Why is lift perpendicular to the wing?

It is said that lift on a wing always acts perpendicular to the relative wind.
Why is this so? Is it because we arbitrarily choose to analyze its components with one purposely chosen to be perpendicular or is it a matter of physics that the lift component is naturally perpendicular. If so why is that?

tex

Isn’t this based more on the changing pressure between air flow under the wing vs air flow over the wing and the pressure gradient pushes the wing upward?

https://en.wikipedia.org/wiki/Lift_(force)

thetexan said:

It is said that lift on a wing always acts perpendicular to the relative wind. Why is this so?

It's defined that way.

thetexan said:

If so why is that?

In many applications we want to maximize lift and minimize drag, so that decomposition is practical there.

PS: Your thread title is wrong. Lift is not always perpendicular to the wing.

A.T. said:

PS: Your thread title is wrong. Lift is not always perpendicular to the wing.

I , on the contrary, think this is the way it is defined. More specifically the lift in a wing is the following integral
$$\oint_S p\cdot \vec{n}dS$$ where ##p## the pressure in the surface of the wing and ##\vec{n}## the normal at the surface of wing at that point.

Delta2 said:

I , on the contrary, think this is the way it is defined. More specifically the lift in a wing is the following integral
$$\oint_S p\cdot \vec{n}dS$$ where ##p## the pressure in the surface of the wing and ##\vec{n}## the normal at the surface of wing at that point.

These are the local pressure forces, not lift. Local pressure forces are obviously perpendicular to the local wing surface, per definition of "pressure".

But "lift" is defined as perpendicular to the freestream relative flow, which is different than perpendicular to the local wing surface.

Difference between lift (L) and drag (D) versus normal force (N) and axial force (A)
From: http://www.aerospaceweb.org/question/aerodynamics/q0194.shtml

Also remember that space has three dimensions. Does it mean that lift can have, and normally does have, out-of-plane component when freestream velocity and therefore drag are defined as in-plane forces?

@A.T. if we define $$\vec{F}=\oint_S p\cdot\hat{n}dS$$ can we say that lift is the component of ##\vec{F}## perpendicular to the freestream velocity while drag is the component of ##\vec{F}## in the same direction to the freestream velocity?

Delta2 said:

@A.T. if we define $$\vec{F}=\oint_S p\cdot\hat{n}dS$$ can we say that lift is the component of ##\vec{F}## perpendicular to the freestream velocity while drag is the component of ##\vec{F}## in the same direction to the freestream velocity?

Lift and drag are components of the total aerodynamic force on the object.

What you have written down doesn't include friction parallel to the local surface (which is part of the aerodynamic force) and includes buoyancy (which is not part of the aerodynamic force)

A.T. said:

Lift and drag are components of the total aerodynamic force on the object.

What you have written down doesn't include friction parallel to the local surface (which is part of the aerodynamic force) and includes buoyancy (which is not part of the aerodynamic force)

Yes, i had in mind that i didn't include the tangential friction force(which i considered negligible for some reason, maybe because air doesn't have high viscosity) but i didnt had in mind that F includes the buoyancy force and that buoyancy isn't part of the aerodynamic force...

Thanks ,i learned something new today !

Delta2 said:

Yes, i had in mind that i didn't include the tangential friction force(which i considered negligible for some reason, maybe because air doesn't have high viscosity) but i didnt had in mind that F includes the buoyancy force and that buoyancy isn't part of the aerodynamic force...

Thanks ,i learned something new today !

You're welcome. But in the end it's just a convention. People doing ballooning might say "lift" to buoyancy.

Think about this practically: the purpose of the wing is to keep the airplane from falling out of the sky. In level flight, lift is equal to the weight of the airplane(caveat about the direction of thrust and its lifting effect). If lift wasn't defined perpendicular to the flow direction, then that force wouldn't tell you if the plane can fly!

This does create a an issue on the other side of the coin though: thrust does not directly oppose drag.

It's an arbitrary choice, but as chosen it works better for what is most important. You can, however, also think of it as a single total force:
https://en.m.wikipedia.org/wiki/Aerodynamic_force

To determine the perpendicular respect to a chambered airfoil (irregular shape) is harder than using the direction of the approaching non-disturbed airstream.
Once the wing is close to a specific volume of air, that perpendicularity is lost (think upwash and downwash).

The wing creates a reactive pressure differential and force vectors, which average direction is off the direction of that approaching non-disturbed airstream.
Then we decide splitting that force as convenient for the analysis we need for the most common condition, which is horizontal level flight: how much weight can the airplane carry (hence the convention for "lift") and how much fuel is going to cost (hence the convention for drag).

