Questions about density and being in the air

In summary, the conversation is about the concept of lift and how it is generated in different scenarios such as with a helicopter, a box being held in the air, and a piece of paper being blown into the air. The argument is whether or not density plays a role in generating lift, with one person believing that it is the main factor while the other argues that it is not always the case.
  • #36
Student100 said:
I still don't understand how you apply N3 to lift and not call it an idealization.
I never claimed that you could calculate or model lift solely based on Newton's third law, only that it applies to all forces, since all forces only exist as one part of a Newton third law pair of forces.
 
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  • #37
rcgldr said:
I never claimed that you could calculate or model lift solely based on Newton's third law, only that it applies to all forces, since all forces only exist as one of a Newton third law pair of forces.

In inertial frames. This isn't an inertial frame of reference.
 
  • #38
rcgldr said:
... all forces only exist as one part of a Newton third law pair of forces.

Student100 said:
In inertial frames. This isn't an inertial frame of reference.
If the ambient air (the air unaffected by an aircraft) is used as a frame of reference, it is an inertial frame. If an aircraft is not accelerating, then the wing as a frame of reference is an inertial frame.
 
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  • #39
rcgldr said:
If the is air is used as a frame of reference, it is an inertial frame. If an aircraft is not accelerating, then using the wing as a frame of reference is an inertial frame.

Then why introduce fictitious forces into the proper calculations of lift if we could simply agree on an inertial frame of reference?

Flows I thought could never be considered inertial frames.

Inertial frames are all just approximately inertial frames correct?
 
  • #40
Student100 said:
Flows I thought could never be considered inertial frames.
I updated my prior post. It's the ambient air (air unaffected by an aircraft) that can be used as an inertial frame of reference, or the ground in a no-wind situation.
 
  • #41
rcgldr said:
I updated my prior post. It's the ambient air (air unaffected by an aircraft) that can be used as an inertial frame of reference, or the ground in a no-wind situation.

Even that air is subject to the fictitious forces to apply N1 & N2, is this not correct?
 
  • #42
rcgldr said:
... the ambient air (air unaffected by an aircraft) that can be used as an inertial frame of reference, or the ground in a no-wind situation.
Student100 said:
Even that air is subject to the fictitious forces to apply N1 & N2, is this not correct?
Ignoring issues related to the rotation of the earth, the ambient air (the air unaffected by the aircraft), is not accelerating. The only idealization here is assuming that ambient air would not be accelerating. The same idealization is also made when using the wing as a frame of reference, as the oncoming flow approaching a wing is considered to have constant velocity. I don't see how any fictitious forces could exist from the perspective of an inertial frame.
 
  • #43
rcgldr said:
Ignoring issues related to the rotation of the earth, the ambient air (the air unaffected by the aircraft), is not accelerating. The only idealization here is assuming that ambient air would not be accelerating. The same idealization is also made when using the wing as a frame of reference, as the oncoming flow approaching a wing is considered to have constant velocity. I don't see how any fictitious forces could exist from the perspective of an inertial frame.

But you have to make assumptions about the fluid to make it inertial. That's my entire confusion here, is that not an idealization? Would such an event ever occur in nature? Then is it not an idealization to apply N3 to lift?

That's all I'm trying to clear up.
 
  • #44
Student100 said:
you have to make assumptions about the fluid to make it inertial.
I'm not sure I understand your point here, but I'm going to guess that you're thinking that if a fluid is not a Newtonian fluid, then it doesn't follow Newtonian physics, which isn't true. A Newtonian fluid is a fluid in which the viscous stresses arising from its flow, at every point, are linearly proportional to the local strain rate. It's a idealization of viscosity, not about Newton's laws of physics. Wiki article

http://en.wikipedia.org/wiki/Newtonian_fluid

Regardless of viscous effects, in an inertial frame of reference, then all fluids would be "inertial". A force exerted onto the air, results in a change in pressure and/or temperature and/or acceleration, and the reaction follows Newton's laws of physics.

Getting back to the original posters question, a change in pressure coexists with a change in density. Wiki article:

http://en.wikipedia.org/wiki/Compressibility_factor#Compressibility_of_air

In the case of a wing, a density differential coexists with the pressure differential, but it's the net pressure differential on a wing that corresponds to lift and drag forces exerted by the air onto the wing. For level flight, you can calculate the net pressure differential by the wing loading divided by the wing area.

http://en.wikipedia.org/wiki/Wing_loading

Take the example of a large commercial aircraft, the MD-11F, at 173 lbs / ft^2 wing loading, this is equal to about 1.2 psi (lbs / in^2) net pressure differential. Say this is distributed as +0.4 psi below the wing and -0.8 psi above. At near sea level where ambient pressure is about 14.7 psi, these differences are relatively small: +0.4/14.7, -0.8/14.7. At 35,000 feet, ambient pressure is only 3.2 psi, so the relative differences in pressure (and density) are greater, but the net pressure differential is the same. At higher altitudes, due to the lower density of the air, a greater speed and/or greater angle of attack is needed to maintain level flight.
 
