Ground Effect....yes again.

[Champcar]

I was reading "Fly the Wing" lastnight and in the aero chapter there is a interesting part were it talks about how the earth supports the weight of a aircraft in flight. It goes on to state that the high pressure is spread accross the ground in a colum of air, but it is so spread out that it is not felt on the ground.

It goes on to state that this works with ground effect where that high pressure area doesnt get spread out as much and is "holding up" the aircraft.

It is a interesting thought that had me thinking for awhile. Any thoughts.

[tgrayson]

Originally Posted by Champcar

It goes on to state that this works with ground effect where that high pressure area doesnt get spread out as much and is "holding up" the aircraft..

That strikes me as meaningless.

Ground effect is a reduction in induced drag when flying close to the ground, because the ground interferes with the wingtip vortices. The phenomenon, as shown in a number of aerodynamics books, is similar to an increase in aspect ratio for the airplane.

Lift is not changed, because in unaccelerated flight, lift = weight. If you had an increase in lift, then the aircraft would be accelerating vertically, which would be unforunate. Sometimes happens with student pilots though.

Ground effect, therefore, is a reduction in drag, which increases the time and distance needed to slow the airplane to the proper landing attitude.

[Champcar]

I understand that part of ground effect. It was an interesting theory though. Which makes me wonder if a airplane with winglets has a reduced ground effect.

The weight of the plane has to be held up by something. So when an aircraft is close to the ground it could have a stronger force pushing back up on it than when at altitude.

[tgrayson]

Originally Posted by Champcar

I understand that part of ground effect. It was an interesting theory though. Which makes me wonder if a airplane with winglets has a reduced ground effect.

The more induced drag the aircraft has, the more pronounced the ground effect, so, yes, I'd *expect* that ground effect would be less for an aircraft with winglets.

<<
The weight of the plane has to be held up by something. So when an aircraft is close to the ground it could have a stronger force pushing back up on it than when at altitude.
>>

That cannot possibly be the case. The "push" force is lift and lift = weight in unaccelerated flight. If you had a greater "push" force, the aircraft would be accelerating vertically. This whole subject is a red herring, though, which is why I brought up the issue of drag. It's a drag reduction that characterizes ground effect, so it's pointless to speculate about an imaginary increase in "push" force. Because there isn't one.

I will offer this caveat though. One view of induced drag is that the lift vector is tilted backwards. When you enter ground effect, this lift vector straightens up and becomes more vertical, so there is a ballooning effect unless you stop it. You will and you do, automatically. The moment you push forward on the yoke to stop the balloon, the increased lift goes away, and all you're left with is a drag reduction, which is what causes all the problems with floating.

My view is the author's theory is silly. The ground doesn't support the aircraft....the aircraft has no idea what happens to the air once it departs the wing. In theory, if the air had no viscosity, yes, it would hit the ground, but long after the aircraft has passed by. But real air has viscosity and the air movement dies out after a while, due to turbulence.

Consider the vertical stabilizer, which generates lift horizontally. What does the air push against here? How about the horizontal stabilizer, which generates downward lift. Does the sky push against the airplane?

Unless something modifies the pressure distribution around the wing, it cannot have *any* effect on aircraft lift, because it's the pressure distribution that holds the airplane up.

[Realms09]

Originally Posted by tgrayson

The ground doesn't support the aircraft....the aircraft has no idea what happens to the air once it departs the wing. In theory, if the air had no viscosity, yes, it would hit the ground, but long after the aircraft has passed by. But real air has viscosity and the air movement dies out after a while, due to turbulence.

Shear forces from velocity gradients will convert kinetic energy of the air into thermal energy, and they will distribute the downward momentum amongst a greater and greater number of air particles. Since viscous forces in this case are just between air particles (no surfaces), the equal and opposite mentality says that downward momentum remains unchanged for the air as a whole. We just go from a low mass/high velocity/high energy to high mass/low velocity/low energy. So, viscosity or no viscosity, the air that was acted on by the wing is going down until something else chanes its direction, such as the ground, a bird, or another airplane. The pressure distribution on these objects will change as a result. The air could also encounter a thermal updraft, in which case the air start moving up but the updraft velocity will be diminished.

But as tgrayson said, the airplane could care less. If there existed a mass of air without the Earth or some other gravitational body, you could still generate lift in it - the air would just keep going "down" for eternity (even with viscosity).

[tgrayson]

Originally Posted by Realms09

Shear forces from velocity gradients will convert kinetic energy of the air into thermal energy, and they will distribute the downward momentum amongst a greater and greater number of air particles.

