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Question:

I'd appreciate it a lot if anyone could explain how the wings can generate lift that is 3.3x more than engine thrust? Is there any experimental proof that this? Or is the assumption that Lift = Weight wrong?

Given that in stable flight, lift equals the weight of an airplane (Lift = Weight); Some commercial airliners have thrust-to-weight ratios of about 0.3; Here the wings generate Lift that is 3.3 times the force generated by the engine thrust (Lift = 3.3 x Thrust); And do so without using any of the force generated by the engines.

Background:

Commercial airliners such as Airbus 320 and Boeing 747, have similar thrust-to-weight ratios of about 0.3. Rearranging this ratio shows that the weight of the airplane (Weight) is about 3.3x engine thrust (Thrust):

Thrust / Weight = 0.3

Rearrange this equation: Weight = Thrust/0.3 = 3.3 x Thrust

=> Weight = 3.3 x Thrust

In stable flight if lift is assumed to equal the weight of the airplane (Lift = Weight) according to Newtons laws (F = ma).

Then combining the two equations (Lift = Weight) and (Weight = 3.3x Thrust):

Lift = Weight = 3.3 x Thrust

=> Lift = 3.3 x Thrust

enter image description here

Then, according to NASA and other advocates of fluid mechanics, the wings generate lift due to the airflow over the wings. Engine thrust is used to overcome drag (Thrust = Drag) and does not contribute towards lift in stable flight.

This means that for every 1 N of Thrust, then 3.3 N of Lift would be created from wings airflows. This is implausible also as it violates the principle of conservation of momentum and energy to produce a net gain of a force (Lift); and so much Lift (3.3x) compared to Thrust. Where does the extra energy used for lift come from?

enter image description here

This proportion (Lift = 3.3 x Thrust) is implausible for many reasons. For example, raising the nose of the airplane (pitch) would provide a net loss of the vertical upward force pushing the airplane up. As the wings generate 3.3x the force of the engine thrust, after the airplane’s nose is raised the upward force from the wings would decline more than the slight increase in upward force from the engines. As Lift = 3.3 x Thrust is implausible, then Lift = Weight is also implausible.

enter image description here

Another example; a Boeing 787-8’s maximum range of about 14,000 km (e.g. New York to New Zealand); It uses about 100,000 kg of fuel to carry a max. payload (aircraft and passengers) of 128,000 kg. [14] It is implausible that on a 14,000 km flight, the wings create an additional upward force (Lift) that is 3.3x greater than the engine thrust (that used 100,000 kg of fuel on this flight). Where exactly does this extra force and energy come from? Remember, fluid mechanics claims that lift is created without using Thrust (as Thrust = Drag). The force and energy for this lift is a net gain to the system; which violates the principles of conservation of momentum and energy. Therefore, Lift = Weight is implausible.

Thanks in advance!

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    $\begingroup$ For us to be able to answer this question, you need to give us more context. You say that the amount of lift is "extremely unlikely" -- but why do you think that? You ask for experimental proof -- so why isn't the very fact that these plane work not experimental proof? $\endgroup$ – knzhou Dec 30 '19 at 8:19
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    $\begingroup$ Gliders have 0 thrust, but still have lift. $\endgroup$ – Adrian Howard Dec 30 '19 at 8:54
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Jan 11 at 21:13
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The thrust is used to keep the plane at a constant speed $v$ in horizontal direction. It compensates the friction, which would slow the plane down. However, the lift is due to Bernoulli's equation $p + \frac{\rho}{2}v^2 + \rho g h = const$ which for constant height becomes $p + \frac{\rho}{2}v^2 = const$. Thus the thrust is given by $$F = A \cdot \Delta p = A \cdot \frac{\rho}{2} [v_{below}^2 - v_{above}^2]$$ where $A$ is the area of the wings, $\rho$ is the mass density of the air at height $h$, and the two velocities are the wind speeds above and below the wings -- since the pressure difference is not constant along the wing, we would obtain an integral, if we would do the calculation properly.

From this simple equation we see, that the thrust is proportional to the area of the wings. If the friction was arbitrary small, the thrust/weight ratio could become arbitrary large, by increasing the area of the wings.

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  • $\begingroup$ Good answer, backed up by lift/drag wind tunnel studies. A flat plate indeed has a lift maximum at 45 degrees deflection, but with huge drag. Same lift at around 15 degrees deflection (much less drag) is obtained by Bernoulli as described so well here. $\endgroup$ – Robert DiGiovanni Jan 2 at 18:25
  • $\begingroup$ @RobertDiGiovanni What force is your F? Lift, or thrust? $\endgroup$ – enbin zheng Jan 11 at 6:36
  • $\begingroup$ This is false. There's not one scientific experiment on a real airplane that proves what you say to be true. NASA says it false, many academics say its false ...... $\endgroup$ – Nick Landell Jan 11 at 6:36
  • $\begingroup$ @NickLandell go to the playground and try the see-saw with one 3.3 times heavier than you. How do you explain this "magic"? $\endgroup$ – Robert DiGiovanni Jan 11 at 7:45
  • $\begingroup$ Sure, but to state the obvious problem with the leverage argument: An airplane wing has no fulcrum, so no leverage is possible. Worse, an airplane is moving not static, so the playground analogy fails. $\endgroup$ – Nick Landell Jan 12 at 14:43
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That's FULL thrust, used when taking off (climbing the hill, so to speak). It is much larger than drag. Typical drag in a jet transport is more like 4% of weight, not 30%.

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    $\begingroup$ This is probably better suited as a comment on the question, though it is a good point. $\endgroup$ – JMac Dec 30 '19 at 16:29
  • $\begingroup$ What do you want to say? $\endgroup$ – enbin zheng Jan 11 at 6:44

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