For flying conditions that deviate from horizontal level, we still use that convention, even though "lift" is not a vertical vector force (respect to the natural horizon) anymore and has to be reduced (via angle of attack) in order to keep a desired more or less vertical trajectory of the airplane.

snorkack said:

Also remember that space has three dimensions. Does it mean that lift can have, and normally does have, out-of-plane component when freestream velocity and therefore drag are defined as in-plane forces?

I don't know what you mean by "defined as in-plane forces".

You project the aerodynamic force vector onto the freestream velocity vector to get the drag vector, and lift is simply the remaining component of aerodynamic force.

I appreciate the replies. I teach basic aerodynamics at college and understand the general principles but have never fully understood this one point. I had several students stop me when I tell them “remember, lift acts perpendicular to the relative wind”. And one will say “ why is that?”
I don’t know why.
My guess is that the only point along the top surface where 100% of the free stream is parallel to the surface at the tangential point.

At all points either in front of or behind that point there will be some additional component normal component that reduces component of parallel flow. Therefore the greatest dynamic pressure occurs at the tangential point...thus the least static pressure.
That’s just my guess.

I need a good explanation I can use in class that doesn’t require me or to explain not them to understand the rigorous Math involved.
tex

thetexan said:

... I need a good explanation I can use in class that doesn’t require me or to explain not them to understand the rigorous Math involved.
tex

This website has excellent explanations that are easy to understand by most students:
https://www.av8n.com/how/

thetexan said:

I need a good explanation I can use in class that doesn’t require me or to explain not them to understand the rigorous Math involved.

It's simply defined this way. To see why it's a useful quantity, you have to go into the applications.

I suspected it might be as simple as an arbitrary definition. But why define it that way? What is the advantage of defining it that way?
For example, it could have just as easily been defined as 10 degrees aft of perpendicular. So why use 90 degrees. Why not 90 degrees to the chord rather than relative wind?

I can see that by using relative wind AOA is more obviously taken into account. If it was based on chord then AOA would not be a part of the calculation.

Any help about this?

tex

thetexan said:

I suspected it might be as simple as an arbitrary definition. But why define it that way? What is the advantage of defining it that way?
For example, it could have just as easily been defined as 10 degrees aft of perpendicular. So why use 90 degrees. Why not 90 degrees to the chord rather than relative wind?

See my post #11 for a logical reason why.

russ_watters said:

See my post #11 for a logical reason why.

Absolutely.
I think this must really be the silly season - or perhaps it's the 'virus'. How can anyone ask why the force that keeps a plane up is defined as acting upwards? I think people must have forgotten how to look at a diagram and think about it before asking questions. Drag, otoh, could be said to act in the reverse direction of the velocity but then you wouldn't have two orthogonal vectors. That would be very inconvenient - but no harder than the rest of aerodynamics. Personally, I wouldn't go for that option.

sophiecentaur said:

Absolutely.
I think this must really be the silly season - or perhaps it's the 'virus'. How can anyone ask why the force that keeps a plane up is defined as acting upwards? I think people must have forgotten how to look at a diagram and think about it before asking questions. Drag, otoh, could be said to act in the reverse direction of the velocity but then you wouldn't have two orthogonal vectors. That would be very inconvenient - but no harder than the rest of aerodynamics. Personally, I wouldn't go for that option.

well, perpendicular to the relative wind is not necessarily upward. I appreciate any help but I don’t appreciate being talked down to. I’ve been a commercial pilot and instructor for 40 years. If I haven’t found a good explanation in that time it’s not because I’m stupid.

tex

thetexan said:

So why use 90 degrees.

To have two linearly independent components. Since drag is 0° the remaining component is 90°.

thetexan said:

Why not 90 degrees to the chord rather than relative wind?

You can do that too, it's called normal and axial force. See diagram in post #5.

thetexan said:

well, perpendicular to the relative wind is not necessarily upward. I appreciate any help but I don’t appreciate being talked down to. I’ve been a commercial pilot and instructor for 40 years. If I haven’t found a good explanation in that time it’s not because I’m stupid.

tex

I didn’t mean to offend but I read a title question that’s basically flawed. A wing has a curved surface so which particular ‘perpendicular’ are you choosing? As the question was so loose, I assumed that the word was “vertical”.
it’s a shame that your training course wasn’t more helpful. Those diagrams can be very basic and imprecise.
But I guess you flew ok and that’s what counts. Knowing the Physics of the things we use everyday often doesn’t affect how well we use them.