But thermal motion is, by definition, random.

[Realms09]

Originally Posted by tgrayson

But thermal motion is, by definition, random.

Yes, the random motion of the air will increase and the temperature will rise. The kinetic energy of the air will approach zero as more and more of the momentum is stored in mass rather than velocity. Nonetheless, the downward movement of the air, in the momentum sense, will remain unchanged, and it's the momentum that matters for talking about "pushing on the Earth."

[tgrayson]

Originally Posted by Realms09

Yes, the random motion of the air will increase and the temperature will rise. The kinetic energy of the air will approach zero as more and more of the momentum is stored in mass rather than velocity. Nonetheless, the downward movement of the air, in the momentum sense, will remain unchanged, and it's the momentum that matters for talking about "pushing on the Earth."

Momentum, however, is always conserved, whereas kinetic energy is not. If the momentum of the downward moving air remains the same, then there can be no momentum embodied in the thermal motion. I don't buy it.

[Realms09]

If the momentum of the downward moving air remains the same, then there can be no momentum embodied in the thermal motion. I don't buy it.

The net momentum of the downward moving air remains the same, and there is no net momentum embodied in the thermal motion. There is no problem with this statement. The net momentum of the random motion embodied by thermal energy must be zero, because it is random. Directional quantities, like momentum, cancel out in this situation. Scalar quantities, like energy, do not. A box of air at rest at some temperature has plenty of energy, but no net momentum. The conservation laws dictate that downward momentum of the air remains unchanged even as it heats up. There is simply no net force acting on the air as a whole to cause its net momentum to change. The shear force that slows down one particle speeds up another.

It's like the classic physics text example of an inelastic collision between two clay balls. The net momentum of the balls remains unchanged, but their temperature increases because kinetic energy is converted to thermal energy as they stick together.

[tgrayson]

Originally Posted by Realms09

there is no net momentum embodied in the thermal motion.

Yes, that occurred to me after I posted.

Originally Posted by Realms09

There is simply no net force acting on the air as a whole to cause its net momentum to change. The shear force that slows down one particle speeds up another.

I do see your point. However, I'm operating on my own momentum due to a number of aerodynamics books stating that the viscosity of the fluid causes these motions to die out. However, I cannot come up with a justification for it, in light of the unavoidable logic of your position. I will have to review these texts to see if they offer any insight. In the end, they could simply be incorrect.

[BeechBoy]

Can anyone explain precisely HOW wingtip vortices create induced drag. I know that they do I just don't know how and have never seen any text that explains it.

p.s. Be kind to my brain in your replies - I don't have a PhD.

[tgrayson]

Originally Posted by BeechBoy

Can anyone explain precisely HOW wingtip vortices create induced drag. I know that they do I just don't know how and have never seen any text that explains it.

p.s. Be kind to my brain in your replies - I don't have a PhD.

1. Draw an airfoil with the leading edge to the left
2. Draw the relative wind coming horizontally from the left.
3. Draw a lift vector perpendicular to the relative wind.

That's life without wing tip vortices.

Now,

1. Modify the relative wind so that it comes from slightly above the first one you drew and angle it downwards.
2. Draw a lift vector perpendicular to *this* relative wind.

Notice that the new lift vector is pointed slightly backwards. That's the origin of induced drag.

The modification you made to where the relative wind was coming from is due to the wingtip vortices; they tend to make the air around the wing move downward.

[UAL747400]

Originally Posted by tgrayson

1. Draw an airfoil with the leading edge to the left
2. Draw the relative wind coming horizontally from the left.
3. Draw a lift vector perpendicular to the relative wind.

That's life without wing tip vortices.

Now,

1. Modify the relative wind so that it comes from slightly above the first one you drew and angle it downwards.
2. Draw a lift vector perpendicular to *this* relative wind.

Notice that the new lift vector is pointed slightly backwards. That's the origin of induced drag.

The modification you made to where the relative wind was coming from is due to the wingtip vortices; they tend to make the air around the wing move downward.

Exactly. That is all the "floating" part of ground effect is. The ground interferes with the vorticies, reducing them if not removing them completely. This causes that lift vector to move more vertical, while reducing induced drag it is also making the lift the wing is making more effective since more of it is being applied directly into lifting the airplane. This gives you the floating effect. You're not riding on a cushion of air like you might think rather the lift the wing is making is being more effective.

Sorry if I repeated what's already been said, I didn't read the longer posts.

[BeechBoy]

Thank you both very much. I was familiar with this explanation for induced drag but I did not know that it was wingtip vortices that caused it. Thanks again.