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Question:

I'd appreciate it a lot if anyone could explain how lift can equal the weight of an airplane in flight (Lift = Weight), if these airplanes have thrust-to-weight ratios of 0.3 (e.g. Boeing 747 or Airbus A320)?

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In stable flight, commercial airliners with thrust-to-weight ratios of about 0.3; mean that the thrust from the engines is just 0.3x the weight of the airplane. For example, if thrust is 1.0 N, then the weight is 3.3 N. (1.0 / 3.3 = 0.3).

Then, if Lift = Weight was true, lift would also be 3.3 N, the same as the weight 3.3 N. BUT this 3.3 N for lift is also 3.3x the thrust from the engines!! This is impossible. The lift generated by the wings is not observed to be 3.3x greater than the thrust produced by the en engines of an airliner. see image below of airliners - the backwash from the engines is observed to be a lot more than the downwash from the wings. Albeit this is comparing backwash form engines on take-off, to downwash from wings landing. Nonetheless.....

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Thrust-to-weight ratios are known and based on verifiable data of engine thrust and aircraft weight. Whereas, Lift = Weight an unproven assumption in almost every textbook (based on Newtons 2nd law of motion F = ma). Therefore, Lift = Weight is wrong: Thrust-to-eight ratios demonstrate that the lift required by an airliner to fly, cannot equal its weight.

Is there any experimental proof that Lift = Weight?

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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
  • $\begingroup$ This question is essentially just asking if Newton's laws are true. $\endgroup$ – D. Halsey Sep 21 at 14:19
  • $\begingroup$ "The lift generated by the wings is not observed to be 3.3x greater than the thrust produced by the en engines of an airliner." Where are you getting this information? I'm pretty sure this is exactly what aircraft designers find to be the case. They get the ratio based on what is realistically going to happen; they don't just make them up. $\endgroup$ – JMac Sep 21 at 14:20
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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 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 Jan 11 at 6:44

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