The lift component (sum of all lift vectors across the top of the wing as viewed from the end of the wing across the airfoil profile) is said to be 90 degrees relative to the relative wind. And the relative wind can be from any angle creating any AOA possible...and it makes no difference what the orientation is to the ground. Assuming the AOA is less than the critical AOA and regardless of the wing’s orientation to the ground, the lift vector through the center of lift is said to always be 90 degrees to the relative wind, which, by the way, can be from any direction. The question is why is the summed lift vector always perpendicular to the relative wind? What about the relative wind over the wing always cause a vector 90 degrees to the relative wind?

I would like a simple enough answer, if that’s possible, that a 40 year commercial pilot veteran can explain to a class of aerodynamic newbies. If I don’t understand it then there’s no way I can explain it to students.

thanks in advance

tex

thetexan said:

The question is why is the summed lift vector always perpendicular to the relative wind?

Asked and answered: convention. It is helpful to break the aerodynamic force into perpendicular components. So we break the aerodynamic force into perpendicular components. If one is going to do this, it is good to have an axis in mind. The velocity of the free stream is a handy and useful reference.

It is exactly the same reason that we talk about normal force and frictional force: because it is a useful way to break the contact force between two objects into perpendicular components.

Aerodynamics people like to use the wind axis to define their major forces. It certainly works well with wind tunnel data. There are other axis systems that can be used, so there is no mathematical or physics reason for the convention other than convenience.
There is no single coordinate system that is convenient for everyone. Aerodynamics people like the wind axis. Propulsion people like the body axis. Navigation people like the Earth axis. Others like the aircraft station system (fuselage station, water line, butt line). The entire field is full of constant coordinate system conversions.

Thank you.

I wondered if that was it. The convention could have just as easily been some other axis But we choose to use a frame of reference that is based on relative wind.

thank all of you. That helps

tex

A.T. said:

I don't know what you mean by "defined as in-plane forces".

You project the aerodynamic force vector onto the freestream velocity vector to get the drag vector, and lift is simply the remaining component of aerodynamic force.

Well, if you look at the drawing where freestream velocity is along the plane of the drawing, then drag is exactly along the freestream velocity, because drag is the only force that does work.
Lift is simply the remaining component which means that it has to be at 90 degrees to freestream velocity. Anywhere along the plane perpendicular to freestream velocity - including out of drawing plane.

snorkack said:

Well, if you look at the drawing where freestream velocity is along the plane of the drawing, then drag is exactly along the freestream velocity, because drag is the only force that does work.
Lift is simply the remaining component which means that it has to be at 90 degrees to freestream velocity. Anywhere along the plane perpendicular to freestream velocity - including out of drawing plane.

You are free to choose your drawing plane. Why would you choose it to have out-of-plane components?

thetexan said:

I wondered if that was it. The convention could have just as easily been some other axis ...

No, not "just as easily". We pick our conventions to make life (the math) easier. To understand why this quantities are useful for this, you have to look at the derivations of formulas describing the efficiency/performance of: aircraft, gliders, sailcraft, turbines, propellers etc.

thetexan said:

I would like a simple enough answer, if that’s possible, that a 40 year commercial pilot veteran can explain to a class of aerodynamic newbies. If I don’t understand it then there’s no way I can explain it to students.

If you don't want to use formulas, and just give general conceptual explanation, then you should start with explaining drag, as the resistance to movement through a fluid. Then lift pops out automatically, as the sometimes existing other component of the fluid force, that is not opposed to the movement.

thetexan said:

Thank you.

I wondered if that was it. The convention could have just as easily been some other axis But we choose to use a frame of reference that is based on relative wind.
...

You are welcome

Copied from
https://www.av8n.com/how/htm/4forces.html

The relative wind acting on the airplane produces a certain amount of force which is called (unsurprisingly) the total aerodynamic force. This force can be resolved into components, called lift and drag, as shown in figure 4.1.


Figure 4.1: Total Aerodynamic Force = Lift + Drag
Here are the official, conventional definitions of the so-called four forces:

• Drag is a component of the aerodynamic force, namely the projection onto the direction parallel to the relative wind.
• Lift is another component of the aerodynamic force, namely the projection onto the two directions perpendicular to the relative wind.

... It is ironic that according to convention, the total aerodynamic force is not listed among the four forces.

... You may think lift, thrust, weight, and drag are defined in a crazy way, but the definitions aren’t going to change anytime soon. They have too much history behind them, and they actually have advantages when analyzing complex situations.