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  • #45
Going back to the original point made:

Air is not a perfect fluid, and the concept of pressure is an idealization. But Newton's third law is not an idealization. Push air down and you go up. That's the ONLY way you can generate lift with a wing.

My point, Newtons third law is an idealization.

To apply it we have to consider an inertial frame of reference. This doesn't exist in lift calculations. Therefore, to apply N3 we make assumptions about the fluid and the plane, or an idealization of an otherwise complex interaction.

The discounting of pressure because it's an idealization, is therefore, flawed.
 
  • #46
I must be missing something -- what is the difficulty here?

An airfoil and a typical (box) fan blade and a ship's propeller all operate on essentially the same principle. Toy helicopters are in fact found with blades very similar to house fans, and they fly. There are also sails, which have basically no thickness at all and are found in all sorts of widths and aspect ratios.

Is there some controversy about whether Newton's 3rd law is true?
 
  • #47
Student100 said:
My point, Newtons third law is an idealization. To apply it we have to consider an inertial frame of reference.
Newton's third law is not something you apply as part of a calculation, it simply exists. As stated before all forces only exist as one of a pair of Newton third law forces. In this case, the aircraft exerts a force onto the air, and the air exerts an equal in magnitude but opposing force onto the aircraft.

Student100 said:
This doesn't exist in lift calculations.
Generally 3d lift calculations divide the upper and lower surfaces of a wing into a very large number of tiny squares, and calculate the pressure at each square, then sum up these pressures to determine the net pressure differential that corresponds to the force that the air exerts onto the wing. Newton's third law still applies, it would be the force that the wing exerts onto the air.
 
  • #48
Student100 said:
Going back to the original point made:
My point, Newtons third law is an idealization.

To apply it we have to consider an inertial frame of reference. This doesn't exist in lift calculations. Therefore, to apply N3 we make assumptions about the fluid and the plane, or an idealization of an otherwise complex interaction.

The discounting of pressure because it's an idealization, is therefore, flawed.

You mean that the Earth is not good enough as an inertial system for lift calculation? What assumption do you need to make about fluid in order to apply Newton's third law?
Are you thinking about the effect of Coriolis forces in lift calculations?

It's not clear why did you even bring in the discussion the inertial systems. It looked like you did it just to keep arguing about the "idealization" of Newton third law.

This seem to go around in loops. :)
Or rather in spiral, getting more and more remote from the original question.
 
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  • Like
Likes Chestermiller
  • #49
I like the N3 explanation better than the pressure difference explanation simply because my daily observation said so

Why does a toy helicopter goes higher when I put my palm about 5 inches under it ? Saying that the pressure difference becomes higher as the volume of air under the heli becomes smaller can't be true as my hand won't change the volume. It goes higher because it exerts force on my palm, which will push the toy upward.

And why can't helicopters climb as high as aircraft ? Some may say that the air density would be too low for it to generate lift, but hey, the decrease of air density under the rotor should be directly proportional with the decrease of air density above the rotor, it should still be able to generate lift right ?

thus I concluded at too high of altitude, the air under the rotor becomes too light to exert force on, which result in the helicopter unable to climb no more

And excuse me for my English.. Its not my first language
 
  • #50
Ariel24K said:
And why can't helicopters climb as high as aircraft ? Some may say that the air density would be too low for it to generate lift, but hey, the decrease of air density under the rotor should be directly proportional with the decrease of air density above the rotor, it should still be able to generate lift right ?
How much lift do you think a helicopter could generate on the moon where the air density is zero?

Density is not a force, so it can't be directly associated with lift (which is a force). On the other hand, pressure is a force per unit area normal to surfaces, and so pressure differences are the direct cause of lift.

Chet
 
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  • #51
Chestermiller said:
How much lift do you think a helicopter could generate on the moon where the air density is zero?

Density is not a force, so it can't generate lift. Pressure, or rather pressure differences, generate lift. This is a force.

Chet

Ah sorry I got the density and pressure difference mixed up after reading the original post, thanks for clearing up
 
  • #52
Some people here are making this way too complicated. Air flow around helicopter blades is approximately incompressible. You can assume incompressible flow and get very close to the right aerodynamics, good enough unless you are actually designing a high performance helicopter blade, in which case you wouldn't be asking this question.

Incompressible means density doesn't change. Lift derives from the difference in pressure, in simple terms the Bernoulli equations with constant density (rho).
https://en.wikipedia.org/wiki/Bernoulli's_principle

Compressible flow occurs as the flow approaches the speed of sound, so don't worry about it, and it doesn't change one's understanding of the answer. For buoyancy to matter you need something the size of a blimp, not a blade.
 

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