowing over the wing change direction and this causes low pressure and that results in lift.

enter image description here

Improve this answer
$\endgroup$
6
  • 2
    $\begingroup$ There is no reason why the air flowing over the top of the wing has to "join up" with the air that passed under it. In fact, the airflow over the top gets to the trailing edge before the air going under the wing. Please see av8n.com/how/htm/airfoils.html#sec-airplane-air. $\endgroup$
    – BillDOe
    Mar 3, 2016 at 6:14
  • 3
    $\begingroup$ The theory of equal transit time has been busted for quite some time. en.wikipedia.org/wiki/… $\endgroup$
    – Simon
    Mar 3, 2016 at 7:15
  • 1
    $\begingroup$ @Simon True but it is still widely taught and even appears in some countries syllabi IIRC. It's a very common misconception that sparked the question in the first place and an answer like this one would be good if it explained such. $\endgroup$
    – Ben
    Mar 3, 2016 at 7:38
  • 1
    $\begingroup$ Nasa says: This is an incorrect theory. $\endgroup$
    – mins
    Mar 3, 2016 at 9:05
  • 1
    $\begingroup$ Also, the aerofoil in the diagram doesn't actually seem to be generating any lift since there's no reaction on the air mass. $\endgroup$
    – David Richerby
    Mar 3, 2016 at 9:12

Not the answer you're looking for? Browse other questions tagged .