The good news is that these subtleties usually don’t bother you. First of all, the angles in figure 4.2 are greatly exaggerated. In ordinary transportation (as opposed to aerobatics), even in climbs and descents, the pitch angle is always rather small, so thrust is always nearly horizontal. Also, the relative wind differs from horizontal by only a few degrees, so drag is always nearly horizontal, and lift is nearly vertical except in turns.

If we don’t like the technical definitions of lift, drag, thrust and weight, we are free to use other terms.

• For example, we can make the following sweeping statement: in unaccelerated flight, the upward forces balance the downward forces, and the forward forces balance the rearward forces. This statement is true whether or not we calculate separately the contributions of lift, drag, thrust and weight. This statement is simple because it expresses all the forces in terms of the horizontal and vertical components, without reference to the direction of flight.
• Alternatively, we can choose, if we want, express all the forces relative to the direction of flight, with minimal references to horizontal and vertical. An example of this is shown in figure 4.7. We resolve the weight into two components: The parallel component of weight is parallel to the direction of flight, while the perpendicular component of weight is at right angles to the direction of flight. These two components added together are equal to the total weight. (The total weight is directly downward, but we don’t care about that in figure 4.7, because we are using the components instead.)
  
  
  
  Figure 4.7: Force Resolved into Components
  This description is nice because it shows things from the pilot’s point of view, which is consistent with the general spirit of this book. Note that in figure 4.7 the parallel component of weight is pulling the aircraft forward along the path of flight. Indeed, in this situation, this component of weight is a larger contribution than engine thrust.

snorkack said:

Well, if you look at the drawing where freestream velocity is along the plane of the drawing, then drag is exactly along the freestream velocity, because drag is the only force that does work.

Are you assume straight and level flight? Otherwise, lift is doing work.

Lift is simply the remaining component which means that it has to be at 90 degrees to freestream velocity.

So you are assuming the lift and drag are orthogonal. That is how the convention defines it, but it is just a convention.

thetexan said:

well, perpendicular to the relative wind is not necessarily upward.

No, but straight and level flight is the most basic and common flight situation. Every other situation is an added complexity.

thetexan said:

I suspected it might be as simple as an arbitrary definition. But why define it that way? What is the advantage of defining it that way?
For example, it could have just as easily been defined as 10 degrees aft of perpendicular. So why use 90 degrees. Why not 90 degrees to the chord rather than relative wind?

Well think about how it is actually used. Here's a basic question: how much lift is required for an airplane to fly straight and level? With the current definition, the answer is simply W: the weight of the plane. If you define "lift" to be perpendicular to the chord line, then the answer depends on angle of attack and to calculate it, you still need the vector opposite the weight vector to use in the calculation. Since that's a useful vector it should probably have a name of its own...

Now take, for example, load factor in a level turn. The graph of load factor is showing the additional lift required to keep the plane level at increasing angles of bank. It is based solely on the angle of bank. It doesn't vary with airspeed or angle of attack. But if you define lift to be perpendicular to the chord line, then in addition to the load factor calculation itself, you also have a similar lift vs angle of attack calculation to do.

Defining the lift vector a different way makes for more work in many common situations. Can you think of a common aviation situation where a different definition such as perpendicular to the chord line can be used "just as easily" (or easier) than the current one?

sophiecentaur said:

Knowing the Physics of the things we use everyday often doesn’t affect how well we use them.

Yeah, I have to keep telling me that, every time I see alleged long time pilots arguing for things like the downwind turn myth (basically against Galilean invariance).

A.T. said:

Yeah, I have to keep telling me that, every time I see alleged long time pilots arguing for things like the downwind turn myth (basically against Galilean invariance).

[googles] wow, never heard of that one. That could actually be dangerous if extrapolated to the other legs.

I'm learning to fly right now and my first CFI (mid-20s, recent grad of a major college flight program) invoked the "equal transit time" myth when explaining how airplanes fly. That was disappointing. He's an airline pilot now. Great CFI.

snorkack said:

Well, if you look at the drawing where freestream velocity is along the plane of the drawing, then drag is exactly along the freestream velocity, because drag is the only force that does work. Lift is simply the remaining component which means that it has to be at 90 degrees to freestream velocity. Anywhere along the plane perpendicular to freestream velocity - including out of drawing plane.

Using the wing as a frame of reference with the x-axis parallel to the freestream velocity, the freestream velocity is diverted downwards, and during the interaction related to lift, the wing exerts a downwards force onto the air (from both above and below), with a component of that force in the same direction as the diverted stream, so the lift force is also doing work. Drag could be considered to be doing negative work, since it decreases the freestream velocity (wrt wing). The net work done would be related to the net change in freestream velocity (wrt wing).